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Physics Notes

[Last Updated 5 May 2014]


General Physics:
Measurement
One-Dimensional Kinematics
Vectors
Two-Dimensional Kinematics
Relative Velocity
Force and Acceleration
Centripetal Force and Acceleration
Work and Energy
Power
Linear Momentum
Rotational Kinematics
Rotational Dynamics
Static Equilibrium
Gravitation
Fluid Statics
Fluid Dynamics
Electric Fields
Gauss’s Law
Capacitors
DC Circuits
Magnetic Force on a Charged Particle
Biot-Savart Law
Faraday's Law
Magnetic Properties
LRC Circuits
Practical Electricity
Heat
Kinetic Theory of Gases
Second Law of Thermodynamics
Sound
Geometric Optics
Optical Instruments
Double-Slit Interference
Diffraction Gratings
Interference by Thin Films
Interferometers
Blackbody Radiation
Scattering of Charged Particles
X-Ray Production
Compton Scattering
Pair Production
Atoms
Molecules
Semiconductor Physics
Nuclear Reactions
Special Relativity:
Relativistic Kinematics
Relativistic Dynamics
Relativistic Doppler Effect
Advanced Classical Mechanics:
Planetary Motion
Motion of a System of Particles
Rigid Bodies
Lagrange's Equations
Quantum Mechanics:
Fourier Analysis
Schrodinger Wave Equation
Nuclear Physics:
Nuclear Reactions


Measurement

PROBLEM

A box of powdered plaster weighs 4.5 lb when full and measures 8" x 5.25" x 3.75". According to the mixing instructions, 1 lb of plaster should be mixed with 6 oz of water. How many cups of plaster should be mixed with 1 cup of water?

SOLUTION

The volume of the box is

V = (8 in)(5.25 in)(3.75 in) = 157.5 in3

The volume of 1 lb of plaster is

V1 = V(1/4.5) = 35 in3 = (35 in3)(1 gal / 231 in3)(128 oz/gal) = 19.393939 oz
=> 19.393939 oz = 2.424242 cups of plaster should be mixed with 6 oz = 3/4 cup of water
=> 3.232323 cups of plaster should be mixed with 1 cup of water

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PROBLEM

Estimate the value and weight of books that can be shipped in a cardboard box whose dimensions are 28 1/2" long x 17 1/4" wide x 18 5/8" high. These are the dimensions of a standard UMAC Express Cargo box.

SOLUTION

By determining the volume, value, and weight of a number of individual books, we can extrapolate to estimate the value and weight of books filling an entire box as described.

Book No. Dimensions Volume (in3) Estimated Value (USD)
1
7/8" x 6" x 9"
47.25
$15.00
2
2" x 6 1/2" x 9 1/4"
120.25
$25.00
3
1" x 5 1/2" x 8 1/2"
46.75
$12.00
4
3/8" x 8 1/4" x 11 1/8"
34.42
$7.50
5
5/8" x 8 1/2" x 9"
47.81
$7.50
6
5/8" x 8 3/8" x 8 1/4"
43.18
$7.50
7
3/8" x 5 1/4" x 7 3/8"
14.52
$5.00
8
1/2" x 5 1/4" x 8"
21.00
$2.00
9
1 1/2" x 5 1/4" x 8"
63.00
$12.00
10
3/8" x 8" x 10"
30.00
$7.50
11
1" x 9" x 10 1/2"
94.50
$7.50
12
3/8" x 8 3/4" x 11"
36.09
$5.00
Total
598.78
$113.50

The total weight of the above books was measured to be 13 lb. The total volume of the box is 28 1/2" x 17 1/4" x 18 5/8" = (28.5 in)(17.25 in)(18.625 in) = 9156.52 in3. Assuming that only 80% of the volume of the box is actually filled with books, the total volume of the enclosed books is (0.8)(9156.52 in3) = 7325.21 in3. Dividing this by the volume of the measured books, we get (7325.21 in3) / (598.78 in3) = 12.23. Thus, the entire box contains 12.23 times as much volume of books or (12.23)(13 lb) = 159.04 lb in weight and (12.23)($113.50) = $1388.52 in value.

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PROBLEM

A "six ounce" yogurt container consists of three sections. The main section has the shape of the frustum of a cone, with height 2 7/16", bottom diameter 2 1/8", and top diameter 2 11/16". The top section has the shape of a cylinder with diameter 2 13/16" and height 1/4". The bottom section has the shape of a cylindrical shell with inner diameter 1 1/2", outer diameter 2 1/8", and height 1/8". Find the volumes of the (a) main section, (b) top section, (c) bottom section, and (d) entire yogurt container in in3, cm3, and ounces.

SOLUTION

(a) The volume of the main section is obtained by dividing it into infinitesimal disks and integrating. Each disk has volume

dV = pr2 dy = p[a + (b - a)y/h]2 dy

where

a = (1/2)(2 1/8") = 1 1/16" = 1.0625"

is the bottom radius,

b = (1/2)(2 11/16") = 1 11/32" = 1.34375"

is the top radius, and

h = 2 7/16" = 2.4375"

is the height. Integrating, we get

Vmain = ∫ dV = ∫0h p[a + (b - a)y/h]2 dy = p0h [a2 + 2a(b - a)y/h + (b - a)2(y/h)2] dy
= p[a2y + 2a(b - a)(y2/2h) + (b - a)2(y3/3h2)] |0h = p[a2h + 2a(b - a)(h2/2h) + (b - a)2(h3/3h2)]
= p[a2h + 2a(b - a)(h/2) + (b - a)2(h/3)] = ph[a2 + a(b - a) + (1/3)(b - a)2]
= ph[a2 + ab - a2 + b2/3 - (2/3)ab + a2/3] = ph[a2/3 + (1/3)ab + b2/3]
= p(2.4375 in)[(1.0625 in)2/3 + (1/3)(1.0625 in)(1.34375 in) + (1.34375 in)2/3] = 11.13 in3
= (11.13 in3)(2.54 cm/in)3 = 182.5 cm3 = (11.13 in3)(1 gal / 231 in3)(128 oz/gal) = 6.170 oz

(b) The volume of the top section is

Vtop = prtop2htop

where

rtop = (1/2)(2 13/16") = 1 13/32" = 1.40625"

is the radius and

htop = 1/4" = 0.25"

is the height. Thus,

V = p(1.40625 in)2(0.25 in) = 1.553 in3 = (1.553 in3)(2.54 cm/in)3 = 25.45 cm3
= (1.553 in3)(1 gal / 231 in3)(128 oz/gal) = 0.8606 oz

(c) The volume of the bottom section is

Vbot = p(rout2 - rin2)hbot

where

rout = (1/2)(2 1/8") = 1 1/16" = 1.0625"

is the outer radius,

rin = (1/2)(1 1/2") = 0.75"

is the inner radius, and

hbot = 1/8" = 0.125"

is the height. Thus,

Vbot = p[(1.0625 in)2 - (0.75 in)2](0.125 in) = 0.2224 in3 = (0.2224 in3)(2.54 cm/in)3 = 3.645 cm3
= (0.2224 in3)(1 gal / 231 in3)(128 oz/gal) = 0.1233 oz

(d) The total volume of the yogurt container is

Vtot = Vmain + Vtop + Vbot = 11.13 in3 + 1.553 in3 + 0.2224 in3 = 12.91 in3 = 211.6 cm3 = 7.154 oz

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PROBLEM

A container of unmixed tile grout is 4.75" high and, as viewed from above, has the shape of a 5" x 5" square with four right triangles with 1" legs cut off from the corners. The grout weighs 7 lb and is to be mixed with one pint of water.
(a) What is the volume of the grout in in3?
(b) What is the volume of the grout in cm3?
(c) What is the mass of the grout in g?
(d) What is the density of the grout in g/cm3?
(e) What volume of water should be mixed with each cm3 of grout?

[based on Polyblend sanded grout]

SOLUTION

(a) The volume of the grout is

V = [(5 in)(5 in) - (4)(1/2)(1 in)(1 in)](4.75 in) = 109.25 in3

(b) V = (109.25 in3)(2.54 cm/in)3 = 1790.287 cm3

(c) The mass of the grout is

m = (7 lb)(1 kg / 2.205 lb)(1000 g/kg) = 3174.603 g

(d) The density of the grout is

r = m/V = (3174.603 g) / (1790.287 cm3) = 1.773 g/cm3

(e) 1 pint = (1 pint)(0.125 gal/pint)(231 in3 / gal)(2.54 cm/in)3 = 473.1765 cm3

This amount of water should be mixed with 1790.287 cm3 of grout. So for each cm3 of grout, (473.1765/1790.287)(1 cm3) = 0.2643 cm3 of water is needed.

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PROBLEM

The Cosmic Calendar (Sagan 1977, Sagan et al. 1980) represents the entire history of the Universe as a single year. Assuming that the Universe is 15 billion years old, how much time in the history of the Universe is represented by (a) one month, (b) one day, (c) one hour, (d) one minute, and (e) one second in the Cosmic Calendar?

SOLUTION

(a) If we take one month to be 1/12 of a year, one month in the Cosmic Calendar corresponds to (1/12)(15 billion years) = 1.25 billion years.

(b) If we take one day to be 1/365 of a year, one day in the Cosmic Calendar corresponds to (1/365)(15 billion years) = 41,095,890 years.

(c) Since there are 24 hours in one day, one hour in the Cosmic Calendar corresponds to (1/24)(41,095,890 years) = 1,712,329 years.

(d) Since there are 60 minutes in one hour, one minute in the Cosmic Calendar corresponds to (1/60)(1,712,329 years) = 28,539 years.

(e) Since there are 60 seconds in one minute, one second in the Cosmic Calendar corresponds to (1/60)(28,539 years) = 476 years.

Note that the currently accepted age of the Universe is 13.73 ± 0.12 billion years (Chang 2008).

Chang, Kenneth 2008, "Gauging Age of Universe Becomes More Precise," New York Times, 9 March 2008.

Sagan, Carl 1977, The Dragons of Eden: Speculations on the Evolution of Human Intelligence (New York: Ballantyne Books).

Sagan, Carl, Ann Druyen, and Steven Soter 1980, "The Shores of the Cosmic Ocean," Cosmos: A Personal Voyage, thirteen-part television series first broadcast by the Public Broadcasting Service (PBS) in 1980.

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One-Dimensional Kinematics

The position x(t) of an object moving with constant acceleration a is given by

x(t) = xi + vit + (1/2)at2

where xi is the initial position, vi is the initial velocity, and t is the elapsed time. The final velocity vf is related to the initial velocity according to

vf = vi + at

and by

vf2 = vi2 + 2a(x - xi)

If an object moves with constant acceleration, its average velocity is

vavg = (vi + vf)/2

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PROBLEM

According to Consumer Reports (2010), "If a driver [of a car] brakes suddenly from about 30 mph, items in the back seat could hit with the force they'd have when falling from a two-story building." Explain.

Consumer Reports 2010, "Pack Your Car Safely," Consumer Reports, August 2010, 10-11.

SOLUTION

The initial and final velocities of an object undergoing uniformly accelerated motion in the vertical direction are related by

vf2 = vi2 + 2a(yf - yi)

where vf is the final velocity, vi is the initial velocity, a is the acceleration, yf is the final y coordinate, and yi is the initial y coordinate.

Consider an object which falls a vertical distance h from rest under the force of gravity. In this case, we have a = - g, where g = 32 ft/s2, yf - yi = - h, and vi = 0. Thus,

vf2 = 2gh => h = vf2 / 2g

If the final velocity of the falling object is vf = 30 mph, which is the assumed initial velocity of the car, then

vf = 30 mph = (30 mph)(5280 ft/mi)(1 hr / 3600 s) = 44 ft/s

and

h = (44 ft/s)2 / [(2)(32 ft/s2)] = 30.25 ft

The object must fall from a height of 30.25 ft above the ground to hit the ground at a speed of 30 mph. If we assume that a typical story in a building spans 10 ft in the vertical direction, and that the first-story floor is 5 ft above the ground, and neglect the distance between the ceiling of a given story and the floor of the next story above, 30.25 ft would be about halfway between the third-story floor and ceiling, or about 5 ft above the roof of a two-story building.

+-------+ 35 ft
|  3rd  |
| story |
+-------+ 25 ft
|  2nd  |
| story |
+-------+ 15 ft
|  1st  |
| story |
+-------+ 5 ft
|       |
+-------+ ground level

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PROBLEM

A dog sees a flowerpot sail up and then back past a window 5.0 ft (= 1.524 m) high. If the total time the pot is in sight is 1 s, find the height above the top of the window to which the pot rises.

[from Halliday and Resnick 1988, Fundamentals of Physics, Third Edition (New York: John Wiley & Sons), problem 2.85]

SOLUTION

Consider the upward part of the motion as the flowerpot goes past the window. The elapsed time is 0.5 s, which is half of the total time that the pot is in sight. Use the kinematic equation

y = vit + (1/2)at2 = vit - (1/2)gt2

Here y = 5.0 ft is the displacement, t = 0.5 s, vi is the unknown initial velocity of the pot at the bottom of the window, a = - g is the acceleration of the pot, and g = 32 ft/s2 = 9.8 m/s2 is the acceleration of gravity. Note that we used a = - g since we are defining the upward direction to be positive, and the acceleration of gravity is directed downward. Everything is known except for vi. We solve for vi.

vi = [y + (1/2)gt2] / t = [5 ft + (1/2)(32 ft/s2)(0.5 s)2] / (0.5 s) = 18 ft/s

This is the velocity of the pot at the bottom of the window. Now we will determine how high the pot goes, relative to the bottom of the window, before it stops and starts going back down. Here we use the kinematic equation

vf2 = vi2 + 2ay = vi2 - 2gh = 0

where vf is the velocity at the top of its motion, which is zero, vi is the velocity at the bottom of the window, which we just found, and y = h is the distance the pot rises above the bottom of the window. Solve for h.

h = vi2 / 2g = (18 ft/s)2 / [(2)(32 ft/s2)] = 5.0625 ft

The height above the top of the window to which the pot rises is 5.0625 ft - 5 ft = 0.0625 ft = 0.75 in.

In MKS units, we would get

vi = [y + (1/2)gt2] / t = [1.524 m + (1/2)(9.8 m/s2)(0.5 s)2] / (0.5 s) = 5.498 m/s

h = vi2 / 2g = (5.498 m/s)2 / [(2)(9.8 m/s2)] = 1.542 m

The height above the top of the window to which the pot rises is 1.542 m - 1.524 m = 0.018 m = 1.8 cm.

Note: 1.8 cm is not exactly the same as 0.75 in. 1.8 cm = (1.8 cm)(1 in / 2.54 cm) = 0.708661 in. The difference arises because the two values of g that were used, 32 ft/s2 and 9.8 m/s2, are not exactly equal, and because there is some roundoff in the above results. In fact, 9.8 m/s2 = (9.8 m/s2)(100 cm/m)(1 in / 2.54 cm)(1 ft / 12 in) = 32.15223 ft/s2. If this value is used in the above calculation, and we retain all digits that our calculator shows, we get, in English units,

vi = [y + (1/2)gt2] / t = [5 ft + (1/2)(32.15223 ft/s2)(0.5 s)2] / (0.5 s) = 18.03806 ft/s

h = vi2 / 2g = (18.03806 ft/s)2 / [(2)(32.15223 ft/s2)] = 5.059859 ft

The height above the top of the window to which the pot rises is 5.059859 ft - 5 ft = 0.059859 ft = 0.718311 in. If we repeat the calculation in MKS units and retain all digits that our calculator shows, we get

vi = [y + (1/2)gt2] / t = [1.524 m + (1/2)(9.8 m/s2)(0.5 s)2] / (0.5 s) = 5.498 m/s

h = vi2 / 2g = (5.498 m/s)2 / [(2)(9.8 m/s2)] = 1.542245 m

The height above the top of the window to which the pot rises is 1.542245 m - 1.524 m = 0.018245 m = 1.8245 cm = 0.718311 in.

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PROBLEM

An elevator ascends with an upward acceleration of 4.0 ft/s2. At the instant its upward speed is 8.0 ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor. Calculate (a) the time of flight of the bolt from the ceiling to the floor and (b) the distance it has fallen relative to the elevator shaft.

[from Resnick, Robert, and Halliday, David 1966, Physics (New York: Wiley), problem 3.30]

SOLUTION

(a) Suppose that at time t = 0 the bottom of the elevator is at y = 0. Then the y coordinate of the bottom of the elevator at time t is

ybot(t) = (1/2)at2

where a = 4.0 ft/s2 is the acceleration of the elevator. The y coordinate of the top of the elevator at time t is

ytop(t) = h + (1/2)at2

where h = 9.0 ft is the distance from the bottom to the top of the elevator. The elevator speed at time t is given by

v(t) = at => t = v/a

The bolt drops when the elevator speed is 8.0 ft/s, which happens at time

t = v/a = (8.0 ft/s) / (4.0 ft/s2) = 2 s

when the top of the elevator is at

ytop(2) = 9.0 ft + (1/2)(4.0 ft/s2)(2 s)2 = 17 ft

The y coordinate of the bolt at time t is

ybolt(t) = ytop(2) + v(2)(t - 2) - (1/2)g(t - 2)2

The bolt hits the floor of the elevator when

ybolt(t) = ybot(t) => ytop(2) + v(2)(t - 2) - (1/2)g(t - 2)2 = (1/2)at2
=> 17 + (8)(t - 2) - (1/2)(32)(t - 2)2 = (1/2)(4)t2 => 17 + 8t - 16 - (16)(t2 - 4t + 4) = 2t2
=> 1 + 8t - 16t2 + 64t - 64 = 2t2 => 18t2 - 72t + 63 = 0 => 2t2 - 8t + 7 = 0
=> t = {8 ± sqrt[64 - (4)(2)(7)]} / [(2)(2)] = [8 ± sqrt(8)] / 4 = 2 ± (1/2) sqrt(2)
=> [2 + (1/2) sqrt(2)] s

The time of flight of the bolt is (1/2) sqrt(2) s = 0.7071 s.

(b) The y coordinate of the bolt when it hits the elevator floor is

ybolt[2 + (1/2) sqrt(2)] = ytop(2) + v(2)[(1/2) sqrt(2)] - (1/2)g[(1/2) sqrt(2)]2
= 17 + (8)[(1/2) sqrt(2)] - (1/2)(32)[(1/2) sqrt(2)]2 = 17 + 4 sqrt(2) - 8 = [9 + 4 sqrt(2)] ft

The distance the bolt has fallen relative to the elevator shaft is

ybolt(2) - ybolt[2 + (1/2) sqrt(2)] = 17 ft - [9 + 4 sqrt(2)] ft = [8 - 4 sqrt(2)] ft = 2.343 ft

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PROBLEM

An elevator ascends from the ground with uniform speed. At time T1 a boy drops a marble through the floor. The marble falls with uniform acceleration g = 9.8 m/s2, and hits the ground T2 seconds later. Find the height of the elevator at time T1.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 1.13]

SOLUTION

Let v equal the elevator's speed and h equal its height when the marble is dropped. Then

h = vT1

The displacement of the marble is

- h = vT2 - (1/2)gT22 = - vT1 => v = (1/2)gT22 / (T1 + T2) => h = (1/2)gT1T22 / (T1 + T2)

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Vectors

PROBLEM

Find a unit vector perpendicular to

A = (i + j - k)

and

B = (2i - j + 3k)

where i, j, and k are unit vectors in the +x, +y, and +z directions.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 1.8]

SOLUTION

The vectors A x B and B x A = - A x B, where "x" denotes a cross product, are perpendicular to both A and B.

A x B = (i + j - k) x (2i - j + 3k)

  | i  j  k |
= | 1  1 -1 | = i(3 - 1) - j(3 + 2) + k(-1 - 2) = 2i - 5j - 3k
  | 2 -1  3 |


The magnitude of A x B is

|A x B| = sqrt[22 + 52 + 32] = sqrt(38)

Thus, the two unit vectors

± (A x B) / |A x B| = ± (2i - 5j - 3k) / sqrt(38)

are perpendicular to A and B.

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Two-Dimensional Kinematics

PROBLEM

A gun crew observes a remotely controlled balloon launching an instrumented spy package in enemy territory. When first noticed the balloon is at an altitude of H = 800 m and moving vertically upward at a constant velocity of vb = 5 m/s. It is D = 1600 m down range. Shells fired from the gun have an initial velocity of v0 = 400 m/s at a fixed angle q (sin q = 3/5 and cos q = 4/5). The gun crew waits and fires so as to destroy the balloon. Assume g = 10 m/s2. Neglect air resistance.
(a) What is the flight time of the shell before it strikes the balloon?
(b) What is the altitude of the collision?
(c) How long did the gun crew wait before they fired?

[from 8.01 Physics I, Fall 2003, Exam 1, Massachusetts Institute of Technology]

SOLUTION

(a) The flight time is the time it takes the shell to travel the distance D in the horizontal (x) direction, which is equal to the distance divided by the horizontal component of the velocity, which is constant. The horizontal component of the velocity is vx = v0 cos q.

D = vxt => t = D/vx = D/(v0 cos q) = (1600 m) / [(400 m/s)(4/5)] = 5 s

(b) The altitude of the shell at a time t after being launched is

y = vy0t + (1/2)at2 = vy0t - (1/2)gt2

vy0 is the initial velocity of the shell in the vertical (y) direction, which is

vy0 = v0 sin q = (400 m/s)(3/5) = 240 m/s

Substituting into the above equation for y, we get

y = (240 m/s)(5 s) - (1/2)(10 m/s2)(5 s)2 = 1200 m - 125 m = 1075 m

(c) Let t0 be the time the gun crew waits, after first seeing the balloon, before firing a shell at it. The altitude of the balloon at the time of the collision is

y = H + vb(t0 + t)

Solving for t0, we get

t0 = (y - H)/vb - t = (1075 m - 800 m)/(5 m/s) - 5 s = 50 s

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PROBLEM

A baseball player hits a large number of baseballs at various initial speeds v0 and initial angles q with the horizontal. Which of the following must be true of the baseball which is in the air for the longest amount of time before touching the ground? Justify your answer.
(a) It has the longest range R.
(b) It has the greatest maximum height h.
(c) It has the highest initial velocity v0.
(d) It was launched at an initial angle of q = 45 deg.

SOLUTION

First we determine the amount of time t the baseball is in the air for a given value of v0 and q. The time it takes for the baseball to reach the highest point of its motion is before it hits the ground is t/2. At this point, its velocity in the vertical (y) direction is zero.

vyf = vyi + ay(t/2) = v0 sin q - gt/2 = 0

t = (2v0/g) sin q (I)

The range R is equal to the product of the horizontal (x) component of the velocity, which is constant, and the total time t.

R = vxt = v0t cos q = (2v02/g) sin q cos q

The maximum height is obtained using

vyf2 = vyi2 + 2ah = (v0 sin q)2 - 2gh = 0 => h = (v0 sin q)2 / 2g (II)

From the expression for t (I) above, we see that

v0 sin q = gt/2

which we can substitute into (II) to get

h = (g2t2/4) / 2g = gt2/8 => t = sqrt(8h/g) = 2 sqrt(2h/g)

We see that t can be expressed as a function of h alone; the rest of the terms in the expression are constants. From this we can conclude that in order for t to be maximized, h must also be maximized. Statements (a), (c), and (d) can only be true if (b) is also true. Thus, the only statement above which must be true is (b). Intuitively, this is because the baseball which reaches the highest elevation is the one which has the largest initial velocity in the y direction and which therefore requires the longest amount of time to decelerate to zero speed.

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PROBLEM

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below.
(a) If the plane is traveling horizontally with a speed of vx = 250 km/h (69.4 m/s), how far in advance of the recipients (horizontal distance) must the goods be dropped?
(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position?
(c) With what speed do the supplies land in the latter case?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 3.38]

SOLUTION

(a) The time t it takes for the goods to fall a distance h = 235 m to the ground is given by

h = (1/2)gt2 => t = sqrt(2h/g)

In this amount of time, the plane travels a horizontal distance

x = vxt = vx sqrt(2h/g) = (69.4 m/s) sqrt[(2)(235 m)/(9.8 m/s2)] = 480.6 m

The goods must therefore be dropped when the plane is a horizontal distance of 480.6 m from the recipients.

(b) The time it takes the goods to reach the horizontal position of the recipients is

t = d/vx = (425 m)/(69.4 m/s) = 6.124 s

In this same amount of time, the goods must have a vertical displacement of - h = - 235 m. If v0 is the initial velocity in the y direction, we have

- h = v0t - (1/2)gt2
=> v0 = [(1/2)gt2 - h] / t = [(1/2)(9.8 m/s2)(6.124 s)2 - 235 m] / (6.124 s) = - 8.367 m/s

The goods should be released with a vertical velocity of 8.367 m/s directed downward.

(c) When the goods reach the recipients, the vertical component of the velocity is

vy = v0 - gt = - 8.367 m/s - (9.8 m/s2)(6.124 s) = - 68.38 m/s

and the horizontal component of the velocity is the same as it was initially, vx = 69.4 m/s. The speed of the goods when they land is

v = sqrt(vx2 + vy2) = sqrt[(69.4 m/s)2 + (- 68.38 m/s)2] = 97.46 m/s

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PROBLEM

A projectile situated on the ground is aimed at an object which is hanging a distance h above the ground and a horizontal distance L from the projectile. The projectile is launched with velocity v0 directly towards the object. If the object is dropped from rest at the instant the projectile is launched, will the projectile hit the object, pass above it, or pass below it?

                +--------
                |
                O
              \-|-/   ^
               / \    |
                      |
       _              |
       /|             h
  v0 /                |
   /                  |
 / q                  v
+----------------------
|<------L------>|


SOLUTION

The horizontal speed of the projectile, which remains constant, is

vx = v0 cos q

The time it takes for the projectile to travel the horizontal distance L is

t = L/vx = L / (v0 cos q)

The initial vertical speed of the projectile is

vy0 = v0 sin q

At time t, the vertical position of the projectile is

y1 = vy0t - (1/2)gt2 = (v0 sin q)[L / (v0 cos q)] - (1/2)gt2 = L tan q - (1/2)gt2 = h - (1/2)gt2

since

tan q = h/L

Also at time t, the vertical position of the object is

y2 = h - (1/2)gt2 = y1

Thus, the projectile hits the object.

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PROBLEM

A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on top of the cliff 7.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 3.73]

SOLUTION

The horizontal distance traveled by the projectile is

L = vxt = v0t cos q (1)

where

vx = v0 cos q

is the horizontal component of the velocity, which is constant, v0 is the initial speed, t = 7.6 s is the travel time, and q is the angle between the initial velocity vector and the horizontal. The vertical distance traveled is

H = vy0t - (1/2)gt2 = v0t sin q - (1/2)gt2 => v0t sin q = H + (1/2)gt2 (2)

where

vy0 = v0 sin q

is the initial vertical component of the velocity. Dividing (2) by (1), we get

tan q = [H + (1/2)gt2] / L => q = tan-1{[H + (1/2)gt2] / L}
= tan-1{[155 m + (1/2)(9.8 m/s2)(7.6 s)2] / (195 m)} = 66.00°

From (1), we get

v0 = L / (t cos q) = (195 m) / [(7.6 s) cos 66.00°] = 63.09 m/s

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PROBLEM

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. The bullet puts a hole in a window of another building and hits the wall that faces the window. The point where the bullet hits the wall is 6.9 m behind the window and 0.50 m below the bullet hole in the window. Determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

v0 = 340 m/s
-----+---->
XXXXX| ^                  +-----+ |
XXXXX| H                  |     | V
XXXXX| V                  o     |---
XXXXX|<---------D-------->|     x---
XXXXX|                    |6.9 m| ^
XXXXX|                    |     | | 0.50 m


[from Cutnell, John D., and Johnson, Kenneth W. 2004, Physics, Sixth Edition (New York: Wiley), problem 3.43]

SOLUTION

Define the following quantities:

t1 = time needed for bullet to travel from window to wall
dx = 6.9 m = horizontal distance between window and wall
dy = 0.50 m = vertical distance between bullet hole in window and point of impact in wall
v0 = 340 m/s = initial velocity of bullet
vx = v0 = 340 m/s = x component of bullet velocity (constant)
v1y = vertical component of bullet velocity when it passes through window
v2y = vertical component of bullet velocity when it hits wall

The horizontal distance the bullet travels between the window and the wall is

dx = vxt1 => t1 = dx/vx = (6.9 m) / (340 m/s) = 2.029 x 10-2 s

v1y and v2y are related by

v2y = v1y + gt1 => v1y = v2y - gt1 (1)
v2y2 = v1y2 + 2gdy (2)

Substituting (1) into (2), we get

v2y2 = (v2y - gt1)2 + 2gdy = v2y2 - 2v2ygt1 + g2t12 + 2gdy
=> 0 = - 2v2ygt1 + g2t12 + 2gdy => 0 = - 2v2yt1 + gt12 + 2dy
=> v2y = (gt12 + 2dy) / 2t1 = [(9.8 m/s2)(2.029 x 10-2 s)2 + (2)(0.50 m)] / [(2)(2.029 x 10-2 s)]
= 24.74 m/s

This is the y velocity of the bullet when it hits the wall, which is related to the total travel time t according to

v2y = gt => t = v2y/g = (24.74 m/s) / (9.8 m/s2) = 2.524 s

The total vertical distance the bullet travels before hitting the wall is

H = (1/2)gt2 = (1/2)(9.8 m/s2)(2.524 s)2 = 31.22 m

The total horizontal distance the bullet travels before hitting the wall is

D + dx = vxt => D = vxt - dx = (340 m/s)(2.524 s) - 6.9 m = 851.3 m

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PROBLEM

A boy stands at the peak of a hill which slopes downward uniformly at angle f. At what angle q from the horizontal should he throw a rock so that it has the greatest range?

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 1.21]

SOLUTION

^ y
|    _
|    /| v0
|  /
|/ q
+----------> x
|\ f
|  \
|    \
|      \
|        \


If we choose the origin of our coordinate system to be at the launch point, the equation of the slope is

yslope = - x tan f (1)

The x coordinate of the rock as a function of time is

xrock = vxt = v0xt = v0t cos q (2)

where vx = v0x = v0 cos q is the x velocity of the rock, which is constant. The y coordinate of the rock as a function of time is

yrock = vy0t - (1/2)gt2 = v0t sin q - (1/2)gt2 (3)

where vy0 = v0 sin q is the initial y velocity of the rock. From (2), we see that

t = xrock / v0 cos q

Substituting into (3), we get

yrock = xrock tan q - (1/2)g(xrock / v0 cos q)2

or simply

yrock = x tan q - (1/2)g(x / v0 cos q)2 (4)

The point at which the rock hits the slope is given by equating yslope and yrock, and solving for x. Using (1) and (4),

yslope = yrock => - x tan f = x tan q - (1/2)g(x / v0 cos q)2

Dividing both sides by x,

- tan f = tan q - (1/2)g(1 / v0 cos q)2x

Solving for x,

x = (2/g)(v02 cos2 q)(tan q + tan f) = (2v02/g)(cos2 q)(tan q + tan f)

To maximize x, we take the derivative of x with respect to q, set it equal to zero, and solve for x.

dx/dq = - (2v02/g)(2 cos q sin q)(tan q + tan f) + (2v02/g)(cos2 q)(sec2 q) = 0
=> - (2 cos q sin q)(tan q + tan f) + 1 = 0
=> - 2 sin2 q - 2 cos q sin q tan f + 1 = 0 (5)

We now make use of the following trigonometric identities:

sin 2q = 2 sin q cos q (6)
cos 2q = cos2 q - sin2 q (7)
1 = cos2 q + sin2 q (8)

Subtracting (7) from (8), we get

1 - cos 2q = 2 sin2 q => sin2 q = (1 - cos 2q) / 2 (9)

Using (6) and (9), we can rewrite (5) as

- 1 + cos 2q - sin 2q tan f + 1 = 0 => cos 2q - sin 2q tan f = 0 => 1 - tan 2q tan f = 0
=> q = (1/2) tan-1(1 / tan f)

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PROBLEM

A projectile is launched at an angle of 40° with an initial velocity of 100 m/s. One hundred meters away is the beginning of a hill that slopes upward at an angle of 20°. The projectile strikes the hill a distance L up the slope. What is the value of this distance up the slope?

[from Wolf, Jonathan S. 1999, Barron's How to Prepare for the Advanced Placement Exam Physics B, 2nd Edition (Hauppauge, New York: Barron's Educational Services), free response problem 8.2]

SOLUTION


                              +
     _                      / |
     /| v0 = 100 m/s    L /   |
   /                    /     | L sin f
 / q=40°              / f=20° |
+--------------------+--------+
    d = 100 m         L cos f


The horizontal distance traveled by the projectile before impact is

d + L cos f = vxt = v0t cos q => t = (d + L cos f) / v0 cos q (1)

where

vx = v0 cos q

is the horizontal component of the velocity, a constant, and t is the time elapsed until impact.

The vertical displacement until impact is

L sin f = vy0t - (1/2)gt2 = v0t sin q - (1/2)gt2 (2)

where

vy0 = v0 sin q

is the initial velocity in the vertical direction and g = 9.8 m/s2 is the acceleration due to gravity.

Combining (1) and (2), we get

L sin f = v0t sin q - (1/2)gt2
=> L sin f = (v0 sin q)(d + L cos f) / v0 cos q
- (1/2)g[(d + L cos f) / v0 cos q]2
=> L sin f = d tan q + L cos f tan q
- (1/2)g[(d2 + 2dL cos f + L2 cos2 f) / (v02 cos2 q)]
=> L sin f = d tan q + L cos f tan q - (1/2)gd2 / (v02 cos2 q)
- (gdL cos f) / (v02 cos2 q) - (1/2)(gL2 cos2 f) / (v02 cos2 q)
=>(1/2)(gL2 cos2 f) / (v02 cos2 q) + L sin f - L cos f tan q
+ (gdL cos f) / (v02 cos2 q) + (1/2)gd2 / (v02 cos2 q) - d tan q = 0
=> aL2 + bL + c = 0 (3)

where

a = (1/2)(g cos2 f) / (v02 cos2 q)
b = sin f - cos f tan q + (gd cos f) / (v02 cos2 q)
c = (1/2)gd2 / (v02 cos2 q) - d tan q

Now

v02 cos2 q = (100 m/s)2 cos2 40° = 5868.2409 m2/s2

Thus,

a = (1/2)(9.8 m/s2)(cos2 20°) / (5868.2409 m2/s2) = 7.373 x 10-4 m-1
b = sin 20° - cos 20° tan 40° +
(9.8 m/s2)(100 m)(cos 20°) / (5868.2409 m2/s2) = - 0.2895463
c = (1/2)(9.8 m/s2)(100 m)2 / (5868.2409 m2/s2) - (100 m)(tan 40°)
= - 75.559931 m

Finally, we solve (3) using the quadratic formula. We get

L = [- b ± sqrt(b2 - 4ac)] / 2a
= [0.2895463 ± sqrt(0.28954632 - (4)(7.373 x 10-4)(- 75.559931)] /
(2)(7.373 x 10-4 m-1)
= (0.2895463 ± 0.5537855) / (1.4746 x 10-3 m-1) => 571.9 m

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Relative Velocity

PROBLEM

A train travels due south at 88.2 ft/s (relative to the ground) in a rain that is blown south by the wind. The path of each raindrop makes the angle 21.6° with the vertical, as measured by an observer stationary on the earth. An observer seated in the train, however, sees perfectly vertical tracks of rain on the window pane. Determine the speed of each raindrop relative to the earth.

[from Resnick, Robert, and Halliday, David 1966, Physics (New York: Wiley), problem 4.36]

SOLUTION

Since the raindrops appear, to the observer on the train, to be traveling in a vertical path, the horizontal component of the raindrop velocity is vx = 88.2 ft/s. The vertical and horizontal components of the raindrop velocity are related to the angle q = 21.6° by

tan q = vx/vy => vy = vx / tan q

The raindrop speed relative to the earth is

v = sqrt(vx2 + vy2) = sqrt(vx2 + vx2 / tan2 q) = vx sqrt(1 + 1 / tan2 q) = vx sqrt(1 + cot2 q)
= vx sqrt(csc2 q) = vx csc q = vx / sin q = (88.2 ft/s) / (sin 21.6°) = 239.6 ft/s

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Force and Acceleration

PROBLEM

Block C is placed on top of block A on a table. A string is attached to block A and runs to the edge of the table, over a small frictionless pulley, and is attached to block B, which hangs from the string. The weights of block A and block B are WA = 44 N and WB = 22 N.

      +---+
      | C |
      |   |
     +-----+    T
     |  A  |____>____o
     |     |        /|
--<--+-----+-----+/  |
  fs             |   |
                 |   ^ T
                 |   |
                 |   |
                 |+-----+
                 ||     |
                 ||  B  |
                 ||     |
                 |+-----+


(a) Determine the minimum weight of block C so that block A does not slide if the coefficient of static friction between block A and the table is ms = 0.20.
(b) Block C is suddenly lifted off block A. What is the acceleration of block A if the coefficient of kinetic friction between block A and the table is mk = 0.15?

[from Halliday and Resnick 1988, Fundamentals of Physics, Third Edition (New York: John Wiley & Sons), problem 6.25]

SOLUTION

(a) Block A will not slide if the force of static friction fs between the table and block A is at least as large as the tension T in the string.

fs > T (I)

The normal force between the table and block A is equal to the combined weight of blocks A and C.

N = WA + WC

The minimum weight of C so that block A does not slide occurs when the force of static friction is equal to the coefficient of static friction times the normal force, which is the maximum value of fs.

fs = msN = ms(WA + WC)

The tension in the string is equal to the weight of block B.

T = WB

Substituting into (I), we get

ms(WA + WC) > WB

Solving for WC,

WC > WB/ms - WA = (22 N) / (0.20) - 44 N = 66 N

(b) If block C is lifted off block A, the normal force between the table and block A is

N = WA

The force of kinetic friction between the table and block A is

fk = mkN = mkWA

Newton's second law (F = ma) for block A is

T - fk = (WA/g)a

or

T - mkWA = (WA/g)a (II)

Newton's second law for block B is

WB - T = (WB)/g)a (III)

Adding (II) and (III), we get

WB - mkWA = [(WA/g) + (WB/g)]a

Solve for a.

a = (WB - mkWA)g / (WA + WB) = [22 N - (0.15)(44 N)](9.8 m/s2) / (44 N + 22 N) = 2.287 m/s2

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PROBLEM

(a) What minimum force F is needed to lift the piano (mass M) using the pulley apparatus shown in the figure below? Assume that the rope and pulleys are massless and neglect friction.
(b) Determine the tension in each section of the rope.

_____________
      |
      |T3
     _|_
    / | \
   /  |  \
  |   o   |
  |\  |  /|
  | \_|_/ |
  |   |   |
T2|   |T1 |
  |   |   |
  | _ |   |
  |/ \|   |
  | + |   |
   \|/    |
    |     |
    |T4   |
    |     |
  +-+-+   |
  |   |   |
  | M |   |
  |   |   |
  +---+   v F
  piano


[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 4.80]

SOLUTION

(a) The minimum force occurs when the piano is lifted at constant speed, in which case the total forces on each pulley and on the piano are all zero. For the upper pulley, we have

T3 - T2 - T1 - F = 0 (1)

For the lower pulley, we have

T2 + T1 - T4 = 0 (2)

For the piano, we have

T4 - Mg = 0 => T4 = Mg

From (2), we get

T1 + T2 = T4 = Mg

Since there is no friction,

F = T2 = T1 => T1 + T2 = 2F = Mg => F = Mg/2

(b) T1 = T2 = Mg/2

From (1), we get

T3 = T1 + T2 + F = 3Mg/2

In part (a) we found that T4 = Mg.

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PROBLEM

Jean, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from Dulles Airport. She notices that the string makes an angle of 25° with respect to the vertical while the aircraft accelerates for takeoff, which takes about 18 seconds. Estimate the takeoff speed of the aircraft.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 4.79]

SOLUTION

+
|\
|q \
|    \ _T (tension)
|     |\
|        \
|          \
|            O
|            |
|            |
|            |
|            V mg (gravitational force)


Write Newton's second law, F = ma, in the x and y directions. In the x direction,

T sin q = ma (1)

In the y direction,

T cos q - mg = 0 => T cos q = mg (2)

Dividing (1) by (2), we get

tan q = a/g => a = g tan q

The takeoff speed vf of the aircraft is

vf = at = gt tan q = (9.8 m/s2)(18 s) tan 25° = 82.26 m/s

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PROBLEM

The system shown below uses massless pulleys and rope. The coefficient of friction between the masses and horizontal surfaces is m. Assume that M1 and M2 are sliding. Gravity is directed downward.

      +-----+               +-----+
      | M1  |----O     O----|  M2 |
______|_____|___/|     |\___|_____|______
                 |     |
                 |     |
                 |     |
                 | ___ |
                 |/   \|
                 |  o  |
                  \_|_/
                    |
                    |
                    |
                 +-----+
                 | M3  |
                 |     |
                 +-----+


(a) Draw force diagrams for each mass.
(b) How are the accelerations related?
(c) Find the tension in the rope connecting M1 and M2

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 2.15]

SOLUTION

(a) Let T1 be the tension in the rope connecting M1 and M2 and let T3 be the tension in the rope supporting M3. Then the forces on each mass are as shown in the diagram below.

         ^                     ^
         |                     |
       N1|                     | N2
         |                     |
      +-----+ T1         T1 +-----+
 f1<--| M1  |->--O     O--<-|  M2 |-->f2
______|_____|___/|     |\___|_____|______
         |       |     |       |
         |    T1 ^     ^ T1    |
     M1g v       |     |       v M2g
                 | ___ |
                 |/   \|
                 |  o  |
                  \_|_/
                    |
                    v
                    ^ T3
                    |
                 +-----+
                 | M3  |
                 |     |
                 +-----+
                    |
                    |
                    v M3g


(b) The movements of M1 to the right and of M2 to the left release rope allowing M3 to move down. M3 moves down by a distance which is half of the total length of rope made available by M1 and M2. Let (x1, x2) be the displacement of (M1, M2) to the (right, left). Let y3 be the displacement of M3 downward. Then

(x1 + x2) / 2 = y3

If we take the second time derivatives of both sides, we get the relationship between the accelerations,

(a1 + a2) / 2 = a3

(c) M1 is constrained to move in the x (horizontal) direction. If we write Newton's second law (F = ma) for both x and y directions for M1, we get

T1 - f1 = M1a1
N1 - M1g = 0

where

f1 = mN1

is the frictional force and N1 is the normal force on M1. If we write Newton's second law for both x and y directions for M2, we get

T1 - f2 = M2a2
N2 - M2g = 0

where

f2 = mN2

is the frictional force and N2 is the normal force on M2. There are only vertical forces on M3, so we write Newton's second law for M3 in the vertical direction only,

M3g - T3 = M3a3

Newton's second law for the pulley above M3 results in

2T1 - T3 = 0

since the pulley is massless.

If we replace f1 with mN1 and f2 with mN2 in the above equations, we have the following seven equations in seven unknowns a1, a2, a3, N1, N2, T1, and T3:

(a1 + a2) / 2 = a3
T1 - mN1 = M1a1
N1 - M1g = 0 => N1 = M1g (1)
T1 - mN2 = M2a2
N2 - M2g = 0 => N2 = M2g (2)
M3g - T3 = M3a3
2T1 - T3 = 0 => T3 = 2T1 (3)

Using (1), (2), and (3), we can eliminate N1, N2, and T3, and are left with four equations in four unknowns, a1, a2, a3, and T1:

(a1 + a2) / 2 = a3 (4)
T1 - mM1g = M1a1 => a1 = T1/M1 - mg (5)
T1 - mM2g = M2a2 => a2 = T1/M2 - mg (6)
M3g - 2T1 = M3a3 => a3 = g - 2T1/M3 (7)

Substituting (5), (6), and (7) into (4), we get

T1/2M1 - mg/2 + T1/2M2 - mg/2 = g - 2T1/M3
=> T1 = (g + mg) / (1/2M1 + 1/2M2 + 2/M3) = M3g(1 + m) / [(M3/2)(1/M1 + 1/M2) + 2]

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PROBLEM

A 45° wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find its acceleration.

|\ /\ m
|  \/
|    \----> A
|    q \        q = 45°
+--------\-----


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 2.16]

SOLUTION

The forces acting on m are as shown below. The normal force N is directed away from the wedge surface. The gravitational force mg is directed downward.

      _
      /| N
|\ /\/
|  \/m
|mg| \----> A
|  v q \
+--------\-----


Let a be the acceleration of m down the wedge relative to the wedge (i.e., as viewed by an observer traveling with the wedge). The acceleration of m in the +x and -y directions (i.e., we take the +x and -y directions as the directions in which ax and ay are considered positive quantities) is

ax = A + a cos q
ay = a sin q

where q = 45°. Write Newton's second law (F = ma) for the +x and -y directions.

N sin q = max = m(A + a cos q) (1)
mg - N cos q = may = ma sin q => N cos q = m(g - a sin q) (2)

Divide (1) by (2).

tan q = (A + a cos q) / (g - a sin q)
=> (tan q)(g - a sin q) = A + a cos q
=> g tan q - a sin q tan q = A + a cos q
=> a = (g tan q - A) / (sin q tan q + cos q)

Since q = 45°, sin q = cos q = 1 / sqrt(2) and tan q = 1, so

a = (g - A) / sqrt(2)

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Centripetal Force and Acceleration

PROBLEM

A jet pilot takes his aircraft in a vertical loop.
(a) If the jet is moving at a speed of 1500 km/h = 416.7 m/s at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's.
(b) Calculate the 80 kg pilot's effective weight (the force with which the seat pushes on him towards the center of the circle) at the bottom of the circle and at the top of the circle, assuming the same speed and the minimum radius found in part (a).

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 5.70]

SOLUTION

(a) The centripetal acceleration is

ac = v2/R => R = v2/ac > v2/6g = (416.7 m/s)2 / [(6)(9.8 m/s2)] = 2953 m

Thus, in order for the centripetal acceleration to not exceed 6.0 g's at the bottom of the loop, the radius of the circle must be at least 2953 m.

(b) At the bottom of the loop, the pilot experiences two forces, the force of gravity mg downward, and the normal force N that his seat exerts on him upward. According to Newton's second law, F = ma,

N - mg = mac = mv2/R

The force that the seat exerts on the pilot is

N = mg + mv2/R = m(g + v2/R) = (80 kg)[9.8 m/s2 + (416.7 m/s)2/(2953 m)] = 5488 N

At the top of the loop, the pilot experiences two forces, the force of gravity mg downward, and the normal force N that his seat exerts on him downward. According to Newton's second law,

N + mg = mac = mv2/R

The force that the seat exerts on the pilot is

N = mv2/R - mg = m(v2/R - g) = (80 kg)[(416.7 m/s)2/(2953 m) - 9.8 m/s2] = 3920 N

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PROBLEM

A circular curve of highway is designed for traffic moving at v = 60 km/h = 16.67 m/s.
(a) If the radius of the curve is R = 150 m, what is the correct angle of banking of the road?
(b) If the curve were not banked, what would be the minimum coefficient of friction between the tires and the road that would keep traffic from skidding at this speed?

[from Halliday, David, and Resnick, Robert 1988, Fundamentals of Physics, Third Edition Extended (New York: John Wiley & Sons), problem 6.52]

SOLUTION

(a) When a car is traveling at the design speed, the only two forces acting on it are the force of gravity mg and the normal force N.

     _  |    /
   N|\ q|  /
       \|/
       /|
     /  |mg
   /    |
 / q    v
+-------------
<---R--->


Writing Newton's second law (F = ma) for both the horizontal and vertical directions, we get

N sin q = mac = mv2/R (1)

where

ac = v2/R

is the centripetal acceleration, which is directed towards the center of the circle or towards the left in the above diagram, for the horizontal direction and

N cos q - mg = 0 => N cos q = mg (2)

for the vertical direction. Dividing (1) by (2), we get

tan q = v2/Rg => q = tan-1(v2/Rg) = tan-1{(16.67 m/s)2 / [(150 m)(9.8 m/s2)]} = 10.70°

(b) For an unbanked curve, we have three forces acting on a car: the force of gravity mg, the normal force N, and the friction f directed towards the center of the circle or to the left in the diagram.

       ^
       |N
       |
     +---+
  f  |   |
<----+---+-----
       |
       |mg
       v


Writing Newton's second law for the horizontal and vertical directions, we get

f = mac = mv2/R

for the horizontal direction and

N - mg = 0 => N = mg

for the vertical direction. The friction force is

f < mN = mmg => mv2/R < mmg => m > v2/Rg = (16.67 m/s)2/[(150 m)(9.8 m/s2)] = 0.1890

The coefficient of friction between the tires and the road must be at least 0.1890.

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PROBLEM

A curve of radius 68 m is banked for a design speed of 85 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve?

[from Giancoli, Douglas C. 2008, Physics for Scientists and Engineers, Fourth Edition (Upper Saddle River, New Jersey: Pearson Prentice Hall), problem 5.59]

SOLUTION

Along the curve, the car is undergoing uniform circular motion with centripetal acceleration

ac = v2/r

towards the center of the circle, or towards the right in the diagram below, where v is the velocity, and r = 68 m is the radius of curvature.

At the design speed v0 = 85 km/h = 23.61 m/s, no frictional force is needed to keep the car moving in uniform circular motion. Equivalently, the frictional force is zero, and the only forces acting on the car are the normal force N and the force of gravity mg, as shown.

|\       |    _
|  \     | q  /| N
|    \   |  /
|      \ / \
|        \ /_ _ _
|        | \ q
|      mg|   \
|        v     \
|              q \
+------------------\


Applying Newton's second law in the horizontal and vertical directions, we get

N sin q = mv02/r (1)
N cos q - mg = 0 => N cos q = mg (2)

Dividing (1) by (2),

tan q = v02/rg

When v differs from v0, a frictional force is needed to keep the car from slipping up or down the slope. When v (>, <) v0, the car would have a tendency to slip (up, down) the slope in the absence of friction, and the frictional force points (down, up) the slope. In particular, when v = (vmax, vmin), the (maximum, minimum) speed at which the car can drive on the curve without slipping, the frictional force is equal to its maximum value

f = mN

where m = 0.30 is the coefficient of static friction.

The force diagram when v > v0 is as follows:

|\       |    _
|  \     | q  /| N
|    \   |  /
|      \ / \
|        \ /_ _ _
|        | \ q
|      mg|   \| f
|        v     \
|              q \
+------------------\


The force diagram when v < v0 is as follows:

|\       |    _
|  \ _   | q  /| N
| f |\   |  /
|      \ / \
|        \ /_ _ _
|        | \ q
|      mg|   \
|        v     \
|              q \
+------------------\


When v = (vmax, vmin), Newton's second law can be written in the horizontal direction as

N sin q (+, -) f cos q = mv2/r => N sin q (+, -) mN cos q = mv2/r (1)

and in the vertical direction as

N cos q - mg (-, +) f sin q = 0 => N cos q - mg (-, +) mN sin q = 0
=> N cos q (-, +) mN sin q = mg (2)

Dividing (1) by (2),

[N sin q (+, -) mN cos q] / [N cos q (-, +) mN sin q] = v2/rg
=> [sin q (+, -) m cos q] / [cos q (-, +) m sin q] = v2/rg
=> [tan q (+, -) m] / [1 (-, +) m tan q] = v2/rg
=> v2 = rg[tan q (+, -) m] / [1 (-, +) m tan q]
= rg[v02/rg (+, -) m] / [1 (-, +) mv02/rg]
= [v02 (+, -) mrg] / [1 (-, +) mv02/rg]
= v02[1 (+, -) mrg/v02] / [1 (-, +) mv02/rg]
= v02[1 (+, -) m/(v02/rg)] / [1 (-, +) mv02/rg]
=> v = v0{[1 (+, -) m/(v02/rg)] / [1 (-, +) mv02/rg]}1/2

Now

v02/rg = (23.61 m/s)2 / [(68 m)(9.8 m/s2)] = 0.8366

Thus,

v = (23.61 m/s){[1 (+, -) 0.3/0.8366] / [1 (-, +) (0.3)(0.8366)]}1/2
= (31.80, 16.91) m/s = (vmax, vmin)

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PROBLEM

A curve of radius R = 60 m is banked for a design speed of v = 100 km/h = 27.78 m/s.
(a) What is the angle of banking?
(b) If the coefficient of static friction is m = 0.30 (wet pavement), in what range of speeds can a car safely make the curve?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 5.73]

SOLUTION

(a) If the curve is banked for a design speed of 100 km/h, a car will execute uniform circular motion at that speed in the absence of friction, and the only two forces acting on it will be the force of gravity mg and the normal force N.

     _  |    /
   N|\ q|  /
       \|/
       /|
     /  |mg
   /    |
 / q    v
+-------------
<---R--->


We set F = ma in both the horizontal and vertical directions. In the horizontal direction, we have

N sin q = mac = mv2/R (1)

where

ac = v2/R

is the centripetal acceleration, which is directed towards the center of the circle or to the left in the above diagram. In the vertical direction, we have

N cos q - mg = 0 => N cos q = mg (2)

Dividing (1) by (2), we get

tan q = v2/Rg => q = tan-1(v2/Rg) = tan-1{(27.78 m/s)2 / [(60 m)(9.8 m/s2)]} = 52.69°

The angle of banking is q = 52.69°. In the absence of friction, the car must travel at exactly 100 km/h in order to undergo circular motion. If the speed is (higher, lower) than this, the car will slide (up, down) the incline.

(b) Now if friction is present, the car can go faster or slower than 100 km/h and still undergo circular motion. The (minimum, maximum) speed at which circular motion can occur, (vmin, vmax), happens when the friction force is f = mN, its maximum value, and is directed (up, down) the incline.

In the case of the minimum speed, vmin, we have the following situation:

             _
     _  |    /|f
   N|\ q|  /
       \|/
       /|
     /  |mg
   /    |
 / q    v
+-------------
<---R--->


We set F = ma in both the horizontal and vertical directions. In the horizontal direction, we have

N sin q - f cos q = mac = mvmin2/R => N sin q - mN cos q = mvmin2/R
=> N(sin q - m cos q) = mvmin2/R (3)

In the vertical direction, we have

N cos q + f sin q - mg = 0 => N(cos q + m sin q) = mg (4)

Dividing (3) by (4), we get

(sin q - m cos q) / (cos q + m sin q) = vmin2/Rg
=> vmin = sqrt[Rg(sin q - m cos q) / (cos q + m sin q)] = sqrt[Rg(tan q - m) / (1 + m tan q)]

where the last expression is obtained by dividing top and bottom by cos q. From above, we know that tan q = v2/Rg, so

vmin = sqrt[Rg(v2/Rg - m) / (1 + mv2/Rg)] = sqrt[(v2 - mRg) / (1 + mv2/Rg)]
= sqrt{[(27.78 m/s)2 - (0.3)(60 m)(9.8 m/s2)] / [1 + (0.3)(27.78 m/s)2/(60 m)(9.8 m/s2)]}
= 20.67 m/s = 74.40 km/h

In the case of the maximum speed, vmax, we have the following situation:

             
     _  |    /
   N|\ q|  /
       \|/
       /|
   f|/  |mg
   /    |
 / q    v
+-------------
<---R--->


We set F = ma in both the horizontal and vertical directions. In the horizontal direction, we have

N sin q + f cos q = mac = mvmax2/R => N sin q + mN cos q = mvmax2/R
=> N(sin q + m cos q) = mvmax2/R (5)

In the vertical direction, we have

N cos q - f sin q - mg = 0 => N(cos q - m sin q) = mg (6)

Dividing (5) by (6), we get

(sin q + m cos q) / (cos q - m sin q) = vmax2/Rg
=> vmax = sqrt[Rg(sin q + m cos q) / (cos q - m sin q)] = sqrt[Rg(tan q + m) / (1 - m tan q)]
= sqrt[Rg(v2/Rg + m) / (1 - mv2/Rg)] = sqrt[(v2 + mRg) / (1 - mv2/Rg)]
= sqrt{[(27.78 m/s)2 + (0.3)(60 m)(9.8 m/s2)] / [1 - (0.3)(27.78 m/s)2/(60 m)(9.8 m/s2)]}
= 39.54 m/s = 142.35 km/h

So the car can safely make the curve at any speed between 74.40 km/h and 142.35 km/h.

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PROBLEM

A 1250-kg car rounds a curve of radius 72 m banked at an angle of 14°.
(a) If the car is traveling at 85 km/h, will a friction force be required? If so, how much and in what direction? Solve taking the x axis to be horizontal and the y axis to be vertical.
(b) Repeat (a) taking the x axis to be along the slope and the y axis to be perpendicular to the slope.
(c) What is the normal force that the surface exerts on the car? Under what conditions is the normal force greater than the weight of the car?
(d) If the car stops, what is the normal force that the surface exerts on the car? Under what conditions is the normal force greater than the weight of the car?

[(a) from Giancoli, Douglas C. 2008, Physics for Scientists and Engineers, Fourth Edition (Upper Saddle River, New Jersey: Pearson Prentice Hall), problem 5.84]

SOLUTION

(a) The car is undergoing uniform circular motion with centripetal acceleration

ac = v2/r

towards the center of the circle, or towards the right in the diagram below, where v = 85 km/h = 23.61 m/s is the velocity, and r = 72 m is the radius of curvature.

Assume that a friction force f is required and that it is directed down the slope.

|\       |    _
|  \     | q  /| N
|    \   |  /
|      \ / \
|        \ /_ _ _
|        | \ q
|      mg|   \| f
|        v     \
|              q \
+------------------\


From Newton's second law of motion, the sum of the vertical forces on the car is zero, so

N cos q - mg - f sin q = 0 => N cos q = mg + f sin q (1)

where N is the normal force between the road surface and the car, m = 1250 kg is the mass of the car, g = 9.8 m/s2 is the acceleration of gravity, and q = 14°.

From Newton's second law of motion, the sum of the horizontal forces on the car is equal to the mass m times the centripetal acceleration ac, or

N sin q + f cos q = mv2/r => N sin q = - f cos q + mv2/r (2)

Dividing (2) by (1),

tan q = (- f cos q + mv2/r) / (mg + f sin q)
=> mg tan q + f sin q tan q = - f cos q + mv2/r
=> mg sin q / cos q + f sin2 q / cos q = - f cos q + mv2/r
=> mg sin q + f sin2 q = - f cos2 q + (mv2/r) cos q
=> f sin2 q + f cos2 q = (mv2/r) cos q - mg sin q
=> f(sin2 q + cos2 q) = (mv2/r) cos q - mg sin q
=> f = (mv2/r) cos q - mg sin q = mg[(v2/rg) cos q - sin q]
= (1250 kg)(9.8 m/s2){[(23.61 m/s)2 / (72 m)(9.8 m/s2)] cos 14° - sin 14°}
= 6428 N

Thus, friction is required and is equal to 6428 N down the slope. Note that if we had assumed that the friction force f was directed up the slope, we would have obtained a negative value for f, indicating that it was in the opposite direction.

(b) Now suppose we choose the +x direction to be down the slope and the +y direction to be perpendicular to the slope, in the same direction as the normal force.

|\       |    _
|  \     | q  /| N
|    \   |  /
|      \ / \
|        \ /_ _ _
|        | \ q
|      mg|   \| f
|        v     \
|              q \
+------------------\


Newton's second law of motion in the x direction is

mg sin q + f = (mv2/r) cos q
=> f = (mv2/r) cos q - mg sin q = mg[(v2/rg) cos q - sin q]
= (1250 kg)(9.8 m/s2){[(23.61 m/s)2 / (72 m)(9.8 m/s2)] cos 14° - sin 14°}
= 6428 N

Note that this choice of axes allowed us to write a single equation where f was the only unknown and then solve for f by plugging in the given values.

(c) Using the same choice of axes as in part (b), Newton's second law of motion in the y direction is

N - mg cos q = (mv2/r) sin q
=> N = mg cos q + (mv2/r) sin q = mg[cos q + (v2/rg) sin q]
= (1250 kg)(9.8 m/s2){cos 14° + [(23.61 m/s)2/(72 m)(9.8 m/s2)] sin 14°}
= 14,228 N

(d) The weight of the car is mg. The normal force will be greater than mg if

cos q + (v2/rg) sin q > 1

With the given values of q, v, and r,

cos q + (v2/rg) sin q = cos 14° + [(23.61 m/s)2/(72 m)(9.8 m/s2)] sin 14°
= 1.161 > 1

so here N > mg.

(d) If the car stops, v = 0, and

N = mg cos q < mg

so the normal force can never be greater than the weight of the car. This situation is the same as an ordinary inclined plane, where there is no circular motion.

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PROBLEM

A mass m is connected to a vertical revolving axle by two strings of length L, each making an angle of 45° with the axle, as shown. Both the axle and mass are revolving with angular velocity w. Gravity is directed downward.

||   ^
||   | w
||\  |
|| \
||q \ L
||   \
||    \
||     \
||      o m     q = 45°
||     /
||    /
||   /
||q / L
|| /
||/
||
||


(a) Draw a clear force diagram for m.
(b) Find the tension in the upper string, Tup, and the lower string, Tlow.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 2.11]

SOLUTION

(a) The forces on the mass are as shown below. There are the two tensions, Tup and Tlow, and the gravitational force mg.

||   ^
||   | w
||\  |
|| \
||q \ 
||  |\ Tup
||   L\
||     \
||    m o       q = 45°
||     /| mg
||   L/ v
||  |/ Tlow
||q /
|| /
||/
||
||


(b) Write Newton's second law (F = ma) in the radial and vertical directions.

Tup sin q + Tlow sin q = mac = mv2/R = mRw2 = mLw2 sin q => Tup + Tlow = mLw2 (1)
Tup cos q - Tlow cos q - mg = 0 => Tup - Tlow = mg / cos q (2)

where ac is the centripetal acceleration and

R = L sin q

is the distance from the axle to the mass. Add (1) and (2).

2Tup = mLw2 + mg / cos q => Tup = (m/2)(Lw2 + g / cos q) = (m/2)[Lw2 + g sqrt(2)]

Use (1) to solve for Tlow.

Tlow = mLw2 - Tup = (m/2)[Lw2 - g sqrt(2)]

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Work and Energy

Suppose that forces F1(r), F2(r), F3(r),…, FN(r) act on an object located at position r, where r varies (i.e., the object is moving). The work done by each force on the object is

W1 = ∫ F1 • dr
W2 = ∫ F2 • dr
W3 = ∫ F3 • dr

WN = ∫ FN • dr

where each integral is over the path taken by the object. The total or net work done on the object is

Wtotal = Wnet = W1 + W2 + W3 + … + WN

When we have a system consisting of a number of objects, we can consider internal forces between the objects within the system, and external forces exerted between something outside of the system and the objects in the system. An external force could be due to a field, such a gravitational or electric field. The work done on the objects by the external forces is called the external work done on the system. A system could consist of a single object.

If work is done on an object by a conservative force such as gravity, the change in the potential energy related to the conservative force is equal to the negative of the work done on the object by the conservative force. Thus, for example, if an object of mass m falls freely a distance h near the surface of the earth under the influence of gravity, the work done by gravity is

Wg = mgh

and the change in the gravitational potential energy is

DPEg = - Wg = - mgh

The work-energy principle states that the net work done on an object is equal to the change in the object’s kinetic energy.

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PROBLEM

(a) Determine the work a hiker must do on a 15.0 kg backpack to carry it up a hill of height h = 10.0 m. Determine also (b) the work done by gravity on the backpack, and (c) the net work done on the backpack. (d) What is the change in the potential energy of the backpack? (e) Show that the work-energy principle is satisfied.

[(a), (b), and (c) from Giancoli, Douglas C. 2008, Physics for Scientists and Engineers, Fourth Edition (Upper Saddle River, New Jersey: Pearson Prentice Hall), example 7.2]

SOLUTION

Let q be the angle between the vertical and the slope.

           /|
         / q|
       /    |
     /      | h
   /        |
 /          |
-------------


The component of the gravitational force along the slope is

Fg = mg cos q

which is directed down the slope. The hiker must exert the same force up the slope on the backpack to keep it moving at constant velocity up the slope. Thus, the hiker’s force on the backpack is

Fh = mg cos q

The distance the backpack moves up the slope is related to h according to

cos q = h/d => d = h / cos q

Thus, the work done by the hiker is

Wh = Fhd = (mg cos q)(h / cos q) = mgh = (15.0 kg)(9.8 m/s2)(10.0 m) = 1470 J

(b) The work done by gravity on the backpack is

Wg = - Fgd = - (mg cos q)(h / cos q) = - mgh = - 1470 J

(c) The net work done on the backpack is

Wnet = Wh + Wg = 0

(d) The change in potential energy is

DPE = mgh = 1470 J

which is the negative of the work done by gravity.

(e) According to the work-energy principle, the net work done on an object is equal to the change in its kinetic energy. The initial and final kinetic energies of the backpack are both zero since the backpack begins and ends at rest. Since the net work done on the backpack is zero, the work-energy principle is satisfied.

Note that we have not explicitly considered the work done by the hiker to accelerate the backpack at the beginning and decelerate it at the end. In fact the work done by the hiker to accelerate the backpack to some constant velocity at the beginning is equal and opposite to the work done by the hiker to decelerate the backpack to rest at the end, so the two amounts of work cancel out.

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PROBLEM

A paratrooper fell from a height of H = 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater h = 1.1 m deep, but survived with only minor injuries. Assuming the paratrooper's mass was m = 80 kg and his terminal velocity was v = 30 m/s, estimate: (a) the work done by the snow in bringing him to rest; (b) the average force exerted on him by the snow to stop him; and (c) the work done on him by air resistance as he fell.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 6.71]

SOLUTION

(a) The paratrooper's kinetic energy just before hitting the ground is

KE = (1/2)mv2 = (1/2)(80 kg)(30 m/s)2 = 36000 J

The paratrooper's potential energy just before hitting the ground, relative to the bottom of the crater, is

PE = mgh = (80 kg)(9.8 m/s2)(1.1 m) = 862.4 J

The paratrooper's total energy just before hitting the ground is

E = KE + PE = 36000 J + 862.4 J = 36862.4 J

When the paratrooper reaches the bottom of the crater and comes to rest, his total energy is zero. The work done by the snow in bringing him to rest is Wsnow = - 36862.4 J. If we neglect the potential energy that the paratrooper had just before he hit the ground, the work done by the snow in bringing him to rest is Wsnow = - 36000 J.

(b) The work done by the snow in bringing the paratrooper to rest is

Wsnow = - Fsnowh

where Fsnow is the average force exerted on the paratrooper by the snow. The negative sign is used because the force and the displacement are in opposite directions.

Fsnow = - Wsnow/h = (36000 J) / (1.1 m) = 32727 N

(c) At the instant the paratrooper begins to fall, his potential energy relative to the ground is

PE = mgH = (80 kg)(9.8 m/s2)(370 m) = 290,080 J

At the instant he hits the ground, his kinetic energy is 36000 J. The work done on him by air resistance as he fell is

Wair = 36000 J - 290,080 J = - 254,080 J

Air resistance did - 254,080 J of work on him, reducing his energy from 290,080 J to 36000 J.

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PROBLEM

A student whose mass is m = 75 kg runs at a speed of vi = 5.0 m/s, grabs a rope of length L = 10.0 m hanging from a tree, and swings out over a lake. He releases the rope when his velocity is zero.
(a) What is the angle q that the rope makes with the vertical when he releases the rope?
(b) What is the tension in the rope just before he releases it?
(c) What is the maximum tension in the rope?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 6.85]

SOLUTION

(a) The student's energy is his initial kinetic energy,

E = KEi = (1/2)mvi2

When he releases the rope, his energy is all potential energy,

E = PEf = mgL(1 - cos q)

where L(1 - cos q) is the height of the student when he releases the rope. Because energy is conserved, we have

KEi = PEf => (1/2)mvi2 = mgL(1 - cos q) => q = arccos[1 - (vi2/2gL)]
= arccos{1 - (5.0 m/s)2/[(2)(9.8 m/s2)(10.0 m)]} = arccos(0.8724) = 29.26 deg

(b) Just before the student releases the rope, the centripetal acceleration is zero because the velocity is zero, and the tension T must balance the component of gravity along the direction of the rope.

Fc = T - mg cos q = 0 => T = mg cos q = (75 kg)(9.8 m/s2) cos 29.26 = 641.25 N

(c) In general, the centripetal force is equal to the tension T minus the component of gravity along the direction of the rope.

Fc = T - mg cos q

The centripetal force is also equal to the mass times the centripetal acceleration.

Fc = mac = mv2/L

Equating the two expressions, we have

T - mg cos q = mv2/L

Solving for T,

T = mg cos q + mv2/L

Both the first and the second terms are maximized when the student initially grabs the rope, where q = 0 deg and v = 5.0 m/s. Thus, the maximum tension is

Tmax = (75 kg)(9.8 m/s2) cos 0 + (75 kg)(5.0 m/s)2/(10.0 m) = 922.5 N

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PROBLEM

A chain of mass M and length L is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, x, has fallen? Neglect the size of the individual links.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 4.11]

SOLUTION

The force exerted on the scale by the chain is of the form

F = F1 + F2

where

F1 = xlg

is the weight of the part of the chain that has already fallen onto the scale,

l = M/L

is the mass per unit length,

F2 = dp/dt = v dm/dt = vl dx/dt = v2l

is the rate dp/dt at which momentum is imparted to the scale by the falling chain, and v is the velocity of the falling part of the chain. From one-dimensional kinematics,

v2 = 2gx

so

F = xlg + v2l = xlg + 2gxl = 3xlg = 3(x/L)Mg

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Power

PROBLEM

A truck can move up a road having a grade of 1.0 ft rise every 50 ft with a speed of 15 miles/hr. The resisting force is equal to 1/25 the weight of the truck. How fast will the same truck move down the hill with the same horse power?

[from Resnick, Robert, and Halliday, David 1966, Physics (New York: Wiley), problem 7.22]

SOLUTION

            _ Fup
            /|  /
          /   /
        / \ /
      /   / truck moving up the slope
    / \ /|
f|/   /  |
    /    V mg
  /  q
/_________________


            _ f
            /|  /
          /   /
        / \ /
Fdown /   / truck moving down the slope
    / \ /|
 |/   /  |
    /    V mg
  /  q
/_________________


The resisting force f is the frictional force that would slow down the truck even if it was moving on level ground and acts in the direction opposite the direction of motion. Let Fup be the force that the motor exerts on the truck when it is moving up the slope. Then the sum of the forces on the truck along the slope is zero:

Fup - f - mg sin q = 0 => Fup = f + mg sin q = mg(1/25 + sin q) = mg(1/25 + 1/50) = 3mg/50

Now when the truck is moving down the slope, the resisting force is directed up the slope. Let Fdown be the force that the motor exerts on the truck when it is moving down the slope. Then the sum of the forces on the truck along the slope is again zero:

Fdown + mg sin q - f = 0 => Fdown = f - mg sin q = mg(1/25 - sin q) = mg(1/25 - 1/50) = mg/50

In each case, the power of the motor is the same and is given by the product of the force and the velocity:

Pup = Pdown => Fupvup = Fdownvdown => (3mg/50)vup = (mg/50)vdown
=> vdown = 3vup = (3)(15 miles/hr) = 45 miles/hr

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Linear Momentum

PROBLEM

A ball of mass m traveling at speed v0 (call this ball 1) collides head-on with another ball, of mass M, which is initially at rest (call this ball 2).
(a) Calculate the final velocities of the two balls if the collision is completely elastic.
(b) Show that your result reduces to the expected answers when (i) m << M, (ii) m >> M, and (iii) m = M.

SOLUTION

Assume that ball 1 is initially traveling in the +x direction and that after the collision it moves in the -x direction with speed v1 and ball 2 moves in the +x direction with speed v2. The initial momentum and kinetic energy are

pi = mv0

and

KEi = (1/2)mv02

The final total momentum and total kinetic energy are

pf = Mv2 - mv1

and

KEf = (1/2)mv12 + (1/2)Mv22

By conservation of momentum, we have

pi = pf => mv0 = Mv2 - mv1 => v0 = (M/m)v2 - v1 => v1 = (M/m)v2 - v0

Because the collision is completely elastic, the initial and final kinetic energies are equal.

KEi = KEf => (1/2)mv02 = (1/2)mv12 + (1/2)Mv22 => v02 = v12 + (M/m)v22

Replacing v1 with the above expression in terms of v0 and v2, we get

v02 = [(M/m)v2 - v0]2 + (M/m)v22 = (M/m)2v22 - 2(M/m)v0v2 + v02 + (M/m)v22

Subtracting v02 from both sides,

0 = (M/m)2v22 - 2(M/m)v0v2 + (M/m)v22 => 0 = (M/m)v2 - 2v0 + v2 = (1 + M/m)v2 - 2v0

Solving for v2,

v2 = 2v0/(1 + M/m) => v1 = 2(M/m)v0/(1 + M/m) - v0 = v0[2/(1 + m/M) - 1]

(b) (i) When m << M, we can neglect m/M relative to 1 and 1 relative to M/m, resulting in v1 ≈ v0 and v2 ≈ 2v0(m/M) << v0. Ball 1 rebounds with approximately the same speed it had and ball 2 barely moves.

Since ball 2 is much more massive, this is like having ball 1 hit a wall and bounce off elastically with the same speed it had initially.

(ii) When m >> M, we can neglect M/m relative to 1 and 1 relative to m/M, resulting in v1 ≈ - v0 and v2 ≈ 2v0. Ball 1 continues to move in the +x direction at approximately the same speed and ball 2 moves in the same direction at approximately twice the speed that ball 1 initially had.

In the reference frame moving in the +x direction at speed v0 (call this the moving reference frame), ball 1 is initially at rest and ball 2 is moving in the -x direction at speed v0. Since ball 1 is much more massive, this is like having ball 2 hit a wall and bounce off elastically with the same speed it had initially. Thus, in the moving reference frame, ball 2 moves in the +x direction at speed v0 after the collision, and in the laboratory frame, ball 2 moves in the +x direction at speed v0 + v0 = 2v0.

(iii) When m = M, v1 = 0 and v2 = v0. Ball 1 stops and ball 2 moves in the +x direction with the same speed ball 1 had before the collision.

In the center-of-mass frame, which is moving in the +x direction at speed v0/2, before the collision ball 1 moves in the +x direction at speed v0/2 and ball 2 moves in the -x direction at speed v0/2. After the collision, ball 1 moves in the -x direction at speed v0/2 and ball 2 moves in the +x direction at speed v0/2. Thus, in the lab frame, after the collision ball 1 stops and ball 2 moves in the +x direction at speed v0.

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PROBLEM

Estimate the terminal velocity of a person of mass m who jumps off an airplane with a massless parachute of radius R. Assume that m = 70 kg, R = 4 m, and the air density is 1.21 kg/m3.

SOLUTION

The terminal velocity is attained when the weight mg of the person equals the force of air resistance on the parachute. Assume that the parachute is circular and that the air which hits it is deflected sideways. Let r be the density of air. The force of air resistance is

Fair = v dm/dt = v rpR2 dy/dt = v rpR2v = v2rpR2

where

dm/dt = rpR2 dy/dt = rpR2v

is the rate at which the air hits the parachute. We have

mg = Fair = v2rpR2
=> v = sqrt(mg/rpR2) = sqrt{(70 kg)(9.8 m/s2) / [(1.21 kg/m3)p(4 m)2]} = 3.358 m/s

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PROBLEM

Material is blown into cart A from cart B at a rate b kilograms per second. The material leaves the chute vertically downward, so that it has the same horizontal velocity as cart B, u. At the moment of interest, cart A has mass M and velocity v. Find dv/dt, the instantaneous acceleration of A.

              +-------------------+
              |ooooooo <-- ooooooo|
    v         |o+---------------+o|        u
 A --->       |o|               |o|       ---> B
+-------------|o|----+     +----|o|-------------+
|             ooo    |     |    |o|             |
|     M      ooooo   |     |oooooooooooooooooooo|
|           ooooooo  |     |oooooooooooooooooooo|
+--------------------+     +--------------------+
     O          O               O          O     
/////////////////////////////////////////////////


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 3.11]

SOLUTION

The total momentum of the system at time t is

p(t) = MA(t)v(t) + MB(t)u

where MA(t) is the mass of car A, MB(t) is the mass of car B, and v(t) is the velocity of car A, all at time t. The total momentum of the system at time t + dt is

p(t + dt) = [MA(t) + dm][v + dv] + [MB(t) - dm]u

Because the total momentum of the system is conserved,

p(t) = p(t + dt)
=> MAv + MBu = (MA + dm)(v + dv) + (MB - dm)u
= MAv + MAdv + v dm + dm dv + MBu - u dm
=> 0 = MAdv + v dm + dm dv - u dm

Neglecting the term dm dv and using

dm = b dt

we get

0 = MAdv + vb dt - ub dt => 0 = MAdv/dt + vb - ub => dv/dt = (u - v)b/MA = (u - v)b/M

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PROBLEM

A sand-spraying locomotive sprays sand horizontally into a freight car situated ahead of it. The locomotive and freight car are not attached. The engineer in the locomotive maintains his speed so that the distance to the freight car is constant. The sand is transferred at a rate dm/dt = 10 kg/s with a velocity of 5 m/s relative to the locomotive. The car starts from rest with an initial mass of 2000 kg. Find its speed after 100 s.

|-----\
|      \----------+ ---> +-----------------+
|   locomotive    |::::::|   freight car   |
|               +-+      +-+               |
+---------------+          +---------------+
  OO    OO   OO              OO         OO  
////////////////////////////////////////////


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 3.12]

SOLUTION

Since the locomotive and freight car remain the same distance apart, they must travel at the same speed. Define

v(t) = velocity of locomotive and freight car at time t
u = 5 m/s = velocity of sprayed sand relative to locomotive
b = dm/dt = 10 kg/s = rate at which sand is transferred
m(t) = m0 + bt = mass of freight car at time t
m0 = 2000 kg = mass of freight car at time t = 0

The momentum of the freight car at time t is

p(t) = m(t)v(t) = (m0 + bt)v(t)

The momentum of the freight car at time t + dt is

p(t + dt) = m(t + dt)v(t + dt) = (m0 + bt + b dt)v(t + dt)

The rate at which the freight car momentum changes is

dp/dt = lim(dt => 0) {[p(t + dt) - p(t)] / dt}
= lim(dt => 0) {[m0v(t + dt) + btv(t + dt) + b dt v(t + dt) - m0v(t) - btv(t)] / dt}
= m0 dv/dt + bt dv/dt + bv(t)

But since the momentum change is due to the sand being sprayed onto the freight car,

dp/dt = (v + u)b

Equating the two expressions for dp/dt, we have

m0 dv/dt + bt dv/dt + bv(t) = (v + u)b
=> m0 dv/dt + bt dv/dt = ub
=> dv/dt = ub / (m0 + bt) = (ub/m0) / (1 + bt/m0)
=> dv = (ub/m0) dt / (1 + bt/m0)
=> ∫ dv = ∫ (ub/m0) dt / (1 + bt/m0) = u ∫ (b/m0) dt / (1 + bt/m0)
=> v(t) = u ln(1 + bt/m0) + C

Setting t = 0,

v(0) = u ln(1) + C = C = 0
=> v(t) = u ln(1 + bt/m0)
=> v(100 s) = (5 m/s) ln[1 + (10 kg/s)(100 s)/(2000 kg)] = 2.027 m/s

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PROBLEM

An inverted garbage can of mass M is suspended in air by water from a geyser. The water shoots up from the ground with a speed v0, at a constant rate dm/dt.
(a) Find a general formula for the maximum height hmax at which the garbage can rides. What assumption must be fulfilled for the maximum height to be reached?
(b) Find hmax if v0 = 20 m/s, M = 10 kg, and dm/dt = 6 kg/s.
(c) Find hmax if v0 = 20 m/s, Mg = 8 N, and dm/dt = 0.5 kg/s.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 3.17]

SOLUTION

Assume that when the water hits the can it is traveling straight up and is reflected directly downward. The rate at which momentum is imparted to the can by the water hitting it is

dp/dt = 2v dm/dt

where v is the velocity of the water when it hits the can. Let y be the height of the bottom of the can, which is actually at the top here because the can is inverted. Then

v2 = v02 - 2gy => v = sqrt(v02 - 2gy) => dp/dt = 2(dm/dt)sqrt(v02 - 2gy)

The can will ride at the height for which the weight Mg is balanced by the force of the water on the can, so

Mg = dp/dt = 2(dm/dt)sqrt(v02 - 2gy) => y = (1/2g)(v02 - {Mg / [2(dm/dt)]}2) = hmax

(b) hmax = {1/[(2)(9.8 m/s2)]}{(20 m/s)2 - [(10 kg)(9.8 m/s2) / (2)(6 kg/s)]2} = 17.01 m

(c) hmax = {1/[(2)(9.8 m/s2)]}{(20 m/s)2 - [(8 N) / (2)(0.5 kg/s)]2} = 17.14 m

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Rotational Kinematics

PROBLEM

An insect of mass 8.0 x 10-2 g walks out with a constant speed of 1.6 cm/s along a radial line marked on a phonograph turntable rotating at a constant angular velocity of 33 1/3 rev/min.
(a) Find the velocity and acceleration of the insect as seen by the ground observer when the insect is 12 cm from the axis of rotation.
(b) What must the minimum coefficient of friction be to allow the insect to get all the way to the edge of the turntable (radius = 16 cm) without slipping?
(c) In what way are the results of (a) and (b) changed if the turntable is given an angular acceleration that increases its angular velocity?

[from Resnick, Robert, and Halliday, David 1966, Physics (New York: Wiley), problem 11.20]

SOLUTION

(a) The velocity of the insect is

v = vrr + vqq = vrr + rwq

where vr is a radial component of the linear velocity, r is the radial coordinate,

vq = rw

is the angular component of the linear velocity, and

w = (33 1/3 rev/min)(2p rad/rev)(1 min / 60 s) = 3.491 rad/s

is the angular velocity. At the instant described, vr = 1.6 cm/s and

vq = rw = (12 cm)(3.491 rad/s) = 41.89 cm/s

The total linear velocity is

v = sqrt(vr2 + vq2) = sqrt[(1.6 cm/s)2 + (41.89 cm/s)2] = 41.92 cm/s

The linear acceleration of the insect is

a = arr + aqq

where

ar = - vq2/r = - rw2 = - (12 cm)(3.491 rad/s)2 = - 146.2 cm/s2

is the radial component of the linear acceleration and

aq = (d/dt)rw = w(dr/dt) = (3.491 rad/s)(1.6 cm/s) = 5.585 cm/s2

is the angular component of the linear acceleration. The total acceleration is

a = sqrt(ar2 + aq2) = sqrt[(- 146.2 cm/s2)2 + (5.585 cm/s2)2] = 146.3 cm/s2

(b) The general formula for the acceleration is

a = sqrt(ar2 + aq2) = sqrt[(rw2)2 + w2(dr/dt)2] = sqrt[r2w4 + w2(dr/dt)2]

The maximum acceleration occurs when r = 16 cm and is

amax = sqrt[(16 cm)2(3.491 rad/s)4 + (3.491 rad/s)2(1.6 cm/s)2] = 195.0 cm/s2

Assume that the bug walks without slipping. The static friction force fs must be sufficient to produce this acceleration, so

fs = mamax < mN = msmg => ms > amax/g = (195.0 cm/s2) / (980 cm/s2) = 0.1990

where

N = mg

is the normal force between the turntable and the bug, m is the mass of the bug, and ms is the coefficient of static friction, which must be at least 0.1990.

(c) If the turntable is given an angular acceleration that increases its angular velocity, the bug's linear acceleration increases at each value of r, both in the radial and the angular directions, and the minimum coefficient of static friction that is necessary also increases.

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Rotational Dynamics

Motorized Car Accelerating. A motorized car is a self-propelled vehicle, in contrast with something like a cart which is pulled by a person or another external force. When a car is driven forward from rest, as viewed from the right side of the car, the motor is attempting to make the wheels turn clockwise. If the car has front-wheel drive, the front wheels are being made to turn clockwise. If the car has four-wheel drive, all four wheels are being made to turn clockwise. If there were no friction between the tires and the road, these wheels would simply spin clockwise, and the car would otherwise remain at rest. Because there is friction, the tires exert a backward force on the road. By Newton's third law, the road exerts an equal and opposite force on the tires in the forward direction. If the car has two-wheel drive, the undriven wheels also turn clockwise, but solely from the friction force between the road and the wheels, which is in the backward direction.

Motorized Car Braking. A typical car has brake pads on all four wheels even if the car has only two-wheel drive. When a car is moving forward and braking or decelerating, as viewed from the right side of the car, the brakes are attempting to make the wheels reduce their clockwise angular velocity. The wheels transfer the linear momentum of the car to the road by exerting a forward force on the road. By Newton's third law, the road exerts an equal and opposite force on the tires in the backward direction.

Cylinder or Sphere Rolling without Slipping down an Incline. A cylinder or sphere which is released at the top of an incline, under the influence of gravity, starts rolling down the incline. The rolling is caused by the friction force between the cylinder or sphere and the incline, which is directed up the slope.

Relationship between Angular Momentum and Angular Velocity. Finite rotations do not commute and cannot be represented by vectors. Infinitesimal rotations do commute and can be represented by vectors. Angular velocity is a result of infinitesimal rotations and can be represented by a vector. The relationship between the linear velocity and angular velocity of a particle is

v = w x r

The relationship between the angular momentum and angular velocity of a rotating rigid body is

+--+   +-         -+ +--+
|Lx|   |Ixx Ixy Ixz| |wx|
|Ly| = |Iyx Iyy Iyz| |wy|
|Lz|   |Izx Izy Izz| |wz|
+--+   +-         -+ +--+


where

Ixx = ∫ (y2 + z2) dm
Iyy = ∫ (x2 + z2) dm
Izz = ∫ (x2 + y2) dm

are moments of inertia and

Ixy = Iyx = - ∫ xy dm
Iyz = Izy = - ∫ yz dm
Ixz = Izx = - ∫ xz dm

are products of inertia.

Torque. The torque on a particle at position r, acted on by a force F, is defined as

t = r x F

  | i  j  k  |
= | x  y  z  |
  | Fx Fy Fz |


= i(yFz - zFy) - j(xFz - zFx) + k(xFy - yFx)

If z = 0 (the particle is in the x-y plane), and Fz = 0 (the force has no z component),

t = k(xFy - yFx)

and the torque only has a component in the z direction. If the particle was initially at rest, its resulting motion will be in the x-y plane (z = 0).

Suppose the force F is applied to a rigid body which is constrained to rotate about the z axis, for example, by an axle running through the rigid body and lying along the z axis. If the force is applied to the rigid body at position r, the torque on the rigid body is calculated to be

t = i(yFz - zFy) - j(xFz - zFx) + k(xFy - yFx)

Because the rigid body is constrained to rotate about the z axis, the only component of the torque which can do anything is the z component,

tz = xFy - yFx

Regardless of the position r where the force F is applied, the only components of r and F which affect the resulting motion are their projections onto the x-y plane. Thus, we can treat the system as if the rigid body and the force F (or any other forces acting on the rigid body) are collapsed onto the x-y plane, and the system behaves like a two-dimensional problem, at least as far as the resulting rotational motion is concerned.

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PROBLEM

Two masses, m1 = 18.0 kg and m2 = 26.5 kg, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius R = 0.260 m and mass M = 7.50 kg. The two masses hang vertically from opposite sides of the pulley. Initially, m1 is on the ground and m2 rests h = 3.00 m above the ground. If the system is now released, use conservation of energy to determine the speed of m2 just before it strikes the ground. Assume the pulley is frictionless so that no energy is lost from the system, but that the pulley turns as the rope moves without any slipping between the pulley and the rope.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 8.55]

SOLUTION

Initially, the kinetic energy of the system is

KEi = 0

and the potential energy is

PEi = m2gh

Let v be the velocity of m2 just before it hits the ground. Since m1 and m2 are connected by a rope, v is also the velocity of m1. Since the rope and pulley move together without slipping, v is also the linear velocity at the outer edge of the pulley.

The kinetic energy of the system is

KEf = (1/2)m1v2 + (1/2)m2v2 + (1/2)Iw2

where I is the moment of inertia of the pulley about its axis and w is its angular velocity. The angular velocity is related to the linear velocity according to

w = v/R

The moment of inertia of the pulley about its axis is

I = (1/2)MR2

Thus, the kinetic energy of the system is

KEf = (1/2)m1v2 + (1/2)m2v2 + (1/2)(1/2)MR2v2/R2 = [(1/2)m1 + (1/2)m2 + (1/4)M]v2

The potential energy of the system is

PEf = m1gh

Since the total initial and final energies of the system are equal,

KEi + PEi = KEf + PEf
=>0 + m2gh = [(1/2)m1 + (1/2)m2 + (1/4)M]v2 + m1gh
=>(m2 - m1)gh = [(1/2)m1 + (1/2)m2 + (1/4)M]v2

Solving for v, we get

v = sqrt{(m2 - m1)gh / [(1/2)m1 + (1/2)m2 + (1/4)M]}
= sqrt{(26.5 kg - 18.0 kg)(9.8 m/s2)(3.00 m) / [(1/2)(18.0 kg) + (1/2)(26.5 kg) + (1/4)(7.50 kg)]}
= 3.218 m/s

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PROBLEM

A wheel of mass M has radius R. It is standing vertically on the floor, and we want to exert a horizontal force F at its axle so that it will climb a step against which it rests. The step has height h, where h < R. What minimum force F is needed?
       ___
     /     \
   /         \
  |           |
  |     o-----|--->F |
  |     |q+.+.|      v
   \   R|    /+---------
_____\__|__/__|      h
        |            ^
        v Mg         |


[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 8.81]

SOLUTION

In order for the wheel to climb the step, the net torque t about the point of contact at the edge of the step must be greater than zero, into the page. The two forces which contribute to the torque are the applied force F and the gravitational force Mg:

t = RF sin qF - RMg sin qMg = RF cos q - RMg sin q

where q is the angle between the radius to the point of contact on the floor and the radius to the point of contact at the edge of the step. We have

cos q = (R - h) / R => sin q = sqrt(1 - cos2 q) = sqrt[1 - (R - h)2/R2]
=> tan q = (sin q) / (cos q) = sqrt[1 - (R - h)2/R2] / [(R - h) / R] = sqrt[R2 / (R - h)2 - 1]

Thus,

t = RF cos q - RMg sin q > 0 => F > Mg tan q = Mg sqrt[R2 / (R - h)2 - 1]

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PROBLEM

A force F is applied to a dumbbell for a time interval t, first as in (i) and then as in (ii). In which case does the dumbbell acquire the greater center-of-mass speed?

(i)          (ii)
      _              _
 F   / \            / \
--->|   |          |   |
     \_/            \_/
      |              |
      |           F  |
      |          --->|
      |              |
      |              |
     / \            / \
    |   |          |   |
     \_/            \_/


[from Physics E-1a, Principles of Physics I: Mechanics, Final Exam, Fall 2001, Harvard Extension School]

SOLUTION

We assume that the time interval t is small enough so that the dumbbell doesn't change its position or orientation significantly while the force is acting on it.

In both cases, the total momentum imparted to the dumbbell is the same, equal to Ft. The momentum of the dumbbell is equal to the product of its mass and its center-of-mass velocity and is independent of how fast the dumbbell is rotating. Therefore, the center-of-mass velocity is the same in both cases.

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PROBLEM

A force F is applied to a dumbbell over the same distance d, first as in (i) and then as in (ii). In which case does the dumbbell acquire the greater center-of-mass speed?

(i)          (ii)
      _              _
 F   / \            / \
--->|   |          |   |
     \_/            \_/
      |              |
      |           F  |
      |          --->|
      |              |
      |              |
     / \            / \
    |   |          |   |
     \_/            \_/


[from Physics E-1a, Principles of Physics I: Mechanics, Final Exam, Fall 2003, Harvard Extension School]

SOLUTION

We assume that the distance d is small enough so that the dumbbell doesn't change its position or orientation significantly while the force is acting on it.

In both cases, the total energy imparted to the dumbbell is the same, equal to Fd, which is the work done by the force. The total energy of the dumbbell is equal to the sum of its translational kinetic energy and its rotational kinetic energy. By symmetry, the dumbbell doesn't rotate when the force is applied at its center. But when the force is applied at one of the ends of the dumbbell, it starts to rotate. Since, in case (i), some of the kinetic energy is rotational, the translational kinetic energy is less than in case (ii), so the center-of-mass velocity is greater in case (ii).

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PROBLEM

A solid disk of mass M and radius R is free to roll on a horizontal surface. A force F is applied at the center of the disk in the +x direction. Determine (a) the acceleration of the disk, (b) the frictional force on the disk, and (c) the maximum value of F for which the disk will not slip if the coefficient of static friction is ms.

SOLUTION

        ___
       /   \
     /       \
    |         |
    |    +----|----> F
    |         |
     \       /
___/___\___/________
   \     C
   f


(a) There are two forces in the horizontal (x) direction, the applied force F and the static friction f. If we write Newton's second law in the x direction, we get

F - f = Ma (1)

where a is the linear acceleration in the x direction. The torque about the point of contact C is

tC = RF = ICaC = [(1/2)MR2 + MR2]aC = (3/2)MR2aC => aC = 2F/3MR

where

IC = [(1/2)MR2 + MR2] = (3/2)MR2

is the moment of inertia of the disk about an axis through C, and aC is the angular acceleration about the same axis. aC and a are related by

a = RaC = 2F/3M

(b) From (1), the frictional force is

f = F - Ma = F - 2F/3 = F/3 (2)

(c) From (2), we get

F = 3f

The maximum value of F for which the disk will not slip is

Fmax = 3fmax = 3msN = 3msMg

where

N = Mg

is the normal force between the surface and the disk.

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PROBLEM

A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at a distance 5R from the center of the planet, the ship fires an instrument package with speed v0 in a direction making an angle q with the direction to the center of the planet. The package has mass m, which is much smaller than the mass of the spaceship. For what angle q will the package just graze the surface of the planet?

             _ v0             ___
             /|              /   \ M
           /               /       \
+-----\  / q              |      R  |
|      >O-----------------|    +--->|
+-----/ m                 |         |
        |                  \       /
        |                    \___/
        |<---------5R--------->|


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.4]

SOLUTION

Angular momentum is conserved. The initial angular momentum of the package, measured relative to the center of the planet, is

Li = ri x pi = ripi sin q k = 5Rmv0 sin q k

where k is the unit vector into the page. The angular momentum of the package when it reaches its closest approach to the planet is

Lf = rf x pf = Rmvf k

where vf is the velocity of the package, which is entirely in the angular direction. Equating the initial and final angular momenta, we get

Li = Lf => 5Rmv0 sin q = Rmvf => vf = 5v0 sin q

Total energy is conserved. The initial energy of the package is

Ei = (1/2)mv02 - GMm/5R

The energy of the package when it reaches its closest approach to the planet is

Ef = (1/2)mvf2 - GMm/R

Equating the initial and final energies, we get

Ei = Ef => (1/2)mv02 - GMm/5R = (1/2)mvf2 - GMm/R => (1/2)v02 = (1/2)vf2 - 4GM/5R
=> (1/2)v02 = (1/2)(25 sin2 q)v02 - 4GM/5R => 1 = 25 sin2 q - 8GM/5Rv02
=> q = sin-1[(1/5) sqrt(1 + 8GM/5Rv02)]

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PROBLEM

A disk of mass M and radius R unwinds from a tape wrapped around it. The tape passes over a frictionless pulley, and a mass m is suspended from the other end. Assume that the disk drops vertically.

////////////////////////
      |
     _|_
    / | \
  /   |   \
 |    |    |
 |    +    |
 |         |
 |\       /|
 |  \___/  |
 |         |
 |         |   ___
 |         |  /   \
 |         |/       \
 |         |      R  |
+-+        |  M +--->|
|m|        |         |
+-+         \       /
              \___/


(a) Relate the accelerations of m and the disk, a and A, respectively, to the angular acceleration a of the disk.
(b) Find a, A, and a.
(c) Explain what happens if m << M or if m >> M.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.23]

SOLUTION

////////////////////////
      |
     _|_
    / | \
  /   |   \
 |    |    |
 |    +    |
 |         |
 |\       /|
 |  \___/  |
 |         ^ T
 ^ T       |   ___
 |         |  /   \
 |         |/       \
 |         |      R  | |
+-+ |      |  M +--->| | A
|m| | a    |    |    | v
+-+ v       \   |   /
 |            \_|_/
 | mg           |
 |              |
 v              v Mg


(a) a, A, and a are related by

A = Ra - a (1)

(b) Write Newton's second law for m and M.

mg - T = ma (2)
Mg - T = MA (3)

The torque on M about its center is

t = RT = Ia = (1/2)MR2a => T = (1/2)MRa (4)

where

I = (1/2)MR2

is the moment of inertia of M about an axis through its center. Substituting (4) into (2) and (3), we get

mg - (1/2)MRa = ma => (1/2)MRa = mg - ma (5)

and

Mg - (1/2)MRa = MA = M(Ra - a) = MRa - Ma
=> Mg = (3/2)MRa - Ma = 3(mg - ma) - Ma = 3mg - (3m + M)a
=> a = (3m - M)g / (3m + M) = (1 - M/3m)g / (1 + M/3m)

From (5) we get

a = (mg - ma) / (1/2)MR
= 2m(g - a) / MR
= 2m[g - (1 - M/3m)g / (1 + M/3m)] / MR
= 2mg[1 - (1 - M/3m) / (1 + M/3m)] / MR
= 2mg[(2M/3m) / (1 + M/3m)] / MR
= (4g/3R) / (1 + M/3m)

From (1) we get

A = Ra - a
= (4g/3) / (1 + M/3m) - (1 - M/3m)g / (1 + M/3m)
= (4/3 - 1 + M/3m)g / (1 + M/3m)
= (1/3)g(1 + M/m) / (1 + M/3m)

(c) If m << M,

a ≈ - g
a ≈ (4g/R)(m/M) ≈ 0
A ≈ g

The tension is very small compared to the weight Mg. M falls almost as if there were no tape attached to it, with very little rotation. As M accelerates downward with A ≈ g, m is pulled upward by the tape with acceleration of magnitude ≈ g.

If m >> M,

a ≈ g
a ≈ 4g/3R
A ≈ g/3

The tension is very small compared to the weight mg. m falls almost as if there were no tape attached to it. M falls with acceleration A ≈ g/3 and angular acceleration a ≈ 4g/3R.

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PROBLEM

Drum A of mass M and radius R is suspended from drum B, also of mass M and radius R, which is free to rotate about its axis. The suspension is in the form of a massless metal tape wound around the outside of each drum, and free to unwind. Gravity is directed downward. Both drums are initially at rest. Find the initial acceleration of A, assuming that it moves straight down.

----+-------
    |
 /->|
   /|\
B | + |
   \_/|
      |
      |
      | _
      |/ \ \
    A | + | |
       \_/  v


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.24]

SOLUTION

----+-------
    |
 /->|
   /|\
B | + |
   \_/|
      v
    T |
      ^ _
      |/ \ \
    A | + | |
       \|/  v
        |
        v Mg


The torques about the centers of drums A and B are

tA = RT = IaA = (1/2)MR2aA => aA = 2T/MR (1)
tB = RT = IaB = (1/2)MR2aB => aB = 2T/MR (2)

where T is the tension in the tape,

I = (1/2)MR2

is the moment of inertia of each drum about its axis, and aA and aB are the angular accelerations of each drum about its axis. Newton's second law for drum A is

Mg - T = MaA => T = Mg - MaA (3)

where aA is the linear acceleration of drum A. aA, aA, and aB are related by

aA = RaA + RaB (4)

Substituting (1), (2), and (3) into (4), we get

aA = 4T/M = 4(g - aA) = 4g - 4aA => 5aA = 4g => aA = 4g/5

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PROBLEM

A marble of mass M and radius R is rolled up a plane of angle q. If the initial velocity of the marble is v0, what is the distance L it travels up the plane before it begins to roll back down?

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.25]

SOLUTION

The initial energy of the marble is all kinetic and consists of translational and rotational kinetic energy.

Ei = KEi = (1/2)Mv02 + (1/2)Iw02 = (1/2)Mv02 + (1/2)(2/5)MR2v02/R2 = (1/2)Mv02 + (1/5)Mv02
= (7/10)Mv02

where

I = (2/5)MR2

is the moment of inertia of the marble about an axis through its center and

w0 = v0/R

is the initial angular velocity of the marble. The final energy of the marble is all potential.

Ef = PEf = MgL sin q

where L is the distance the marble has rolled up the plane by the time it stops. Equating the initial and final energies, we get

Ei = Ef => (7/10)Mv02 = MgL sin q => L = (7/10)v02 / g sin q

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PROBLEM

A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2. The yo-yo is placed upright on a table and the string is pulled with a horizontal force F as shown. The coefficient of friction between the yo-yo and the table is m. What is the maximum value of F for which the yo-yo will roll without slipping?

        ___
       /   \
     /       \
    |    -    |
    |   |+|   |
    |    -----|----> F
     \       /
_______\___/________
         


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.27]

SOLUTION

        ___
       /   \
     /       \
    |    -    |
    |   |+|   |
    |    -----|----> F
     \       /
___/___\___/________
   \     C
   f


The torque about the point of contact C is

t = (R - b)F = ICaC = (3/2)MR2aC => aC = (2/3)(R - b)F/MR2 (1)

where

IC = (1/2)MR2 + MR2 = (3/2)MR2

is the moment of inertia of the yo-yo about an axis through C and aC is the angular acceleration about the same axis. The yo-yo will roll to the right. The linear acceleration of the yo-yo is

a = RaC (2)

Newton's second law in the horizontal direction is

F - f = Ma (3)

where f is the frictional force between the yo-yo and the table. Using (1), (2), and (3), we get

F - f = MR(2/3)(R - b)F/MR2 = (2/3)(R - b)F/R = (2/3)(1 - b/R)F
=> f = F - (2/3)(1 - b/R)F = F(1 - 2/3 + 2b/3R) = F(1/3 + 2b/3R) = (F/3)(1 + 2b/R)
=> F = 3f / (1 + 2b/R)

The maximum value of F for which the yo-yo rolls without slipping is

Fmax = 3fmax / (1 + 2b/R) = 3mN / (1 + 2b/R) = 3mMg / (1 + 2b/R)

where

N = Mg

is the normal force between the table and the yo-yo.

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PROBLEM

A bowling ball is thrown down the alley with speed v0. Initially it slides without rolling, but due to friction it begins to roll. Show that its speed when it rolls without sliding is (5/7)v0.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.30]

SOLUTION

       /->
        ___
       /   \
     /       \
    |         |
    |    +    | ---> v0
    |         |
     \       /
___/___\___/__________
   \
   f


The friction between the bowling ball and the surface causes the angular velocity of the ball to increase, from zero, but at the same time causes the center of mass to decelerate. Slipping stops when the center of mass velocity vf and the angular velocity wf are related by

vf = Rwf

The torque about the center of mass is

t = Rf = Ia = (2/5)MR2a => a = 5f/2MR

where

I = (2/5)MR2

is the moment of inertia about the ball's (a solid sphere) axis and a is the angular acceleration about the same axis. The linear acceleration of the ball is given by

- f = Ma => a = - f/M

where f is the frictional force. Let t be the time elapsed until the ball starts rolling without slipping. Then

wf = at = 5ft/2MR => ft = (2/5)MRwf = (2/5)Mvf (1)

and

vf = v0 + at = v0 - ft/M (2)

Substituting (1) into (2), we get

vf = v0 - (2/5)vf => (7/5)vf = v0 => vf = (5/7)v0

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PROBLEM

A cylinder of radius R spins with angular velocity w0. When the cylinder is laid on a plane, it skids for a short time and eventually rolls without slipping. What is the final angular velocity, wf?

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.31]

SOLUTION

      w0
       /->
        ___
       /   \
     /       \
    |         |
    |    +    |
    |         |
     \       /
_______\___/___\____
               /
               f


The friction between the cylinder and the surface causes the angular velocity of the cylinder to decrease but at the same time causes the center of mass to accelerate to the right. Slipping stops when the center of mass velocity vf and the angular velocity wf are related by

vf = Rwf

The torque about the center of mass is

t = - Rf = Ia = (1/2)MR2a => a = - 2f/MR

where

I = (1/2)MR2

is the moment of inertia about the cylinder's axis and a is the angular acceleration about the same axis. The linear acceleration of the cylinder is given by

f = Ma => a = f/M

where f is the frictional force. Let t be the time elapsed until the cylinder starts rolling without slipping. Then

wf = w0 + at = w0 - 2ft/MR (1)

and

vf = at = ft/M => ft = Mvf = MRwf (2)

Substituting (2) into (1), we get

wf = w0 - 2wf => wf = w0/3

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PROBLEM

A marble of radius b rolls back and forth in a shallow dish of radius R. Find the frequency of small oscillations. R >> b.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.34]

SOLUTION

Suppose that, at some instant of time, the marble is positioned as defined by the following parameters:

s = distance marble has rolled along the dish from its equilibrium position at the very bottom of the dish
q = angle through which marble has turned about axis through center of curvature of dish

When the marble is positioned at angle q relative to its equilibrium position, it experiences a normal force N acting at the point of contact towards the center of curvature of the dish. The marble also experiences the force of gravity acting downward at its center. The force of gravity produces a torque, about the point of contact, given by

t = - bmg sin q ≈ - bmgq (1)

for small oscillations (q << 1). There is a negative sign because the torque tries to move the marble back to its equilibrium position. The moment of inertia of the marble about an axis through its point of contact is, by the parallel axis theorem,

I = I0 + mb2 = (2/5)mb2 + mb2 = (7/5)mb2 (2)

where

I0 = (2/5)mb2

is the moment of inertia of the marble about an axis through its center. The rotational form of Newton's second law is

t = Ia = I d2f/dt2 (3)

where

a = d2f/dt2

is the angular acceleration about the point of contact. Now suppose the marble has rotated an infinitesimal amount df about its point of contact with the bowl. The corresponding linear displacement of its center is b df, which is also equal to the distance R dq that the marble has rolled along the surface of the bowl. Thus,

b df = R dq => df = (R/b) dq => d2f/dt2 = (R/b) d2q/dt2 (4)

Combining (1), (2), (3), and (4), we get

- bmgq = (7/5)mb2(R/b) d2q/dt2 = (7/5)mbR d2q/dt2 => - gq = (7/5)R d2q/dt2
=> d2q/dt2 = - (5g/7R)q = - w2q => d2q/dt2 + w2q = 0 (5)

where

w2 = 5g/7R => w = sqrt(5g/7R)

Equation (5) is the equation for simple harmonic motion with angular frequency w.

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PROBLEM

A cubical block of side L rests on a fixed cylindrical drum of radius R. Find the largest value of L for which the block is stable.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.35]

SOLUTION

The block will be stable if, when it is tilted to the side a small amount and released, its center of mass is positioned so that gravity causes the block to tilt back towards its original position.

Let (x, y) = (0, 0) be at the center of the cylinder, and let the block be resting on top of the cylinder so that its center is at (0, R + L/2). The block will be stable if, when it is tilted towards the +x direction, its center of mass is to the left (i.e., at a smaller value of x) than its new point of contact with the cylinder.

Suppose the block has tilted towards the +x direction a small amount so that its point of contact with the cylinder is at angle q with respect to the vertical, as measured about the cylinder's axis. Then the point of contact is at (xc, yc) = (R sin q, R cos q).

The block has "rolled" along the surface of the cylinder a distance Rq. The x coordinate of the center of mass of the block is xcom = (R + L/2) sin q - Rq cos q. If xcom < xc, the block will tilt back towards its original position when released.

xcom < xc => (R + L/2) sin q - Rq cos q < R sin q => (R + L/2)q - Rq < Rq => L < 2R

where we have made the small angle approximations sin qq and cos q ≈ 1. Thus, as long as L < 2R, the block will be stable.

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PROBLEM

A boy of mass m runs on ice with velocity v0 and steps on the end of a plank of length L and mass M which is perpendicular to his path.
(a) Describe quantitatively the motion of the system after the boy is on the plank. Neglect friction with the ice.
(b) One point on the plank is at rest immediately after the collision. Where is it?

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.39]

SOLUTION

(a)

     +-+ ^
     | | |
     | | |
     | | |
     | | |
     | | |
   M | | L
     | | |
     | | |
     | | |
     | | |
m    | | |
o--->+-+ v
v0


Let (x, y) = (0, 0) be at the bottom of the plank, as viewed in the above diagram, before the collision. After the collision, the center of mass of the system is located at

ycm = (ML/2) / (M + m) = (L/2) / (1 + m/M)

The initial and final momenta, pi and pf, are both entirely in the x direction and are equal.

pi = pf => mv0 = (M + m)vcm => vcm = mv0 / (M + m)

where vcm is the center-of-mass velocity. After the collision, the center of mass moves to the right at speed vcm. The boy-plus-plank system rotates counterclockwise, at angular velocity w, about its center of mass. Calculate the initial and final angular momenta, Li and Lf about the center of mass.

Li = mv0(L/2) / (1 + m/M)
Lf = Iw

where

I = Iboy + Iplank

is the moment of inertia of the boy-plus-plank about an axis through the center of mass,

Iboy = m(L/2)2 / (1 + m/M)2

is the moment of inertia of the boy about the axis,

Iplank = Iplank0 + M[L/2 - (L/2) / (1 + m/M)]2 (1)

is the moment of inertia of the plank about the same axis, and Iplank0 is the moment of inertia of the plank about an axis through its center. Equation (1) follows from the parallel axis theorem. The moment of inertia of a thin stick of mass M and length L about an axis through its center of mass is

Istick = ∫ r2 dm = 2 ∫0L/2 r2 l dr = 2l(r3/3)|0L/2 = 2(M/L)(L3/24) = ML2/12

where

l = M/L

is the linear density of the stick. Thus,

Iplank0 = ML2/12

and

I = m(L/2)2 / (1 + m/M)2 + ML2/12 + M[L/2 - (L/2) / (1 + m/M)]2
= m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2[1 - 1 / (1 + m/M)]2
= m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2(m/M)2 / (1 + m/M)2
=> Lf = {m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2(m/M)2 / (1 + m/M)2}w

Equating the initial and final angular momenta, we get

Li = Lf
=> mv0(L/2) / (1 + m/M) = {m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2(m/M)2 / (1 + m/M)2}w
=> w = [mv0(L/2) / (1 + m/M)] / {m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2(m/M)2 / (1 + m/M)2}

(b) The point which is at rest immediately after the collision is at a distance r above the center of mass such that

rw = vcm = mv0 / (M + m)
=> r = (mv0/w) / (M + m)
= {[mv0 / (M + m)] / [mv0(L/2) / (1 + m/M)]}
· {m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2(m/M)2 / (1 + m/M)2}
= {[(m/M)v0 / (1 + m/M)] / [mv0(L/2) / (1 + m/M)]}
· {m(L/2)2 / (1 + m/M)2 + ML2/12 + M(L/2)2(m/M)2 / (1 + m/M)2}
= (1/M)[(mL/2) / (1 + m/M)2 + ML/6 + M(L/2)(m/M)2 / (1 + m/M)2]
= [(m/M)(L/2) / (1 + m/M)2 + L/6 + (L/2)(m/M)2 / (1 + m/M)2]
= [(L/6) / (1 + m/M)2][(3m/M) + (1 + m/M)2 + 3(m/M)2]
= [(L/6) / (1 + m/M)2][1 + 5m/M + 4(m/M)2]

The distance from the bottom of the plank to the point which is at rest immediately after the collision is

r + ycm = [(L/6) / (1 + m/M)2][1 + 5m/M + 4(m/M)2] + (L/2) / (1 + m/M)
= [(L/6) / (1 + m/M)2][1 + 5m/M + 4(m/M)2 + 3(1 + m/M)]
= [(L/6) / (1 + m/M)2][4 + 8m/M + 4(m/M)2]
= [(2L/3) / (1 + m/M)2](1 + m/M)2
= 2L/3

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PROBLEM

A wheel with fine teeth is attached to the end of a spring with constant k and unstretched length l. For x > l, the wheel slips freely on the surface, but for x < l the teeth mesh with the teeth on the ground so that it cannot slip. Assume that all the mass of the wheel is in the rim.

|
|
|                  ___
|       k         /   \
|----/\/\/\/-----|--+  |
|                 \   /
+^^^^^^^^^^--------------
|<---l--->|<---b--->|


(a) The wheel is pulled to x = l + b and released. How close will it come to the wall on its first trip?
(b) How far out will it go as it leaves the wall?
(c) What happens when the wheel next hits the gear track?

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.40]

SOLUTION

(a) The initial energy of the system is all potential energy and given by

E1 = PE1 = (1/2)kb2

The wheel moves without rolling until its bottom reaches x = l. When it reaches this point, the energy of the system is all kinetic,

E2 = KE2 = (1/2)mv22

where m is the mass of the wheel and v2 is its velocity when it reaches x = l. By conservation of energy,

E1 = E2 => (1/2)kb2 = (1/2)mv22 => v2 = b sqrt(k/m)

When the wheel reaches the gear track, the teeth impart an impulse Dp to the wheel in the +x direction, causing it to start rotating counterclockwise with angular velocity w3 and rolling with velocity

v3 = Rw3

where R is the wheel radius. Conservation of linear momentum requires that

mv3 = mv2 - Dp (1)

Conservation of angular momentum about the wheel axis requires that

RDp = Iw3 = mR2w3 => Dp = mRw3 = mv3 (2)

where

I = mR2

is the moment of inertia of the wheel about its axis. Substituting (2) into (1), we get

mv3 = mv2 - mv3 => v3 = v2/2 = (b/2) sqrt(k/m) => w3 = (b/2R) sqrt(k/m)

The energy of the system immediately after the wheel starts rotating is all kinetic and is equal to

E3 = KE3 = (1/2)mv32 + (1/2)Iw32 = (1/2)m(b2/4)(k/m) + (1/2)mR2(b2/4R2)(k/m) = kb2/4

In its first approach to the wall, when the wheel comes to rest all the energy of the system is potential energy,

E4 = PE4 = (1/2)k(l - x)2

By energy conservation,

E3 = E4 => kb2/4 = (1/2)k(l - x)2 => b2/2 = (l - x)2 => l - x = b / sqrt(2) => x = l - b / sqrt(2)

When the wheel comes to rest, its distance from the wall is l - b / sqrt(2).

(b) When the wheel moves back out and reaches x = l its linear and angular velocities are the same as when it first entered the area where x < l, except that now the wheel is rotating clockwise because it is rolling away from the wall. The energy of the system is all kinetic,

E5 = KE5 = (1/2)mv52 + (1/2)Iw52 = (1/2)mv32 + (1/2)Iw32 = kb2/4

When it reaches its furthest point from the wall, its angular velocity and rotational kinetic energy are the same as when it was at x = l, but its translational kinetic energy has been converted completely into potential energy. The energy of the system is

E6 = (1/2)k(x - l)2 + (1/2)Iw32 = (1/2)k(x - l)2 + kb2/8 = kb2/4 => x = l + b/2

Thus, the center of the wheel goes out to x = l + b/2.

(c) The instant before the wheel reaches the gear track again, it is still rotating clockwise with angular velocity w3 and its center of mass is moving in the -x direction with velocity v3. When it makes contact with the gear track, the teeth impart an impulse Dq to the wheel in the +x direction, which causes it to start rotating counterclockwise with angular velocity w7 and rolling with velocity

v7 = Rw7

Conservation of linear momentum requires that

mv7 = mv3 - Dq => mRw7 = mv3 - Dq => Dq = mv3 - mRw7

Conservation of angular momentum requires that
Iw7 = RDq - Iw3 => mR2w7 = RDq - mR2w3 => mRw7 = Dq - mRw3
=> mRw7 = mv3 - mRw7 - mRw3 => w7 = v3/2R - w3/2 = (1/2)(v3/R - w3) = 0

The wheel stops moving when it hits the teeth.

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PROBLEM

A plank of length 2L leans against a wall. It starts to slip downward without friction. Show that the top of the plank loses contact with the wall when it is at two-thirds of its initial height.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 6.41]

SOLUTION

  Nw
|---->
|\ q
| \
|  \
|   \ 2L
|   |\
|   | \  ^ Nf
| mg|  \ |
|   v q \|
+----------
         P


The torque about the point of contact with the floor is

t = 2LNw sin q - Lmg cos q = IPa = IP d2q/dt2 = (4mL2/3) d2q/dt2 (1)

where Nw is the normal force that the wall exerts on the plank,

IP = ∫ r2 dm = ∫02L r2 l dr = l(r3/3)|02L = (m/2L)(8L3/3) = 4mL2/3

is the moment of inertia of the plank about an axis through the point of contact with the floor,

a = d2q/dt2

is the angular acceleration about the same axis, and

l = m/2L

is the linear density of the plank. If the origin is at the bottom of the wall, the x and y coordinates of the center of mass are

x = L cos q
y = L sin q

The x and y components of the center-of-mass velocity are

dx/dt = - L sin q dq/dt
dy/dt = L cos q dq/dt

The x component of the center-of-mass acceleration is

d2x/dt2 = - L cos q (dq/dt)2 - L sin q d2q/dt2 (2)

Newton's second law (F = ma) in the horizontal direction is

Nw = max = m d2x/dt2 = - mL cos q (dq/dt)2 - mL sin q d2q/dt2

The initial energy of the plank is all potential and is equal to mg times the initial height of the center of mass,

Ei = mgL sin q0

where q0 is the initial angle the plank makes with the horizontal. When the plank has slipped so that it makes an angle q < q0 with the horizontal, the energy is the sum of the potential energy and the translational and rotational kinetic energies,

Ef = mgL sin q + (1/2)m[(dx/dt)2 + (dy/dt)2] + (1/2)I0(dq/dt)2
= mgL sin q + (1/2)mL2(dq/dt)2 + (1/2)I0(dq/dt)2 = mgL sin q + (1/2)(mL2 + I0)(dq/dt)2

where

I0 = ∫ r2 dm = 2 ∫0L r2 l dr =2l(r3/3)|0L = (2)(m/2L)L3/3 = mL2/3

is the moment of inertia of the plank about a horizontal axis through its center. The translational kinetic energy is calculated assuming that all of the mass is concentrated at the center of mass and that it moves at the speed of the center of mass. The rotational kinetic energy is the energy of rotation about the center of mass. Thus,

Ef = mgL sin q + (1/2)(mL2 + mL2/3)(dq/dt)2 = mgL sin q + (2mL2/3)(dq/dt)2

Equating the initial and final energies,

Ei = Ef => mgL sin q0 = mgL sin q + (2mL2/3)(dq/dt)2 => L sin q0 = L sin q + (2L2/3g)(dq/dt)2 (3)

At the instant when the plank loses contact with the wall, we have

Nw = 0 => d2x/dt2 = 0

From (1) and (2) we get

d2q/dt2 = - (3g/4L) cos q
(dq/dt)2 = - tan q d2q/dt2 = (3g/4L) sin q

Substituting into (3),

L sin q0 = L sin q + (2L2/3g)(3g/4L) sin q = (3/2)L sin q => L sin q = (2/3)L sin q0
=> 2L sin q = (2/3)(2L sin q0)

But 2L sin q is the y coordinate of the top of the plank when it loses contact with the wall and 2L sin q0 is the initial y coordinate of the top of the plank. Thus, the plank loses contact with the wall when it is at two-thirds of its initial height.

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PROBLEM

A thin hoop of mass M and radius R rolls without slipping about the z axis. It is supported by an axle of length R through its center. The hoop circles around the z axis with angular speed W.
(a) What is the instantaneous angular velocity w of the hoop?
(b) What is the angular momentum L of the hoop? Is L parallel to w?
[Note: The moment of inertia of a hoop for an axis along its diameter is (1/2)MR2.]

z ^ wz = W
  |
  |       ++
  |       ||
  |       ||
  |     R ||
  |       ||
  |   R   ||
  +=======|| hoop instantaneously rolling into page
  |       ||
  |       ||
  |       ||
  |       ||
  |       ||
  |       ++
////////////


[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 7.1]

SOLUTION

(a) Choose the coordinate system so that the origin is at the point where the axle intersects with the z axis.

z ^ wz = W
  |
  |       ++
  |       ||
  |       ||
  |     R ||
  |       ||
  |   R   ||
  +=======||-----> y, x out of page
  |       ||
  |       ||
  |       ||
  |       ||
  |       ||
  |       ++
////////////


The z component of the angular velocity is

wz = W

The time required for the hoop to roll once around the z axis is

T = 1/n = 2p/W

where n is the frequency at which the hoop circles the z axis. The distance that the hoop rolls around the z axis is 2pR, which is the same as the circumference of the hoop, so the hoop rotates exactly once each time it circles the z axis. The linear velocity of the hoop is

v = 2pR/T = 2pR/(2p/W) = RW

and the y component of the angular velocity is

wy = - v/R = - W

The instantaneous angular velocity vector is

w = wxx + wyy + wzz = - Wy + Wz

It has magnitude

w = sqrt(wx2 + wy2 + wz2) = sqrt[02 + (-W)2 + W2] = W sqrt(2)

and makes an angle of

tan-1(|wy| / |wz|) = tan-1(1) = p/4 rad = 45°

with the z axis.

(b) We calculate L using

+--+   +-         -+ +--+
|Lx|   |Ixx Ixy Ixz| |wx|
|Ly| = |Iyx Iyy Iyz| |wy|
|Lz|   |Izx Izy Izz| |wz|
+--+   +-         -+ +--+


or

Lx = Ixxwx + Ixywy + Ixzwz
Ly = Iyxwx + Iyywy + Iyzwz
Lz = Izxwx + Izywy + Izzwz

where

Ixx = ∫ (y2 + z2) dm
Iyy = ∫ (x2 + z2) dm
Izz = ∫ (x2 + y2) dm

are moments of inertia and

Ixy = Iyx = - ∫ xy dm
Iyz = Izy = - ∫ yz dm
Ixz = Izx = - ∫ xz dm

are products of inertia. By symmetry,

Ixy = Iyx = 0
Ixz = Izx = 0
Iyz = Izy = 0

The moment of inertia about the y axis is

Iyy = MR2

The moment of inertia about the z axis is

Izz = (1/2)MR2 + MR2 = (3/2)MR2

Since wx = 0, we have

Lx = 0
Ly = Iyywy = - MR2W
Lz = Izzwz = (3/2)MR2W

The instantaneous angular momentum vector is

L = Lxx + Lyy + Lzz = - MR2Wy + (3/2)MR2Wz

It has magnitude

L = sqrt(Lx2 + Ly2 + Lz2) = sqrt{02 + (-MR2W)2 + [(3/2)MR2W]2} = (1/2)MR2W sqrt(13)

and makes an angle of

tan-1(|Ly| / |Lz|) = tan-1(2/3) = 0.5580 rad = 33.69°

with the z axis. So L and w are not parallel.

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PROBLEM

If you start a coin rolling on a table with care, you can make it roll in a circle. The coin "leans" inward, towards the center of the circle, with its axis tilted. If the radius of the coin is b, the radius of the circle it follows on the table is R, and its velocity is v, find the angle f that the axis makes with the horizontal. Assume that there is no slipping.

[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGraw-Hill), problem 7.6]

SOLUTION
        _
        /| z
       /
     c//
    o//_ _ _ coin instantaneously rolling out of page in +x direction
   i//\ f
  n//   \       table top
   \     \|
     \
       \| y


Choose the origin of the coordinate system to be at the point of contact between the coin and the table top. Choose the y axis to be parallel to the coin's spin axis. Choose the z axis to be aligned with the coin. The x axis points out of the page.

Neglecting the motion of the coin in a circle on the table top, its spin angular velocity is

w = wyy = (v/b)y

The spin angular momentum of the coin is given by

+--+   +-         -+ +--+
|Lx|   |Ixx Ixy Ixz| |wx|
|Ly| = |Iyx Iyy Iyz| |wy|
|Lz|   |Izx Izy Izz| |wz|
+--+   +-         -+ +--+


Since wx = wz = 0, this reduces to

Lx = Ixywy
Ly = Iyywy
Lz = Izywy

Because of the way we defined our coordinate system,

Ixy = - ∫ xy dm = 0
Izy = - ∫ yz dm = 0

so the spin angular momentum is

L = Iyywyy

Using the parallel axis theorem, the moment of inertia of the coin about the y axis is

Iyy = (1/2)mb2 + mb2 = (3/2)mb2

Thus,

L = (3/2)mb2(v/b)y = (3/2)mbvy

Now because the coin is rolling in a circle, the spin angular momentum vector rotates about the vertical axis and sweeps out a circle of radius

Lh = L cos f = (3/2)mbv cos f

The torque on the coin is due to the force of gravity acting at the coin's center and is equal to

t = r x F = - bmg sin f x

The magnitude of the torque is equal to the time derivative of the angular momentum, so

t = dL/dt => bmg sin f = 2pLh/(2pR/v) = vLh/R = (3mbv2/2R) cos f => tan f = 3v2/2gR

Note that we did not include the angular momentum due to the coin's circular motion along the table top because this angular momentum points in the upward direction and is constant.

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Static Equilibrium

PROBLEM

A 50 kg uniform square sign, 2.0 m on a side, is hung from a 3.0 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.0 m above the point where the rod is fixed to the wall.
(a) What is the tension in the cable?
(b) What are the horizontal and vertical components of the force exerted by the wall on the rod?
(c) Solve the problem if the sign is attached to the rod along the entire length of the sign by a hinge, rather than at two points.

    |
    |\
    | \
    |  \
4 m |   \
    |    \
    |     \
   O|     q\3
    | 1+----+2
    |  |    |
    |  |    | 2 m
    |  +----+
    |   2 m
    |
    <--3 m-->


[from Halliday, David, and Resnick, Robert 1988, Fundamentals of Physics, Third Edition Extended (New York: John Wiley & Sons), problem 13.25]

SOLUTION

(a) The torque about O must be zero. There are three contributions to the torque about O: the weight of the sign supported at points 1 and 2, with half the weight supported at each point, and the tension of the cable at point 3, which is really the same as point 2. The torque is

t = t1 + t2 + t3 = r1 x F1 + r2 x F2 + r3 x F3

If O is the origin of our coordinate system, with +x to the right, +y up, and +z out of the page, and we define a = 2.0 m, b = 3.0 m, and M = 50 kg, then

r1 = (b - a)x
r2 = bx = r3
F1 = - (Mg/2)y = F2
F3 = - (T cos q)x + (T sin q)y

and

t = - (b - a)(Mg/2)z - (bMg/2)z + (bT sin q)z = - (Mg/2)(2b - a)z + (bT sin q)z = 0
=> T = (Mg/2)(2b - a) / (b sin q) = Mg(1 - a/2b) / (sin q)

where T is the tension in the cable. The triangle formed by the wall, the rod, and the cable is a 3-4-5 right triangle, so

sin q = 4/5 = 0.8

Thus,

T = (50 kg)(9.8 m/s2)[1 - (2 m) / (2)(3 m)] / (0.8) = 408.3 N

(b) Let Fh and Fv be the horizontal and vertical components of the force of the wall on the rod. Since the sum of the horizontal and vertical forces on the rod must be zero,

Fh - T cos q = 0 => Fh = T cos q = (408.3 N)(3/5) = 245 N (to the right)
Fv + T sin q - Mg/2 - Mg/2 = 0 => Fv = Mg - T sin q = (50 kg)(9.8 m/s2) - (408.3 N)(4/5)
= 163.3 N (upward)

(c) If the sign is attached continuously along the rod, the torque on the rod due to the weight of the sign can be calculated by doing an integral. We divide the section of the rod between x = 1 m = b - a and x = 3 m = b into infinitesimal segments dx. The torque due to each of these infinitesimal segments is

dtsign = r x dFsign = xx x [- (Mg/a) dx y] = - (Mg/a) x dx z
=> tsign = - (Mg/a)zb-ab x dx = - Mgz [(1/a) ∫b-aa x dx]

Now the average value of x between x = b - a and x = a is

xavg = (1/a) ∫b-aa x dx = (1/2)[(b - a) + b] = (1/2)(2b - a) => tsign = - (Mg/2)(2b - a)z

This is the same as the torque, due to the sign, that was calculated in part (a), so T, Fh, and Fv will be the same.

Alternatively, we could have treated the rod and the sign as if they were a single object, with the weight of the sign being applied at its center of mass. We would obtain

tsign = rsign x Fsign = {(1/2)[(b - a) + b]x - (a/2)y} x (- Mgy) = - (Mg/2)(2b - a)z

which yields the same result.

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PROBLEM

Four similar, uniform bricks of length L are stacked on a table as shown. We seek to maximize the overhang distance h, measured from the edge of the table. Find the optimum indicated distances and calculate h.

               +------+
               |brick1|
           +---+--+---+
           |brick2| a1
        +--+---+--+
        |brick3|a2
      +-+----+-+
      |brick4| |<- a3
------+----+-+
table    ->| |<- a4
           |
           |<---h---->|


[from Halliday, David, and Resnick, Robert 1988, Fundamentals of Physics, Third Edition Extended (New York: John Wiley & Sons), problem 13.37(a)]

SOLUTION

Brick 1 will be stable if its center of mass is above the corresponding pivot point, which is at the upper right corner of brick 2. Thus,

a1 = L/2

Brick 2 will be stable if the center of mass of bricks 1 and 2 is above the corresponding pivot point, which is at the upper right corner of brick 3. Let M be the mass of each brick. Then the x coordinate of the center of mass of brick 2 relative to the upper right corner of brick 3 is
- (L/2 - a2) and the x coordinate of the center of mass of brick 1 relative to the upper right corner of brick 3 is a2. The x coordinate of the center of mass of bricks 1 and 2 relative to the upper right corner of brick 3 is

[- M(L/2 - a2) + Ma2] / (2M) = 0 => - L/2 + a2 + a2 = 0 => a2 = L/4

Brick 3 will be stable if the center of mass of bricks 1, 2, and 3 is above the corresponding pivot point, which is at the upper right corner of brick 4. The x coordinate of the center of mass of brick 3 relative to the upper right corner of brick 4 is - (L/2 - a3) and the x coordinate of the center of mass of bricks 1 and 2 relative to the upper right corner of brick 4 is a3. The x coordinate of the center of mass of bricks 1, 2, and 3 relative to the upper right corner of brick 4 is

[- M(L/2 - a3) + 2Ma3] / (3M) = 0 => - L/2 + a3 + 2a3 = 0 => a3 = L/6

Brick 4 will be stable if the center of mass of bricks 1, 2, 3, and 4 is above the corresponding pivot point, which is at the corner of the table. The x coordinate of the center of mass of brick 4 relative to the corner of the table is - (L/2 - a4) and the x coordinate of the center of mass of bricks 1, 2, and 3 relative to the corner of the table is a4. The x coordinate of the center of mass of bricks 1, 2, 3, and 4 relative to the corner of the table is

[- M(L/2 - a4) + 3Ma4] / (4M) = 0 => - L/2 + a4 + 3a4 = 0 => a4 = L/8

Thus,

h = a1 + a2 + a3 + a4 = L/2 + L/4 + L/6 + L/8 = 25L/24

Alternate Method:

The nth brick will be stable if the center of mass of bricks 1 through n is above the corresponding pivot point. The x coordinate of the center of mass of the nth brick relative to the pivot point is - (L/2 - an) and the x coordinate of the center of mass of bricks 1 through n - 1 relative to the pivot point is an. The x coordinate of the center of mass of bricks 1 through n relative to the pivot point is

[- M(L/2 - an) + (n-1)Man] / (nM) = 0 => - L/2 + an + (n-1)an = 0 => an = L/2n

Thus,

a1 = L/2
a2 = L/4
a3 = L/6
a4 = L/8
h = a1 + a2 + a3 + a4 = L/2 + L/4 + L/6 + L/8 = 25L/24

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PROBLEM

Four similar, uniform bricks of length L are stacked on a table as shown. Assume that brick 2 overhangs brick 4 by the same amount that brick 3 overhangs brick 4, and that brick 1 is directly above brick 4. We seek to maximize the overhang distance h, measured from the edge of the table. Find the optimum indicated distances and calculate h.

              L
        +----------+
        | 1        |
        |       A  |
+-------+--+----+--+-------+
| 2        |    | 3        |
|          |    |  B       |
+-------+--+----+--+-------+
        | 4        |
        |          |
--------+---+------+<--c1->|
    table   |
            |<-c2->|
            |
            |<------h----->|


[from Halliday, David, and Resnick, Robert 1988, Fundamentals of Physics, Third Edition Extended (New York: John Wiley & Sons), problem 13.37(c)]

SOLUTION

Half of the weight brick 1 is supported by each of the two bricks (2 and 3) on which it rests. When brick 3 is on the verge of falling, bricks 1 and 3 make contact at point A and bricks 3 and 4 make contact at point B. Calculate the torque on brick 3 relative to point B. There are two contributions to the torque on brick 3. One contribution is from the force Mg/2 of brick 1 on brick 3 at point A. The second contribution is from the force of gravity Mg acting at the center of mass of brick 3. The total torque must be zero.

t = t1 + t2 = r1 x F1 + r2 x F2 (1)

Define the coordinate system so that +x points to the right, +y points up, and +z points out of the page. Let (M, H) be the (mass, height) of each brick. Then

r1 = Hy - (L - c1)x
r2 = (H/2)y + (c1 - L/2)x
F1 = - (Mg/2)y
F2 = - Mgy
r1 x F1 = (L - c1)(Mg/2)z
r2 x F2 = - (c1 - L/2)Mgz

Substituting into (1), we have

t = (L - c1)(Mg/2)z - (c1 - L/2)Mgz = 0 => (L - c1)(1/2) - (c1 - L/2) = 0
=> L/2 - c1/2 - c1 + L/2 = 0 => L - 3c1/2 = 0 => c1 = 2L/3

The entire assembly of four bricks will be stable if its center of mass is directly above the edge of the table. By symmetry, the center of mass is a horizontal distance L/2 from the right end of brick 4. Thus,

c2 = L/2

and

h = c1 + c2 = 2L/3 + L/2 = 7L/6

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PROBLEM

A massless plank of length L is attached to a rod whose end is embedded in a wall and free to rotate without friction. A small mass M is attached to the plank at its midpoint. An upward force F is applied to the end of the plank opposite the rod and prevents the plank from rotating.

             
          |M|
o---------+-+---------
r                    ^
o                    |
d                    | F
                     |


(a) Find F and the vertical and horizontal components of the force the rod exerts on the plank.
(b) If the force F is removed, find the angular acceleration of the plank about the rod the instant after F is removed.
(c) Find the linear acceleration of the center of the plank the instant after F is removed.
(d) Find the vertical and horizontal components of the force the rod exerts on the plank the instant after F is removed.
(e) Find the angular velocity of the plank about the rod after the plank has rotated through angle q following the removal of F.
Repeat parts (a)-(e) if the plank has mass m.

SOLUTION

(a) The torque about the rod is zero. Define z to be the unit vector pointing out of the page. Then

t = - (MgL/2)z + FLz = 0 => - MgL/2 + FL = 0 => F = Mg/2

Let (Fv, Fh) be the (vertical, horizontal) component of the force the rod exerts on the plank. The sum of the vertical forces on the plank is zero, so

Fv - Mg + F = 0 => Fv = Mg - F = Mg - Mg/2 = Mg/2

The sum of the horizontal forces on the plank is zero. Since the only horizontal force on the plank is Fh, Fh must be zero.

(b) When the force F is removed, the net torque on the plank is due to the weight of the mass M.

t = MgL/2 = Ia = (ML2/4)a => a = 2g/L

where

I = ML2/4

is the moment of inertia about the rod, and a is the angular acceleration of the plank about the rod.

(c) The linear acceleration of the center of the plank is

a = (L/2)a = g

(d) The sum of the vertical forces is

Fv - Mg = - Ma = - Mg => Fv = 0

The only horizontal force is Fh, and this must be zero because the acceleration in the horizontal direction is zero.

(e) Use energy conservation. The instant the force F is removed, the plank is at rest and its kinetic energy is zero. We can define its potential energy in this position to be zero. Thus, the total initial energy is

Ei = 0

After the plank has rotated through an angle q, its kinetic energy is

KEf = (1/2)Iw2 = (1/2)(ML2/4)w2 = (ML2/8)w2

and its potential energy is

PEf = - (MgL/2) sin q

Thus the total energy is

Ef = KEf + PEf = (ML2/8)w2 - (MgL/2) sin q = Ei = 0 => w = sqrt[(4g/L) sin q]

Now, if the plank isn't massless and has mass m:

(a) The torque about the rod is zero. Define the z to be the unit vector pointing out of the page. Then

t = - [(M+m)gL/2]z + FLz = 0 => - (M+m)gL/2 + FL = 0 => F = (M+m)g/2

Let (Fv, Fh) be the (vertical, horizontal) component of the force the rod exerts on the plank. The sum of the vertical forces on the plank is zero, so

Fv - (M+m)g + F = 0 => Fv = (M+m)g - F = (M+m)g - (M+m)g/2 = (M+m)g/2

The sum of the horizontal forces on the plank is zero. Since the only horizontal force on the plank is Fh, Fh must be zero.

(b) When the force F is removed, the net torque on the plank is due to the weight of the mass M and the weight of the plank.

t = (M+m)gL/2 = Ia = [(ML2/4) + (mL2/3)]a = (M/4 + m/3)L2a
=> a = [(M+m) / (M/4 + m/3)]g/2L

where

I = ML2/4 + mL2/3

is the moment of inertia about the rod, and a is the angular acceleration of the plank about the rod.

(c) The linear acceleration of the center of the plank is

a = (L/2)a = [(M+m) / (M/4 + m/3)]g/4

(d) The sum of the vertical forces is

Fv - (M+m)g = - (M+m)a = - (M+m)[(M+m) / (M/4 + m/3)]g/4
=> Fv = (M+m)g - (M+m)[(M+m) / (M/4 + m/3)]g/4 = (M+m)g{1 - [(M+m) / (M/4 + m/3)]/4}

The only horizontal force is Fh, and this must be zero because the acceleration in the horizontal direction is zero.

(e) Use energy conservation. The instant the force F is removed, the plank is at rest and its kinetic energy is zero. We can define its potential energy in this position to be zero. Thus, the total initial energy is

Ei = 0

After the plank has rotated through an angle q, its kinetic energy is

KEf = (1/2)Iw2 = (1/2)(ML2/4 + mL2/3)w2 = (1/2)(M/4 + m/3)L2w2

and its potential energy is

PEf = - [(M+m)gL/2] sin q

Thus the total energy is

Ef = KEf + PEf = (1/2)(M/4 + m/3)L2w2 - [(M+m)gL/2] sin q = Ei = 0
=> w = sqrt({[(M+m)g/L] / (M/4 + m/3)} sin q)

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Gravitation

PROBLEM

Calculate the radius of a geosynchronous orbit.

SOLUTION

The sidereal day is T = 23 h 56 m 4.1 s = 86164.1 s. This is the time it takes the Earth to make one rotation about its axis. A satellite in a geosynchronous orbit must orbit above the equator in a circular orbit with the same period. Equating the gravitational force to the centripetal force, we have

GMem/r2 = mv2/r => GMe/r = v2

where G = 6.67 x 10-11 N m2/kg2 is the gravitational constant, Me = 5.98 x 1024 kg is the mass of the Earth, m is the mass of the satellite, r is the radius of the orbit, and v is the orbital speed. The orbital speed is

v = 2pr/T

Thus,

GMe/r = 4p2r2/T2
=> r = (GMeT2/4p2)1/3
= [(6.67 x 10-11 N m2/kg2)(5.98 x 1024 kg)(86164.1 s)2/4p2]1/3
= 4.217 x 107 m = 4.217 x 104 m = 42,170 km

This is about 1.05 times the circumference of the Earth.

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PROBLEM

The Hubble Space Telescope (HST) orbits the Earth about 600 km above the ground.
(a) Assuming a circular orbit, what is the period of a complete revolution around the Earth?
(b) If the HST was deployed on 25 April 1990, estimate the date on which it completed its 100,000th orbit.

SOLUTION

(a) The gravitational force FG on the HST must be equal to its mass m times its centripetal acceleration ac.

FG = mac

The gravitational force is

FG = GMem / r2

where G = 6.67 x 10-11 N m2 kg-2 is the gravitational constant, Me = 5.98 x 1024 kg is the mass of the Earth, and r is the orbital radius. The centripetal acceleration is

ac = v2 / r

where v is the orbital velocity. Thus,

GMem / r2 = mv2 / r => GMe / r = v2 => v = sqrt(GMe / r) = 2pr / T

where T is the orbital period. Solving for T,

T = 2pr / sqrt(GMe / r) = 2p sqrt(r3 / GMe) (1)

The orbital radius is

r = Re + h

where Re = 6.37 x 106 is the radius of the Earth and h = 600 km = 0.6 x 106 m is the altitude of the HST above the ground. So

r = 6.37 x 106 m + 0.6 x 106 m = 6.97 x 106 m

Substituting into (1), we get

T = 2p{(6.97 x 106)3 / [(6.67 x 10-11 N m2 kg-2)(5.98 x 1024 kg)]}1/2 = 5789 s
= (5789 s)[(1 min) / (60 s)] = 96.48599 min = 96.49 min

(b) The time required to complete 100,000 orbits is

Ttot = 100,000T = (100,000)(96.48599 min) = (9648599 min)[(1 day) / (1440 min)]
= 6700.416 days ≈ (6700.416 days)[(1 yr) / (365 days)] = 18.3573 years

Since the HST was deployed on 25 April 1990, it completed its 100,000th orbit sometime in 2008 after 25 April 2008. Since 1992, 1996, 2004, and 2008 were leap years (2000 was not a leap year), the number of days elapsed between 25 April 1990 and 25 April 2008 is (18)(365) + 4 = 6574. The number of days elapsed between 25 April 1990 and 29 August 2008 is 6574 + 5 (for the remainder of April 2008) + 31 (for May) + 30 (for June) + 31 (for July) + 29 (for August) = 6700. So the date on which the HST completed its 100,000th orbit was approximately 29 August 2008. The actual date was 11 August 2008.

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PROBLEM

(a) How much work would be required to move a satellite of mass m from a circular orbit of radius r1 = 2rE about the earth to another circular orbit of radius r2 = 3rE, where rE is the radius of the earth?
(b) How much work is done by gravity?
(c) What is the change in the potential energy of the satellite?
(d) Show that the work-energy principle is satisfied.

[(a) from Giancoli, Douglas C. 2008, Physics for Scientists and Engineers, Fourth Edition (Upper Saddle River, New Jersey: Pearson Prentice Hall), problem 8.57]

SOLUTION

(a) For a satellite of mass m in a circular orbit of radius r around the earth, the gravitational force Fg must equal the mass times the centripetal acceleration ac, or

Fg = mac

The gravitational force is

Fg = GMm/r2

where G is the gravitational constant, and M is the mass of the earth. The centripetal acceleration is

ac = v2/r

where v is the orbital velocity of the satellite. Thus,

GMm/r2 = mv2/r => GM/r = v2

The kinetic energy of the satellite is

KE = (1/2)mv2 = (1/2)m(GM/r) = GMm/2r

The gravitational potential energy of the satellite is

PE = - GMm/r

Thus, the total mechanical energy of the satellite is

Etotal = KE + PE = GMm/2r – GMm/r = - GMm/2r

Moving the satellite from a circular orbit of radius 2rE to a circular orbit of radius 3rE increases the energy from

E1 = - GMm/4rE

to

E2 = - GMm/6rE

The change in energy is

DE = E2 - E1 = - GMm/6rE + GMm/4rE = (GMm/rE)(- 1/6 + 1/4)
= (GMm/rE)(- 2/12 + 3/12) = (GMm/rE)(1/12) = GMm/12rE

The work required to move the satellite is equal to the change in energy,

W = DE = GMm/12rE

This is work done by some external agent, such as the Space Shuttle, to move the satellite to its new orbit and give it the proper orbital velocity so that it stays there.

(b) The work done by gravity is

Wg = ∫ Fg • dr

where

Fg = - (GMm/r2) r

is the force of gravity and dr is an infinitesimal element of the path taken by the satellite in moving from r = 2rE to r = 3rE.

Since the force of gravity is a radial force,

Wg = ∫ Fg • dr = ∫ - (GMm/r2) r • dr = ∫ – (GMm/r2) rr dr

where the integral is from r = 2rE to r = 3rE.

Note that the vector dr is an actual path element. Since only the component of dr in the radial direction contributes to the integral, we have replaced it with r dr, where r is the unit vector which points radially outward, and dr is the change in the radial coordinate r that corresponds to the path element dr.

Carrying out the integral,

Wg = - ∫ (GMm/r2) dr = GMm/3rE - GMm/2rE = (GMm/rE)(1/3 – 1/2)
= (GMm/rE)(2/6 – 3/6) = - GMm/6rE

(c) The gravitational potential energy at radial distance r is

PE(r) = - GMm/r

Thus the change in potential energy is

DPE = PE2 - PE1 = PE(3rE) – PE(2rE) = - GMm/3rE + GMm/2rE
= (GMm/rE)(- 1/3 + 1/2) = (GMm/rE)(- 2/6 + 3/6) = GMm/6rE

which is the negative of the work done by gravity. To answer part (b), we could have just calculated the change in potential energy and taken the negative.

(d) The change in the kinetic energy of the satellite is

DKE = KE2 - KE1 = GMm/6rE - GMm/4rE = (GMm/rE)(1/6 – 1/4)
= (GMm/rE)(2/12 – 3/12) = - GMm/12rE

The total work done on the satellite is the sum of the work done by the external agent, calculated in part (a), and the work done by gravity, calculated in part (b). Thus,

Wtotal = W + Wg = GMm/12rE – GMm/6rE = - GMm/12rE

The total work done on the satellite is equal to the change in its kinetic energy, so the work-energy principle is satisfied.

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Fluid Statics

PROBLEM

(a) Calculate the fraction of an iceberg which is underwater.
(b) If an iceberg has a mass of M = 1000 kg, how much weight must be placed on top of it for the iceberg to be totally submerged, without causing it to sink further?

SOLUTION

(a) The buoyant force Fbuoy on the iceberg is equal to the weight of the water that the iceberg displaces.

Fbuoy = rwaterVsubg (I)

where rwater is the density of the water, Vsub is the volume of the submerged part of the object, and g = 9.8 m/s2 is the acceleration of gravity. Let f be the fraction of the iceberg's volume V which is submerged. Then

Vsub = fV

The buoyant force supports the weight of the iceberg.

Fbuoy = riceVg (II)

Equating (I) and (II), we get

rwaterVsubg = riceVg => rwaterfV = riceV
=> f = rice/rwater = (0.917 g/cm3) / (1.00 g/cm3) = 0.917

(b) The weight W placed on top of the iceberg must be large enough so that the iceberg must be fully submerged in order for the buoyant force to support the combined weight of the iceberg and the additional weight.

Fbuoy = rwaterVg = rwater(M/rice)g = (rwater/rice)Mg = Mg + W
=> W = [(rwater/rice) - 1]Mg = [(1.00 g/cm3) / (0.917 g/cm3) - 1](1000 kg)(9.8 m/s2) = 887.0 N

Alternate Method:

From part (a), we know that, without any additional weight placed on top of the iceberg, 917 kg of the iceberg is submerged and 83 kg is above the water. If placing a weight W on top of the iceberg is just sufficient to submerge the rest of the iceberg, the additional buoyant force due to the part of the iceberg that was above water equals the weight W.

Fbuoy = rwaterVaboveg = rwater(Mabove/rice)g = (rwater/rice)Maboveg = W
=> W = [(1.00 g/cm3) / (0.917 g/cm3)](83 kg)(9.8 m/s2) = 887.0 N

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PROBLEM

You place a glass beaker, partially filled with water, in a sink. The beaker has a mass of mb = 390 g and an interior volume of Vin = 500 cm3. You now start to fill the sink with water and you find, by experiment, that if the beaker is less than half full, it will float; but if it is more than half full, it remains on the bottom of the sink as the water rises to its rim. What is the density of the material of which the beaker is made?

[from Halliday, David, and Resnick, Robert 1988, Fundamentals of Physics, Third Edition Extended (New York: John Wiley and Sons), problem 16.49]

SOLUTION

When the beaker is exactly half filled with water, then, when the water in the sink reaches the rim of the beaker, the buoyant force will just balance the weight of the beaker and the water it contains.

The buoyant force is equal to the weight of the displaced water. Let Vb be the volume of the beaker itself. Then the volume of the displaced water is Vin + Vb and the buoyant force is

Fbuoy = rw(Vin + Vb)g

where rw = 1 g/cm3 is the density of water and g = 9.8 m/s2 is the acceleration of gravity.

The volume of the water inside the beaker is

Vw = (1/2)Vin

The weight of the beaker plus the water it contains is

W = mbg + rwVwg = mbg + (1/2)rwVing

Equating this to the buoyant force, we get

rw(Vin + Vb)g = mbg + (1/2)rwVing => rw(Vin + Vb) = mb + (1/2)rwVin
=> Vb = mb/rw + (1/2)Vin - Vin = mb/rw - (1/2)Vin

The density of the beaker is

rb = mb/Vb = mb / [mb/rw - (1/2)Vin] = rw / [1 - (1/2)Vinrw/mb]
= (1 g/cm3) / [1 - (1/2)(500 cm3)(1 g/cm3)/(390 g)] = 2.786 g/cm3

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PROBLEM

An ice cube is placed in a partially filled glass of water in which it floats. When the ice cube has completely melted, will the water level be higher, lower, or the same? Neglect evaporation.

|         |
|  +---+  |
|--|ice|--|
|  |   |  |
|  +---+  |
|         |
|  water  |
|         |
+---------+


SOLUTION

Let

V0 = volume of ice cube
r0 = density of ice
rw = density of water
Vd = volume of water displaced by ice cube.

The buoyant force which supports the ice cube is equal to its weight, so

FB = miceg = r0V0g (1)

where mice = r0V0 is the mass of the ice cube, and g is the acceleration of gravity. But the buoyant force is also equal to the weight of the displaced water, so

FB = rwVdg (2)

Equating (1) and (2), we get

r0V0g = rwVdg => Vd = (r0/rw)V0

When the ice has completely melted, the mass of the water produced is equal to the mass of the ice.

mw = mice => rwVw = r0V0 => Vw = (r0/rw)V0 = Vd

Thus, the volume of the water produced by the melting ice cube is equal to the volume of the water that was displaced by the ice cube, and the water level stays the same.

Because the weight of the ice cube is equal to the weight of the displaced water, we could have seen immediately, without doing any calculation, that when the ice cube melts, the resulting water must occupy the same volume that was displaced by the ice cube, so the water level remains the same after the ice cube melts.

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PROBLEM

Suppose that three blocks of ice, each measuring 25 cm x 25 cm x 25 cm, have been prepared such that there is something different in the middle of each one. Each block of ice is floating in a bucket of water, filled to the brim. When the ice blocks melt, determine if the top of the water in the bucket will spill over, stay at the brim but not spill over, or move below the brim for the ice block that contains
(a) 10 cm3 of iron
(b) 10 cm3 of air
(c) 10 cm3 of water

[from Physics E-1a, Principles of Physics I: Mechanics, Final Exam, Fall 2003, Harvard Extension School]

SOLUTION

(a) Let mice be the mass of the ice in the ice block and mFe be the mass of the iron. The weight of the ice block is

W = (mice + mFe)g

The buoyant force is equal to the weight of the ice block and is also equal to the weight of the displaced water,

Fbuoy = rwVg

where rw is the density of water and V is the volume of the displaced water. Thus,

(mice + mFe)g = rwVg => V = (mice + mFe) / rw = mice/rw + mFe/rw

When the ice block melts, the volume added to the water is

V' = mice/rw + mFe/rFe < V

where rFe is the density of iron, which is greater than the density of water. Since V' < V, the water level drops.

(b) The weight of the ice block is

W = miceg

The buoyant force is equal to the weight of the ice block and is also equal to the weight of the displaced water,

Fbuoy = rwVg

Thus,

miceg = rwVg => V = mice/rw

When the ice block melts, the volume added to the water is

V' = mice/rw = V

Since V' = V, no water spills over and the bucket stays filled to the brim.

(c) Let mice be the mass of the ice in the ice block and mw be the mass of the water inside the ice block. The weight of the ice block is

W = (mice + mw)g

The buoyant force is equal to the weight of the ice block and is also equal to the weight of the displaced water,

Fbuoy = rwVg

Thus,

(mice + mw)g = rwVg => V = (mice + mw) / rw = mice/rw + mw/rw

When the ice block melts, the volume added to the water is

V' = mice/rw + mw/rw = V

Since V' = V, no water spills over and the bucket stays filled to the brim.

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PROBLEM

A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 7.00 kg. This cylinder is floating in water. Then oil (r = 725 kg/m3) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in the oil?

|             |
|-------------|
|     oil     |
|   +-----+   |
|   | mg  |y  |
|---|-----|---|
|   |     |H-y|
|   +-----+   |
|    water    |
+-------------+


[from Cutnell, John D., and Johnson, Kenneth W. 2004, Physics, Sixth Edition (New York: Wiley), problem 11.47]

SOLUTION

Let H = 0.120 m be the total height of the cylinder and let y be the height of the cylinder in the oil. Then H - y is the height of the cylinder in the water. The buoyant force is equal to the weight of the displaced liquid and also equal to the weight of the cylinder.

Fbuoy = Fbuoywater + Fbuoyoil = pR2(H - y)rwaterg + pR2yroilg = pR2g[(H - y)rwater + yroil] = mg
=> m/pR2 = Hrwater - yrwater + yroil => y = (Hrwater - m/pR2) / (rwater - roil)
= [(0.120 m)(1000 kg/m3) - (7.00 kg) / p(0.150 m)2] / (1000 kg/m3 - 725 kg/m3) = 7.626 x 10-2 m

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PROBLEM

A polar bear partially supports herself by pulling part of her body out of the water and onto a rectangular slab of ice. The ice sinks down so that only half of what was once exposed is now exposed, and the bear has 70 percent of her volume (and weight) out of the water. Estimate the bear’s mass, assuming that the total volume of the ice is 10 m3, and the bear’s specific gravity is 1.0.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 10.31]

SOLUTION

Let mbear be the bear’s mass and mice be the mass of the ice slab. Then the buoyant force has to equal the combined weight of the bear and the ice slab, or

Fbuoy = mbearg + miceg (1)

But, by Archimedes’ principle, the buoyant force is also equal to the weight of the displaced water. The volume of the displaced water is equal to the sum of the volumes displaced by the bear and the ice slab. If Vbear is the total volume of the bear, Vice is the total volume of the ice slab, and f2 is the fraction of the ice slab which is exposed, then the volume of the displaced water after the bear has partially climbed onto the slab is

Vw = 0.3Vbear + (1 – f2)Vice

The buoyant force is

Fbuoy = rwVwg = 0.3rwVbearg + (1 – f2)rwViceg

where rw = 1.00 x 103 kg/m3 is the density of water.

Since the specific gravity of the bear is 1.0, the bear’s density is the same as that of water, and

mbear = rwVbear => Fbuoy = 0.3mbearg + (1 – f2)rwViceg (2)

Equating (1) and (2), and dividing by g, we have

mbear + mice = 0.3mbear + (1 – f2)rwVice
=> mbear = [(1 – f2)rwVice - mice] / 0.7

Since

mice = riceVice

where rice = 0.917 x 103 kg/m3 is the density of ice,

mbear = [(1 – f2)rw - rice]Vice / 0.7 (3)

Before the bear climbs onto the ice slab, the fraction of the ice slab which is exposed is f1 = 2f2, and the buoyant force on the ice slab is

Fbuoy = miceg = (1 – f1)rwViceg
=> f1 = 1 - mice/rwVice = 1 - rice/rw = 1 – 0.917 = 0.083
=> f2 = f1/2 = 0.083/2 = 0.0415

Substituting into (3), we get

mbear = [(1 – 0.0415)(1.0 x 103 kg/m3) - 0.917 x 103 kg/m3](10 m3) / 0.7
= 592.9 kg

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PROBLEM

Intravenous infusions are often made under gravity. Assuming the fluid has a density of 1.00 g/cm3, at what height h above the needle should the top of the fluid in the container be placed so the liquid pressure at the needle is (a) 65 mm Hg, (b) 550 mm H2O? (c) If the blood pressure is 18 mm Hg above atmospheric pressure, how high should the bottle be placed so that the fluid just barely enters the vein?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 10.61]

SOLUTION

(a) Assume that the pressure at the top of the fluid in the container is atmospheric pressure, P0 = 760 mm Hg = 1.013 x 105 N/m2 = 1.013 x 106 dyne/cm2. Then the pressure at the needle is

P = P0 + rgh => h = (P – P0)/rg

The stated pressure of 65 mm Hg is a gauge pressure, specified relative to atmospheric pressure. Thus, P – P0 = 65 mm Hg, which should be converted to dyne/cm2, and

h = (P – P0)/rg
= (65 mm Hg)[(1.013 x 106 dyne/cm2) / (760 mm Hg)] / [(1.00 g/cm3)(980 cm/s2)] = 88.41 cm

(b) Since the density of mercury is 13.6 g/cm3, while that of water is 1.0 g/cm3, a pressure of 550 mm H2O is equivalent to (550/13.6) mm Hg = 40.44 mm Hg.

The required value of h is

h = (P – P0)/rg
= (40.44 mm Hg)[(1.013 x 106 dyne/cm2) / (760 mm Hg)] / [(1.00 g/cm3)(980 cm/s2)] = 55.00 cm

(c) The fluid will just barely enter the vein if the gauge pressure at the needle is equal to the specified blood pressure. Thus,

h = (P – P0)/rg = (18 mm Hg)[(1.013 x 106 dyne/cm2) / (760 mm Hg)] / [(1.00 g/cm3)(980 cm/s2)] = 24.48 cm

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PROBLEM

Suppose a person can reduce the pressure in the lungs to – 80 mm-Hg gauge pressure. How high can water then be sucked up a straw?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 10.71]

SOLUTION

When the person sucks on the straw, the pressure at the end of the straw inside the person’s mouth is reduced to 80 mm-Hg below atmospheric pressure, and the water rises inside the straw. Assume that the straw is oriented vertically. If the pressure at the water surface inside the container is P0 = 760 mm-Hg, and the pressure at the end of the straw inside the person’s mouth is P = 760 mm-Hg – 80 mm-Hg = 680 mm-Hg, P0 and P are related by

P = P0 - rgh

where r = 1.0 g/cm3 is the density of water, g = 980 cm/s2 is the acceleration of gravity, and h is the height of the water inside the straw. Thus,

h = (P0 - P) / rg
= (760 mm-Hg – 680 mm-Hg)(1 atm / 760 mm-Hg)(1.013 x 106 dyne/cm2 atm) / [(1.0 g/cm3)(980 cm/s2)]
= 108.81 cm = 1.0881 m

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PROBLEM

A patient is being fed through a nasogastric tube. A flow rate of 100 cm3/min is observed when the food is a height h = 45 cm above the level of the stomach and the pressure in the stomach is 10 mm Hg. Calculate the new flow rate if the food is raised to a new height of 75 cm. The food density is r = 1.2 g/cm2.

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 7.9]

SOLUTION

Assume that the pressure at the top of the food is P0 = 760 mm Hg = 1.013 x 105 N/m2 = 1.013 x 106 dyne/cm2. Then the pressure at the end of the tube at the level of the stomach is

P1 = P0 + rgh => P1 – P0 = rgh
= (1.2 g/cm2)(980 cm/s2)(45 cm) = 5.292 x 104 dyne/cm2

If the food is raised to a new height of 75 cm, the pressure at the end of the tube at the level of the stomach is

P2= P0 + rgh => P2 - P0 = rgh
= (1.2 g/cm2)(980 cm/s2)(75 cm) = 8.820 x 104 dyne/cm2

The gauge pressure of the stomach is

Ps - P0 = (10 mm Hg)[(1.013 x 106 dyne/cm2) / (760 mm Hg)]
= 1.333 x 104 dyne/cm2

The flow rate is proportional to the difference between the pressure at the end of the tube and the stomach pressure. Therefore, the new flow rate F2 is related to the original flow rate F1 according to

F2 = F1(P2 - Ps) / (P1 - Ps) = F1[(P2 – P0) – (Ps - P0)] / [(P1 - P0) – (Ps - P0)]
= (100 cm3/min)(8.820 x 104 dyne/cm2 - 1.333 x 104 dyne/cm2) / (5.292 x 104 dyne/cm2 - 1.333 x 104 dyne/cm2) = 189.1 cm3/min

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PROBLEM

(a) A fluid mass is rotating at constant angular velocity w about the central vertical axis of a cylindrical container. Show that the variation of pressure in the radial direction is given by

dp/dr = rw2r

(b) Take p = pc at the axis of rotation (r = 0) and show that the pressure p at any point r is

p = pc + (1/2)rw2r2

(c) Show that the variation of pressure with depth h is

dp = rg dh

(d) Show that the liquid surface is of paraboloidal form; that is, a vertical cross section of the surface is the curve

y = w2r2/2g

[from Resnick, Robert, and Halliday, David 1966, Physics (New York: Wiley), problem 17.22]

SOLUTION

(a) Consider an infinitesimal fluid element contained within an infinitesimal volume element located between radial coordinates r and r + dr, of small angular width Dq, and extending over a small distance Dy parallel to the axis of the cylindrical container. Here, dr is infinitesimal and much less than either rDq or Dy.

Pressure is exerted on the fluid element, along the radial direction, by the parts of the fluid adjacent to it at radial distances r and r + dr. The fluid at radial distances r and r + dr exerts pressures p(r) and p(r + dr). The pressure at r acts over an area

A(r) = r Dq Dy

The pressure at r + dr acts over an area

A(r + dr) = (r + dr) Dq Dy = r Dq Dy + dr Dq Dy = A(r) + dr Dq Dy => A(r)

in the limit dr => 0.

The total force, in the radial direction, on the fluid element is

dF = p(r) A(r) r - p(r + dr) A(r) r = [p(r) - p(r + dr)] A(r) r

In the limit dr => 0, the mass of the fluid element is

dm = r dV = r A(r) dr

The centripetal acceleration of the fluid element is

ac = - (v2/r) r = - rw2 r

Writing Newton's second law, in the radial direction, for the fluid element, we get

dF = dm ac
=> [p(r) - p(r + dr)] A(r) r = - r A(r) dr rw2 r
=> p(r) - p(r + dr) = - r dr rw2
=> [p(r + dr) - p(r)] / dr = rrw2
=> lim(dr => 0) {[p(r + dr) - p(r)] / dr} = lim(dr => 0) rrw2
=> dp/dr = rw2r (1)

(b) Multiplying both sides of (1) by dr, we get

dp = rw2r dr => ∫pcp(r) dp = ∫0r rw2r dr => p(r) - pc = rw2r2/2 => p(r) = pc + (1/2)rw2r2 (2)

(c) Consider an infinitesimal fluid element extending from depth h to depth h + dh with uniform cross sectional area dA, so that its volume is

dV = dA dh

The mass of the fluid element is

dm = r dV = r dA dh

The pressure exerted on the fluid element at depths h and h + dh is p(h) and p(h + dh). The total force on the fluid element in the vertical direction is

dF = - p(h) dA y + p(h + dh) dA y - g dm y = 0
=> - p(h) dA + p(h + dh) dA - gr dA dh = 0
=> - p(h) + p(h + dh) - gr dh = 0
=> [p(h + dh) - p(h)] / dh = rg
=> lim(dh => 0) {[p(h + dh) - p(h)] / dh} = lim(dh => 0) rg
=> dp/dh = rg (3)

(d) Multiplying both sides of (3) by dh, we get

dp = rg dh => ∫p0p(h) dp = rg ∫0h dh => p(h) - p0 = rgh => p(h) = p0 + rgh (4)

where p0 is atmospheric pressure. So the fluid pressure increases linearly with the distance from the fluid's surface. Define y = 0 to be the level of the fluid's surface at r = 0, and define p(r, y) to be the fluid pressure as a function of r and y, with y measured positive upward. Then from (2) we can write

p(r, 0) = p0 + (1/2)rw2r2 (5)

and from (4) we can write

p(r, y) = p0 + rg(ys[r] - y) => p(r, 0) = p0 + rgys(r) (6)

where ys(r) is the level of the fluid's surface at radial coordinate r. Equating the expressions for p(r, 0) from (5) and (6), we get

p0 + (1/2)rw2r2 = p0 + rgys(r) => ys(r) = w2r2/2g

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Fluid Dynamics

PROBLEM

A given pressure difference produces a flow rate of 2.0 liters/min through one tube and 3.0 liters/min through another tube. What would the flow rate be if these tubes were hooked together so that the fluid has to flow through them sequentially? The pressure difference remains the same.

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 6.37]

SOLUTION

The flow rate F through a tube is related to the pressure at the endpoints by

F = (P1 - P2) / R

where P1 is the pressure where the fluid enters the tube, P2 is the pressure where the fluid exits the tube, and R is the resistance.

For the two individual tubes, we have the flow rates

F1 = (P1 - P2) / R1 = 2.0 L/min

and

F2 = (P1 - P2) / R2 = 3.0 L/min

where R1 and R2 are the resistances of the tubes,

R1 = (P1 - P2) / F1

and

R2 = (P1 - P2) / F2

Now suppose that we connect the tubes in series, with the fluid flowing through tube 1 first and then tube 2. Assume that P1 is the pressure at the entrance of tube 1, P2 is the pressure at the end of tube 2, and P’ is the pressure at the junction between tubes 1 and 2. Then the flow rate through the connected tubes is

F = (P1 - P’) / R1 = (P’ – P2) / R2

because the flow rate must be the same through both tubes. The flow rate through the connected tubes is also

F = (P1 - P2) / R

where R is the resistance of the combination. Thus,

(P1 - P’) / R1 = (P1 - P2) / R (1)

and

(P’ – P2) / R2 = (P1 - P2) / R => P’ = (P1 - P2)(R2 / R) + P2 (2)

Rewriting (1), we get

P1 / R1 - P’ / R1 = (P1 - P2) / R
=> P1 / R1 - [(P1 - P2) / R](R2 / R1) - P2 / R1 = (P1 - P2) / R
=> (P1 - P2) / R1 = [(P1 - P2) / R](1 + R2 / R1)
=> (P1 - P2) / R = (P1 - P2) / [R1(1 + R2 / R1)] = (P1 - P2) / (R1 + R2)

Thus the resistance of the combination is

R = R1 + R2 = (P1 - P2) / F1 + (P1 - P2) / F2 = (P1 - P2)(1 / F1 + 1 / F2)

and the flow rate through the combination is

F = (P1 - P2) / R = (P1 - P2) / [(P1 - P2)(1 / F1 + 1 / F2)] = 1 / (1 / F1 + 1 / F2)
= 1 / [(F1 + F2) / F1F2] = F1F2 / (F1 + F2)
= (2.0 L/min)(3.0 L/min) / (2.0 L/min + 3.0 L/min)
= 1.2 L/min

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PROBLEM

A Pitot tube (named after Henri Pitot in 1732) is used to measure the airspeed of an airplane. It consists of an outer tube with a number of small holes, B, that is connected to one arm of a U-tube. The other arm of the U-tube is connected to a hole, A, at the front end of the device, which points in the direction the plane is headed. At A the air becomes stagnant so that vA = 0. At B, however, the speed of the air presumably equals the airspeed of the aircraft. A streamline connects A and B. Assume that A and B are at the same height.
(a) Use Bernoulli's equation to show that the airspeed is given by

v = sqrt(2rgh/rair)

where r is the density of the liquid in the U-tube, g is the acceleration of gravity, h is the difference in the height of the liquid in the two arms of the U-tube, and rair is the density of the air.
(b) If rair = 1.03 kg/m3, the U-tube contains alcohol with density r = 0.81 x 103 kg/m3, and the height difference is h = 26 cm, what is the plane's airspeed?

[from Halliday, David, and Resnick, Robert 1988, Fundamentals of Physics, Third Edition Extended (New York: John Wiley and Sons), problems 16.68 and 16.69]

SOLUTION

(a) Bernoulli's equation states that

p + (1/2)rv2 + rgy = constant

Thus,

pA + (1/2)rairvA2 + rairgyA = pB + (1/2)rairvB2 + rairgyB

Solving for pA - pB, we get

pA - pB = (1/2)rair(vB2 - vA2) + rairg(yB - yA) = (1/2)rairvB2

But

pA - pB = rgh

Equating the two expressions for pA - pB,

(1/2)rairvB2 = rgh => vB = sqrt(2rgh/rair) = v

(b) v = sqrt[(2)(0.81 x 103 kg/m3)(9.8 m/s2)(0.26 m)/(1.03 kg/m3)] = 63.31 m/s

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PROBLEM

To fight a fire on the fourth floor of a building, firemen want to use a hose of diameter 6.35 cm (2 1/2 in) to shoot 950 L/min of water to a height of 12 m.
(a) With what minimum speed must the water leave the nozzle of the fire hose if it is to ascend 12 m?
(b) What pressure must the water have inside the fire hose? Ignore friction.

[from Ohanion, Hans C. 1985, Physics, Second Edition (New York: W. W. Norton & Company), problem 18.42]

SOLUTION

(a) By conservation of energy, the kinetic energy of the water when it leaves the nozzle must equal its potential energy when it reaches its highest point.

(1/2)mv2 ≥ mgh => v ≥ sqrt(2gh) = sqrt[(2)(9.8 m/s2)(12 m)] = 15.34 m/s

(b) Consider a streamline which passes from the inside of the hose to the outside through the nozzle. Call point (1, 2) a point on the streamline (inside the hose, just outside the nozzle). Using Bernoulli's equation, we have

(1/2)rv12 + rgz1 + p1 = (1/2)rv22 + rgz2 + p2

Take z1 = z2. Then

p1 - p2 = (1/2)r(v22 - v12)

The volume of water flowing through the hose is

dV/dt = p(d/2)2v1 => v1 = (dV/dt) / [p(d/2)2]
= (950 L/min)(103 mL/L)(1 cm3/mL)(10-6 m3/cm3)(1 min / 60 s) / [p(6.35 x 10-2 m / 2)2]
= 5.000 m/s
=> p1 - p2 = (1/2)(103 kg/m3)[(15.34 m/s)2 - (5.000 m/s)2] = 1.051 x 105 N/m2 = 1.041 atm

The water inside the hose must be at a pressure of 2.041 atm, or 1.041 atm above atmospheric pressure.

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PROBLEM

What gauge pressure in the water mains is necessary if a fire hose is to spray water to a height of 12.0 m?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 10.37]

SOLUTION

The velocity of the water v0 when it emerges in the upward direction from the nozzle is related to its final velocity vf = 0 when it reaches the maximum height y = h = 12.0 m according to

vf2 = v02 + 2ay => 0 = v02 - 2gh => v02 = 2gh

where a = - g is the acceleration due to gravity.

Now consider what happens as the water passes from the water main to the nozzle and exits through the nozzle.

water main
----------            ^
           \          |
             \ nozzle |
               ------  |
   --->                |
               --------+
             /       2
           /
----------
    1


Consider a streamline that passes from the water main to the nozzle and exits. Bernoulli’s equation applied to the streamline from point 1 to point 2 is

P1 + (1/2)rv12 + rgh1 = P2 + (1/2)rv22 + rgh2

If we assume that h1 = h2 (i.e., neglect any difference in height between the water main and the nozzle), this becomes

P1 + (1/2)rv12 = P2 + (1/2)rv22

The rate of flow through the water main and the nozzle must be the same. Thus, if A1 is the cross sectional area of the water main, and A2 is the cross sectional area of the nozzle, the rate at which water flows is

A1v1 = A2v2 => v1 = (A2/A1)v2

The pressure at point 2 is equal to atmospheric pressure. Thus, the gauge pressure in the water main is

P1 - P2 = (1/2)rv22 - (1/2)rv12 = (1/2)rv22 - (1/2)r(A2/A1)2v22
= (1/2)rv22[1 – (A2/A1)2]

Since A2 << A1,

P1 - P2 ≈ (1/2)rv22

But v2 = v0, because point 2 is the exit point, so

P1 - P2 ≈ (1/2)rv02 = (1/2)r(2gh) = rgh = (103 kg/m3)(9.8 m/s2)(12 m)
= 1.176 x 105 N/m2 = (1.176 x 105 N/m2)(1 atm / 1.013 x 105 N/m2)
= 1.161 atm

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PROBLEM

A cylindrical tank has a base area A and a height H; it is initially full of water. The tank has a small hole of area A' at its bottom. Calculate how long it will take all the water to flow out of this hole.

[from Ohanion, Hans C. 1985, Physics, Second Edition (New York: W. W. Norton & Company), problem 18.52]

SOLUTION

Consider a streamline from the top of the tank to the hole at the bottom. Call point (1, 2) the position along the streamline at the (top, bottom). According to Bernoulli's equation,

(1/2)rv12 + rgz1 + p1 = (1/2)rv22 + rgz2 + p2

We have v1 ≈ 0, p1 = p2 = 1 atm, z1 = h, the height of the water level above the bottom of the tank, and z2 = 0.

rgh = (1/2)rv22 => v2 = sqrt(2gh)

Using the equation of continuity, we have

- A dh/dt = A'v2 => dh/dt = - (A'/A) sqrt(2gh) => dh / sqrt(h) = - (A'/A) sqrt(2g) dt
=> ∫ dh / sqrt(h) = - (A'/A) sqrt(2g) ∫ dt => 2h1/2 = C - (A't/A) sqrt(2g)

where C is a constant. When t = 0, we have h = H, so

C = 2H1/2

and

2h1/2 = 2H1/2 - (A't/A) sqrt(2g)

The time at which h = 0 is given by

0 = 2H1/2 - (A't/A) sqrt(2g) => t = 2H1/2(A/A') / sqrt(2g) = (A/A') sqrt(2H/g)

This is the time it takes for all the water to flow out of the tank through the hole.

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PROBLEM

You need to siphon water from a clogged sink. The sink has an area of 0.375 m2 and is filled to a height of 4.0 cm. Your siphon tube rises 50 cm above the bottom of the sink and then descends 100 cm to a pail. The siphon tube has a diameter of 2.0 cm.
(a) Assuming that the water enters the siphon tube with almost zero velocity, calculate its velocity when it enters the pail.
(b) Estimate how long it will take to empty the sink.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 10.80]

SOLUTION

(a)

  -         +----------+   -
  ^         |+--------+|   ^
  | ---+    ||  +---  ||   |
  |    |    ||  |     ||   |
50 cm  |    ||  |     ||   |
  |    |    ||  |     ||   |
  |    |----||--|     ||   |
  v    |4 cm||  |     ||100 cm
  -    +--------+     ||   |
          sink        ||   |
                      ||   |
                      ||   |
                      ||   |
                      ||   |
                      ||   v
                   +------+
                   |      |
                   | pail |
                   |      |
                   |      |
                   +------+


Consider a streamline running from one end of the siphon tube to the other end. Let point (1, 2) be at the (sink, pail) end. Bernouilli's equation is

P1 + (1/2)rv12 + rgy1 = P2 + (1/2)rv22 + rgy2

Both pressures P1 and P2 are approximately equal to atmospheric pressure, so we eliminate them and the equation becomes

(1/2)rv12 + rgy1 = (1/2)rv22 + rgy2

Since v1 ≈ 0, we can write

rgy1 = (1/2)rv22 + rgy2 => gy1 = (1/2)v22 + gy2
=> v2 = sqrt[2g(y1 - y2)] = sqrt[(2)(9.8 m/s2)(0.5 m)] = 3.130 m/s

(b) The rate at which water is siphoned out of the sink is

dV/dt = pr2v2

where r = 1 cm = 0.01 m is the radius of the siphon tube. The total volume of water that has to be siphoned out is

Vwater = Asinkd

where Asink = 0.375 m2 is the area of the sink and d = 0.04 m is the initial depth of the water. The time required to empty the sink is

t = Vwater / (dV/dt) = Asinkd / pr2v2 = (0.375 m2)(0.04 m) / p(0.01 m)2(3.130 m/s) = 15.25 s

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Electric Fields

An electric dipole consists of two charges of equal magnitude but opposite sign separated by some distance. The electric dipole moment is defined as a vector whose magnitude is equal to the product of the magnitude of each charge and the separation distance, and which points from the negative charge to the positive charge.

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PROBLEM

Identical charges q1 = q and q2 = q are located at (x, y) = (-a, 0) and (a, 0), at the endpoints of the base of an isosceles triangle whose base angles are q = 30°. Where on the y axis should a third charge q3 = 2q be placed so that the electric field at the third vertex of the triangle, at (x, y) = (0, d), is zero?
             ^ y
             |
     q3 = 2q o y = ?
             |
             |
            /|\
           / | \
          /  |  \
         /   |   \
        /    |    \
       /    d|     \
      /      |      \
     /       |       \
    / q      |      q \
---o---------+---------o--->x
q1 = q              q2 = q
(-a,0)               (a,0)


SOLUTION

The tangent of q is

tan q = d/a => d = a tan q = a tan 30° = a/sqrt(3)

The locations of the three charges are

r1 = -a i
r2 = a i
r3 = y j

where i is the unit vector in the +x direction and j is the unit vector in the +y direction. The location of the apex of the triangle is

r = d j = [a/sqrt(3)] j

The electric field at r is

E(r) = Si=13 kqi(r - ri) / |r - ri|3
= kq{[a/sqrt(3)] j + a i} / |[a/sqrt(3)] j + a i|3 + kq{[a/sqrt(3)] j - a i} / |[a/sqrt(3)] j - a i|3
+ 2kq{[a/sqrt(3)] j - y j} / |[a/sqrt(3)] j - y j|3

Now the distances from the apex of the triangle to r1 and r2 are the same,

|[a/sqrt(3)] j + a i| = |[a/sqrt(3)] j - a i| = sqrt(a2/3 + a2) = 2a/sqrt(3)

and

|[a/sqrt(3)] j - y j| = y - a/sqrt(3)

so

E(r) = kq[2a/sqrt(3)] j / [2a/sqrt(3)]3 - 2kq[y - a/sqrt(3)] j / [y - a/sqrt(3)]3
= kq j / [2a/sqrt(3)]2 - 2kq j / [y - a/sqrt(3)]2 = kq j (3)/(4a2) - 2kq j / [y - a/sqrt(3)]2 = 0
=> (3)/(4a2) - 2 / [y - a/sqrt(3)]2 = 0 => 8a2 = 3[y - a/sqrt(3)]2 => 2a sqrt(2) = sqrt(3)[y - a/sqrt(3)] => y = 2a sqrt(2/3) + a/sqrt(3) = [a/sqrt(3)][2 sqrt(2) + 1]

The third charge must be placed at y = [a/sqrt(3)][2 sqrt(2) + 1].

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PROBLEM

The electric dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule has a dipole moment p which can be considered as the vector sum of two dipole moments, p1 and p2. The distance between each hydrogen (H) and the oxygen (O) is about 0.96 x 10-10 m. The lines joining the center of the O atom with each H atom make an angle of 104°, and the net dipole moment has been measured to be p = 6.1 x 10-30 C m. Determine the charge q on each H atom.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 17.27]

SOLUTION

Because oxygen is more electronegative than hydrogen, the electron distributions around the hydrogens are drawn towards the oxygen. The result is that there is effectively a positive charge of +q at each of the hydrogens and a negative charge of -2q at the oxygen.

         +q H
          /
        / L
      /
-2q O 2q
      \
        \ L
          \
         +q H


The system is equivalent to two electric dipoles, one with electric dipole moment p1 = qL pointing from the oxygen to the upper hydrogen, and the other with electric dipole moment p2 = qL pointing from the oxygen to the lower hydrogen. The total electric dipole moment points to the right and has magnitude

p = p1 cos q + p2 cos q = qL cos q + qL cos q = 2qL cos q

where 2q = 104° or q = 52°. Solving for q,

q = p / 2L cos q = (6.1 x 10-30 C m) / [(2)(0.96 x 10-10 m) cos 52°]
= 5.160 x 10-20 C

Note that q < e = 1.6 x 10-19 because the probability distribution of the single electron around each hydrogen nucleus isn’t drawn completely to the oxygen, so the proton in each hydrogen nucleus is still partially shielded.

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Gauss’s Law

Gauss’s law relates the electric charge at a certain location to an integral over a surface enclosing the charge. Specifically, Gauss’s law states that

e0E · dS = qenc

where e0 = 8.85 x 10-12 C2/N m2 is the permittivity of free space, the integral is over a surface enclosing charge qenc, E is the electric field at each infinitestimal element dS of the surface, dS = dS n, dS is the area of the surface element, and n is the unit vector pointing outward normal to the surface.

Gauss’s law can be used to calculate the electric field due to infinite sheets of charge which are parallel to each other. There are two ways to do this.

Method 1: Determine the electric field magnitude and direction due to each sheet of charge.  Then in each region (to the left of all sheets, in between the sheets, and to the right of all sheets) add up the contributions from each sheet to get the total electric field magnitude and direction.  

Method 2: In cases where the electric field can be deduced to be zero at some locations, for example, within the plates of a parallel plate capacitor, consider a Gaussian pill box enclosing the surface of one of the plates, with one end of the pill box inside the conductor and the other end outside, calculate the electric field magnitude and direction outside of the conductor, and ignore the other plate.  In this case, where you calculate the total electric field by considering only one of the charged plates, the electric field values you use in Gauss's law are the total electric field, not just due to one sheet of charge.

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PROBLEM

Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities ± s. Ignore edge effects and use Gauss’s law to show (a) that for points far from the edges, the electric field between the plates is E = s/e0 and (b) that outside the plates on either side the field is zero. (c) How would your results be altered if the two plates were nonconductors? (d) How would your results for (a), (b), and (c) change if the surface charge density was +s on both plates?

[from Giancoli, Douglas C. 2008, Physics for Scientists and Engineers, Fourth Edition (Upper Saddle River, New Jersey: Pearson Prentice Hall), problems 22.24 and 22.25]

SOLUTION

(a) Since the plates are conducting, the charge is drawn to the sides of the plates which face each other.
      +s     -s
   +---+      +---+
   |  +|      |-  |
   |  +|      |-  |
+--|---|--+   |-  |
|  |  +|  |   |-  |
|  |  +|  |   |-  |
+--|---|--+   |-  |
   |  +|      |-  |
   |  +|      |-  |
I  +---+  II  +---+  III  --->
                           x


Construct a Gaussian pill box with one end to the left of the positively charged plate (region I) and the other end to the right of the plate (region II), both parallel to the plate. Apply Gauss’s law,

e0E+ · dS = qenc => ∫ E+ · dS = qenc/e0

where E+ is the electric field due to the positively charged plate. We break up the integral into contributions from the left end, the side, and the right end, and get

left E+ · dS + ∫side E+ · dS + ∫right E+ · dS = qenc/e0 (1)

To the left of the positively charged plate, the electric field due to the positive charge points to the left. To the right of the positively charged plate, the electric field of the positive charge points to the right. On both sides of the positively charged plate, the electric field due to the positive charge has the same magnitude, which we can call E+.

On the left end of the Gaussian pill box, dS points to the left. On the right end of the Gaussian pill box, dS points to the right. Along the side of the pill box, E+ is perpendicular to dS, so E+ · dS = 0. Thus, if the area of each end of the pill box is A, we have

left E+ · dS = ∫left E+ dS = E+A
side E+ · dS = 0
right E+ · dS = ∫left E+ dS = E+A
qenc = sA

Substituting into (1),

E+A + 0 + E+A = sA/e0 => 2E+A = sA/e0 => E+ = s/2e0

Thus, the electric field due to the positively charged plate is

E+ = - (s/2e0)x (region I)
E+ = + (s/2e0)x (regions II and III)

Similarly, we can show that the electric field due to the negatively charged plate is

E- = + (s/2e0)x (regions I and II)
E- = - (s/2e0)x (region III)

Between the plates (region II), the total electric field is the sum of the fields due to each plate,

E = E+ + E- = + (s/2e0)x + (s/2e0)x = + (s/e0)x

(b) The electric field to the left of both plates (region I) is

E = E+ + E- = - (s/2e0)x + (s/2e0)x = 0

The electric field to the right of both plates (region III) is

E = E+ + E- = + (s/2e0)x - (s/2e0)x = 0

(c) If the plates were nonconductors, the results would be the same.

(d) Since the plates are conducting, the charge is drawn to the sides of the plates which face away from each other.
   +s           +s
   +---+      +---+
   |+  |      |  +|
   |+  |      |  +|
+--|---|--+   |  +|
|  |+  |  |   |  +|
|  |+  |  |   |  +|
+--|---|--+   |  +|
   |+  |      |  +|
   |+  |      |  +|
I  +---+  II  +---+  III  --->
     1          2          x


Using the same approach as above, the electric field due to the plate on the left is

E1 = - (s/2e0)x (region I)
E1 = + (s/2e0)x (regions II and III)

The electric field due to the plate on the right is

E2 = - (s/2e0)x (regions I and II)
E2 = + (s/2e0)x (region III)

The total electric field is

E = E1 + E2 = - (s/2e0)x - (s/2e0)x = - (s/e0)x (region I)
E = E1 + E2 = + (s/2e0)x - (s/2e0)x = 0 (region II)
E = E1 + E2 = + (s/2e0)x + (s/2e0)x = + (s/e0)x (region III)

The results are unchanged if the plates are nonconducting.

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Capacitors

PROBLEM

A capacitor C1 carries a charge Q0. It is then connected directly to a second, uncharged, capacitor C2, as shown. What charge will each carry now? What will be the potential difference across each?

      C1     C2
      ||     ||
+-----||-----||-----+
|     ||     ||     |
|                   |
|                   |
+-------------------+


[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 17.65]

SOLUTION

The circuit is equivalent to the following diagram.

        |
        |
    +---+---+
    |       |
   +|       |+
C1 ---     --- C2
   ---     ---
   -|       |-
    |       |
    +---+---+
        |
        |


After the two capacitors are connected together, the potential differences across capacitors C1 and C2 are

V1 = Q1/C1
V2 = Q2/C2

where Q1 and Q2 are the charges on each capacitor. Because the capacitors are connected in parallel,

V1 = V2 => Q1/C1 = Q2/C2 => Q2 = Q1(C2/C1)

The charge Q0 that was on C1 is now distributed between C1 and C2. From charge conservation,

Q0 = Q1 + Q2 = Q1 + Q1(C2/C1) = Q1(1 + C2/C1) => Q1 = Q0/(1 + C2/C1)
=> Q2 = (C2/C1)Q0/(1 + C2/C1) = Q0C2/(C1 + C2)

The potential differences across the capacitors are

V1 = Q1/C1 = Q0/(C1 + C2)
V2 = Q2/C2 = Q0/(C1 + C2) = V1

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DC Circuits

PROBLEM

(a) Determine the current through each of the resistors in the diagram if R1 = 22 W, R2 = 25 W, R3 = 15 W, R4 = 12 W, R5 = 14 W, and E = 6.0 V.
(b) What is the net resistance of the circuit connected to the battery?

            ^
           /|\
          / | \
        R1  |  R4
        /   |   \
       /    |    \
+-----<    R3     >-----+
|      \    |    /      |
|       \   |   /       |
|       R2  |  R5       |
|         \ | /         |
|          \|/          |
|           v           |
|                       |
|          |            |
+----------||-----------+
           |
           E


[from Giancoli, Douglas C. 2008, Physics for Scientists and Engineers, Fourth Edition (Upper Saddle River, New Jersey: Pearson Prentice Hall), problem 26.38]

SOLUTION

(a) Let Ii be the current through resistor Ri. Assume I1, I2, I4, and I5 flow towards the right, and I3 flows downward. Let I be the current through the battery.

Write the loop theorem for the loop which passes through the battery, R2, and R4, in that order. The sum of the potential differences encountered in going around a complete loop must be zero. Thus,

6 – 25I2 - 14I5 = 0 (1)

Write the loop theorem for the loop which passes through R1, R3, and R2.

- 22I1 - 15I3 + 25I2 = 0 (2)

Write the loop theorem for the loop which passes through R3, R4, and R5.

15I3 - 12I4 + 14I5 = 0 (3)

Applying current conservation at each of the four junctions, we get

I = I1 + I2 = I4 + I5 (4)
I1 = I3 + I4 (5)
I5 = I2 + I3 (6)

Use (6) to substitute for I5 in (1) through (5). We get

6 – 25I2 - 14(I2 + I3) = 0 => 6 – 39I2 - 14I3 = 0 (7)
- 22I1 - 15I3 + 25I2 = 0 (8)
15I3 - 12I4 + 14(I2 + I3) = 0 => 29I3 - 12I4 +14I2 = 0 (9)
I = I1 + I2 = I4 + I2 + I3 => I1 = I3 + I4 (10)

Use (10) to substitute for I1 in (7) through (9).

6 – 39I2 - 14I3 = 0 => I3 = 3/7 – (39/14)I2 (11)
- 22(I3 + I4) – 15I3 + 25I2 = 0 => - 37I3 - 22I4 + 25I2 = 0 (12)
29I3 - 12I4 + 14I2 = 0 (13)

Use (11) to substitute for I3 in (12) and (13).

- 37[3/7 – (39/14)I2] – 22I4 + 25I2 = 0 (14)
29[3/7 – (39/14)I2] – 12I4 + 14I2 = 0 (15)

Multiply (14) and (15) by 14.

- 37(6 – 39I2) – (22)(14)I4 + (25)(14)I2 = 0 (16)
29(6 – 39I2) – (12)(14)I4 + (14)(14)I2 = 0 (17)

Simplifying (16) and (17), we get

- 222 + 1443I2 - 308I4 + 350I2 = 0 => - 222 + 1793I2 - 308I4 = 0 (18)
174 – 1131I2 - 168I4 + 196I2 = 0 => 174 - 935I2 - 168I4 = 0 (19)

Factoring the coefficients of I4 in (18) and (19), we get

308 = 22 x 7 x 11
168 = 23 x 3 x 7

The lowest common multiple of 308 and 168 is the product of the highest power of each prime factor of either number or 23 x 3 x 7 x 11 = 1848. Thus, multiply (18) by 2 x 3 = 6 and multiply (19) by 11.

- 1332 + 10758I2 - 1848I4 = 0 (20)
1914 - 10285I2 - 1848I4 = 0 (21)

Subtract (21) from (20).

- 3246 + 21043I2 = 0 => I2 = 3246/21043

Solving (18) for I4,

I4 = (- 222 + 1793I2)/308 = - 222/308 + (1793/308)I2
= - 222/308 + (1793/308)(3246/21043)
= [- (222)(21043) + (1793)(3246)] / [(308)(21043)]
= 1148532 / [(308)(21043)] = 339/1913

Using (11), we get

I3 = 3/7 – (39/14)I2 = 3/7 – (39/14)(3246/21043) = 3/7 – (39/7)(1623/21043) = [(3)(21043) – (39)(1623)] / [(7)(21043)] = - 168 / [(7)(21043)] = - 24/21043

Thus, I3 actually flows upward through R3.

Using (6),

I5 = I2 + I3 = 3246/21043 – 24/21043 = 3222/21043

Using (5),

I1 = I3 + I4 = - 24/21043 + 339/1913 = [- 24 + (339)(11)]/21043 = 3705/21043

So we have I1 = 3705/21043 A, I2 = 3246/21043 A, I3 = - 24/21043 A, I4 = 339/1913 A, and I5 = 3222/21043 A. Note that the magnitude of I3 is much smaller than the magnitudes of the other currents. To confirm that these are correct, it can be shown that they satisfy the original equations (1) through (6).

(b) The total current through the battery is

I = I1 + I2 = 3705/21043 A + 3246/21043 A = 6951/21043 A

The net resistance of the circuit is

R = E / I = (6 V) / (6951/21043 A) = (2 V) / (2317/21043 A) = 42086/2317 W
= 18.16 W

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Magnetic Force on a Charged Particle

A charged particle of charge q moving with velocity v in a magnetic field B experiences a magnetic force

F = qv x B

If the particle is moving in a plane perpendicular to B, the force will be perpendicular to the particle's direction of motion, and the particle will move in a circle. We can then write

F = - qvB r

where r is the radial unit vector. The force F is equal to the time derivative of the particle momentum, or

F = dp/dt = - qvB r (1)

Multiplying (1) by dt, we get

dp = - qvB dt r (2)

Equating the magnitudes of both sides of (2),

dp = qvB dt (3)

During an infinitesimal time interval dt, the particle's momentum will change from some initial value p1 to another value p2 such that

p2 = p1 + dp

where dp is the change in the momentum during the time interval dt. p1 and p2 have the same magnitudes but point in slightly different directions due to the curvature of the circular path. The angle between the directions of p1 and p2 is dq, which is also the angular distance that the particle moves along its circular path during the time interval dt. Thus, the magnitude of dp is

dp = p dq (4)

Combining (3) and (4) yields

p dq = qvB dt

Dividing both sides by dt,

p dq/dt = qvB (5)

Now dq/dt is the angular velocity of the particle in its circular path. Since the particle traverses 2p radians in the time T (its period) that it takes to complete one circle,

dq/dt = 2p/T

But if R is the radius of the path, then the period is

T = 2pR/v

so

dq/dt = 2p / (2pR/v) = v/R

Substituting this result into (5), we get

pv/R = qvB => p/R = qB => R = p/qB

This result is valid even if the particle is relativistic, but if the particle is relativistic, we must use

p = gmv

where g = 1 / sqrt(1 - b2), b = v/c, c is the speed of light in a vacuum, and m is the particle's rest mass. For a nonrelativistic particle, g => 1 and p => mv, the classical result.

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Biot-Savart Law

According to the Biot-Savart law, the magnetic field dB(r) at point r due to a current I running through an infinitesimal wire segment dl at position r' is

dB(r) = (m0/4p) I dl x (r - r') / |r - r'|3

where dB(r), r, r', and dl are all vectors, m0 = 4p x 10-7 T m/A is the permeability constant, and "x" indicates a vector cross product. r - r' is the vector pointing from r' to r. dl points in the direction of current flow in the wire at r'. The current

I = dq/dt

is the charge per unit time flowing through the wire segment dl. The product

I dl = (dq/dt) dl = dq (dl/dt) = v dq

can be thought of as representing an amount of charge dq which flows through the wire segment dl in time dt, or an amount of charge dq which flows through the wire segment dl with velocity

v = dl/dt

Thus, the Biot-Savart law may be rewritten as

dB(r) = (m0/4p) dq v x (r - r') / |r - r'|3

to give the magnetic field due to a charge dq which moves with velocity v. Eliminating the differentials, we have

B(r) = (m0/4p) q v x (r - r') / |r - r'|3

We can use this formula to calculate the magnetic field B(r) due a charge q moving with velocity v at r'.

PROBLEM

Particle 1, with charge q0 > 0 is located a distance d above particle 2, with charge - q0 < 0. The direction from particle 2 to particle 1 is the +y direction. Both are moving in the +x direction with velocity v0.

^ 1 o---->
|   q0  v0
|          y^
d           |
|           +--->
|               x
v 2 o---->
   -q0  v0


(a) Determine the magnitude and direction of the magnetic force acting on particle 1. Ignore electric forces and gravity.
(b) What is the magnitude and direction of the magnetic field at the midpoint between the particles?

[from Physics 2212, Intro Physics II, Quiz 4, Fall 2005, Georgia Institute of Technology]

SOLUTION

(a) First determine the magnetic field at the location of particle 1 due to particle 2. We use the form of the Biot-Savart law for the magnetic field due to a moving point charge.

B(r) = (m0/4p) q v x (r - r') / |r - r'|3

where r is the position of particle 1, r' is the position of particle 2, q = - q0 is the charge on particle 2, and v = v0 i is the velocity of particle 2. We have

r - r' = d j

and

|r - r'|3 = d3

The magnetic field at the location of particle 1 due to particle 2 is

B(r) = (m0/4p)(-q0)(v0 i) x (d j) / d3 = (m0/4p)(-q0v0/d2) k = (m0/4p)(q0v0/d2) (-k)

The magnetic field at the location of particle 1, due to particle 2, points in the -k or -z direction, which is into the screen. Now use the magnetic force equation to determine the magnetic force on particle 1.

F = qv x B = q0(v0 i) x (m0/4p)(q0v0/d2) (-k) = q0v0(m0/4p)(q0v0/d2) j = (m0q02v02/4pd2) (+j)

The magnetic force on particle 1 has magnitude m0q02v02/4pd2 and points in the +j or +y direction.

(b) By symmetry, both particles produce the same magnetic field at the point halfway between them, so it is sufficient to calculate the magnetic field due to one of them and double it to get the result. We will calculate the magnetic field due to particle 2 and call this (1/2)B(r1/2). The magnetic field due to particle 2 is calculated the same way as in part (a) except that we replace d with d/2 everywhere. Thus,

(1/2)B(r1/2) = (m0/p)(q0v0/d2) (-k) = (m0q0v0/pd2) (-k)

and the magnetic field due to both particles at the point halfway between them is

B(r1/2) = (2m0q0v0/pd2) (-k)

It has magnitude 2m0q0v0/pd2 and points in the -k or -z direction, which is into the screen.

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Faraday's Law

PROBLEM

A horizontal conducting rod of mass M and length D can slide freely in the vertical direction between vertical conducting rails with which it makes electrical contact at its ends. The rails are connected to each other at ground level via a resistance R, and a uniform magnetic field B0 is oriented horizontally through the rectangular loop formed by the rod, the rails, and the ground.

x | x x x x x | x
x | x x D x x | x
x ============= x
x | x x | x x | x
x | x x | x x | x B0 into screen
x | x x v x x | x
x | x x x x x | x
x | x x x x x | x
x +-----R-----+ x


(a) Use Faraday's Law to determine the induced emf in the circuit at a moment when the rod is at height h and falling with speed v. Does the induced current flow clockwise or counterclockwise?
(b) Find an expression (in terms of the given parameters) for the power dissipated in the resistor when the rod is at height h and falling with speed v.
(c) If the rod is released from rest at some height h0, it will fall and eventually reach a terminal speed vmax. By considering the rate at which gravitational potential energy is lost by the rod, find an expression for the terminal speed in terms of the given parameters.

[from Physics 2212, Intro Physics II, Quiz 5, Fall 2004, Georgia Institute of Technology]

SOLUTION

(a) The induced emf is

E = - dFB/dt

where FB is the magnetic flux enclosed by the loop.

FB = ∫ B · dS = B0Dh

where the integral is done over the area enclosed by the loop and h is the height of the rod.

E = - (d/dt) B0Dh = - B0D dh/dt = B0Dv

The induced current flows in the direction which tries to increase the enclosed magnetic flux into the screen, so the direction is clockwise.

(b) The current through the resistor is

I = E/R = B0Dv/R

The power dissipated in the resistor is

P = I2R = B02D2v2/R

(c) The gravitational potential energy of the rod is

U = Mgh

The rate at which the gravitational potential energy of the rod changes is

dU/dt = Mg dh/dt = - Mgv

After the rod reaches terminal velocity vmax the rate at which its kinetic energy K increases is zero. Thus the rate at which the total mechanical energy of the rod changes is

dE/dt = dK/dt + dU/dt = 0 - Mgvmax = - Mgvmax (I)

This is negative and equal to the rate at which the magnetic field does work on the rod. The magnetic field exerts an upward force on the rod due to the clockwise current which is induced in the loop. By the right hand rule, the magnetic force on the rod is

FB = IL x B => FB = IDB0 = B02D2v/R

The rate at which the magnetic field does work on the rod is

PB = dWB/dt = - FBv = IDB0 = - B02D2v2/R

After the rod reaches terminal velocity,

PB = - B02D2vmax2/R (II)

Equating the expressions in (I) and (II), we get

- Mgvmax = - B02D2vmax2/R => vmax = MgR/B02D2

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PROBLEM

A square loop of aluminum wire is 20.0 cm on a side. It is to carry a constant 25.0 A current and rotate in a 1.65 T magnetic field.
(a) Determine the minimum diameter of the wire so that it will not fracture from tension or shear. Assume a safety factor of 10.
(b) What is the resistance of a single loop of this wire?
(c) Calculate the magnetic field at the center of the loop due to the current in the loop.
(d) Now assume that a current whose peak value is 25.0 A is to be generated in the loop by rotating the loop in the magnetic field. How fast would the loop have to be rotated?

[parts (a) and (b) from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 20.71]

SOLUTION

(a) The greatest stress on the loop will occur when the face of the loop is perpendicular to the direction of the external magnetic field, with the direction of the current such that the magnetic field due to the current is in the same direction as that of the external field. Then the magnetic force on each side of the loop is

F = ILB = (25.0 A)(0.20 m)(1.65 T) = 8.25 N

directed away from the center of the loop. We neglect the magnetic field of the current in calculating the magnetic force on the loop.

           ^
           |
           | F
           |
    +------>------+
    |      I      |
    |             |
  F |             |
<---^ B into page v--->
    |             | F
    |             |
    |             |
    +------<------+
           |
           | F
           |
           v


Each corner of the loop experiences simultaneously a shearing force of F/2 and a tension of F/2. In order for the loop to withstand these forces without fracture, with a safety factor of 10, the loop must be able to withstand a shearing force and tension both equal to (10)(F/2) = 5F. The tensile strength and shear strength of aluminum are both equal to S = 200 x 106 N/m2. Thus,

5F = Spr2 = Spd2/4

where r is the radius of the wire and d is the diameter of the wire. Solving for d, we get

d = sqrt(20F/pS) = sqrt[(20)(8.25 N)/p(200 x 106 N/m2)] = 0.0005125 m = 0.5125 mm

(b) R = r(4L)/p(d/2)2 = (2.65 x 10-8 W m)(4)(0.20 m)/{p[(0.0005125 m)/2]2} = 0.1028 W

where r = 2.65 x 10-8 W m is the resistivity of aluminum.

(c) Use the Biot-Savart law. By symmetry we only need to calculate the magnetic field at the center of the loop due to one side and multiply by four. Choose a coordinate system so that the center of the loop is at the origin, one side of the loop extends from (x, y) = (- 0.10 m, 0.10 m) to (x, y) = (0.10 m, 0.10 m), and the current in this side of the loop flows in the +x (+i) direction. The magnetic field due to an infinitesimal segment of the side of the loop is

dB(r) = (m0I/4p) dl x (r - r') / |r - r'|3

We have dl = dx i, r = 0, and r' = x i + a j, where a = 0.10 m. Substituting into the above equation, we get

dB(0) = (m0I/4p)(dx i) x (- x i - a j) / |- x i - a j|3 = - (m0I/4p)(dx a k) / (x2 + a2)3/2

Integrating from x = - a to x = + a, we get

B(0) = (m0Ia/4p) k-aa dx / (x2 + a2)3/2 = (m0Ia/2p) k0a dx / (x2 + a2)3/2

Define x = a tan q => dx = a sec2 q dq, x2 + a2 = a2(tan2 q + 1) = a2 sec2 q

B(0) = - (m0Ia/2p) k0p/4 a sec2 q dq / (a2 sec2 q)3/2 = - (m0I/2pa) k0p/4 cos q dq
= - (m0I/2pa) k sin(p/4) = - (1.26 x 10-6 H/m)(25.0 A)/[(2p)(0.10 m)sqrt(2)] k = 3.545 x 10-5 T (-k)

which is much smaller than the external field and directed in the -z (-k) direction.

(d) The electromotive force (EMF) generated in the loop is

E = - dFB/dt = - (d/dt) B · A = - (d/dt) BL2 cos wt = BL2w sin wt = E0 sin wt

where A is a vector whose magnitude is the area of the loop and which points perpendicular to the face of the loop, and

E0 = BL2w

is the amplitude of the induced EMF, which is related to the peak value of the induced current according to

E0 = I0R => BL2w = I0R => w = I0R/BL2 = (25.0 A)(0.1028 W/[(1.65 T)(0.20 m)2)] = 38.94 rad/s = (38.94 rad/s)(1 rev / 2p rad) = 6.197 rev/s

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PROBLEM

A transformer has 420 turns in the primary coil and 120 in the secondary coil. What kind of a transformer is this and, assuming 100 percent efficiency, by what factor does it change the voltage? By what factor does it change the current?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 21.31]

SOLUTION

This is a step-down transformer because the output voltage is lower than the input voltage.

VS/VP = NS/NP = 120/420 = 2/7

The output voltage is 2/7 times the input voltage. Assuming 100 percent efficiency, the power output is equal to the power input.

Pout = Pin => VPIP = VSIS => IS/IP = VP/VS = 7/2 = 3.5

The output current is 3.5 times the input current.

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PROBLEM

Calculate the peak output voltage of a simple generator whose square armature windings are L = 6.60 cm on a side if the armature contains N = 125 loops and rotates in a field of B = 0.200 T at a rate of w = 120 rev/s.

[from Giancoli, Douglas C. 1998, Physics, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall, problem 21.95]

SOLUTION

The electromotive force generated is

E = - N dFB/dt

where dFB/dt is the rate at which the magnetic flux through the armature is changing. The magnetic flux through the armature is

FB = AB cos wt

where

A = L2

is the area of the armature and wt is the angle between the normal to the area enclosed by the armature and the direction of the magnetic field. So

E = - N d/dt (L2B cos wt) = NL2Bw sin wt = E0 sin wt

where

E0 = NL2Bw = (125)(6.60 x 10-2 m)2(0.200 T)(120 rev/s)(2p rad/rev) = 82.11 V

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PROBLEM

A coil has 2.25 W resistance and 440 mH inductance. If the current is 3.00 A and is increasing at a rate of 3.50 A/s, what is the potential difference across the coil at this moment?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 21.47]

SOLUTION

The potential drop across the resistor is

VR = IR

If the current is increasing, the inductance causes a back emf of magnitude

VL = L dI/dt

The potential drop across the coil is

V = VR + VL = IR + L dI/dt = (3.00 A)(2.25 W) + (0.440 H)(3.50 A/s) = 8.29 V

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Magnetic Properties

When a (paramagnetic, diamagnetic) substance is placed in an external magnetic field, the field induces a magnetic dipole moment whose direction is (parallel to, opposite) that of the external field.  If the field is nonuniform, such as that above the north pole of a strong bar magnet or a solenoid, the substance will be (attracted to, repelled from) the pole because the substance effectively becomes a magnet whose polarity is (the same as, opposite) that of the external magnet or solenoid.  The (attraction, repulsion) can be understood by considering the (paramagnetic, diamagnetic) substance to be equivalent to two magnetic poles, one of which is further away from the pole of the external magnet.  Both magnetic poles experience a magnetic force from the external field, in opposite directions, but the magnetic pole which is closer to the external magnet or solenoid is subjected to a larger external field and thus experiences a larger force.  Therefore the (paramagnetic, diamagnetic) substance is (attracted, repelled) because the closer pole is (attracted, repelled).

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LRC Circuits

PROBLEM

In the circuit shown below, resistors 1 and 2 of resistance R1 and R2, respectively, and an inductor of inductance L are connected to a battery of emf E and a switch S. The switch is closed at time t = 0. Express all algebraic answers in terms of the given quantities and fundamental constants.

          /
         /
   +----o S o---R1---+------+
   |                 |      |
   |                 |      |
E----                R2     L
  --                 |      |
   |                 |      |
   |                 |      |
   +-----------------+------+


(a) Determine the current through resistor 1 immediately after the switch is closed.
(b) Determine the magnitude of the initial rate of change of current, dI/dt, in the inductor.
(c) Determine the current through the battery a long time after the switch has been closed.
(d) Sketch a graph of the current through the battery as a function of time.
(e) Some time after steady state has been reached, the switch is opened. Determine the voltage across resistor 2 just after the switch has been opened.

[from AP Physics C Exam, Electricity and Magnetism, 2005]

SOLUTION

(a) Immediately after the switch is closed, the current through the inductor is zero, and all the current flows through R1 and R2. Applying the loop theorem around the left-hand loop,

E - IR1 - IR2 = 0 => I = E/(R1 + R2)

(b) Applying the loop theorem around the right-hand loop, starting at the bottom of R2, we have

IR2 - L dI/dt = 0 => dI/dt = IR2/L = (E/L)/(1 + R1/R2)

(c) A long time after the switch has been closed, steady state has been reached, and we can neglect the inductor, which is effectively a short. All the current flows through R1 and the inductor, and none through R2. The total resistance seen by the battery is R1, and the current through the battery is

I = E/R1

(d) The current through the battery starts at E/(R1 + R2) and increases asymptotically to E/R1.

(e) Just before the switch is opened, the current through the inductor is E/R1, flowing downward. Just after the switch is opened, the current through the inductor is still E/R1, flowing downward, but now the current flows upward through R2. The voltage across R2 is ER2/R1 with the bottom of R2 at the higher potential.

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PROBLEM

The potential difference across a given coil is V1 = 22.5 V at an instant when the current is I1 = 860 mA and is increasing at a rate of dI/dt = 3.40 A/s. At a later instant, the potential difference is V2 = 16.2 V whereas the current is I2 = 700 mA and is decreasing at a rate of dI/dt = - 1.80 A/s. Determine the inductance L and resistance R of the coil.

[from Giancoli, Douglas C. 1998, Physics, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall, problem 21.53]

SOLUTION

At the first instant, we have V = V1 = 22.5 V, I = I1 = 860 mA, and dI/dt = 3.40 A/s. Since the current is increasing, the EMF generated in the coil is in the direction opposite the current. This means that the potential differences across the coil due to the resistance and inductance both have the same sign; they both decrease in the direction of the current. We treat V as the magnitude of the potential difference across the coil.

V = RI + L dI/dt => V1 = RI1 + L dI/dt => 22.5 = 0.860R + 3.40L (I)

At the second instant, we have V = V2 = 16.2 V, I = I2 = 700 mA, and dI/dt = - 1.80 A/s. Since the current is decreasing, the EMF generated in the coil is in the direction of the current. The potential differences across the coil due to the resistance and inductance have opposite signs; the potential difference due to the resistance decreases in the direction of the current whereas the potential difference due to the inductance increases in the direction of the current. Assume the potential difference due to the inductance has a smaller magnitude than the potential difference due to the resistance, so that RI + L dI/dt is a positive quantity.

V = RI + L dI/dt => V2 = RI2 + L dI/dt => 16.2 = 0.700R - 1.80L (II)

Solve (II) for L in terms of R.

L = (0.700R - 16.2) / 1.80 = (7/18)R - 9 (III)

Substitute into (I) and solve for R.

22.5 = 0.860R + 3.40[(7/18)R - 9] => 2250 = 86R + 340[(7/18)R - 9] => 2250 = 86R + (1190/9)R - 3060 => 5310 = (1964/9)R => R = (5310)(9)/1964 = (2655)(9)/982 = 24.33 W

Now use (III) to solve for L.

L = (7/18)R - 9 = (7/18)(24.33) - 9 = 0.4628 H

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PROBLEM

A series RLC circuit consists of an 81 W resistor, a 110 mH inductor, and a 47 mF capacitor. It is attached to a power line in Algeria, operating at Vrms = 230 V and 50 Hz. What is the average power supplied by the power line?

[from Physics 2212, Intro Physics II, Quiz 5, Summer 2005, Georgia Institute of Technology]

SOLUTION

The total impedance of the circuit is

Z = sqrt[R2 + (wL - 1/wC)2] = sqrt[R2 + (2pfL - 1/2pfC)2]
=sqrt{(81 W)2 + [2p(50 Hz)(110 x 10-3 H) - 1/2p(50 Hz)(47 x 10-6 F)]2} = 87.53 W

The rms current in the circuit is

Irms = Vrms / Z = (230 V) / (87.53 W) = 2.628 A

The phase angle by which the voltage leads the current is

f = tan-1[(wL - 1/wC) / R] = tan-1[(2pfL - 1/2pfC) / R]
= tan-1{[2p(50 Hz)(110 x 10-3 H) - 1/2p(50 Hz)(47 x 10-6 F)] / (81 W)} = - 0.3887 rad = - 22.27°

The power factor is

cos f = cos(-22.27°) = 0.9254

The average power supplied by the power line is

Pavg = IrmsVrms cos f = (230 V)(2.628 A)(0.9254) = 559.3 W

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PROBLEM

A circuit contains two elements, but it is not known if they are L, R, or C. The current in this circuit when connected to a 120 V 60 Hz source is 5.6 A and lags the voltage by 50 deg. What are the two elements and what are their values?

[from Giancoli, Douglas C. 1998, Physics, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall, problem 21.99]

SOLUTION

The amplitude of the current is

I0 = V0/Z

where

Z = sqrt[R2 + (wL - 1/wC)2]

is the impedance. The voltage is of the form

V(t) = V0 cos(wt + f)

where f is the phase by which the voltage leads the current. The tangent of the phase is

tan f = (wL - 1/wC) / R

Since the phase is 50 deg, one of the components must be a resistor. Otherwise, the phase would be ± 90 deg. The voltage leads the current by 50 deg. Since the voltage leads the current, the other component must be an inductor. Thus, we have

Z = sqrt[R2 + (wL)2] = V0/I0

and

tan f = wL/R = tan 50 => wL = R tan 50 => sqrt[R2 + (R tan 50)2] = V0/I0
=> R = V0/[I0 sqrt(1 + tan2 50)] = (120 V) / [(5.6 A) sqrt(1 + tan2 50)] = 13.77 W
=> L = (R tan 50) / w = (R tan 50) / 2pf = (13.77 W)(tan 50) / [2p(60 Hz)] = 43.54 mH

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PROBLEM

In a series LRC circuit, the inductance is L = 20 mH, the capacitance is C = 50 nF, and the resistance is R = 200 W. At what frequencies is the power factor equal to 0.17?

[from Giancoli, Douglas C. 1998, Physics, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall, problem 21.103]

SOLUTION

In a series LRC circuit, the average power Pavg can be expressed in terms of the rms current Irms and the rms voltage Vrms as

Pavg = IrmsVrms cos f

where

cos f = R/Z

is the power factor,

Z = sqrt[R2 + (wL - 1/wC)2]

is the impedance, and w is the angular frequency of the driving voltage, which is related to the frequency f by

w = 2pf

The power factor is maximized, at one, when Z is minimized, which occurs when

wL - 1/wC = 0 => w = w0 = 1/sqrt(LC) = 1/sqrt[(20 x 10-3 H)(50 x 10-9 F)] = 31623 rad/s => f = f0 = w0/2p = 5033 Hz

On either side of the resonant frequency f0, the power factor decreases. We require that cos f = 0.17. We have

cos f = R / sqrt[R2 + (wL - 1/wC)2] = 1 / sqrt[1 + (wL - 1/wC)2/R2]

Squaring both sides,

cos2 f = 1 / [1 + (wL - 1/wC)2/R2]

Solving for (wL - 1/wC)/R,

(wL - 1/wC) / R = ± sqrt(1 / cos2 f - 1) = ± sqrt(1 / 0.172 - 1) = ± 5.797

Multiply by wCR.

w2LC - 1 = ± 33.60wCR => - LCw2 ± 33.60CRw + 1 = 0

This is a quadratic equation of the form

aw2 + bw + c = 0

where

a = - LC
b = ± 5.797CR
c = 1

The solution is

w = [- b ± sqrt(b2 - 4ac)] / 2a

Using b = + 5.797CR, we get

w = {-(5.797)(50 x 10-9 F)(200 W) ± sqrt[(5.797)2(50 x 10-9 F)2(200 W)2 + (4)(20 x 10-3 H)(50 x 10-9 F)]} / (2)(- 20 x 10-3 H)(50 x 10-9 F) = ± 13912 rad/s => f = w/2p = ± 2214 Hz => 2214 Hz

Using b = - 5.797CR, we get

w = {-(- 5.797)(50 x 10-9 F)(200 W) ± sqrt[(- 5.797)2(50 x 10-9 F)2(200 W)2 + (4)(20 x 10-3 H)(50 x 10-9 F)]} / (2)(- 20 x 10-3 H)(50 x 10-9 F) = ± 71879 rad/s => f = w/2p = ± 11440 Hz => 11440 Hz

The two frequencies that result in a power factor of 0.17 are 2214 Hz and 11440 Hz.

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Practical Electricity

Standard electricity available in the United States is in the form of sinusoidal alternating current (AC) with a root-mean-square voltage of 120 V and a frequency of 60 Hz (http://www.electricityforum.com/basic-electricity.html). The current drawn from this power source by an electrical device is also sinusoidal. The current rating of a device is its root-mean-square current. The power rating of a device refers to the product of the root-mean-square voltage and the root-mean-square current drawn.

Circuit breakers are designed to limit the current passing through the wiring to a particular area of a building. Circuit breakers are rated at the highest root-mean-square current which the circuit breaker will allow to pass without tripping or breaking the circuit. If the maximum allowed current is exceeded, the circuit breaker shuts off electricity to the section of the building that is drawing the current.

Electric power usasge is measured in kilowatt-hours. By estimating the energy used by various devices in your home, you can verify that the amount of electricity usage you are billed for seems reasonable. If your estimated usage and the amount that your bill says you use are very different, this might indicate either that something in your home is using a very different amount of electricity than you think, or that someone is tapping into your electric power. For example, someone may be using your exterior electrical receptacles without your knowing it.

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PROBLEM

An electrical device operated in the United States is rated at 1500 W. What electrical current is drawn by the device?

SOLUTION

The power used by the device is given by

P = VI

where V is the voltage supplied, and I is the current through the device. Thus,

I = P / V = (1500 W) / (120 V) = 12.5 A

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Heat

PROBLEM

A 15 g lead bullet traveling at 220 m/s passes through a thin iron wall and emerges at a speed of 160 m/s. If the bullet absorbs 50 percent of the heat generated,
(a) what will be the temperature rise of the bullet?
(b) If the ambient temperature is 20°C, will any of the bullet melt, and if so, how much?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 14.59]

SOLUTION

(a) The kinetic energy lost by the bullet is

KElost = KEi - KEf = (1/2)m(vi2 - vf2) = (1/2)(0.015 kg)[(220 m/s)2 - (160 m/s)2] = 171 J

The amount of heat Q absorbed by the bullet is half of this. The temperature rise DT of the bullet is given by

Q = mcDT => DT = Q/mc = (0.5)(171 J) / [(0.015 kg)(130 J/kg C°)] = 43.85°C

This assumes that the temperature increase doesn't cause the temperature of the bullet to exceed the melting point of lead (327 K).

(b) If the ambient temperature is 20°C = 293 K, an increase of 43.85°C would bring the temperature of the bullet to 336.85 K, which is above the melting point of lead, so some of it will melt. If m is the total mass of the bullet, mmelt is the mass which melts, and Lf = 0.25 x 105 J/kg is the heat of fusion,

Q = mcDT + mmeltLf => mmelt = (Q - mcDT) / Lf = [(0.5)(171 J) - (0.015 kg)(130 J/kg C°)(327 K - 293 K)] / (0.25 x 105 J/kg) = 0.768 g

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PROBLEM

A house has well-insulated walls 17.5 cm thick (assume conductivity of air) and area 410 m2, a roof of wood 6.5 cm thick and area 280 m2, and uncovered windows 0.65 cm thick and total area 33 m2.
(a) Assuming that the heat loss is only by conduction, calculate the rate at which heat must be supplied to this house to maintain its temperature at 23° C if the outside temperature is -10° C.
(b) If the house is initially at 10° C, estimate how much heat must be supplied to raise the temperature to 23° C within 30 min. Assume that only the air needs to be heated and that its volume is 750 m3.
(c) If natural gas costs $0.080 per kilogram and its heat of combustion is 5.4 x 107 J/kg, how much is the monthly cost to maintain the house as in part (a) for 24 h each day assuming 90 percent of the heat produced is used to heat the house? Take the specific heat of air to be 0.24 kcal/kg C°.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 14.58]

SOLUTION

(a) The rate at which heat is conducted out of the house through the walls, roof, and windows can be calculated using the equation

dQ/dt = kA dT/dx

walls: (dQ/dt)walls = (0.055 x 10-4 kcal/s m C°)(410 m2)(33 C°)/(0.175 m) = 0.4252 kcal/s

roof: (dQ/dt)roof = (0.3 x 10-4 kcal/s m C°)(280 m2)(33 C°)/(0.065 m) = 4.2646 kcal/s

windows: (dQ/dt)windows = (2.0 x 10-4 kcal/s m C°)(33 m2)(33 C°)/(0.0065 m) = 33.5077 kcal/s

The total rate of heat loss through conduction is

(dQ/dt)total = (dQ/dt)walls + (dQ/dt)roof + (dQ/dt)windows = 0.4252 kcal/s + 4.2646 kcal/s + 33.5077 kcal/s = 38.1975 kcal/s

(b) The total rate of heat loss through conduction if the inside temperature is 10° C instead of 23° C is

(dQ/dt)total = (38.1975 kcal/s)(20 C°)/(33 C°) = 23.1500 kcal/s

The average rate of heat loss during the 30 minutes that the house is heated is

(dQ/dt)avg = (0.5)(38.1975 kcal/s + 23.1500 kcal/s) = 30.6738 kcal/s

The total heat lost through conduction during the 30 minutes that the house is heated is

Q1 = (30.6738 kcal/s)(30 min)(60 s/min) = 55212.8 kcal

The total mass of the air in the house is

m = rV = (1.29 kg/m3)(750 m3) = 967.5 kg

The total heat needed to heat the air from 10° C to 23° C is

Q2 = (967.5 kg)(0.24 kcal/kg C°)(13 C°) = 3018.6 kcal

The total heat that must be supplied during the 30 minutes that the house is heated from 10° C to 23° C is

Q = Q1 + Q2 = 55212.8 kcal + 3018.6 kcal = 58231.4 kcal

(c) The cost per month of keeping the house at 23° C is (38.1975 kcal/s)(4.186 kJ/kcal)(3600 s/hr)(24 hr/day)(30 day/month)(1 kg / 5.4 x 104 kJ)($0.080/kg)/0.9 = $682.22/month

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Kinetic Theory of Gases

PROBLEM

An ideal gas undergoes an adiabatic expansion followed by an isothermal compression back to its original pressure.
(a) Is the final volume higher or lower than the initial volume?
(b) Is the final temperature higher or lower than the initial temperature?
(c) Sketch the two processes on a pressure vs. volume diagram.

SOLUTION

(a) Since the first process is adiabatic, the pressure p and volume V during this process are related by

pVg = constant

where g = Cp/Cv, Cp is the molar heat capacity at constant pressure, and Cv is the molar heat capacity at constant volume.

Thus,

p1V1g = p2V2g => V2 = V1(p1/p2)1/g (1)

where p1 and V1 are the initial pressure and volume, and p2 and V2 are the pressure and volume at the end of the adiabatic expansion.

According to the ideal gas law,

pV = nRT

where n is the number of moles of gas, R is the gas constant, and T is the temperature of the gas.

During the isothermal compression, the temperature of the gas remains constant, so

pV = constant => p2V2 = p3V3

where p3 and V3 are the pressure and volume at the end of the isothermal compression. We are told that the isothermal compression brings the gas back to its initial pressure, so p3 = p1, and we have

p2V2 = p1V3 => V3 = (p2/p1)V2 = V1(p2/p1)(p1/p2)1/g = V1(p2/p1)1-1/g
=> V3/V1 = (p2/p1 )1-1/g (2)

We know that V2 > V1 because the first process is an adiabatic expansion. Since 1/g = Cv/Cp is a positive number, we can deduce, from (1), that

p1/p2 > 1 => p2/p1 < 1

Since Cp is larger than Cv, regardless of whether the gas is monatomic, diatomic, or polyatomic,

g = Cp/Cv > 1 => 1/g < 1 => 1 – 1/g > 0

It follows from (2) that

V3/V1 < 1 => V3 < V1

so the final volume is lower than the initial volume.

(b) From the ideal gas law, we have

p1V1 = nRT1 (3)

and

p2V2 = nRT2 (4)

Dividing (3) by (4),

p1V1/p2V2 = T1/T2 => T2/T1 = p2V2/p1V1 = (V2/V1)(p2/p1)

From (1),

p2/p1 = (V1/V2)g => T2/T1 = (V2/V1)(V1/V2)g = (V1/V2)g-1 < 1 => T2 < T1

But T2 = T3, so T3 < T1, and the final temperature is lower than the initial temperature.

(c) Suppose that the gas is monatomic, V1 = 10 L, V2 = 2V1 = 20 L, and p1 = 1 atm. From (1), we have

p2 = p1(V1/V2)g = (1 atm)(10 L / 20 L)5/3 = 0.3150 atm

From (2), we get

V3 = V1(p2/p1)1-1/g = (10 L)(0.3150 atm / 1 atm)1-3/5 = 6.300 L

The two processes are sketched in GasProblem110821.xls. The adiabatic expansion is in blue, and the isothermal compression is in pink.

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PROBLEM

A two-story house has length L = 43' 3/16" (43 feet and 3/16 inch) and width W = 24' 5/16". Each floor has height H = 8', excluding the ceilings and floors. The exterior wall thickness is a = 7 1/2". The ceiling thickness on the second floor is b = 5 11/16". With the heat shut off, the temperature on the first floor is observed to decrease from Ti = 70 °F to T1f = 65 °F in two hours when the temperature outside is Tout = 5 °F. The air pressure is p = 1 atm. Neglect any interior walls or other structure in the house in performing the following calculations.
(a) What is the volume of the air filling each floor?
(b) How many moles of air molecules are on the first floor at the average temperature during the two hours?
(c) How much heat is lost by the air on the first floor? Neglect heat lost by any other objects on the first floor including the walls, floor, and ceiling.
(d) Estimate the thermal conductivity of the exterior walls on the first floor. Assume that the windows and doors have the same thickness and thermal conductivity as the walls.
(e) Estimate the temperature decrease on the second floor in two hours if the heat is shut off and the initial temperature is also Ti = 70 °F, assuming that the exterior walls and ceiling on the second floor have the same thermal conductivity as the exterior walls on the first floor. Neglect heat transfer between the first and second floors, and heat lost by any other objects on the second floor including the walls, floor, and ceiling. Again, assume that the thickness and thermal conductivity of the windows and doors are the same as those of the walls.

SOLUTION

(a) The volume of the interior of each floor is

V = (L - 2a)(W - 2a)H

We have

L - 2a = 43' 3/16" - (2)(7 1/2") = 43 ft + 3/16 in - (2)(7.5 in)
= (43 ft)(12 in/ft)(2.54 cm/in)(1 m / 100 cm) + (3/16 in)(2.54 cm/in)(1 m / 100 cm)
- (2)(7.5 in)(2.54 cm/in)(1 m / 100 cm)
= 13.1064 m + 0.004763 m - 0.3810 m = 12.73 m

W - 2a = 24' 5/16" - (2)(7 1/2") = 24 ft + 5/16 in - (2)(7.5 in)
= (24 ft)(12 in/ft)(2.54 cm/in)(1 m / 100 cm) + (5/16 in)(2.54 cm/in)(1 m / 100 cm)
- (2)(7.5 in)(2.54 cm/in)(1 m / 100 cm)
= 7.3152 m + 0.007938 m - 0.3810 m = 6.94 m

H = 8 ft = (8 ft)(12 in/ft)(2.54 cm/in)(1 m / 100 cm) = 2.44 m

Thus,

V = (12.73 m)(6.94 m)(2.44 m) = 215.5 m3

(b) From the ideal gas law, the number of moles of air molecules on the first floor at the average temperature is

n1 = pV / RT1avg (1)

where

T1avg = (Ti + Tf) / 2 = (70 °F + 65 °F) / 2 = 67.5 °F

To convert from °F to °C, subtract 32 and multiply by 5/9:

T1avg = (5/9 °C/°F)(67.5 °F - 32 °F) = 19.72 °C

The size of a Celsius degree is the same as that of a Kelvin degree, but the zero points differ for the two temperature scales. The zero point of the Celsius scale corresponds to the freezing point of water, while the zero point of the Kelvin scale is absolute zero. To convert from °C to °K, multiply by one and add 273:

T1avg = (19.72 °C)(1 °K/°C) + 273 °K = 292.7 °K

We must use °K in (1), so

n1 = pV / RT1avg = [(1 atm)(1.01 x 105 N/m2) / (1 atm)](215.5 m3) / [(8.315 J/mole °K)(292.7 °K)] = 8942 moles

(c) The heat lost by the air on the first floor is

DQ1 = - n1CpDT1

where

DT1 = T1f - Ti = (65 °F - 70 °F)(5/9 °K/°F) = - 2.778 °K

is the temperature change, and Cp is the molar heat capacity of the air in the room at constant pressure. Air is almost entirely nitrogen and oxygen, which are diatomic gases. For a diatomic ideal gas, the molar heat capacity at constant pressure is

Cp = (5/2)R = (5/2)(8.315 J/mole °K) = 20.79 J/mole °K

Thus,

DQ1 = - (8942 moles)(20.79 J/mole °K)(- 2.778 °K) = 5.163 x 105 J

(d) The surface area of the walls on the first floor is

Awall = 2(L - 2a)H + 2(W - 2a)H = 2H[(L - 2a) + (W - 2a)] = (2)(2.44 m)(12.73 m + 6.94 m)
= 95.94 m2

The average rate at which heat is lost by the first floor is

dQ1/dt = kAwall(dT1/dx)wall (2)

where k is the thermal conductivity,

(dT1/dx)wall = (T1avg - Tout) / a = (67.5 °F - 5 °F)(5/9 °K/°F) / [(7.5 in)(2.54 cm/in)(1 m / 100 cm)]
= 182.3 °K/m

is the average temperature gradient through the wall on the first floor. The rate at which heat is lost by the first floor is

dQ1/dt = (5.163 x 105 J) / [(2 hr)(60 min/hr)(60 s/min)] = 71.71 J/s

Solving for k in (2),

k = (dQ1/dt) / [Awall(dT1/dx)wall] = (71.71 J/s) / [(95.94 m2)(182.3 °K/m)] = 4.101 x 10-3 W/m °K

(e) The surface area of the walls on the second floor is the same as on the first floor. The surface area of the ceiling on the second floor is

Aceil = (L - 2a)(W - 2a) = (12.73 m)(6.94 m) = 88.37 m2

The thickness of the ceiling on the second floor is

b = 5 11/16" = (5 in + 11/16 in)(2.54 cm/in)(1 m / 100 cm) = 0.1445 m

If T2f is the final temperature on the second floor at the end of the two-hour period,

T2avg = (T1 + T2f) / 2 (3)

is the average temperature on the second floor during the two hours, and the average temperature gradients across the walls and ceiling on the second floor are

(dT2/dx)wall = (T2avg - Tout) / a (4)
(dT2/dx)ceil = (T2avg - Tout) / b (5)

The average rate at which heat is lost by the second floor is

dQ2/dt = kAwall(dT2/dx)wall + kAceil(dT2/dx)ceil = k[Awall(dT2/dx)wall + Aceil(dT2/dx)ceil] (6)

The total heat lost by the second floor is

DQ2 = (dQ2/dt)T0 (7)

where T0 = 2 hr = (2 hr)(60 min/hr)(60 s/min) = 7200 s. This is related to the temperature change of the second floor according to

DQ2 = - n2CpDT2 (8)

where n2 is the number of moles of air molecules on the second floor. The average value of n2 is obtained from the ideal gas law,

n2 = pV/RT2avg (9)

Equating the expressions for DQ2 in (7) and (8), we solve for DT2 and get

DT2 = - (dQ2/dt)T0 / n2Cp = T2f - Ti (10)

where T2f is the final temperature,

T2f = Ti + DT2 (11)

Equations (3) through (11) can be used iteratively to obtain a value for T2f. A trial value has to be assumed for T2f. If we assume T2f = 65 °F, and use the above equations, we get

T2avg = 67.5 °F
(dT2/dx)wall = 182.3 °K/m
(dT2/dx)ceil = 240.4 °K/m
dQ2/dt = 158.8 J/s
n2 = 8942 moles
DT2 = - 11.07 °F
T2f = 58.93 °F

If the process is repeated, the eventual self-consistent result is DT2 = - 10.53 °F and T2f = 59.47 °F. See TwoStoryHouseProblem091012.xls.

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Second Law of Thermodynamics

PROBLEM

One mole of an ideal monatomic gas, initially at point A at a pressure of 1.0 x 105 N/m2 and a volume of 25 x 10-3 m3, is taken through a 3-process cycle A => B => C => A (GasCycle060219.01.xls). Each process is done slowly and reversibly. At point B, the pressure is 2.0 x 105 N/m2 and the volume is 25 x 10-3 m3. At point C, the pressure is 1.0 x 105 N/m2 and the volume is 50 x 10-3 m3. For a monatomic gas, the heat capacities for constant volume and constant pressure are, respectively, Cv = (3/2)R and Cp = (5/2)R, where R is the universal gas constant, 8.32 J/mole K. Determine each of the following:
(a) the temperature of the gas at each of the vertices, A, B, and C, of the triangular cycle
(b) the net work done by the gas for one cycle
(c) the net heat absorbed by the gas for one full cycle
(d) the heat given off by the gas for the third process from C to A
(e) the efficiency of the cycle

SOLUTION

(a) For an ideal gas,

pV = nRT => T = pV/nR

where p = pressure, V = volume, n = number of moles, R = 8.32 J/mole K is the universal gas constant, and T = temperature.

TA = pAVA/nR = (1.0 x 105 N/m2)(25 x 10-3 m3)/[(1 mole)(8.32 J/mole K)] = 300.5 K
TB = pBVB/nR = (2.0 x 105 N/m2)(25 x 10-3 m3)/[(1 mole)(8.32 J/mole K)] = 601.0 K
TC = pCVC/nR = (1.0 x 105 N/m2)(50 x 10-3 m3)/[(1 mole)(8.32 J/mole K)] = 601.0 K

(b) The work done by the gas during each part of the cycle is

W = ∫ p dV

which is the area under the graph of pressure vs. volume. If the volume (increases, decreases) during a process, the work done by the gas is (positive, negative). Calculate the work done by the gas for each part of the cycle. We can make use of the fact that this is a triangular cycle (i.e., it has a triangular shape on the pressure vs. volume graph).

W(A=>B) = ∫A=>B p dV = 0 because the change in volume is zero. The gas neither expands nor contracts, so the work done by the gas is zero.

The work done from B=>C is the area under the pressure vs. volume graph from B=>C.

W(B=>C) = (1.0 x 105 N/m2)(25 x 10-3 m3) + (0.5)(1.0 x 105 N/m2)(25 x 10-3 m3) = 3750 J

W(C=>A) = ∫C=>A p dV = (1.0 x 105 N/m2)(25 x 10-3 m3 - 50 x 10-3 m3) = - 2500 J

The net work done by the gas for one cycle is

W = W(A=>B) + W(B=>C) + W(C=>A) = 0 + 3750 J - 2500 J = 1250 J

(c) The heat absorbed by the gas during the constant volume process A=>B is

Q(A=>B) = nCv(TB - TA) = (1 mole)(3/2)(8.32 J/mole K)(601.0 K - 300.5 K) = 3750 J

The heat absorbed by the gas during the constant pressure process C=>A is

Q(C=>A) = nCp(TA - TC) = (1 mole)(5/2)(8.32 J/mole K)(300.5 K - 601.0 K) = - 6250 J

For any part of the cycle, the change in internal energy is equal to the heat absorbed minus the work done by the gas. For process B=>C, the change in internal energy is

DU(B=>C) = Q(B=>C) - W(B=>C)

But the internal energy of a monatomic ideal gas is

U = (3/2)nRT

since TB = TC = 601.0 K, the change in internal energy for process B=>C is zero. Thus,

0 = Q(B=>C) - W(B=>C) => Q(B=>C) = W(B=>C) = 3750 J

The net heat absorbed by the gas for one full cycle is

Q = Q(A=>B) + Q(B=>C) + Q(C=>A) = 3750 J + 3750 J - 6250 J = 1250 J

(d) The heat absorbed by the gas in process C=>A is Q(C=>A) = - 6250 J, so the heat given off by the gas in this process is 6250 J.

(e) The cycle represents an engine which absorbs heat from some source, does work, and exhausts heat. The efficiency of the cycle is the ratio between the work done and the heat absorbed.

e = (1250 J) / (3750 J + 3750 J) = 0.1667

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PROBLEM

A heat engine consists of an oil-fired steam turbine driving an electric power generator with a power output of 120 megawatts. The thermal efficiency of the heat engine is 40 percent.
(a) Determine the time rate at which heat is supplied to the engine.
(b) If the heat of combustion of oil is 4.4 x 107 joules per kilogram, determine the rate in kilograms per second at which oil is burned.
(c) Determine the time rate at which heat is discarded by the engine.
(d) If the discarded heat is continually and completely absorbed by the water in a full tank measuring 200 meters by 50 meters by 10 meters, determine the change in the temperature of the water in 1 hour. (Density of water is r = 1.0 x 103 kg/m3; specific heat of water is c = 4.2 x 103 J/kg °C.)

SOLUTION

(a) Forty percent of the heat supplied to the engine drives the electric power generator. The time rate at which heat is supplied to the engine is (120 MW)/(0.40) = 300 MW = 300 x 106 J/s.

(b) The rate at which oil is burned is (300 x 106 J/s)(1 kg / 4.4 x 107 J) = 6.818 kg/s.

(c) The rate at which heat is discarded by the engine is (300 MW)(1 - 0.4) = 180 MW = 180 x 106 J/s.

(d) The heat Q absorbed by the water and its temperature change DT are related by

Q = mcDT

The temperature change of the water in one hour is

DT = Q/mc = (180 x 106 J/s)(3600 s/hr)/[(1.0 x 103 kg/m3)(200 m)(50 m)(10 m)(4.2 x 103 J/kg °C)] = 1.543 °C

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PROBLEM

One mole of an ideal monatomic gas is taken through the cycle abca (GasCycle060219.02.xls. State a has volume Va = 17 x 10-3 m3 and pressure Pa = 1.2 x 105 N/m2. State b has volume Vb = 51 x 10-3 m3 and pressure Pb = 1.2 x 105 N/m2. State c has volume Vc = 51 x 10-3 m3. Process ca lies along the 250 K isotherm. The molar heat capacities for the gas are Cp = 20.8 J/mole K and Cv = 12.5 J/mole K. Determine each of the following:
(a) the temperature Tb of state b
(b) the heat Qab added to the gas during process ab
(c) the change in internal energy Ub - Ua
(d) the work Wbc done by the gas on its surroundings during process bc
The net heat added to the gas for the entire cycle is 1800 J. Determine each of the following:
(e) the net work done by the gas on its surroundings for the entire cycle
(f) the efficiency of a Carnot engine that operates between the maximum and minimum temperatures in this cycle

SOLUTION

(a) PbVb = nRTb => Tb = PbVb/nR = (1.2 x 105 N/m2)(51 x 10-3 m3)/[(1 mole)(8.32 J/mole K)] = 735.6 K

(b) Qab = nCp(Tb - Ta) = (1 mole)(20.8 J/mole K)(735.6 K - 250 K) = 10100 J

(c) Ub - Ua = (3/2)nR(Tb - Ta) = (3/2)(1 mole)(8.32 J/mole K)(735.6 K - 250 K) = 6060 J

(d) Wbc = 0 because the volume is constant during process bc.

(e) The change in the internal energy of the gas in one cycle is

DU = Q - W = 0 => W = Q = 1800 J

where Q is the net heat absorbed in one cycle and W is the net work done in one cycle.

(f) eCarnot = (TH - TC)/TH = (735.6 K - 250 K)/(735.6 K) = 0.6601

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PROBLEM

A real heat engine working between heat reservoirs at 970 K and 650 K produces 550 J of work per cycle for a heat input of 2200 J.
(a) Compare the efficiency of this real engine to that of an ideal (Carnot) engine.
(b) Calculate the total entropy change of the Universe (i.e., the world outside the engine) per cycle of the real engine.
(c) Calculate the total entropy change of the Universe per cycle of a Carnot engine operating between the same two temperatures.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 15.41]

SOLUTION

(a) The efficiency of a heat engine is defined as (work output) / (heat input). For the real heat engine, the efficiency is

e = Wout/Qin = 550 J / 2200 J = 0.25

For a Carnot engine, the efficiency is

eCarnot = (TH - TC) / TH = (970 K - 650 K) / (970 K) = 0.3299

(b) The internal energy change for one cycle of the real heat engine is

DU = Qin - Qout - Wout = 0 => Qout = Qin - Wout = 2200 J - 550 J = 1650 J

The entropy change of the Universe per cycle of the real heat engine is

DSuniv = - Qin/TH + Qout/TC = - (2200 J) / (970 K) + (1650 J) / (650 K) = + 0.2704 J/K

(c) The efficiency of a Carnot engine is given by

eCarnot = (Qin - Qout) / Qin = (TH - TC) / TH => 1 - Qout/Qin = 1 - TC/TH => Qout/Qin = TC/TH => Qout = QinTC/TH

The entropy change of the Universe per cycle is

DSuniv = - Qin/TH + Qout/TC = - Qin/TH + Qin/TH = 0

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Sound

PROBLEM

An ideal organ pipe resonates at frequencies of 50 Hz, 150 Hz, 250 Hz, etc. Take the velocity of sound to be v = 343 m/s (at 20 °C).
(a) Determine if the pipe is open at both ends or open at one end and closed at the other end.
(b) Determine the length of the pipe.

[from Physics E-1a, Principles of Physics I: Mechanics, Final Exam, Fall 2001, Harvard Extension School]

SOLUTION

(a) The first harmonic (i.e., the fundamental) is f1 = 50 Hz. The second harmonic (i.e., the first overtone) is f2 = 150 Hz = 3f1. The third harmonic (i.e., the second overtone) is f3 = 250 Hz = 5f1. Thus, the pipe is open at one end and closed at the other end.

(b) The fundamental is related to the velocity of sound v and the length of the pipe L according to

f1 = v/4L => L = v/4f1 = (343 m/s)/[(4)(50 Hz)] = 1.715 m

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PROBLEM

You look directly overhead and see a plane exactly h = 1.5 km above the ground, flying faster than the speed of sound. By the time you hear the sonic boom, the plane has traveled a horizontal distance of d = 2.0 km.
(a) Find the angle of the shock cone, q.
(b) Find the speed of the plane, v. Assume the speed of sound is vs = 330 m/s.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 12.66]

SOLUTION

(a) The tangent of the angle of the shock cone is

tan q = h/d => q = tan-1(h/d) = tan-1[(1.5 km) / (2.0 km)] = tan-1(3/4) = 36.87°

(b) The sine of the angle of the shock cone is

sin q = vs/v => v = vs / sin q = vs / sin[tan-1(3/4)] = (330 m/s) / (3/5) = 550 m/s

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Geometric Optics

PROBLEM

A - 6.0 diopter lens is held 14.0 cm from an ant 1.0 mm high. What is the position, type, and height of the image?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.55]

SOLUTION

The power P of the lens is related to its focal length f by

P = 1 / f = - 6.0 D = - 6.0 m-1

Because the power of the lens is negative, it is a diverging lens. According to the lens equation,

1 / do + 1 / di = 1 / f

where do is the object distance, and di is the image distance. Solving for di,

di = (1 / f - 1 / do)-1 = [- 6.0 m-1 - 1 / (14.0 cm)]-1 = [- 6.0 m-1 - 1 / (0.140 m)]-1
= - 0.07609 m = - 7.609 cm

Since di is negative, the image is on the same side of the lens as the object and is therefore virtual. The magnification of the lens is

m = - di / do = - (- 7.609 cm) / (14.0 cm) = 0.5435

The height of the image is

hi = mho = (0.5435)(1 mm) = 0.5435 mm

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PROBLEM

A double concave lens has surface radii of 31.2 cm and 23.8 cm. What is the focal length if n = 1.52?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.68]

SOLUTION

The lensmaker's equation states that

1 / f = (n - 1)(1 / R1 + 1 / R2) (1)

where f is the focal length of the lens, n is the index of refraction, and R1 and R2 are the radii of curvature of the surfaces of the lens. A (positive, negative) radius of curvature corresponds to a (convex, concave) surface.

Since the two surfaces are concave, R1 = - 31.2 cm and R2 = - 23.8 cm. Solving (1) for f,

f = [(n - 1)(1 / R1 + 1 / R2)]-1 = {(1.52 - 1)[1 / (- 31.2 cm) + 1 / (- 23.8 cm)]}-1
= - 25.96 cm

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PROBLEM

A prescription for a corrective lens calls for + 1.50 diopters. The lensmaker grinds the lens from a "blank" with n = 1.56 and a preformed convex front surface of radius of curvature of 20.0 cm. What should be the radius of curvature of the other surface?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.73]

SOLUTION

The lensmaker's equation states that

1 / f = (n - 1)(1 / R1 + 1 / R2) (1)

where f is the focal length of the lens, n is the index of refraction, and R1 and R2 are the radii of curvature of the surfaces of the lens. A (positive, negative) radius of curvature corresponds to a (convex, concave) surface.

Let R1 = 20.0 cm = 0.200 m be the radius of curvature of one of the surfaces. Solving (1) for R2,

R2 = {[1 / f(n - 1)] - 1 / R1}-1

The power of a lens is equal to the reciprocal of its focal length. Thus, if the power of the lens is P = + 1.50 diopters,

1 / f = P = + 1.50 diopters = + 1.50 m-1
=> R2 = [(1.50 m-1) / (1.56 - 1) - 1 / (0.200 m)]-1 = - 0.4308 m = - 43.08 cm

The other surface of the lens is concave with a radius of curvature of - 43.08 cm.

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PROBLEM

We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of 14.0° above the horizontal. Calculate the depth of the pool.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.77]

SOLUTION

Consider a ray which is incident along the 14° line of sight, hits the surface of the water at the near edge of the pool, and is then refracted at the interface. The angle of incidence is q1 = 90° - 14° = 76°. By Snell's law, at the air-water interface,

n1 sin q1 = n2 sin q2

where (n1, n2) = (1.00, 1.33) is the index of refraction of (air, water). Solving for q2, we get

q2 = sin-1[(n1/n2) sin q1]

But the width x and depth d of the pool are related by

tan q2 = x/d => d = x / tan q2 = x / tan{sin-1[(n1/n2) sin q1]}
= (5.50 m) / tan{sin-1[(1.00/1.33) sin 76°]} = 5.156 m

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PROBLEM

A 1.65 m tall person stands 3.25 m from a convex mirror and notices that he looks precisely half as tall as he does in a plane mirror placed at the same distance. What is the radius of curvature of the convex mirror? Assume that sin qq.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.89]

SOLUTION

Let h = 1.65 m be the height of the object and do = d = 3.25 m be the object distance. When the person stands in front of a plane mirror, the image distance is di = - d = - 3.25 m. It is negative because the image is virtual and located behind the mirror.


object             plane
(person)           mirror               image
^                    |                    ^
|h                   |                    |h
|                    |                    |
+--------------------|--------------------+
          d          |          d
                     |
                     |


The angular size of the image can be expressed as the angle q subtended by the image of the person as viewed by the person, the tangent of which is

tan q = h / (do - di) = h / (d + d) = h / 2d

If the person stands the same distance in front of a convex mirror, the object distance is do' = d, the same as in front of the plane mirror, but the image distance and image height are different and can be denoted by di' = - d' and h'.


object             convex
(person)           mirror
^                    |          image
|h                   |            ^
|                    |            |h'
+--------------------|------------+
          d          |      d'
                     |
                     |


The angular size of the image can again be expressed as the angle q' subtended by the image of the person as viewed by the person, the tangent of which is

tan q' = h' / (do' - di') = h' / (d - di') (1)

where di' = - d' (i.e., d' is a positive quantity). We are told that the size of the image as viewed in the convex mirror is half the size of the image as viewed in the plane mirror. We assume that tan qq (i.e., if sin qq, then tan qq is also valid) and tan q' ≈ q'. So

q' = q / 2 => tan q' = (1/2) tan q = h / 4d = (1.65 m) / [(4)(3.25 m)] = 1.65 / 13 = 165 / 1300 = 33 / 260 ≡ a

Now the magnification of the mirror is

m = h' / h = - di' / do' = - di' / d => h' = - di'h / do' = - di'h / d

Substituting into (1), we get

tan q' = h' / (d - di') = - di'h / [d(d - di')] = a => - di'h = ad(d - di') = ad2 - addi'
=> di' = ad2 / (ad - h)
= (33/260)(3.25 m)2 / [(33/260)(3.25 m) - 1.65 m] = (33/260)(325/100)2 m / [(33/ 260)(325/100) - 165/100]
= (33/260)(13/4)2 m / [(33/260)(13/4) - 33/20] = (1/13)(13/4)2 m / (1/4 - 1) = (13/16) m / (- 3/4) = - (13/12) m

Using the mirror equation, we can obtain the focal length f ' of the convex lens.

1 / do' + 1 / di' = 1 / f ' => 1 / d + 1 / di' = 1 / f ' => 1 / f ' = 1 / (3.25 m) - 12 / (13 m) = - 8 / (13 m) => f ' = - 13 m / 8 = - 1.625 m

The radius of curvature is

r = 2f ' = (2)(- 1.625 m) = - 3.25 m

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PROBLEM

A beam of light enters the end of an optic fiber. Show that we can guarantee total internal reflection at the side surface of the material if the index of refraction is greater than about 1.42. In other words, regardless of the angle of incidence at the end of the optic fiber, the light beam reflects back into the material when it hits the side surface.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.46]

SOLUTION

Consider the extreme case where a light beam is incident on the end surface at an angle of incidence of 90° - d, where d is infinitesimal. The light beam is bent at the end surface and hits the side surface, where it is again refracted. If this light beam is refracted with an angle of refraction of 90° at the side surface, then a light beam hitting the end surface with any other angle of incidence will undergo total internal reflection at the side surface.

Let b be the angle of refraction at the end surface. Then the angle of incidence at the side surface is 90° - b. Applying Snell's law at the end surface, we have

n1 sin q1 = n2 sin q2 => sin 90° = n sin b => 1 = n sin b (1)

Applying Snell's law at the side surface, we have

n sin(90° - b) = sin 90° => n cos b = 1 (2)

Combining (1) and (2), we get

n sin b = n cos b => tan b = 1 => b = 45° => n = 1 / sin b = 1 / sin 45° = sqrt(2) = 1.414

Thus, if the index of refraction is greater than sqrt(2) = 1.414, total internal refraction occurs at the side surface for any angle of incidence on the end surface.

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PROBLEM

A converging lens with a focal length of 10.0 cm is placed in contact with a diverging lens with a focal length of 20.0 cm. What is the focal length of the combination and is the combination converging or diverging?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.88]

SOLUTION

Neglect the thicknesses of the lenses. With the Sun or some other distant source as the object, assume the converging lens is closer to the source than the diverging lens. Applying the lens equation, we have

1/do + 1/di = 1/f => di = (1/f - 1/do)-1 = [1/(10.0 cm) - 1/(+∞)]-1 = 10.0 cm

The image of the converging lens is 10.0 cm to the right of the two lenses. Using the image of the converging lens as the object of the diverging lens, use the lens equation to determine the location of the image of the diverging lens.

1/do + 1/di = 1/f => di = (1/f - 1/do)-1 = [1/(-20.0 cm) - 1/(- 10.0 cm)]-1 = 20.0 cm

The final image is 20.0 cm to the right of the lenses, so the focal length of the combination is 20.0 cm. Since the final image is to the right of the lenses, the combination is converging.

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PROBLEM

A double convex lens whose radii of curvature are both 17.5 cm is made of crown glass. Find the distance between the focal points for 400-nm and 700-nm light.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 24.17]

SOLUTION

Use the lensmaker’s equation,

1/f = (n – 1)(1/R1 + 1/R2) => f = [(n – 1)(1/R1 + 1/R2)]-1

where f is the focal length, n is the index of refraction, and R1 and R2 are the radii of curvature of the two surfaces of the lens.

The index of refraction of crown glass is 1.52 at 400 nm and 1.50 at 700 nm. Thus, the focal lengths of the lens for 400 nm and 700 nm light are

f400 = {(1.52 – 1)[(1 / 17.5 cm) + (1 / 17.5 cm)]}-1 = 16.82692 cm
f700 = {(1.50 – 1)[(1 / 17.5 cm) + (1 / 17.5 cm)]}-1 = 17.50000 cm

The distance between the focal points is

f700 - f400 = 17.50000 cm – 16.82692 cm = 0.6731 cm = 6.731 mm

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PROBLEM

A planoconvex lens (convex on the left side and flat on the right side), in the shape of a hemisphere of glass, has index of refraction n = 1.50 and radius of curvature R = 12.0 cm. An incoming ray of light approaches the convex surface traveling parallel to the principal axis and a height h above it. Determine the distance d, from the flat face of the lens, to where the ray crosses the principal axis if (a) h = 1.0 cm, and (b) h = 6.0 cm. (c) How far apart are these "focal points"? (d) How large is the "circle of confusion" at the "focal point" for h = 1.0 cm? Note that this is not a thin lens.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.45]

SOLUTION

(a) The ray bends towards the normal when it passes through the convex side of the lens and away from the normal when it passes through the flat side. Thus, in both cases, the ray is refracted towards the principal axis.

Let qi be the angle of incidence on the convex surface. Then

sin qi = h/R => qi = sin-1(h/R) (1)

According to Snell's law, the angle of refraction qr on the convex surface is related to qi by

sin qi = n sin qr => qr = sin-1[(1/n) sin qi] = sin-1(h/Rn) (2)

Consider the triangle formed by the refracted ray within the lens, the radius from the center of curvature of the convex side to the point where the ray enters the convex side of the lens, and the line segment between the point where the refracted ray exits the lens on the flat surface and the center of curvature of the convex side.

Using the fact that the sum of the angles of this triangle equals p, the angle of incidence qi' at the flat surface is related to qi and qr according to

p/2 - qi + qr + p/2 + qi' = p => - qi + qr + qi' = 0 => qi' = qi - qr (3)

The angle of refraction qr' at the flat surface is obtained from

n sin qi' = sin qr' => qr' = sin-1(n sin qi') (4)

Consider the triangle formed by the ray exiting the lens, the flat surface of the lens, and the principal axis. The angle A between the ray and the principal axis is qr'. The length of the side of the triangle opposite A can be shown to be equal to R sin qi - R cos qi tan qi'. Thus,

tan qr' = (R sin qi - R cos qi tan qi') / d
=> d = (R sin qi - R cos qi tan qi') / tan qr'
= R(sin qi - cos qi tan qi') / tan qr' (5)

Using equations (1), (2), (3), (4), and (5), we get

qi = sin-1(h/R) = sin-1(1.0 cm / 12.0 cm) = sin-1(1/12) = 4.780192°
qr = sin-1(h/Rn) = sin-1{(1.0 cm) / [(12.0 cm)(1.50)]} = sin-1(1/18) = 3.184739°
qi' = qi - qr = 4.780192° - 3.184739° = 1.595453°
qr' = sin-1(n sin qi') = sin-1[(1.50) sin 1.595453°] = 2.392871°
d = R(sin qi - cos qi tan qi') / tan qr'
= (12.0 cm)(sin 4.780192° - cos 4.780192° tan 1.595453°) / tan 2.392871°
= 15.95982 cm

(b) Using equations (1), (2), (3), (4), and (5), we get

qi = sin-1(h/R) = sin-1(6.0 cm / 12.0 cm) = sin-1(1/2) = 30°
qr = sin-1(h/Rn) = sin-1{(6.0 cm) / [(12.0 cm)(1.50)]} = sin-1(1/3) = 19.47122°
qi' = qi - qr = 30° - 19.47122° = 10.52878°
qr' = sin-1(n sin qi') = sin-1[(1.50) sin 10.52878°] = 15.70443°
d = R(sin qi - cos qi tan qi') / tan qr'
= (12.0 cm)(sin 30° - cos 30° tan 10.52878°) / tan 15.70443° = 14.46985 cm

(c) Dd = 15.95982 cm - 14.46985 cm = 1.489966 cm

(d) Determine how far from the principal axis the 6 cm ray passes the focus for the 1 cm ray. The angle between the refracted 6 cm ray and the principal axis is qr'. If Dy is the distance from the principal axis to the refracted 6 cm ray when it passes the focus for the 1 cm ray,

tan qr' = Dy / Dd => Dy = (Dd) tan qr' = (1.489966 cm) tan 15.70443°
= 0.418935 cm

The diameter of the circle of confusion is

2Dy = (2)(0.418935 cm) = 0.83787 cm

Giancoli25.45.110527.xls shows how the above results were obtained.

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PROBLEM

A lens whose index of refraction is n is submerged in a material whose index of refraction is n' (n' ≠ 1). Derive the lensmaker's equation and the lens equation for this lens.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.91]

SOLUTION

We generalize the derivation given by Giancoli for the case where the lens is placed in a medium where the index of refraction is n' ≠ 1. Consider a convex-convex thin lens and a ray incident on the left side of the lens, parallel to the lens axis. Define

q1 = angle of incidence at first surface of lens
q2 = angle of refraction at first surface of lens
q3 = angle of incidence at second surface of lens
q4 = angle of refraction at second surface of lens

At the first surface, by Snell's law,

n' sin q1 = n sin q2

Using the small angle approximation,

n'q1 = nq2 => q2 = (n'/n)q1 (1)

Similarly, at the second surface,

nq3 = n'q4 => q3 = (n'/n)q4 (2)

Define

h1 = vertical distance from axis to point on lens at which ray enters lens
h2 = vertical distance from axis to point on lens at which ray exits lens
a = angle between axis and direction from center of curvature of second surface to point on lens at which ray exits lens
b = angle between axis and direction from focus on exit side of lens to point on lens at which ray exits lens
R1 = radius of curvature of first surface of lens
R2 = radius of curvature of second surface of lens
f = focal length of lens (in the general medium with index of refraction n' ≠ 1)

Then

q1 = sin q1 = h1/R1 (3)
a = sin a = h2/R2 (4)
b = sin b = h2/f (5)

Define

g = q1 - q2 (6)

It can be seen that

a + g = q3 => a = q3 - g (7)

and that

q4 = a + b

Thus, from (1)-(7),

a = q3 - g = (n'/n)q4 - (q1 - q2) = (n'/n)a + (n'/n)b - q1 + q2
=> h2/R2 = (n'/n)(h2/R2) +(n'/n)(h2/f) - h1/R1 + (n'/n)(h1/R1) (8)

In the thin lens approximation, h1 ≈ h2, and we can remove the h1 and h2 from (8) to get

1/R2 = (n'/n)(1/R2) +(n'/n)(1/f) - 1/R1 + (n'/n)(1/R1)
=> (n'/n)(1/f) = (1 - n'/n)(1/R1 + 1/R2) => 1/f = (n/n' - 1)(1/R1 + 1/R2) (9)

This is the lensmaker's equation generalized for the case where the medium in which the lens is placed has an index of refraction n' ≠ 1. The lens equation is

1/do + 1/di = 1/f

where do is the object distance, di is the image distance, and 1/f is given by the generalized lensmaker's equation (9).

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Optical Instruments

PROBLEM

A 35 mm slide (picture size is actually 24 by 36 mm) is to be projected on a screen 1.80 by 2.70 m placed 9.00 m from the projector. What focal length lens should be used if the image is to cover the screen?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 23.81]

SOLUTION

The required magnification is

M = hi / ho = (- 2.70 m) / (36 mm) = (- 2700 mm) / (36 mm) = - 75

The image height hi is negative because the image is inverted relative to the object. The magnification is related to the object distance do and the image distance di by

M = - di / do => do = - di / M = - (9.00 m) / (- 75) = (900 cm) / 75 = 12 cm

The focal length f is related to do and di by

1/f = 1/do + 1/di => f = 1 / (1/do + 1/di) = 1 / (1/12 + 1/900) = 11.84 cm

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PROBLEM

Suppose a woman has normal vision. (a) What is the accommodated strength of her eyes if she has a 10% ability to accommodate? (b) What is the closest object she can see clearly?

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 15.9]

SOLUTION

(a) The eyes of a person with normal vision focus the image of an object at infinity (do = ∞) 2.0 cm beyond the eye lens. The corresponding power is

P = 1/f = 1/do + 1/di = 1/di = 1 / (0.02 m) = 50 D

The accommodated power is

Pacc = 1.1P = (1.1)(50 D) = 55 D

(b) The closest object that she can see clearly corresponds to the accommodated power.

1/do + 1/di = 1/f => do = (1/f – 1/di)-1 = [55 D – 1 / (0.02 m)]-1 = 0.2 m = 20 cm

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PROBLEM

The eyes of a certain myopic administrator have a minimum strength of 52.0 D. (a) What is the accommodated strength of his eyes if he has a normal 8% ability to accommodate? (b) What is the most distant object he can see clearly? (c) What is the closest object he can see clearly?

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 15.10]

SOLUTION

(a) The accommodated strength is

Pacc = 1.08P = (1.08)(52.0 D) = 56.16 D

(b) The most distant object he can see clearly corresponds to the minimum strength.

1/do + 1/di = 1/f => do = (1/f – 1/di)-1 = [52.0 D – 1 / (0.02 m)]-1 = 0.5 m
= 50 cm

(c) The closest object he can see clearly corresponds to the accommodated strength.

1/do + 1/di = 1/f => do = (1/f – 1/di)-1 = [56.16 D – 1 / (0.02 m)]-1 = 0.1623 m
= 16.23 cm

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PROBLEM

(a) What is the accommodated strength of a farsighted man who can see objects clearly that are no closer than 1.50 m? (b) What spectacle lens will allow him to see objects clearly at 25 cm distance?

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 15.17]

SOLUTION

(a) Assume that the distance between the eye lens and the retina is 2.0 cm. Then we calculate the power of the lens, or the inverse of its focal length, if the object distance is do = 1.50 m and the image distance is di = 2.0 cm = 0.02 m.

P = 1/f = 1/do + 1/di = 1 / (1.50 m) + 1 / (0.02 m) = 50.67 D

(b) Neglecting the distance between the spectacle lens and the eye, do = 25 cm = 0.25 m and di = - 1.50 m. We get

P = 1/f = 1/do + 1/di = 1 / (0.25 m) - 1 / (1.50 m) = 3.333 D

Assuming the distance between the spectacle lens and the eye is 2.0 cm, do = 25 cm – 2 cm = 23 cm and di = - (1.50 m – 0.02 m) = - 1.48 m.

P = 1/f = 1/do + 1/di = 1 / (0.23 m) – 1 / (1.48 m) = 3.672 D

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PROBLEM

A mother sees that her child's spectacle prescription is +0.75 D. What is the closest object her child can see clearly with glasses off? Solve (a) neglecting the distance between the glasses and the eyes and (b) assuming the distance between the glasses and the eyes is 2.0 cm.

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 15.21]

SOLUTION

The glasses have a positive focal length and are thus converging, so the child is farsighted. The glasses correct the child's vision so that an object held at a distance of 25.0 cm from the eyes appears to be at the child's near point.

(a) Use the lens equation.

1/d0 + 1/di = 1/f => di = (1/f - 1/do)-1

We have 1/f = +0.75 D = +0.75 m-1. If we neglect the distance between the child's glasses and eyes, do = 25.0 cm = 0.250 m. Then

di = [0.75 m-1 - 1 / (0.250 m)] = - 0.3077 m = - 30.77 cm

The child's near point is |di| = 30.77 cm.

(b) If the distance between the child's glasses and eyes is 2.0 cm, do = 25.0 cm - 2.0 cm = 23.0 cm = 0.230 m and

di = (1/f - 1/do)-1 = [0.75 m-1 - 1 / (0.230 m)]-1 = - 0.2779 m = - 27.79 cm

The child's near point is |di| + 2.0 cm = 27.79 cm + 2.0 cm = 29.79 cm.

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PROBLEM

Sam purchases +3.2 diopter eyeglasses which correct his faulty vision to put his near point at 25 cm. Assume he wears the lenses 2.0 cm from his eyes.
(a) Is Sam nearsighted or farsighted?
(b) Calculate the focal length of Sam's glasses.
(c) Calculate Sam's near point without glasses.
(d) Pam, who has normal eyes with near point at 25 cm, puts on Sam's glasses. Calculate Pam's near point with Sam's glasses on.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.60]

SOLUTION

(a) If the power of a lens is (positive, negative), the lens is (converging, diverging). The +3.2 D eyeglasses are converging. Without the glasses, images are focused behind Sam's retina. This is a characteristic of farsightedness, so Sam is farsighted.

(b) f = 1/P = 1/3.2 = 0.3125 m = 31.25 cm

(c) When an object is placed 25 cm from Sam's eye, or 23 cm from the lens, we want the image to be (x - 2) cm away from the lens on the same side of the lens as the object, where x cm is Sam's near point without glasses. Thus,

1/f = 1/do + 1/di => 1/31.25 = 1/23 - 1/(x-2) => x = 2 + 1/(1/23 - 1/31.25) = 89.12 cm

Sam's near point without glasses is 89.12 cm.

(d) We need to determine the distance x from Pam's eye to the object so that the image is 25 cm from Pam's eye.

1/f = 1/do + 1/di => 1/31.25 = 1/(x-2) - 1/(25-2) => x = 2 + 1/(1/31.25 + 1/23) = 15.25 cm

Pam's near point with Sam's glasses on is 15.25 cm.

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PROBLEM

A woman can see clearly with her right eye only when objects are between 40 cm and 180 cm away. Prescription bifocals should have what powers so that she can see distant objects clearly (upper part) and be able to read a book 25 cm away (lower part)?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.64]

SOLUTION

For viewing distant objects, the upper part of the bifocals should make an object at ∞ appear to be 180 cm from the eye or 178 cm from the lens on the same side of the lens as the object.

1/f = 1/do + 1/di => f = 1/(1/do + 1/di) = 1/(1/∞ - 1/178) = - 178 cm = - 1.78 m
=> P = 1/f = - 1/1.78 = - 0.5618 D

For reading, the lower part of the bifocals should make an object 25 cm from the eye or 23 cm from the lens appear to be 40 cm from the eye or 38 cm from the lens on the same side of the lens as the object.

1/f = 1/do + 1/di => f = 1/(1/do + 1/di) = 1/(1/23 - 1/38) = 58.27 cm = 0.5827 m
=> P = 1/f = 1/0.5827 = +1.716 D

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PROBLEM

A child has a near point of 15 cm. What is the maximum magnification the child can obtain using an 8.0 cm focal length magnifier? Compare to that for a normal eye.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.65]

SOLUTION

The maximum magnification occurs if the eye is focused on the near point, rather than at infinity, and is equal to

M = N/f + 1 = (15 cm) / (8.0 cm) + 1 = 2.875

for the child with a near point of 15 cm. For a normal eye it would be

M = N/f + 1 = (25 cm) / (8.0 cm) + 1 = 4.125

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PROBLEM

What is the magnifying power of a +4.0 D lens used as a magnifier? Assume a relaxed normal eye.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.66]

SOLUTION

The focal length of the lens is

f = 1/P = 1/4.0 = 0.25 m = 25 cm

The magnifying power of the lens is

M = N/f = (25 cm) / (25 cm) = 1.00

where N = 25 cm is the near point of a normal eye.

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PROBLEM

A 50-year-old man uses +2.5 diopter lenses to be able to read a newspaper 25 cm away from the lenses.
(a) What is the man's near point?
(b) Ten years later, he finds that he must hold the paper 35 cm away from the lenses to see clearly with the same lenses. What power lenses does he need now?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.69]

SOLUTION

(a) The focal length of the lens is

f = 1/P = 1/2.5 = 0.4 m = 40 cm

When the man is 50 years old, the object distance (i.e., distance from the lens to the object) is do = 25 cm.

1/f = 1/do + 1/di => di = 1/(1/f - 1/do) = 1/(1/40 - 1/25) = - 66.67 cm

The man's near point is 66.67 cm from the lens.

(b) Ten years later, the same glasses make an object which is do = 35 cm from the lens appear at the man's new near point.

1/f = 1/do + 1/di => di = 1/(1/f - 1/do) = - 280 cm

The man's new near point is 280 cm from the lens. He needs new glasses with focal length f ' which will make an object at do = 25 cm appear at di = - 280 cm.

1/f ' = 1/do + 1/di
=> f ' = 1/(1/do + 1/di) = 1/(1/25 - 1/280) = 27.45 cm = 0.2745 m
=> P' = 1/f ' = +3.643 D

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PROBLEM

A small reflecting telescope has a focal length of 1.0 m. (a) What magnification does it produce when photographing the sun, 1.50 x108 km away? Note that the magnification should be very small since the object being photographed is very large. (b) How large is the image of a sunspot 25,000 km in diameter?

[from Urone, Paul Peter 1986, Physics with Health Science Applications (New York, New York: John Wiley and Sons), problem 14.30]

SOLUTION

(a) Using the mirror equation,

1/do + 1/di = 1/f => di = (1/f – 1/do)-1 = [1 / (1.0 m) – 1 / (1.50 x 1011 m)]-1
= 1.0 m

The magnification is

m = - di/do = - (1.0 m) / (1.50 x 1011 m) = - 6.667 x 10-12

(b) hi = mho = (- 6.667 x 10-12)(25,000 x 103 m) = - 1.667 x 10-4
= - 0.1667 mm

The image is 0.1667 mm in diameter.

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PROBLEM

Suppose that you wish to construct a telescope that can resolve features 10 km across on the Moon, 384,000 km away. You have a 2.2 m focal length objective lens whose diameter is 12 cm. What focal length eyepiece is needed if your eye can resolve objects 0.10 mm apart at a distance of 25 cm? What is the resolution limit set by the size of the objective lens (that is, by diffraction)? Use l = 500 nm.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.71]

SOLUTION

The eye can resolve objects whose angular separation is

q = (0.10 mm) / (25 cm) = (0.10 mm) / (250 mm) = 0.0004 rad = 4 x 10-4 rad

The angular size of the lunar features that are to be resolved is

f = (10 km) / (384,000 km) = 2.604 x 10-5 rad

Thus the magnification of the telescope should be at least

M = q/f = (4 x 10-4 rad) / (2.604 x 10-5 rad) = 15.36

The magnification of a refracting telescope is

M = fo/fe => fe = fo/M = (2.2 m)/15.36 = 0.1432 m = 14.32 cm

According to the Rayleigh criterion, the resolution limit set by the size of the objective lens is

R = 1.22l/D = (1.22)(500 x 10-9 m) / (0.12 m) = 5.083 x 10-6 rad

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PROBLEM

The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is 20 diopters, what is the total magnification of the telescope?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 25.73]

SOLUTION

The focal length of the eyepiece is

fe = 1/Pe = 1/20 = 0.05 m = 5 cm

The distance between the lenses is the sum of the focal lengths of the eyepiece and the objective. Thus the focal length of the objective is fo = 85 cm - 5 cm = 80 cm. The total magnification of the telescope is

M = - fo/fe = - (80 cm) / (5 cm) = - 16

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Double-Slit Interference

PROBLEM

In a double-slit experiment, it is found that blue light of wavelength 460 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 24.10]

SOLUTION

Let l1 = 460 nm. The interference maxima for wavelength l1 occur when

d sin q = ml1

where d is the separation between the slits and m is an integer. For a second-order maximum, m = 2 and

d sin q = 2l1 (1)

Interference minima occur at the same value of q for wavelength l2 if

d sin q = (m + 1/2)l2 (2)

where m is an integer. Equating (1) and (2),

2l1 = (m + 1/2)l2 => l2 = 2l1 / (m + 1/2)

For m = 0,

l2 = 2l1 / (1/2) = 4l1 = (4)(460 nm) = 1840 nm

which is in the infrared.

For m = 1,

l2 = 2l1 / (3/2) = (4/3)l1 = (4/3)(460 nm) = 613.3 nm

which is in the visible and is yellow-orange.

For m = 2,

l2 = 2l1 / (5/2) = (4/5)l1 = (4/5)(460 nm) = 368.0 nm

which is in the ultraviolet.

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Diffraction Gratings

PROBLEM

Two first-order spectrum lines are measured by an 8500-line/cm spectroscope at angles, on each side of center, of + 26°38’, + 41°08’ and – 26°48’, - 41°19’.
(a) What are the wavelengths?
(b) Suppose the angles measured were produced when the spectrometer (but not the source) was submerged in water. What then would be the wavelengths?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problems 24.36 and 24.37]

SOLUTION

(a) For a diffraction grating, maxima occur for a given wavelength l when

d sin q = ml

where d is the line spacing, and m is an integer. For the first-order maxima,

d sin q = l

For each wavelength, there is a maximum on each side of q = 0. The line spacing is

d = 1 cm / 8500 = (1/8500) cm

Thus, considering only the magnitudes of q,

l = d sin q = [(1/8500) cm] sin(26°38’, + 41°08’, 26°48’, 41°19’)
= [(1/8500) cm] sin(26.63°, 41.13°, 26.80°, 41.32°)
= 5.274 x 10-5, 7.739 x 10-5, 5.304 x 10-5, 7.767 x 10-5) cm
= (527.4, 773.9, 530.4, 776.7) nm

The 527.4 and 530.4 nm correspond to the same wavelength; the difference is due to measurement error. The same is true for the 773.9 and 776.7 nm. Averaging the 527.4 and 530.4 nm, and the 773.9 and 776.7 nm, we get l = 528.9 and 775.3 nm for the two wavelengths.

(b) If the spectrometer is submerged in water, the measured wavelengths are reduced by a factor of n = 1.33, the index of refraction, which is assumed the same for both wavelengths. Thus, the true wavelengths are obtained by multiplying the reduced wavelengths by n, or

l = nln = (1.33)(528.9, 775.3) nm = (703.4, 1031.2) nm

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Interference by Thin Films

PROBLEM

A thin film of alcohol (n = 1.36) lies on a flat glass plate (n = 1.51). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for l = 512 nm and a maximum for l = 640 nm. What is the thickness of the film? Assume that n in a given material is independent of wavelength, and that the maximum at 640 nm is the first maximum that occurs as the wavelength is increased from 512 nm.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey), problem 24.45]

SOLUTION

Light incident from above is reflected at the interfaces between the air and the alcohol, and between the alcohol and the glass. The emerging beams of light interfere with each other and produce maxima and minima in the total intensity, depending on the wavelength of the light.

     |  |
     |  |  air (n = 1.00)
-----+--+-----
     |
    t|     alcohol (n = 1.36)
     |
-----+--------
           glass (n = 1.51)


At both interfaces, the index of refraction of the material on the incident side is less than the index of refraction of the material on the other side. Thus, the reflected light undergoes a phase shift of 180° or half a cycle at both interfaces, and the relative phase difference between the two emerging beams depends only on the number of wavelengths along the path of the beam passing through the alcohol.

The wavelength of the light in alcohol is

ln = l/n

where l is the wavelength in a vacuum and n = 1.36 is the index of refraction in the alcohol.

If t is the thickness of the film of alcohol, the number of wavelengths along the path of the light beam passing through the alcohol is

N = 2t / ln = 2t / (l/n) = 2nt / l

If N is an integer, the two beams of light emerging into the air interfere constructively, and there is a maximum. If N is an odd half integer (e.g., 1/2, 3/2, 5/2,…), the two beams of light emerging into the air interfere destructively, and there is a minimum.

Define la = 512 nm and lb = 640 nm. Then

N = 2tn / la (1)

where N is an odd half integer, because there is a minimum for la = 512 nm, and

N’ = 2tn / lb

where N’ is an integer, because there is a maximum for lb = 640 nm. As the wavelength is increased from la to lb, the number of wavelengths along the path through the alcohol must decrease by 1/2. Thus,

N’ = N – 1/2

and

N – 1/2 = 2tn / lb (2)

Subtracting (2) from (1), we get

1/2 = 2tn(1/la - 1/lb)
=> t = (1/4n) / (1/la - 1/lb) = {1/[(4)(1.36)]} / (1 / 512 nm – 1 / 640 nm)
= 470.6 nm

The thickness of the film is 470.6 nm.

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Interferometers

PROBLEM

The yellow sodium D lines have wavelengths of 589.0 and 589.6 nm. When they are used to illuminate a Michelson interferometer, it is noted that the interference fringes disappear and reappear periodically as the movable mirror is moved. Why does this happen? How far must the mirror move between one disappearance and the next?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 24.53]

SOLUTION

There is interference between the two wavelengths which causes minima and maxima in the fringe amplitudes as the difference between path lengths in the two arms changes. When the difference in path length is such that the number of wavelengths of the 589.0 nm line is 1/2 more than the number of wavelengths of the 589.6 nm line in the path length difference, there will be minima in the fringe amplitude. Let the path length difference at which a minimum occurs be d. Let n be the number of wavelengths of the 589.6 nm line. Then

(n + 1)(589.0 nm) = (n)(589.6 nm) => 589n + 589 = 589.6n => 0.6n = 589
=> n = 589 / 0.6 = 981 2/3 and n + 1/2 = 982 1/6

The corresponding path length difference is

d = (981 2/3)(589.6 nm) = 578790.66 nm = 0.57879 mm

This is the minimum path length difference for which a minimum in the fringe amplitude occurs. Every time the path length difference increases by this amount, there will be another minimum. The mirror must move half this amount or 0.2894 mm between minima.

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Blackbody Radiation

PROBLEM

Prove the relation

u(l, T) = 4E(l, T)/c

between the energy density u(l, T) per unit wavelength in a cavity, at temperature T, and the emissive power E(l, T) (the energy emitted by a blackbody per unit time per unit area per unit wavelength at wavelength l).

[from Gasiorowicz, Stephen 1974, Quantum Physics (New York: Wiley), problem 1.1]

SOLUTION

We consider as an ideal blackbody a cavity at temperature T with a small opening of infinitesimal area dA. Let the origin of our coordinate system be placed at the opening with the +z axis pointing directly into the cavity. The radiation exiting the cavity through the hole will come from all directions within 2p steradians inside the cavity.

Consider a hemispherical shell of radius r and infinitesimal thickness dr whose center is at the origin and whose base lies in the xy plane (i.e., along the surface of the cavity). Then any radiation which is within the hemispherical shell, between radial coordinates r and r + dr, at time t = 0, that is traveling towards the opening, will arrive at the opening at time t = r/c, where c is the speed of light.

Consider a particular volume element

dV = r2 dr sin q dq df

located within the shell at spherical coordinates (r, q, f). If u is the energy density, the amount of energy enclosed in dV is

dU = u dV = ur2 dr sin q dq df

This radiation spreads out into 4p steradians. The amount which passes through dA at time t = r/c is

(dU / 4pr2)(dA cos q) = (ur2 dr sin q dq df / 4pr2)(dA cos q)
= (u/4p) dr dA sin q cos q dq df

where dA cos q is the projection of dA in the direction facing the volume element dV. The total energy which emerges from dA from the hemispherical shell is

∫ (dU / 4pr2)(dA cos q) = (u/4p) dr dA ∫0p/2 sin q cos q dq02p df
= (u/2) dr dA ∫0p/2 sin q cos q dq = (u/2) dr dA (1/2) sin2 q |0p/2 = (u/4) dr dA

The emissivity is

E = [(u/4) dr dA] / (dA dt) = (u/4) dr/dt

where dt = dr/c is the time interval over which the radiation from the spherical shell emerges. Thus,

E = uc/4 => u = 4E/c

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PROBLEM

How much of the Sun's energy is radiated in the range of wavelengths 4000-7000 Å? Assume a surface temperature of T = 5800 K.

[from Gasiorowicz, Stephen 1974, Quantum Physics (New York: Wiley), problem 1.4]

SOLUTION

According to Planck's blackbody radiation formula, the energy emitted per unit frequency per unit area per unit time by a surface at temperature T is

E(n, T) = (2ph/c2)n3/[exp(hn/kT) - 1]

where h = 6.63 x 10-34 J s is Planck's constant, c = 3 x 108 m/s is the speed of light, k = 1.38 x
10-23 J/K is Boltzmann's constant, and n is the frequency. We can also write this as

dU(n, T)/dAdtdn = (2ph/c2)n3/[exp(hn/kT) - 1]

The total energy emitted per unit area per unit time is

[dU(n, T)/dAdt]tot = ∫0 dn (2ph/c2)n3/[exp(hn/kT) - 1]

Define

x = hn/kT => dx = (h/kT) dn

Then

(dU/dAdt)tot = ∫0 dx (kT/h)(2ph/c2)x3(kT/h)3/(ex - 1) = (2ph/c2)(kT/h)40 dx x3/(ex - 1)
= (2ph/c2)(kT/h)40 dx x3 e-x /(1 - e-x) (1)

Now we use the fact that, for |r| < 1,

1/(1 - r) = Sn=0 rn

and write

1/(1 - e-x) = Sn=0 e-nx

so

(dU/dAdt)tot = (2ph/c2)(kT/h)40 dx x3 e-x Sn=0 e-nx = (2ph/c2)(kT/h)4 Sn=00 dx x3 e-(n+1)x

Define

y = (n+1)x => dy = (n+1)dx

Then

(dU/dAdt)tot = (2ph/c2)(kT/h)4 Sn=0 [1/(n+1)4] ∫0 dy y3 e-y
= (2ph/c2)(kT/h)4 Sn=1 (1/n4) ∫0 dy y3 e-y

Let u = y3 and dv = e-y dy. Then du = 3y2 dy and v = - e-y, and

0 dy y3 e-y = 3 ∫0 y2 e-y dy

Let u = y2 and dv = e-y dy. Then du = 2y dy and v = - e-y, and

0 y2 e-y dy = 2 ∫0 ye-y dy

Let u = y and dv = e-y dy. Then du = dy and v = - e-y, and

0 ye-y dy = ∫0 e-y dy = - e-y |0 = 1 => ∫0 dy y3 e-y = 6
=> (dU/dAdt)tot = (12ph/c2)(kT/h)4 Sn=1 (1/n4)

It can be shown that

Sn=1 (1/n4) = p4/90

Thus,

(dU/dAdt)tot = (12ph/c2)(kT/h)4(p4/90) = (12p5k4 / 90c2h3)T4 = (2p5k4 / 15c2h3)T4 = sT4

where

s = (2p5k4 / 15c2h3) = 5.642 x 10-8 W/m2K4

is the Stefan-Boltzmann constant. The accepted value is 5.67 x 10-8 W/m2K4.

Now we determine the energy emitted per unit area per unit time in the 4000-7000 Å wavelength range, which is in the visible part of the spectrum. Let lmin = 4000 Å and lmax = 7000 Å. The corresponding frequencies are

nmin = c / lmax = (3 x 108 m/s) / (7 x 10-7 m) = 4.286 x 1014 Hz

and

nmax = c / lmin = (3 x 108 m/s) / (4 x 10-7 m) = 7.500 x 1014 Hz

The corresponding values of the variable

x = hn/kT

are

xmin = hnmin/kT = (6.63 x 10-34 J s)(4.286 x 1014 Hz) / [(1.38 x 10-23 J/K)(5800 K)] = 3.550

and

xmax = hnmax/kT = (6.63 x 10-34 J s)(7.500 x 1014 Hz) / [(1.38 x 10-23 J/K)(5800 K)] = 6.213

We use equation (1) from above, and set the limits of integration equal to xmin and xmax.

(dU/dAdt)vis = (2ph/c2)(kT/h)4xminxmax dx x3/(ex - 1)
= (2ph/c2)(kT/h)43.5506.213 dx x3/(ex - 1) = 2.385(2ph/c2)(kT/h)4 = 4.770pk4T4/c2h3

The fraction of the total energy emitted in the visible range is

(dU/dAdt)vis/(dU/dAdt)tot = 4.770p/(2p5/15) = 0.3673

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PROBLEM

Calculate the fraction of the Sun's energy which is emitted in the following wavelength ranges:
(a) 740 nm - 0.1 cm (infrared)
(b) 380-740 nm (visible)
(c) 10-380 nm (ultraviolet)
(d) 0.01-10 nm (X-ray)
Assume that the Sun is a blackbody with a surface temperature of T = 5800 K.

SOLUTION

For each of the four wavelength ranges lmin < l < lmax, the minimum and maximum frequencies n are

nmin = c / lmax

and

nmax = c / lmin

where c = 3 x 108 m/s is the speed of light. The minimum and maximum values of the parameter

x = hn/kT are

xmin = hnmin/kT

and

xmax = hnmax/kT

where h = 6.63 x 10-34 J s is Planck's constant and k = 1.38 x 10-23 J/K is Boltzmann's constant. The minimum and maximum values of the energy E are

Emin = hnmin

and

Emax = hnmax

These are summarized in the following table.

Part of SpectrumWavelength RangeFrequency Range (Hz)hn/kT RangeEnergy Range
(a) infrared
740 nm - 0.1 cm
3 x 1011 - 4.054 x 1014
2.485 x 10-3 - 3.358
1.243 x 10-3 - 1.680 eV
(b) visible
380-740 nm
(4.054-7.895) x 1014
3.358-6.539
1.680-3.271 eV
(c) ultraviolet
10-380 nm
7.895 x 1014 - 3 x 1016
6.539-248.5
3.271-124.3 eV
(d) X-ray
0.01-10 nm
3 x 1016 - 3 x 1019
248.5-248501
0.1243-124.3 keV

The total energy emitted per unit area per unit time can be shown to be

(dU/dAdt)tot = (2ph/c2)(kT/h)40 dx x3/(ex - 1) = (2p5k4 / 15c2h3)T4

The energy emitted per unit area per unit time in each of the above ranges in x is

(dU/dAdt)range = (2ph/c2)(kT/h)4xminxmax dx x3/(ex - 1)

The above integral can be done numerically. The fraction of the Sun's energy emitted in each wavelength range is

frange = (dU/dAdt)range / (dU/dAdt)tot

The following table lists the values of ∫xminxmax dx x3/(ex - 1) and frange for each of the four wavelength ranges.

Part of Spectrumxminxmax dx x3/(ex - 1)frange
(a) infrared
3.051
0.4698
(b) visible
2.787
0.4292
(c) ultraviolet
0.6555
0.1009
(d) X-ray
0.0
0.0

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Scattering of Charged Particles

PROBLEM

A beam of 10-MeV protons is incident on a copper (Z = 29, A = 65) foil. For protons scattered 90°:
(a) Find the impact parameter.
(b) Using angular-momentum and energy conservation, determine the closest distance of approach of a proton to a copper nucleus for 90° scattering and compare this with the impact parameter.

[from Weidner, Richard T., and Sells, Robert L. 1980, Elementary Modern Physics, Third Edition (Boston, Massachusetts: Allyn and Bacon), problem 6.9]

SOLUTION

(a) For a particle of charge Q1 incident on a nucleus of charge Q2, the impact parameter is

b = (k/2)(Q1Q2/Ek) cot q/2

where k = 1/4pe0 = 8.98755 x 109 N m2/C2, Ek is the kinetic energy of the incident particle, and q is the scattering angle.

We have Q1 = e, Q2 = Ze = 29e, Ek = 10 MeV, and q = 90°. Thus, the impact parameter is

b = (k/2)(29e2/Ek) cot q/2
= [(8.98755 x 109 N m2/C2) / 2]{(29)(1.6 x 10-19 C)2 / [(10 MeV)(1.6 x 10-13 J/MeV)]} cot(90°/2)
= 2.085 x 10-15 m

(b) Consider a single proton incident on a single copper nucleus. Assume that the proton is initially traveling in the +x direction as shown in the diagram below with its distance from the x axis equal to the impact parameter b.

                   ^y     +z out of screen
|                   |
v Q1 = e            |
--o---->T1          |
b                   |Q2 = Ze = 29e
--------------------O-------------------->
^                   |                    x
|                   |


Let position 1 be a position of the proton far away from the nucleus as the proton is approaching the nucleus; in the above diagram, this would be a position to the far left, with the proton’s path still a distance b from the x axis. Let position 2 be the position of the proton at its closest approach to the nucleus.

Use conservation of energy and angular momentum of the proton as it moves from position 1 to position 2 to determine the closest distance of approach rmin.

The initial energy of the proton, at position 1, is

E1 = T1 + V1

where

T1 = 10 MeV = (10 MeV)(1.6 x 10-13 J/MeV) = 1.6 x 10-12 J

is the initial kinetic energy, and V1 is the initial potential energy, which we can take to be equal to zero because the proton is far away from the nucleus.

The final energy of the proton, at position 2, is

E2 = T2 + V2

where T2 and V2 are the kinetic and potential energies of the proton when it reaches its closest distance to the nucleus.

The initial and final kinetic energies can be written in terms of the initial and final velocities, v1 and v2,

T1 = (1/2)mpv12
T2 = (1/2)mpv22

where mp is the proton mass.

The final potential energy of the proton is

V2 = kZe2/rmin

where k = 1/4pe0 = 8.98755 x 109 N m2/C2.

Thus, conservation of energy yields

E1 = E2 => (1/2)mpv12 = (1/2)mpv22 + kZe2/rmin (1)

The initial angular momentum of the proton is

L1 = r1 x p1

where r1 and p1 are the position and momentum vectors of the proton at position 1. Using the right hand rule, L1 points into the screen. The magnitude of the initial angular momentum is

L1 = rmpv1 sin f = bmpv1

where f is the angle between r1 and p1. The final angular momentum, when the proton is closest to the nucleus, is

L2 = r2 x p2

also pointing into the screen, the magnitude of which is

L2 = rminmpv2

By conservation of angular momentum,

L1 = L2 => bmpv1 = rminmpv2 => bv1 = rminv2 => v2 = (b/rmin)v1 (2)

Using (2) to substitute for v2 in (1),

(1/2)mpv12 = (1/2)mp(b/rmin)2v12 + kZe2/rmin => T1 = (b/rmin)2T1 + kZe2/rmin
=> T1rmin2 = T1b2 + kZe2rmin => T1rmin2 - kZe2rmin - T1b2 = 0
=> rmin = {kZe2 + sqrt[(kZe2)2 + (4T1)(T1b2)]} / 2T1
= {kZe2 + sqrt[(kZe2)2 + (2T1b)2]} / 2T1
= [(8.98755 x 109 N m2/C2)(29)(1.6 x 10-19 C)2
+ sqrt{(8.98755 x 109 N m2/C2)2(29)2(1.6 x 10-19 C)4
+ [(2)(1.6 x 10-12 J)(2.085 x 10-15 m)]2}] / (2)(1.6 x 10-12 J)
= 5.034 x 10-15 m ~ 2.4b

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PROBLEM

A hydrogen gas is ionized by projecting an incident beam of particles through the gas. Use the conservation laws of energy and linear momentum to explain why ionization requires a smaller incident-particle energy when (a) photons are used instead of electrons; (b) when electrons are used instead of protons.

[from Weidner, Richard T., and Sells, Robert L. 1980, Elementary Modern Physics, Third Edition (Boston, Massachusetts: Allyn and Bacon), problem 6.29]

SOLUTION

(a) Consider the collision between the incident particle and a hydrogen atom in the center-of-mass frame. The minimum incident-particle energy occurs when the final particles are at rest.

Consider the case where the incident particle is a photon of energy Eg.

Eg               vH
o------>      <------O mH     ----->
                                x


The initial energy of the system is the sum of the photon energy, the kinetic energy of the hydrogen atom (assumed nonrelativistic), and the rest energy of the hydrogen atom,

E1 = Eg + (1/2)mHvH2 + mHc2

where mH = 1.007825 u is the hydrogen mass, vH is the speed of the hydrogen atom, and c = 3 x 108 m/s is the speed of light. The photon is absorbed by the electron in the hydrogen atom, ionizing the atom, and leaving a proton and electron at rest. The final energy of the system is the sum of the rest energies of the proton and electron,

E2 = mpc2 + mec2

where mp = 1.0072765 u is the proton mass, and me = 0.000548624 u is the electron mass.

Equating the initial and final energies,

E1 = E2 => Eg + (1/2)mHvH2 + mHc2 = mpc2 + mec2 (1)

In the center-of-mass frame, the total momentum is zero, so the momentum of the photon is equal to the momentum of the hydrogen atom. The momentum of the photon is

pg = Eg/c (2)

The momentum of the hydrogen atom is

pH = mHvH (3)

Equating (2) and (3),

pg = pH => Eg/c = mHvH => vH = Eg/mHc

Substituting into (1), we get

Eg + (1/2)mH(Eg/mHc)2 + mHc2 = mpc2 + mec2
=> Eg + (1/2mHc2)Eg2 + mHc2 = mpc2 + mec2
=> (1/2mHc2)Eg2 + Eg + (mH - mp - me)c2 = 0
=> Eg = {- 1 + sqrt[1 - (4)(1/2mHc2)(mH - mp - me)c2]} / (1/mHc2)
= mHc2{- 1 + sqrt[1 - (4)(1/2mHc2)(mH - mp - me)c2]}
= mHc2{- 1 + sqrt[1 - (2)(mH - mp - me)/mH]}
= (1.007825 u)(931.502 MeV/c2u){- 1 + sqrt[1 - (2)(1.007825 u - 1.0072765 u - 0.000548624 u) / (1.007825 u)]}
= 115.5 eV

The velocity of the hydrogen atom before the collision is

vH = Eg/mHc
= (115.5 eV)(1.6e-19 J/eV) / (1.007825 u)(931.502 MeV/c2u)(1.6 x 10-13 J/MeV)c
= 1.230 x 10-7c = (1.230 x 10-7)(3 x 108 m/s) = 36.91 m/s

The hydrogen atom was initially moving in the -x direction in the center-of-mass frame. Since the hydrogen atom is at rest in the laboratory frame, the center-of-mass frame is moving in the +x direction relative to the laboratory frame at velocity 36.91 m/s. The energy of the photon in the laboratory frame is obtained using the Lorentz transformation equation,

Eg' = g(Eg + bcpg) = g(Eg + bEg) = gEg(1 + b)

where

b = vH/c = (36.91 m/s) / (3 x 108 m/s) = 1.230 x 10-7
g = 1 / sqrt(1 - b2) ≈ 1

Thus,

Eg' ≈ Eg = 115.5 eV

Consider the case, in the center-of-mass frame, where the incident particle is an electron with kinetic energy Te and velocity ve.

      ve            vH
me o------>      <------O mH     ----->
      Te                           x


The initial energy of the system is the sum of the kinetic energies of the electron and hydrogen atom (both assumed nonrelativistic), and their rest energies,

E1 = Te + mec2 + (1/2)mHvH2 + mHc2
= (1/2)meve2 + mec2 + (1/2)mHvH2 + mHc2

The incident electron collides with the hydrogen atom, ionizing it, resulting in a proton and two electrons. The final energy of the system is the sum of the rest energies of the proton and two electrons,

E2 = mpc2 + 2mec2

Equating the initial and final energies,

E1 = E2 => (1/2)meve2 + mec2 + (1/2)mHvH2 + mHc2 = mpc2 + 2mec2 (4)

In the center-of-mass frame, the total momentum is zero, so the momentum of the incident electron is equal to the momentum of the hydrogen atom. The momentum of the incident electron is

pe = meve (5)

The momentum of the hydrogen atom is

pH = mHvH (6)

Equating (5) and (6),

pe = pH => meve = mHvH => vH = (me/mH)ve

Substituting into (4), we get

(1/2)meve2 + mec2 + (1/2)mH(me/mH)2ve2 + mHc2 = mpc2 + 2mec2
=> (1/2)[me + mH(me/mH)2]ve2 + mHc2 = mpc2 + mec2
=> (1/2)me(1 + me/mH)ve2 + mHc2 = mpc2 + mec2
=> ve = sqrt[2(mp + me - mH)c2 / me(1 + me/mH)]
= c sqrt[2(mp + me - mH) / me(1 + me/mH)]
= c sqrt[2(1.0072765 u + 0.000548624 u – 1.007825 u) / (0.000548624 u)(1 + 0.000548624 u / 1.007825 u)]
= 0.02126c = (0.02126)(3 x 108 m/s) = 6.377 x 106 m/s

The velocity of the hydrogen atom before the collision is

vH = (me/mH)ve = (0.000548624 u / 1.007825 u)(6.377 x 106 m/s)
= 3471 m/s

The hydrogen atom was initially moving in the -x direction in the center-of-mass frame. Since the hydrogen atom is at rest in the laboratory frame, the center-of-mass frame is moving in the +x direction relative to the laboratory frame at velocity 3471 m/s. The velocity of the electron in the laboratory frame is

ve’ = ve + vH = 6.377 x 106 m/s + 3471 m/s = 6.380 x 106 m/s

The speed of the electron is marginally relativistic. Using the nonrelativistic formula for the kinetic energy,

Te’ = (1/2)meve2 = (1/2)(9.11 x 10-31 kg)(6.380 x 106 m/s)2 = 1.854 x 10-17 J
= (1.854 x 10-17 J) / (1.6 x 10-19 J/eV) = 115.9 eV > Eg

Using the relativistic formula,

Te’ = (g - 1)mec2 = {1 / sqrt[1 – (ve’/c)2]}mec2
= {1 / sqrt[1 – (6.380 x 106 m/s / 3 x 108 m/s)2]}(9.11 x 10-31 kg)(3 x 108 m/s)2
= 1.855 x 10-17 J = (1.855 x 10-17 J) / (1.6 x 10-19 J/eV) = 115.9 eV > Eg

So assuming that the electron is nonrelativistic is a good approximation.

(b) Consider the case, in the center-of-mass frame, where the incident particle is a proton with kinetic energy Tp and velocity vp

      vp            vH
mp o------>      <------O mH     ----->
      Tp                           x


The initial energy of the system is the sum of the kinetic energies of the proton and hydrogen atom (both assumed nonrelativistic), and their rest energies,

E1 = Tp + mpc2 + (1/2)mHvH2 + mHc2
= (1/2)mpvp2 + mpc2 + (1/2)mHvH2 + mHc2

The incident proton collides with the hydrogen atom, ionizing it, resulting in two protons and one electron. The final energy of the system is the sum of the rest energies of the two protons and the electron,

E2 = 2mpc2 + mec2

Equating the initial and final energies,

E1 = E2 => (1/2)mpvp2 + mpc2 + (1/2)mHvH2 + mHc2 = 2mpc2 + mec2 (7)

In the center-of-mass frame, the total momentum is zero, so the momentum of the incident proton is equal to the momentum of the hydrogen atom. The momentum of the incident proton is

pp = mpvp (8)

The momentum of the hydrogen atom is

pH = mHvH (9)

Equating (8) and (9),

pp = pH => mpvp = mHvH => vH = (mp/mH)vp

Substituting into (7), we get

(1/2)mpvp2 + mpc2 + (1/2)mH(mp/mH)2vp2 + mHc2 = 2mpc2 + mec2
=> (1/2)[mp + mH(mp/mH)2]vp2 + mHc2 = mpc2 + mec2
=> (1/2)mp(1 + mp/mH)vp2 + mHc2 = mpc2 + mec2
=> vp = sqrt[2(mp + me - mH)c2 / mp(1 + mp/mH)]
= c sqrt[2(mp + me - mH) / mp(1 + mp/mH)]
= c sqrt[2(1.0072765 u + 0.000548624 u – 1.007825 u) / (1.0072765 u)(1 + 1.0072765 u / 1.007825 u)]
= (3.509 x 10-4)c = (3.509 x 10-4)(3 x 108 m/s) = 1.053 x 105 m/s

The velocity of the hydrogen atom before the collision is

vH = (mp/mH)vp = (1.0072765 u / 1.007825 u)(1.053 x 105 m/s)
= 1.052 x 105 m/s

The hydrogen atom was initially moving in the -x direction in the center-of-mass frame. Since the hydrogen atom is at rest in the laboratory frame, the center-of-mass frame is moving in the +x direction relative to the laboratory frame at velocity 1.052 x 105 m/s. The velocity of the proton in the laboratory frame is

vp’ = vp + vH = 1.053 x 105 m/s + 1.052 x 105 m/s = 2.105 x 105 m/s

The kinetic energy of the proton in the laboratory frame is

Tp’ = (1/2)mpvp2 = (1/2)(1.67 x 10-27 kg)(2.105 x 105 m/s)2 = 3.700 x 10-17 J
= (3.700 x 10-17 J) / (1.6 x 10-19 J/eV) = 231.2 eV > Te

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X-Ray Production

PROBLEM

A 5.0 MeV electron traveling east passes near a copper nucleus (rest energy, 6.0 x 104 MeV) and is deflected by the large electrostatic force. The electron is observed to emerge from the collision traveling south with an energy of 4.9 MeV. A single created photon travels away in the northeast direction.
(a) Find the energy and momentum of the created photon.
(b) Show that the recoil copper nucleus has a momentum 76 times that of the photon but a kinetic energy only 1/200 that of the photon.
(c) Determine the direction of the recoil nucleus.

[(a) and (b) from Weidner, Richard T., and Sells, Robert L. 1980, Elementary Modern Physics, Third Edition (Boston, Massachusetts: Allyn and Bacon), problem 4.21]

SOLUTION

         ^    _
         |  g /| pg Eg    ^ north
         |  /             |
e-       |/ q = 45°       |
o--->    O-----           +---> east
T1
      e- o T2
         |
         |
         v


(a) The (initial, final) kinetic energy of the electron is (T1, T2) = (5.0, 4.9) MeV. Neglect the kinetic energy of the recoil copper nucleus because it is much more massive than the electron. Then the energy of the photon is

Eg = T1 - T2 = 5.0 MeV - 4.9 MeV = 0.1 MeV

The momentum of the photon is

pg = Eg/c = 0.1 MeV/c

(b) In the interaction between the copper nucleus and the electron, the electron approaches the nucleus slightly north of the nucleus. The attractive electrostatic force between the electron and nucleus causes the electron to be pulled around the nucleus and end up traveling south. The recoil nucleus travels east of north due to the electrostatic force.

The total initial energy of the electron is

E1 = T1 + mec2

where me is the electron rest mass. The total initial energy of the electron is related to its total initial momentum, p1, and rest mass according to

E12 = p12c2 + me2c4 => (T1 + mec2)2 = p12c2 + me2c4
=> p1 = (1/c) sqrt[(T1 + mec2)2 - me2c4]
= (1/c) sqrt[(5.0 MeV + 0.511003 MeV)2 - (0.511003 MeV)2] = 5.487 MeV

Similarly, the final momentum of the electron is

p2 = (1/c) sqrt[(T2 + mec2)2 - me2c4]
= (1/c) sqrt[(4.9 MeV + 0.511003 MeV)2 - (0.511003 MeV)2] = 5.387 MeV

By conservation of momentum, the x (east-west) component of the momentum of the recoil nucleus, pNx, is related to p1 and the momentum of the photon according to

p1 = pNx + pg cos q => pNx = p1 - pg cos q

where q = 45° because the photon is traveling in the northeast direction. So

pNx = p1 - pg cos q = 5.487 MeV/c - (0.1 MeV/c) cos 45° = 5.417 MeV/c

Also by conservation of momentum, the y (north-south) component of the momentum of the recoil nucleus, pNy, is related to p2 and pg according to

0 = - p2 + pNy + pg sin q => pNy = p2 - pg sin q
= 5.387 MeV/c - (0.1 MeV/c) sin 45° = 5.316 MeV/c

The total momentum of the recoil nucleus is

pN = sqrt(pNx2 + pNy2) = sqrt[(5.417 MeV/c)2 + (5.316 MeV/c)2]
= 7.589 MeV/c => pN / pg = (7.589 MeV/c) / (0.1 MeV/c) = 75.89 ≈ 76

The kinetic energy of the recoil nucleus, assumed nonrelativistic, is

TN = pN2 / 2mN

where mN = 6.0 x 104 MeV/c2 is the mass of the copper nucleus. Thus,

TN = pN2 / 2mN = (7.589 MeV/c)2 / [(2)(6.0 x 104 MeV/c2)]
= 0.00048 MeV => Eg / TN = (0.1 MeV) / (0.00048 MeV) = 208.3 ≈ 200

(c) The direction of the recoil nucleus is given by the angle f where

tan f = pNy / pNx
=> f = tan-1(pNy / pNx) = tan-1[(5.316 MeV/c) / (5.417 MeV/c)]
= 44.46°

The direction of the recoil nucleus is 44.46° north of east.

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Compton Scattering

PROBLEM

A 100 MeV photon collides with a proton that is at rest. What is the maximum possible energy loss for the photon?

[from Gasiorowicz, Stephen 1974, Quantum Physics (New York: Wiley), problem 1.8]

SOLUTION

This is Compton scattering except that a photon scatters off a proton instead of an electron. The change in the photon wavelength is given by

Dl = l' - l = (h/mpc)(1 - cos q)

where l is the initial photon wavelength, l' is the final wavelength, h = 6.63 x 10-34 J s is Planck's constant, mp = 938.2592 MeV/c2 is the proton rest mass, c is the speed of light, and q is the photon scattering angle. The initial and final photon energies are

E = hc/l => l = hc/E

and

E' = hc/l' => l' = hc/E'

Thus,

hc/E' - hc/E = (h/mpc)(1 - cos q) => 1/E' - 1/E = (1/mpc2)(1 - cos q)
=> 1/E' = 1/E + (1/mpc2)(1 - cos q)
=> E' = 1/[1/E + (1/mpc2)(1 - cos q)] = E/[1 + (E/mpc2)(1 - cos q)]
=> E'min = E/[1 + (2E/mpc2)] = (100 MeV)/[1 + (2)(100 MeV)/(938.2592 MeV)] = 82.43 MeV

The maximum possible energy loss for the proton is

E - E'min = 100 MeV - 82.43 MeV = 17.57 MeV

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Pair Production

PROBLEM

A photon collides with a free electron at rest. Show that the minimum photon energy necessary for this interaction to create an electron-positron pair is four times the rest energy of an electron.

[from Weidner, Richard T., and Robert L. Sells 1980, Elementary Modern Physics (Boston, Massachusetts: Allyn and Bacon), problem 4.32]

SOLUTION

laboratory frame
   g       e-
------->   o
 Eg, pg    me


The minimum photon energy Eg in the laboratory frame occurs when, in the center-of-mass frame, the three final particles (original electron, new positron, and new electron) are at rest.

The energy-momentum four-vector has components (ipx, ipy, ipz, E/c), where px is the x component of the total linear momentum, py is the y component, pz is the z component, E is the total energy, c is the speed of light, and i is the square root of - 1.

The quantity (E/c)2 - px2 - py2 - pz2 is invariant and has the same value in the laboratory frame and in the center-of-mass frame. In this problem, we can define the +x direction to be the direction in which the incident photon is traveling and set both py and pz equal to zero, so (E/c)2 - px2 is invariant, or

(E/c)2 - px2 = constant (1)

In the laboratory frame, the energy is

E1 = Eg + mec2

where Eg is the photon energy and me is the electron mass, and the momentum is

p1 = p1x = pg = Eg/c

where pg = Eg/c is the photon momentum.

In the center-of-mass frame, the energy is

E2 = 3mec2

and the momentum is

p2 = 0

Using (1), we have

(E1/c)2 - p12 = (E2/c)2 - p22 => (Eg + mec2)2 / c2 - (Eg/c)2 = (3mec2)2 / c2
=> Eg2/c2 + 2Egme + me2c2 - Eg2/c2 = 9me2c2 => 2Egme + me2c2 = 9me2c2
=> 2Egme = 8me2c2 => Eg = 4mec2

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PROBLEM

A beam of 2.04-MeV photons is incident on a thin copper target. One of the photons passing near the nucleus of a copper atom becomes annihilated and produces an electron-positron pair at a distance of 1.0 x 10-3 Å from the nucleus. The created particles have the same kinetic energy Ek at the point of creation and move away in the direction in which the incident photon is moving. Because of the strong Coulomb force between each particle and the copper nucleus (charge +29e) the two oppositely charged particles will have different kinetic energies Ek+ and Ek- when they have moved far from the nucleus.
(a) Show that after the created particles have moved away a distance much larger than 1.0 x 10-3 Å, the difference in kinetic energies will be Ek+ - Ek- = 0.84 MeV.
(b) If the two particles enter perpendicularly into a uniform magnetic field of 0.15 T, what will the radius of curvature be?

[from Weidner, Richard T., and Robert L. Sells 1980, Elementary Modern Physics (Boston, Massachusetts: Allyn and Bacon), problem 4.35]

SOLUTION

(a) For each of the two particles, energy is conserved between the time they are created and the time they are far from the copper nucleus. When the particles are a large distance from the copper nucleus, their potential energy is negligible. Thus, their kinetic energies at that point will each be equal to the sum of their kinetic and potential energies when they were created. As stated in the problem, their initial kinetic energies are both Ek. Their initial potential energies are

V± = ± (1/4pe0)(29e2/rmin)

where rmin = 1.0 x 10-3 Å = 1.0 x 10-13 m, V+ is the initial potential energy of the positron, and V- is the initial potential energy of the electron. Thus, the kinetic energies of the positron and electron far away from the copper nucleus are

Ek+ = Ek + V+ = Ek + (1/4pe0)(29e2/rmin) (1)
Ek- = Ek + V- = Ek - (1/4pe0)(29e2/rmin) (2)

Subtracting (2) from (1), we get

Ek+ - Ek- = (2)(1/4pe0)(29e2/rmin)
= (2)(9 x 109 N m2 C-2)(29)(1.6 x 10-19 C)2 / (1.0 x 10-13 m) = 1.336 x 10-13 J
= (1.336 x 10-13 J)(1 MeV / 1.6 x 10-13 J) = 0.8352 MeV

(b) The radius of curvature of the circular path of a charged particle with rest mass m, velocity v, momentum p = gmv, and charge q, moving in a plane perpendicular to the direction of a magnetic field B, is

R = p/qB = gmv/qB (3)

where g = 1 / sqrt(1 - b2), b = v/c, and c = 3 x 108 m/s is the speed of light in a vacuum.

From part (a), the kinetic energies of the positron and electron, when they are far from the copper nucleus, are

Ek± = Ek + V± (4)

When the photon is converted into an electron-positron pair, all of its energy is converted into the total energy of the electron-positron pair, with includes rest energy, kinetic energy, and potential energy. Thus,

Eg = mec2 + Ek + V+ + mec2 + Ek + V- = 2mec2 + 2Ek

because

V+ = (1/4pe0)(29e2/rmin) = - V-

Thus,

Ek = (Eg - 2mec2)/2 = [2.04 MeV - (2)(0.511 MeV)] / 2 = 0.509 MeV

The potential energies right after the electron-positron pair is created are

V± = ± (1/4pe0)(29e2/rmin) = ± (9 x 109 N m2 C-2)(29)(1.6 x 10-19 C)2 / (1.0 x 10-13 m)
= ± 6.682 x 10-14 J = ± (6.682 x 10-14 J)(1 MeV / 1.6 x 10-13 J) = ± 0.4176 MeV

From (4), the kinetic energies far from the copper nucleus are

Ek± = Ek + V± = 0.509 MeV ± 0.4176 MeV = (0.9266 MeV, 0.09140 MeV)

The total energies of the positron and electron far from the copper nucleus are

E± = Ek± + mec2 = (0.9266 MeV + 0.511 MeV, 0.09140 MeV + 0.511 MeV)
= (1.438 MeV, 0.6024 MeV)

The relativistic factor g for the two particles is

g± = E± / mec2 = (1.438 MeV / 0.511 MeV, 0.6024 MeV / 0.511 MeV) = (2.813, 1.179)

The relativistic factor b is related to g by

g = 1 / sqrt(1 - b2) => b = sqrt(1 - 1/g2)

Thus,

b+ = sqrt[1 - 1/(g+)2] = sqrt(1 - 1/2.8132) = 0.9347
b- = sqrt[1 - 1/(g-)2] = sqrt(1 - 1/1.1792) = 0.5296

The velocities of the positron and electron are

v+ = b+c = (0.9347)(3 x 108 m/s) = 2.804 x 108 m/s
v- = b-c = (0.5296)(3 x 108 m/s) = 1.589 x 108 m/s

Substituting into (3), and setting q = e = 1.6 x 10-19 C for both, we get

R+ = g+mev+/qB = (2.813)(9.11 x 10-31 kg)(2.804 x 108 m/s) / [(1.6 x 10-19 C)(0.15 T)] = 0.02994 m = 2.994 cm
R- = g-mev+/qB = (1.179)(9.11 x 10-31 kg)(1.589 x 108 m/s) / [(1.6 x 10-19 C)(0.15 T)] = 0.007109 m = 0.7109 cm

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Atoms

PROBLEM

The ionization (binding) energy of the outermost electron in boron is 8.26 eV. (a) Use the Bohr model to estimate the "effective charge," Zeff, seen by this electron. (b) Estimate the average orbital radius.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 28.29]

SOLUTION

(a) According to the Bohr model, the angular momentum L of an electron in an atom can have values of the form

L = Ln = nħ (1)

where n is a positive integer, ħ = h / 2p, and h = 6.63 x 10-34 J s is Planck's constant. The angular momentum of an electron of mass m in a circular orbit of radius rn is

L = Ln = mvrn (2)

where v is the orbital velocity. Combining (1) and (2), we get

mvrn = nħ => v = nħ / mrn (3)

A boron atom contains five electrons. Thus, its ground state electronic configuration is 1s22s22p1. The outermost electron is the 2p electron, which orbits a nucleus of effective charge Zeffe, where = 1.6 x 10-19 C. The effective charge of the nucleus is less than the charge of the five protons it contains, 5e, due to shielding by the other four electrons. The 2p electron orbits the nucleus in a circular orbit. The centripetal force it experiences is

FC = mv2 / rn (4)

This force is supplied by the electrical attraction between the electron and the nucleus, given by Coulomb's law, so

FC = kZeffe2 / rn2 (5)

where k = 9 x 109 N m2 / C2. Equating (4) and (5) gives us

mv2 / rn = kZeffe2 / rn2

If we multiply both sides by rn / 2, we get

(1/2)mv2 = kZeffe2 / 2rn (6)

The total energy of the 2p electron is the sum of its kinetic and potential energies,

En = (1/2)mv2 - kZeffe2 / rn = kZeffe2 / 2rn - kZeffe2 / rn = - kZeffe2 / 2rn (7)

Substituting (3) into (6),

(1/2)m(n2ħ2 / m2rn2) = kZeffe2 / 2rn => (1/2)n2ħ2 / mrn2 = kZeffe2 / 2rn => n2ħ2 / mrn = kZeffe2 => 1 / rn = mkZeffe2 / n2ħ2

Substituting into (7),

En = - (kZeffe2 / 2)(mkZeffe2 / n2ħ2) = - mk2Zeff2e4 / 2n2ħ2

For the 2p electron, n = 2, so its energy is

E2 = - mk2Zeff2e4 / 8ħ2

Solving for Zeff,

Zeff = (- 8ħ2E2 / mk2e4)1/2

We are told that the ionization energy of the electron is 8.26 eV, so its energy when bound is E2 = - 8.26 eV. We can now calculate what Zeff is.

Zeff = {[- (8)(6.63 x 10-34 J s)2(- 8.26 eV)(1.6 x 10-19 J/eV) / 4p2] /
[(9.11 x 10-31 kg)(9 x 109 N m2 C-2)2(1.6 x 10-19 C)4]}1/2
= 1.5602

(b) Solve (7) for rn.

rn = - kZeffe2 / 2En => r2 = - kZeffe2 / 2E2
= - (9 x 109 N m2 C-2)(1.5602)(1.6 x 10-19 C)2 /
[(2)(- 8.26 eV)(1.6 x 10-19 J/eV)]
= 1.36 x 10-10 m = 1.36 Å

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Molecules

PROBLEM

Estimate the binding energy of the H2 molecule by calculating the difference in the kinetic energy of the electrons when they are in separate atoms and when they are in the molecule, using the uncertainty principle. Take Dx for the electrons in the separated atoms to be the radius of the first Bohr orbit, 0.053 nm, and for the molecule take Dx to be the separation of the nuclei, 0.074 nm.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 29.39]

SOLUTION

According to the Heisenberg uncertainty principle, the uncertainties in the position Dx and in the momentum Dp are related by

DxDp > ~ h/2p

where h = 6.63 x 10-34 J s is Planck's constant. From this we get

p ~ Dp ~ h/2pDx

The total kinetic energy of the two electrons is

KE = 2(p2/2m) = p2/m ~ (h/2pDx)2/m

where m = 9.11 x 10-31 kg is the electron mass. The total kinetic energy of the two electrons when they are in separate atoms is

KEatoms = {(6.63 x 10-34 J s)/[(2p)(0.053 x 10-9 m)]}2/(9.11 x 10-31 kg) = 4.351 x 10-18 J = 27.19 eV

The total kinetic energy of the electrons when they are in the H2 molecule is

KEmolecule = {(6.63 x 10-34 J s)/[(2p)(0.074 x 10-9 m)]}2/(9.11 x 10-31 kg) = 2.232 x 10-18 J = 13.95 eV

The change in the total kinetic energy of the electrons is

DKE = KEmolecule - KEatoms = 13.95 eV - 27.19 eV = - 13.24 eV

Therefore, the binding energy of the H2 molecule is 13.24 eV.

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PROBLEM

In the ionic salt KF, the separation distance between ions is about 0.27 nm.
(a) Estimate the electrostatic potential energy between the ions assuming them to be point charges (magnitude 1e).
(b) It is known that F releases 4.07 eV of energy when it "grabs" an electron, and 4.34 eV is required to ionize K. Find the binding energy of KF relative to free K and F atoms, neglecting the energy of repulsion (i.e., activation energy).

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 29.41]

SOLUTION

(a) V = - (1/4pe0)e2/r = - (9 x 109 N m2/C2)(1.6 x 10-19 C)2/(0.27 x 10-9 m) = - 8.533 x 10-19 J
= - 5.333 eV

(b) The amount of energy needed to convert a KF molecule to separate K and F atoms is the sum of 5.333 eV to separate the K+ and the F-, 4.07 eV to remove the extra electron from the F-, and - 4.34 eV released when K+ captures an electron. The sum is 5.333 eV + 4.07 eV - 4.34 eV = 5.063 eV. This is the binding energy of the KF molecule.

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Semiconductor Physics

PROBLEM

For an arsenic donor atom in a doped silicon semiconductor, assume that the "extra" electron moves in a Bohr orbit about the arsenic ion. For this electron in the ground state, take into account the dielectric constant K = 12 of the Si lattice (which represents the weakening of the Coulomb force due to all the other atoms and ions in the lattice), and estimate (a) the orbit radius and (b) the binding energy for this extra electron. [Hint: Substitute e = Ke0 in Coulomb's law.]

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 29.43]

SOLUTION

(a) The atomic number of arsenic is Z = 33. Its ground state electron configuration is 1s22s22p63s23p64s23d104p3. The extra electron orbits the arsenic ion with a net charge of +e. If there was only a single arsenic ion, the electric field experienced by the free electron at radial distance r would be

E = (1/4pe0)e/r2

Because of the presence of all the other atoms and ions in the lattice, the electric field is reduced by a factor of K and is given by

E = (1/4pKe0)e/r2

The Coulomb force experienced by the extra electron is

F = eE = (1/4pKe0)e2/r2

and is equal to the electron mass m times the centripetal acceleration

ac = v2/r

where v is the velocity of the electron in its orbit, so

(1/4pKe0)e2/r2 = mv2/r => (1/4pKe0)e2/r = mv2 (1)

According to Bohr's quantum condition, the angular momentum of the extra electron is

L = mvr = nh/2p => v = nh/2pmr

where n is a positive integer. Substituting this into (1), we get

(1/4pKe0)e2/r = m(nh/2pmr)2 = n2h2/4p2mr2 => r = n2h2Ke0/pme2 = rn

For n = 1, the ground state, we have

r1 = h2Ke0/pme2 = (6.63 x 10-34J s)2(12)(8.85 x 10-12C2/N m2)/[p(9.11 x 10-31 kg)(1.6 x 10-19 C)2] = 6.372 x 10-10 m = 6.372 Å = 0.6372 nm

(b) The energy of the electron is equal to the sum of its kinetic and potential energies,

E = (1/2)mv2 - (1/4pKe0)e2/r = (1/2)(1/4pKe0)e2/r - (1/4pKe0)e2/r = - (1/8pKe0)e2/(n2h2Ke0/pme2)
= - me4/8K2e02n2h2 = En

The ground state energy is

E1 = - me4/8K2e02h2
= - (9.11 x 10-31 kg)(1.6 x 10-19 C)4 / [(8)(12)2(8.85 x 10-12 C2/N m2)2(6.63 x 10-34 J s)2]
= - 1.505 x 10-20 J = - (1.505 x 10-20 J)(1 eV)/(1.6 x 10-19 J) = - 0.09408 eV

The binding energy is 0.09408 eV.

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PROBLEM

The zener diode voltage regulator shown below has R = 1.80 kW and a diode breakdown voltage of 130 V. The diode is rated at a maximum current of 100 mA.
(a) If Rload = 15.0 kW, over what range of supply voltages will the circuit maintain the voltage at 130 V?
(b) If the supply voltage is 200 V, over what range of load resistance will the voltage be regulated?

   +-----R-----+-----o-----+
   |           |     ^     |
   |           |     |     |
  +|     zener _     |     |
Vsupply  diode ^   Vout  Rload
  -|           |     |     |
   |           |     |     |
   |           |     v     |
   +-----------+-----o-----+


[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 29.45]

SOLUTION

(a) Let Vmin and Vmax be the minimum and maximum values of Vsupply for which Vout is regulated at 130 V.

When Vsupply = Vmin, the current through Rload is (130 V) / (15.0 x 103 W) = 8.667 x 10-3 A and the current through the diode is zero. The current through R is therefore also 8.667 x 10-3 A and the voltage drop across R is (8.667 x 10-3 A)(1800 W) = 15.6 V. Thus, Vmin = Vsupply = 130 V + 15.6 V = 145.6 V.

When Vsupply = Vmax, the current through Rload is still 8.667 x 10-3 A. The current through the diode is the maximum current of 100 x 10-3 A. The total current through R is 8.667 x 10-3 A + 100 x 10-3 A = 108.667 x 10-3 A and the voltage drop across R is (108.667 x 10-3 A)(1800 W) = 195.6 V. Thus, Vmax = Vsupply = 130 V + 195.6 V = 325.6 V.

So 195.6 V < Vsupply < 325.6 V.

(b) Let (Rload)min and (Rload)max be the minimum and maximum values of Rload for which Vout is regulated at 130 V.

If Rload = 0, all of the current goes through Rload and none goes through the diode. The potential difference across Rload and across the diode is zero. If Rload is increased, the potential difference across Rload and across the diode increases. The current through the diode is still zero because the potential difference across the resistor and across the diode is still less than 130 V. If I is the current through R and Rload, Kirchoff's second rule, which states that the sum of the potential differences around a loop is zero, yields

Vsupply - IR - IRload = 0 => I = Vsupply/(R + Rload)

The voltage across Rload is

Vload = IRload = Vsupply[Rload/(R + Rload)] => (Vload/Vsupply)(R + Rload) = Rload

=> Rload = (Vload/Vsupply)R/(1 - Vload/Vsupply)

When Rload = (Rload)min, the potential difference across Rload and across the diode is 130 V, the breakdown voltage, and we have

Rload = (Vload/Vsupply)R/(1 - Vload/Vsupply)
= [(130 V)/(200 V)](1800 W)/[1 - (130 V)/(200 V)] = 3343 W = 3.343 kW = (Rload)min

At this value of Rload, the diode switches on in the reverse bias direction and current starts flowing through the diode. As we continue to increase Rload, the diode continues to maintain its voltage at 130 V no matter how large Rload gets because the current through the diode never reaches its maximum current of 100 mA. The potential difference across R is

VR = Vsupply - Vload = 200 V - 130 V = 70 V

The current through R is

I = VR/R = (70 V) / (1800 W) = 38.89 mA

So when Rload gets very large, the maximum current through the diode is equal to 38.89 mA, the current through R. We have (Rload)max = ∞ and 3.343 kW < Rload < ∞.

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PROBLEM

A strip of silicon 1.5 cm wide and 1.0 mm thick is immersed in a magnetic field of strength 1.5 T perpendicular to the strip. When a current of I = 0.20 mA is run through the strip, there is a resulting Hall effect voltage of 18 mV across the strip. How many electrons per silicon atom are in the conduction band? The density of silicon is r = 2330 kg/m3.

         ^
         |
         | B = 1.5 T
         |

+------------------+
| o I out of paper | t = 1.0 mm
+------------------+
     w = 1.5 cm


[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 29.47]

SOLUTION

The presence of the magnetic field causes a Hall effect emf, to the left, of magnitude

EH = vdBw => vd = EH/Bw (1)

where vd is the electron drift velocity. The current density is

j = nevd = I/A => vd = I/Ane = I/wtne (2)

where n is the density of free electrons (i.e., electrons in the conduction band), e = 1.6 x 10-19 C is the elementary charge, and A = wt is the cross-sectional area of the conductor. Equating the expressions for vd in (1) and (2), we get

EH/Bw = I/wtne

Solving for the free electron density, we get

n = IB/teEH = (0.20 x 10-3 A)(1.5 T) / [(1.0 x 10-3 m)(1.6 x 10-19 C)(18 x 10-3 V)]
= 1.042 x 1020 electrons/m3 (3)

The molar mass of silicon is

M = (0.9223)(27.976927 g/mole) + (0.0777)(30.975362 g/mole) = 28.21 g/mole
= 28.21 x 10-3 kg/mole.

The molar density of silicon is

r/M = (2330 kg/m3)/(28.21 x 10-3 kg/mole) = 8.260 x 104 mole/m3

The atomic density of silicon is

NAr/M = (6.022 x 103atoms/mole)(8.260 x 104 mole/m3) = 4.974 x 1028 atoms/m3 (4)

where NA = 6.022 x 103/mole is Avogadro's number. Dividing (3) by (4), the number of conduction band electrons per silicon atom is

n/(NAr/M) = (1.042 x 1020) / (4.974 x 1028) electrons/atom = 2.094 x 10-9 electrons/atom

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Nuclear Reactions

PROBLEM

(a) Show, using the laws of conservation of energy and momentum, that for a nuclear reaction requiring added energy, based on rest mass considerations, the minimum kinetic energy of the bombarding particle (the threshold energy) is equal to Qmpr / (mpr - mb), where Q is the energy corresponding to the difference in total mass between products and reactants, mb is the rest mass of the bombarding particle, and mpr the total rest mass of the products. Assume the target nucleus is at rest before an interaction takes place, and that all particles are nonrelativistic.
(b) Apply the formula derived above to the reaction 6C13(p, n)7N13 to determine the minimum kinetic energy of the bombarding proton that will allow the reaction to proceed.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problems 31.15 and 31.16]

SOLUTION

(a) Assume that there are four particles involved, numbered as follows: 1 = bombarding particle, 2 = target particle, 3 = heavier product particle, and 4 = lighter product particle.

In the center-of-mass reference frame, the minimum initial kinetic energy occurs when the final particles are at rest. In the laboratory reference frame, this corresponds to the final particles moving at the same speed (the center-of-mass velocity) in the direction in which the bombarding particle was moving. Then the situations before and after the collision are as follows:

before collision:
vi
o---> o target
m1    m2


after collision:
vf
oo--->
m3, m4


Conservation of momentum requires that

m1vi = (m3 + m4)vf => vf = vim1 / (m3 + m4) (1)

Conservation of energy requires that the sum of the total rest mass energy and kinetic energy before and after the collision be the same, or

(m1 + m2)c2 + (1/2)m1vi2 = (m3 + m4)c2 + (1/2)(m3 + m4)vf2

Using (1), we get

(1/2)m1vi2 = (m3 + m4)c2 - (m1 + m2)c2 + (1/2)m1vi2[m1 / (m3 + m4)]
= Q + (1/2)m1vi2[m1 / (m3 + m4)] (2)

where

Q = (m3 + m4)c2 - (m1 + m2)c2

Rearranging (2), we get

(1/2)m1vi2{1 – [m1 / (m3 + m4)]} = Q
=> (1/2)m1vi2 = Q / {1 – [m1 / (m3 + m4)]}
= (m3 + m4)Q / [(m3 + m4 - m1) = mprQ / (mpr - mb)

where

mpr = m3 + m4

is the total rest mass of the products and

mb = m1

is the mass of the bombarding particle.

Since the kinetic energy of the incident particle is

K = (1/2)m1vi2

we have

K = mprQ / (mpr - mb)

(b) In the reaction 6C13(p, n)7N13, a proton collides with a 6C13 nucleus and produces a 7N13 nucleus and a neutron. Equivalently, a 1H1 nucleus, which is a proton, collides with a 6C13 nucleus and produces a 7N13 nucleus and a neutron, or

1H1 + 6C13 => 7N13 + 0n1

Using the definitions 1 = bombarding particle, 2 = target particle, 3 = heavier product particle, and 4 = lighter product particle, we have

Q = (m3 + m4)c2 - (m1 + m2)2
= [m(7N13) – 7me + mn]c2 - [m(1H1) – me + m(6C13) – 6me]c2
= [m(7N13) + mn]c2 - [m(1H1) + m(6C13)]c2

where mn is the neutron mass and me is the electron mass. Plugging in values for m(6C13), m(7N13), and m(1H1), which are atomic masses, mn, and me,

Q = (13.005738 u + 1.008665 u)(931.5 MeV/c2u)c2
- (1.007825 u + 13.003355 u)(931.5 MeV/c2u)c2
= 3.002 MeV
mb = m1 = m(1H1) - me = 1.007825 u – 0.000549 u = 1.007276 u
mpr = m3 + m4 = m(7N13) – 7me + mn
= 13.005738 u – (7)(0.000549 u) + 1.008665 u = 14.01056 u

Thus the minimum kinetic energy is

K = mprQ / (mpr - mb) = (14.01056 u)(3.002 MeV) / (14.01056 u – 1.007276 u) = 3.235 MeV

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Relativistic Kinematics

Define the S frame to be the laboratory frame, and define the S' frame to be the frame moving in the +x direction with speed u relative to S. Define b = u/c, where c = 3 x 108 m/s is the speed of light in a vacuum, and g = 1 / sqrt(1 - b2). The Lorentz coordinate transformation equations for converting from coordinates in S to coordinates in S' are

x' = g(x - bct) (1)
y' = y
z' = z
ct' = g(ct - bx) (2)

Taking the differentials of (1) and (2), we get

dx' = g(dx - bc dt) (3)
c dt' = g(c dt - b dx) => dt' = g[dt - (b/c) dx] (4)

Dividing (3) by (4), the x component of the velocity of an object in S' is

vx' = dx'/dt' = (dx - bc dt) / [dt - (b/c) dx] = (dx/dt - u) / [1 - (b/c) dx/dt] = (vx - u) / (1 - uvx/c2)

Thus, if the x component of the velocity of an object in S is vx, the x component of the velocity of the object in S' is given by

vx' = (vx - u) / (1 - uvx/c2)

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PROBLEM

In the laboratory (S) frame, an a particle moving east with a speed of 0.5c is passed by an electron moving west with a speed of 0.95c. Find the speed of the electron relative to the a particle.

[from Richtmyer, F. K., Kennard, E. H., and Cooper, J. N. 1969, Introduction to Modern Physics, Fifth Edition (New York: McGraw-Hill), problem 2.8]

SOLUTION

The velocity transformation equation is

vx' = (vx - u) / (1 - uvx/c2)

where vx is the velocity in the S frame, vx' is the velocity in the S' frame, and u is the velocity of the S' frame relative to the S frame. Let east be in the direction of +x. The velocity of the electron in the S frame is vx = - 0.95c. Let the S' frame be the frame of the a particle. Then u = 0.5c. The velocity of the electron relative to the a particle is

vx' = (- 0.95c - 0.5c) / [1 - (0.5)(- 0.95)] = - 0.98305c

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PROBLEM

Two protons, each having a speed of 0.935c in the laboratory, are moving toward each other. Determine (a) the momentum of each proton in the laboratory, (b) the total momentum of the two protons in the laboratory, and (c) the momentum of one proton as seen by the other proton.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 26.63]

SOLUTION

(a) Define b = v/c, where v is the speed of one of the protons, and g = 1 / sqrt(1 - b2). Then

b = 0.935
g = 1 / sqrt(1 - 0.9352) = 2.8197

The momentum of each proton in the laboratory is

p = gmpv = gmpbc = (2.8197)(938.3 MeV/c2)(0.935c) = 2474 MeV/c = 1.319 x 10-18 kg m/s

(b) The total momentum of the two protons in the laboratory is zero.

(c) The velocity of the two protons in the laboratory can be written as u = ± 0.935c. Let S be the laboratory frame, and let S' be the reference frame of the proton with u = + 0.935c. The velocity of the other proton, as measured in S, is vx = - 0.935c. The velocity measured in S' is

vx' = (vx - u) / (1 - uvx/c2) = (- 0.935c - 0.935c) / [1 - (0.935c)(- 0.935c)/c2] = - 0.99775c

The momentum of the proton measured in S' is

px' = g'mpvx' = g'mpb'c

where

b' = vx'/c = - 0.99775
g' = 1 / sqrt(1 - b'2) = 14.92

Thus,

px' = (14.92)(938.3 MeV/c2)(- 0.99775c) = - 13964 MeV/c = - 7.447 x 10-18 kg m/s

The momentum of one proton as seen by the other proton is 13964 MeV/c = 7.447 x 10-18 kg m/s.

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Relativistic Dynamics

PROBLEM

A 2000-kg spaceship, initially at rest in an inertial system, is accelerated along a straight line for 1.0 yr (3.1 x 107 s) by means of a photon rocket of 300-MW power.
(a) What is the loss in mass of the spaceship during this time?
(b) What is the speed of the spaceship after one year?
(c) What distance does it travel?
(d) What fraction of the energy released is in kinetic energy of the spaceship?

[from Weidner, Richard T., and Robert L. Sells 1980, Elementary Modern Physics (Boston, Massachusetts: Allyn and Bacon), problem 4.17]

SOLUTION

(a) The energy E expended by the photon rocket is equal to the energy equivalent to the change in mass Dm of the spaceship.

E = Dmc2

But the energy expended is equal to the product of the power P and the time t or

E = Pt

Thus,

Pt = Dmc2 => Dm = Pt / c2 = (300 x 106 W)(3.1 x 107 s) / (3 x 108 m/s)2 = 0.1033 kg

(b) The total momentum of the photons emitted by the rocket is

p = E / c = Pt / c

This will be the final momentum of the rocket,

p = (m - Dm)v

where m = 2000 kg is the initial mass and v is the velocity. Solving for v,

v = p / (m - Dm) = Pt / (m - Dm)c = (300 x 106 W)(3.1 x 107 s) / [(2000 kg - 0.1033 kg)(3 x 108 m/s)] = 15,500 m/s

Note that the rocket is still nonrelativistic since

b = v/c = (15,500 m/s) / (3 x 108 m/s) = 5.167 x 10-5 << 1

(c) Since the rocket loses so little of its mass, its mass is approximately constant, and its acceleration is approximately constant. The average velocity is approximately half of the final velocity. The distance traveled is approximately

d = vavgt = vt / 2 = (15,500 m/s)(3.1 x 107 t) / 2 = 2.403 x 1011 m

(d) The kinetic energy of the spaceship is

T = (1/2)(m - Dm)v2 = (1/2)(2000 kg - 0.1033 kg)(15,500 m/s)2 = 2.403 x 1011 J

The fraction of the energy released which goes into the kinetic energy of the spaceship is

f = T / Pt = (2.403 x 1011 J) / [(300 x 106 W)(3.1 x 107 s)] = 2.583 x 10-5

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PROBLEM

An unknown particle is measured to have a negative charge and a speed of 2.24 x 108 m/s. Its momentum is determined to be 3.07 x 10-22 kg m/s. Identify the particle by finding its rest mass.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 26.60]

SOLUTION

The momentum of a relativistic particle is

p = gmv = gmbc

where m is its rest mass, v is its speed, g = 1 / sqrt(1 - b2), and b = v/c. We have

b = v/c = (2.24 x 108 m/s) / (3 x 108 m/s) = 0.7467
g = 1 / sqrt(1 - 0.74672) = 1.503

The particle's rest mass is

m = p / gv = (3.07 x 10-22 kg m/s) / [(1.503)(2.24 x 108 m/s)] = 9.117 x 10-31 kg

The particle is an electron.

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PROBLEM

A proton of rest mass m is accelerated to a speed v ≈ c in a synchrotron. If it undergoes a perfectly inelastic collision with a nucleus at rest of mass M, find the rest mass and the speed of the resulting compound nucleus.

[from Richtmyer, F. K., Kennard E. H., and Cooper, J. N. 1969, Introduction to Modern Physics, Sixth Edition (New York: McGraw-Hill Book Company), problem 3.8]

SOLUTION

Let M1 be the mass of the compound nucleus and v' be its speed. Define

b = v/c
g = 1 / sqrt(1 - b2)
b' = v'/c
g' = 1 / sqrt(1 - b'2)

The initial momentum of the system is

p1 = gmv = gmbc

The final momentum is

p2 = g'M1v' = g'M1b'c

The initial energy is

E1 = gmc2 + Mc2

The final energy is

E2 = g'M1c2

By conservation of momentum and energy, we have

p1 = p2 => gmb = g'M1b' (1)
E1 = E2 => gmc2 + Mc2 = g'M1c2 => gm + M = g'M1 (2)

Dividing (1) by (2),

gmb / (gm + M) = b' = v'/c => v' = gmv / (gm + M)

Using (1), we solve for M1.

M1 = gmb / g'b' = gmb / [b' / sqrt(1 - b'2)] = [gmb sqrt(1 - b'2)] / b'
= [gmb sqrt(1 - b'2] / [gmb / (gm + M)] = (gm + M) sqrt{1 - [gmb / (gm + M)]2}
= sqrt[(gm + M)2 - (gmb)2] = sqrt(g2m2 + 2gmM + M2 - g2m2b2)
= sqrt[g2m2(1 - b2) + 2gmM + M2] = sqrt(m2 + M2 + 2gmM)

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PROBLEM

A positive pion can be produced through the reaction p + p => p + n + p+ by bombarding protons at rest with high-energy protons. If the rest energies for p, n, and p+ are respectively 938, 939.5, and 135 MeV, find the minimum kinetic energy for the incident protons for this reaction. Hint: Work first in a frame in which the center of mass is at rest, the zero-momentum frame.

[from Richtmyer, Kennard, and Cooper 1969, Introduction to Modern Physics, Fifth Edition (New York: McGraw-Hill), problem 3.18]

SOLUTION

In the center-of-mass frame, the minimum energy occurs when the products are at rest. The total energy is thus equal to the sum of the rest energies of the final particles.

E' = mpc2 + mnc2 + mpc2

In the center-of-mass frame, the total momentum is

p' = 0

In the laboratory frame, the total energy is the total energy of the proton which is initially moving plus the rest energy of the proton which is initially at rest.

E = gmpc2 + mpc2 = (g + 1)mpc2

The total momentum is the momentum of the proton that was initially moving.

p = gmpbc

Here,

g = 1 / sqrt(1 - b2)

b = v/c

and v is the initial velocity of the proton that was moving. Now we make use of the fact that, since (cp, E) is a four vector, E2 - (cp)2 is invariant.

E2 - (cp)2 = E'2 - (cp')2 => (g + 1)2mp2c4 - g2b2mp2c4 = E'2 => (g + 1)2 - g2b2 = (E'/mpc2)2

Since

b = sqrt(1 - 1/g2)

we can write

(g + 1)2 - g2(1 - 1/g2) = (E'/mpc2)2 => (g + 1)2 - (g2 - 1) = (E'/mpc2)2
=> g2 + 2g + 1 - g2 + 1 = (E'/mpc2)2 => g = (1/2)(E'/mpc2)2 - 1

Substituting for E', we have

g = (1/2)[(mp + mn + mp) / mp]2 - 1 = (1/2)[(938 + 939.5 + 135) / 938]2 - 1 = 1.302

The kinetic energy of the incident proton is

T = (g - 1)mpc2 = (1.302 - 1)(938 MeV) = 282.9 MeV

This is the minimum kinetic energy that the incident protons must have in order for the reaction to occur.

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PROBLEM

Antiprotons are produced by bombarding protons at rest with high-energy protons through the reaction p + p => p + p + (p + p*-).
(a) In the center-of-mass frame find the minimum kinetic energy of each proton required for the reaction.
(b) In this frame and for this energy what are the values of b and g for each initial proton?
(c) Transform to the laboratory frame and find the minimum kinetic energy of the incident proton for the reaction.

[from Richtmyer, Kennard, and Cooper 1969, Introduction to Modern Physics, Fifth Edition (New York: McGraw-Hill), problem 3.19]

SOLUTION

(a) In the center-of-mass frame, the total momentum is zero, and the total energy Etot' must be sufficient to produce the products at rest. The two protons have equal velocities in opposite directions, so the kinetic energies are equal.

Etot' = 2mpc2 + 2T' = 4mpc2

where mp is the rest mass of each proton and antiproton, and T' is the kinetic energy of each initial proton in the center-of-mass frame. Solving for T', we get

T' = mpc2

(b) The total energy of each proton in the center-of-mass frame is

E' = mpc2 + T' = 2mpc2 = g'mpc2 => g' = 2

g' and b' are related by

g' = 1 / sqrt(1 - b'2) => b' = sqrt(1 - 1/g'2) = sqrt(1 - 1/22) = (1/2) sqrt(3)

(c) We make use of the fact that since (cptot, Etot) is a four vector, whose components are related by the Lorentz transformation, the quantity Etot2 - (cptot)2 is invariant. Thus,

Etot2 - (cptot)2 = Etot'2 - (cptot')2 (I)

where the unprimed quantities are measured in the lab frame and the primed quantities are measured in the center-of-mass frame. In the lab frame, the total energy is equal to the total energy of the proton that was initially moving plus the rest energy of the proton that was initially at rest.

Etot = gmpc2 + mpc2 = (g + 1)mpc2

where

g = 1 / sqrt(1 - b2)

b = v/c

and v is the initial velocity of the proton that was moving in the lab frame. The total momentum in the lab frame is the momentum of the proton that was initially moving, which is equal to its relativistic mass times its velocity.

ptot = gmpv = gmpbc

In the center-of-mass frame, the total energy, from above, is

Etot' = 4mpc2

and the total momentum is zero. Substituting into (I), we get

(g + 1)2mp2c4 - g2mp2b2c4 = 16mp2c4 => (g + 1)2 - g2b2 = 16

Since

b2 = 1 - 1/g2

we can write

(g + 1)2 - g2 + 1 = 16 => g2 + 2g + 1 - g2 + 1 = 16 => g = 7

The initial kinetic energy of the proton that was moving in the lab frame is

T = (g - 1)mpc2 = (7 - 1)mpc2 = 6mpc2

This is the minimum kinetic energy that the incident proton must have in the lab frame in order for the reaction to occur.

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Relativistic Doppler Effect

PROBLEM

Through some coincidence, the Balmer lines from singly ionized helium in a distant star happen to overlap with Balmer lines from hydrogen in the Sun. How fast is the star receding from us?

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 33.45]

SOLUTION

The electron energy levels in hydrogenic atoms (i.e., those with a single electron such as ordinary hydrogen and singly ionized helium) are proportional to Z2, where Z is the atomic number or the number of protons in the nucleus. Thus, the allowed energies in singly ionized helium (Z = 2) are four times the allowed energies in ordinary hydrogen (Z = 1). The difference in energy DE between two different energy levels in a hydrogenic atom is also proportional to Z2. DE is related to the wavelength of the corresponding photon according to

DE = hc/l => l = hc/DE ~ 1/Z2

where h = 6.63 x 10-34 J s is Planck's constant, and c = 3 x 108 m/s is the speed of light in a vacuum. Thus, the wavelength is proportional to 1/Z2. So if the distant star were not receding, its Balmer lines (i.e., the lines corresponding to transitions to the n = 2 level) from singly ionized helium would have wavelengths that are 1/4 of those of hydrogen in the Sun. The distant star must be receding fast enough so that the observed wavelengths are four times the values that would be seen by someone at rest with respect to the star. Then the Balmer lines from singly ionized helium in the star will overlap with the Balmer lines from hydrogen in the Sun. The following diagrams illustrate the situation for wavelengths corresponding to a specific transition (i.e., a particular choice of the initial and final principal quantum numbers).

^
|        |  wavelength of light emitted by hydrogen
|        |  in the Sun as observed on Earth
+---------->
         l0


^
|  |  wavelength of light emitted by singly ionized helium
|  |  in distant star as observed in star's reference frame
+---------->
   l0/4


^
|        |  wavelength of light emitted by singly ionized
|        |  helium in distant star as observed on Earth
+---------->
         l0


The relativistic formula for the Doppler effect for light is

l' = l sqrt[(1 + v/c) / (1 - v/c)] = l sqrt[(1 + b) / (1 - b)]

where l' is the wavelength observed on the earth, l is the wavelength observed in the star's reference frame, v is the recession velocity, and b = v/c. Dividing both sides by l,

l'/l = sqrt[(1 + b) / (1 - b)] = 4 => (1 + b) / (1 - b) = 16 => 1 + b = 16 - 16b
=> 17b = 15 => b = 15/17 = 0.8824 = v/c
=> v = 0.8824c = (0.8824)(3 x 108 m/s) = 2.647 x 108 m/s

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Planetary Motion

PROBLEM

Sputnik I had a perigee (point of closest approach to the Earth) hp = 227 km above the Earth's surface, at which point its speed was vp = 28,710 km/hr. Find its apogee (maximum) distance from the Earth's surface and its period of revolution. (Assume the Earth is a sphere, and neglect air resistance. You need only look up g and the Earth's radius to do this problem.)

[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 3.52]

SOLUTION

The distance from the center of the Earth at perigee is

rp = hp + Re = 227 km + 6370 km = 6597 km

The velocity of Sputnik I at perigee is

vp = (28,710 km/hr)(1000 m/km)(1 hr / 3600 s) = 7975 m/s

The total energy of Sputnik I is

E = (1/2)mvp2 - GMem/rp => E/m = (1/2)vp2 - GMe/rp

On the Earth's surface, we know that for any object of mass m

GMem/Re2 = mg => GMe = gRe2

Thus,

E/m = (1/2)vp2 - gRe2/rp = (1/2)(7975 m/s)2 - (9.8 m/s2)(6.37 x 106 m)2/(6597 x 103 m) = - 2.848 x 107 J/kg

The constant K is

K = - GMem => K/m = - GMe = - gRe2 = - (9.8 m/s2)(6.37 x 106 m)2 = - 3.977 x 1014 m3/s2

The semimajor axis is

a = |K/2E| = |(K/m)/2(E/m)| = [3.977 x 1014 m3/s2 / (2)(2.848 x 107 J/kg)] = 6.982 x 106 m = 6982 km

The eccentricity e of the orbit is related to rp and a according to

rp = a(1 - e) => e = 1 - rp/a = 1 - (6597 km) / (6982 km) = 0.05512

The distance from the center of the Earth at apogee is

ra = a(1 + e) = (6982 km)(1 + 0.05512) = 7367 km

The distance above the surface of the Earth at apogee is

ha = ra - Re = 7367 km - 6370 km = 997 km

The period of revolution is

T = sqrt(4p2a3|m/K|) = sqrt[4p2(6982 x 103 m)3/(3.977 x 1014 m3/s2)] = 5813 s = 1.615 hr

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PROBLEM

Explorer I, an artificial satellite, had a perigee 360 km and an apogee 2549 km above the Earth's surface. Find its distance above the Earth's surface when it passed over a point 90 deg around the Earth from its perigee.

[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 3.53]

SOLUTION

The Earth's radius is Re = 6370 km. At (perigee, apogee), the distance from the center of the Earth to Explorer I is

(rp, ra) = (6370 + 360, 6370 + 2549) km = (6730, 8919) km.

The radial distances at perigee and apogee are related to the semimajor axis a and the eccentricity e of the orbit by

rp = a(1 - e) (I)

and

ra = a(1 + e)

If we add these two equations together, we get

rp + ra = 2a => a = (rp + ra)/2 = (6730 km + 8919 km)/2 = 7824.50 km

From the above equation (I), we can solve for the eccentricity e.

rp = a(1 - e) => e = 1 - rp/a = 1 - (6730 km)/(7824.50 km) = 0.139881

Now we use the equation which gives the radial distance r as a function of the angular position q, for an object in an elliptical orbit.

r(q) = a(1 - e2) / (1 + e cos q)

Here q = (0, p) corresponds to (perigee, apogee). When q = p/2 or 90 deg, we get

r(p/2) = a(1 - e2) / [1 + e cos (p/2)] = a(1 - e2) = (7824.50 km)(1 - 0.1398812) = 7671.40 km

The distance above the Earth's surface when q = p/2 is thus 7671.40 km - 6370 km = 1301 km.

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PROBLEM

A comet is observed a distance of ri = 1.00 x 108 km from the Sun, traveling toward the Sun with a velocity of vi = 51.6 km/s at an angle of 45 deg with the radius from the Sun. Work out an equation for the orbit of the comet in polar coordinates with origin at the Sun and x-axis through the observed position of the comet. (The mass of the Sun is Ms = 2.00 x 1030 kg.)

[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 3.54]

SOLUTION

The equation of the orbit is of the form

1/r = B + A cos (q - q0)

where

A = e / [a(1 - e2)]

and

B = 1 / [a(1 - e2)]

To calculate the eccentricity e and the semimajor axis a, we first calculate the energy E, angular momentum L, and constant K = GMsm. The energy is

E = (1/2)mvi2 - GMsm/ri => E/m = (1/2)vi2 - GMs/ri
= (1/2)(51.6 x 103 m/s)2 - (6.67 x 10-11 N m2 / kg2)(2.00 x 1030 kg) / (1.00 x 1011 m)
= - 2.720 x 106 J/kg

The angular momentum is

L = r x p => L = rimvi sin 45 => L/m = rivi sin 45 = (1.00 x 1011 m)(51.6 x 103 m/s) / sqrt(2) = 3.649 x 1015 m2/s

is

K = - GMsm => K/m = - GMs = - (6.67 x 10-11 N m2/kg2)(2.00 x 1030 kg) = - 1.334 x 1020 N m2/kg

The eccentricity is

e = sqrt[1 + 2EL2/mK2] = sqrt[1 + 2(E/m)(L/m)2/(K/m)2]
= sqrt[1 + (2)(- 2.720 x 106 J/kg)(3.649 x 1015 m2/s)2/(1.34 x 1020 N m2/kg)2] = 0.997963102

The semimajor axis is

a = |K/2E| = |(K/m)/2(E/m)| = (1.34 x 1020 N m2/kg) / [(2)(2.720 x 106)] = 2.452 x 1013 m = (2.452 x 1013 m)[(1 AU) / (1.49 x 1011 m)] = 164.6 AU

The constants A and B are

A = e / [a(1 - e2)] = 1 x 10-11 m-1

and

B = 1 / [a(1 - e2)] = 1.00204 x 10-11 m-1

The reference angle q0 = 135 deg.

From the equation for r(q) used in this problem, we can find the perihelion and aphelion distances, rp and ra.

1/rp = B + A = 1.00204 x 10-11 m-1 + 1 x 10-11 m-1 = 2.00205 x 10-11 m-1 => rp = 4.995 x 1010 m

1/ra = B - A = 1.00204 x 10-11 m-1 - 1 x 10-11 m-1 = 2.041 x 10-14 m-1 => ra = 4.899 x 1013 m

The perihelion and aphelion distances can also be calculated from the formulas in terms of a and e.

rp = a(1 - e) = (2.452 x 1013 m)(1 - 0.997963102) = 4.995 x 1010 m

ra = a(1 + e) = (2.452 x 1013 m)(1 + 0.997963102) = 4.899 x 1013 m

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PROBLEM

Mariner 4 left the Earth on an orbit whose perihelion distance from the Sun was approximately the distance of the Earth (1.49 x 108 km), and whose aphelion distance was approximately the distance of Mars from the Sun (2.2 x 108 km). With what velocity did it leave relative to the Earth? With what velocity must it leave the Earth (relative to the Earth) in order to escape altogether from the Sun's gravitational pull? (You need no further data to answer this problem except the length of the year, if you assume the Earth moves in a circle.)

[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 3.59]

SOLUTION

The (perihelion, aphelion) distance of Mariner 4 from the Sun is (rp, ra) = (1.49, 2.2) x 108 km. The semimajor axis is

a = (1/2)(rp + ra) = (1/2)(1.49 x 108 km + 2.2 x 108 km) = 1.845 x 108 km

The semimajor axis is related to the constant K and the energy E according to

a = |K/2E| (I)

The constant K is

K = - GMsm (II)

where G is the gravitational constant, Ms is the mass of the Sun, and m is the mass of the object in orbit. The energy E at the Earth's orbit is

E = (1/2)mv2 - GMsm/re (III)

Kepler's third law for an object orbiting the Sun is of the form

T2 = 4p2a3/GMs

where T is the period of the orbit and a is the semimajor axis. From this equation, we see that

GMs = 4p2a3/T2

For the planet Earth, T = (365.25 days)(24 hr/day)(3600 s/hr) = 3.156 x 107 s and a = 1.49 x 1011 m. It follows that

GMs = 4p2(1.49 x 1011 m)3/(3.156 x 107 s)2 = 1.311 x 1020 m3/s2

From (I), (II), and (III), we get

a = GMsm / (2GMsm/re - mv2) = GMs / (2GMs/re - v2) => GMs/a = 2GMs/re - v2

Solving for v,

v = sqrt[GMs(2/re - 1/a)] = sqrt{(1.311 x 1020 m3/s2)[2/(1.49 x 1011 m) - 1/(1.845 x 1011 m)]} = 32392 m/s

The velocity of the Earth in its orbit around the Sun is

vearth = 2pre/T = 2p(1.49 x 1011 m)/(3.155 x 107 s) = 29673 m/s

Assuming that Mariner 4 is launched in the direction of the Earth's velocity around the Sun, its velocity relative to the Earth is

vmariner = v - vearth = 32392 m/s - 29673 m/s = 2719 m/s = 2.719 km/s

To escape altogether from the Sun's gravitational pull, Mariner 4's velocity must be such that

E = (1/2)mv2 - GMsm/re = 0 => v = sqrt(2GMs/re) = sqrt[(2)(1.311 x 1020)/(1.49 x 1011 m)] = 41949 m/s

The velocity relative to the Earth is

vmariner = v - vearth = 41949 m/s - 29673 m/s = 12276 m/s = 12.276 km/s

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Motion of a System of Particles

PROBLEM

A lunar landing craft approaches the Moon's surface. Assume that one-third of its weight is fuel, that the exhaust velocity from its rocket engine is v = 1500 m/s, and that the acceleration of gravity at the lunar surface is one-sixth of that at the Earth's surface. How long can the craft hover over the Moon's surface before it runs out of fuel?

[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 4.7]

SOLUTION

Let m0 = initial mass of ship, including fuel => m0/3 = initial mass of fuel, 2m0/3 = mass of ship without fuel.

The force of gravity is balanced by the force imparted to the ship by the mass being ejected.

F = mg/6 = - v dm/dt => dm/m = - (g/6v) dt => ∫ dm/m = - (g/6v) ∫ dt => ln[m(t)/m0] = - gt/6v

m(t) = m0 e-gt/6v

Let tf be the time at which the ship runs out of fuel.

m(tf) = m0 exp(-gtf/6v) = 2m0/3 => exp(-gtf/6v) = 2/3

Solving for tf, we get

tf = - (6v/g) ln(2/3) = - [(6)(1500 m/s)/(9.8 m/s2)] ln(2/3) = 372.4 s

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Rigid Bodies

PROBLEM

How many yards of thread 0.03 inch in diameter can be wound on the spool shown in the figure below? The maximum diameter (top and bottom) is Dmax = 1 inch. The minimum diameter (middle) is Dmin = 0.5 inch. The height is b = 1.5 inches. The angles between the top and bottom and the diagonals are all q = 45 deg.

 Dmax = 1"
+---------+ ^
 \       /  |
  \     /   |
   |   | ^  |
   |   | |  |
   |   | a  b = 1.5"
   |   | |  |
   |   | v  |
  /     \   |
 /       \  |
+---------+ v


[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 5.19]

SOLUTION

To solve this problem, we calculate the total volume of space that can contain the thread and then multiply this volume by the fraction of the volume that the thread can actually occupy. Then we determine the length of thread that can make up this net volume.

We calculate the total volume of space that can contain the thread by using thin cylindrical shells as our volume elements and integrating in the radial direction from the inner radius Rmin = Dmin/2 = 0.25" to the outer radius Rmax = Dmax/2 = 0.5". The volume element is

dV(r) = 2pr h(r) dr

where

h(r) = a + (b - a)(r - Rmin)/(Rmax - Rmin)

and a = b - 0.5" = 1".

dV(r) = 2pr [a + (b - a)(r - Rmin)/(Rmax - Rmin)] dr

Integrating, we get

V = int(r = Rmin, r = Rmax) 2pr [a + (b - a)(r - Rmin)/(Rmax - Rmin)] dr
= 2p int(r = Rmin, r = Rmax) r[a + (b - a)(r - Rmin)/(Rmax - Rmin)] dr
= 2p int(r = Rmin, r = Rmax) r[a + a(r - Rmin)] dr

where

a = (b - a)/(Rmax - Rmin) = (1.5 in - 1 in) / (0.5 in - 0.25 in) = 2

Thus,

V = 2p int(r = Rmin, r = Rmax) r[(a - aRmin) + ar] dr
= 2p[(a - aRmin)(1/2)(Rmax2 - Rmin2) + a(1/3)(Rmax3 - Rmin3)]
= 2p{[(1 - (2)(1/4)](1/2)[(1/2)2 - (1/4)2] + (2)(1/3)[(1/2)3 - (1/4)3]} in3
= 2p[(1/2)(1/2)(1/4 - 1/16) + (2/3)(1/8 - 1/64)] in3
= 2p(9/192 + 14/192) in3
= 2p(23/192) in3
= 23p/96 in3

Now determine the fraction of the volume that the thread can actually occupy. This is equal to the fraction of area that closely packed circles of equal radius can actually fill.

Imagine constructing a horizontal row of circles of equal radius R that lie right next to each other. Now place a second horizontal row of circles on top of the first row so that each of the circles in the second row (1) lies above the point of contact between two circles in the first row and (2) touches these two circles.

Now, for any two adjacent circles in the first row and the one above their point of contact in the second row, draw a line segment connecting the center of each of these three circles to the centers of the other two.

This results in an equilateral triangle whose side is R. What fraction of the area of the equilateral triangle contains parts of the circles? The area of the equilateral triangle is

Atriangle = (2R)(2R) sin 60 = 4R2sqrt(3)/2 = [4 sqrt(3)/2]R2

The area of the circles enclosed by the equilateral triangle is

Acircle = (3)(pR2/6) = pR2/2

The fraction of the area of the triangle that contains parts of the circles is

f = Acircle/Atriangle = (pR2/2) / {[4 sqrt(3)/2]R2} = p / [4 sqrt(3)]

The volume of space actually occupied by thread is

Vthread = Vf = [23p/96 in3]p / [4 sqrt(3)] = {[23p2 / [384 sqrt(3)]} in3

Vthread is also given by

Vthread = pRthread2L

where Rthread = 0.015 inch is the radius of the thread and L is its length. Thus,

L = Vthread / pRthread2 = {[23p2 / [384 sqrt(3)]} in / p(0.015)2 = 482.8 inches = 13.41 yd

If we didn't take into account the fact that the thread can only occupy a fraction of the total volume, our result would have been

L = V / pRthread2 = (23p/96) / p(0.015)2 inches = 1065 inches = 29.58 yd

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PROBLEM

A cable 20 ft long is suspended between two points A and B, 15 ft apart. The line AB makes an angle of 30 deg with the horizontal (B higher). A weight of 2000 lb is hung from a point C 8 ft from the end of the cable at A.
(a) Find the position of point C, and the tensions in the cable, if the cable does not stretch.
(b) If the cable is 1/2 inch in diameter and has a Young's modulus of 5 x 105 lb/in2, find the position of point C and the tensions, taking cable stretch into account. Carry out two successive approximations, and estimate the accuracy of your result.

[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: Addison-Wesley), problem 5.33]

SOLUTION

(a) Assume that the cable is massless and that it deforms into two straight segments, from A to C and from C to B, but doesn't stretch. Define

a = distance between points A and B = 20 ft
b = distance between points A and C = 8 ft
c = distance between points B and C = 20 ft - 8 ft = 12 ft
q = angle between AB and the horizontal = 30 deg
W = weight of hanging mass = 2000 lb
a = angle between AB and AC (unknown)
b = angle between AB and BC (unknown)

Use the law of cosines to solve for a and b.

c2 = a2 + b2 - 2ab cos a

a = arccos[(a2 + b2 - c2) / 2ab] = arccos[(152 + 82 - 122) / (2)(15)(8)] = arccos(0.6042) = 52.83 deg

b2 = a2 + c2 - 2ac cos b

b = arccos[(a2 + c2 - b2) / 2ac] = arccos[(152 + 122 - 82) / (2)(15)(12)] = arccos(0.8472) = 32.09 deg

The angle between AC and the horizontal is

a1 = a - q = 52.83 deg - 30 deg = 22.83 deg

Thus, point C is located 8 ft from point A in the direction 22.83 deg below the horizontal. The angle between BC and the horizontal is

b1 = b + q = 32.09 deg + 30 deg = 62.09 deg

Let (T1, T2) be the tension in (AC, BC) and let T be the tension in the cable that supports W and is connected to C. The tension T must equal the weight W because the hanging mass is at equilibrium.

T = W

The horizontal components of T1 and T2 are equal.

T1 cos a1 = T2 cos b1 => T1 = T2 (cos b1) / (cos a1)

The vertical components of T1 and T2 balance T.

T1 sin a1 + T2 sin b1 = T = W

Substitute for T1.

T2 cos b1 sin a1 / cos a1 + T2 sin b1 = W

T2 cos b1 tan a1 + T2 sin b1 = W

Solve for T2.

T2 = W / (cos b1 tan a1 + sin b1) = (2000 lb) / (cos 62.09 tan 22.83 + sin 62.09) = 1851 lb

T1 = (1851 lb)(cos 62.09) / (cos 22.83) = 940 lb

(b)In part (a), we obtained a first approximation. To obtain a second approximation, we assume the values of T1 and T2 obtained in part (a) and calculate the amount of stretching that would result. If r is the radius of the cable, the stresses in the cable segments AB and BC are

stress1 = T1/pr2

and

stress2 = T2/pr2

The strains in the cable segments AB and BC are

strain1 = Db/b

and

strain2 = Dc/c

The ratio between stress and strain is the Young's modulus Y, so

stress1/strain1 = stress2/strain2 = Y

From the above, we can get estimates for the stretched lengths of cable segments AB and BC.

b' = b(1 + strain1) = b(1 + stress1 / Y) = b(1 + T1/pr2Y) = (8 ft)[1 + (1851 lb)/p(0.25 in)2(5 x 105 lb/in2)] = 8.077 ft

c' = c(1 + strain2) = c(1 + stress2 / Y) = c(1 + T2/pr2Y) = (12 ft)[1 + (940 lb)/p(0.25 in)2(5 x 105 lb/in2)] = 12.226 ft

Using these values for b' and c', we solve for new values of a1, b1, T1, and T2, using the above formulas. The results are

a1 = 24.49 deg
b1 = 62.53 deg
T1 = 924 lb
T2 = 1823 lb

If we calculated the stretched lengths of AB and BC using these new tensions, and did a third approximation, we would get

b' = 8.075 ft
c' = 12.223 ft
a1 = 24.46 deg
b1 = 62.52 deg
T1 = 924 lb
T2 = 1823 lb

So the second approximation is quite accurate. A fourth approximation would yield

b' = 8.075 ft
c' = 12.223 ft
a1 = 24.46 deg
b1 = 62.52 deg
T1 = 924 lb
T2 = 1823 lb

which is the same as the third approximation.

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Lagrange's Equations

In the direct application of Newton's laws, we use Cartesian coordinates when they are practical. Sometimes they are not the most convenient coordinates for solving a problem. Some examples are the following:

(1) central force motion (e.g., planetary motion): Use spherical coordinates r, q, and f.
(2) two-body problem: Use center-of-mass coordinates and relative coordinates.
(3) many-body problem: Use center-of-mass coordinates and relative coordinates.
(4) moving coordinate systems: Newton's laws assume a fixed coordinate system.
(5) rigid bodies: Six coordinates (three translational and three rotational) are generally needed to completely specify the location and orientation of a rigid body.

Generalized coordinates include Cartesian, polar, relative, center-of-mass, and other types of coordinates. Joseph-Louis Lagrange (1736-1813), an Italian mathematician and astronomer who spent most of his life in Prussia and France, developed a method to directly set up equations of motion in terms of generalized coordinates and solve them for the resulting motion.

For an N-particle system in three-dimensional space, we may define generalized coordinates q1, q2, q3,..., q3N. If there are no constraints, there are a total of 3N generalized coordinates describing the system, consistent with the 3N degrees of freedom. Each generalized coordinate can in principle be written as a function of 3N Cartesian coordinates and the time,

q1 = q1(x1, y1, z1, x2, y2, z2,..., xN, yN, zN, t)
q2 = q2(x1, y1, z1, x2, y2, z2,..., xN, yN, zN, t)
q3 = q3(x1, y1, z1, x2, y2, z2,..., xN, yN, zN, t)
...
q3N = q3N(x1, y1, z1, x2, y2, z2,..., xN, yN, zN, t)

A constraint is a restriction on the freedom of motion of a system in the form of a condition that must be satisfied by their coordinates or a restriction on the velocities. A rigid body is a system subject to contraints that restrict the possible values of the coordinates of the individual particles that make up the rigid body. A holonomic constraint is one that can be expressed in the form of an equation relating the coordinates.

Lagrange's equations are

(d/dt)(∂T/∂q1') - ∂T/∂q1 = Q1
(d/dt)(∂T/∂q2') - ∂T/∂q2 = Q2
(d/dt)(∂T/∂q3') - ∂T/∂q3 = Q3
...
(d/dt)(∂T/∂q3N') - ∂T/∂q3N = Q3N

where T is the kinetic energy of the system, qi' = ∂qi/∂t, and Qi is the generalized force corresponding to generalized coordinate qi, defined by

dWi = Qi dqi

where dWi is the work done on the system if qi is increased by an infinitesimal amount dqi, with all other generalized coordinates held constant.

If the forces Qi can be derived from a potential energy function V that depends on the qi but not on the qi', the Lagrangian function

L = T - V

can be defined, and Lagrange's equations can be written in the form

(d/dt)(∂L/∂q1') - ∂L/∂q1 = 0
(d/dt)(∂L/∂q2') - ∂L/∂q2 = 0
(d/dt)(∂L/∂q3') - ∂L/∂q3 = 0
...
(d/dt)(∂L/∂q3N') - ∂L/∂q3N = 0

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PROBLEM

Consider an Atwood's machine, consisting of two masses, m1 and m2, connected by a massless string which passes over a massless and frictionless pulley of radius R.

       ___
     /     \
   /         \
  |        R  |
  |     o-----|
  |           |
  |\         /|
  |  \ ___ /  |
  |           |
  |           |
+-+-+         |
| m1|         |
+---+         |
            +-+-+
            | m2|
            +---+


(a) Uses Lagrange's equations to determine the acceleration of the system.
(b) Use Lagrange's equations to determine the tension in the string.
(c) Use Newtonian mechanics to obtain the acceleration and the tension.

SOLUTION

(a)
           ___
         /     \
       /         \
      |        R  |
---   |     o-----|   ---
 ^    |           |    ^
 |    |\         /|    |
 x1   |  \ ___ /  |    |
 |   ^t          |    x2
 v    |           ^t   |
--- +-+-+         |    |
    | m1|         |    |
    +---+         |    v
      |         +-+-+ ---
      |         | m2|
      v         +---+
     m1g          |
                  |
                  v
                 m2g


Define

x1 = vertical distance from pulley axis to m1
x2 = vertical distance from pulley axis to m2

x1 and x2 are related by the constraint equation

x1 + x2 + pR = L (1)

where L is the length of the string.

The kinetic energy of the system is

T = (1/2)m1x1'2 + (1/2)m2x2'2

where x1' = dx1/dt and x2' = dx2/dt.

The generalized force Q1 associated with x1 is obtained by considering the work dW1 done on m1 if x1 is increased by an infinitesimal amount dx1 while keeping x2 constant. The work is done by the force of gravity m1g and the force of the tension t. Thus,

dW1 = (m1g - t) dx1 = Q1 dx1

where

Q1 = m1g - t

Similarly, the generalized force Q2 associated with x2 is

Q2 = m2g - t

Lagrange's equations for the system are

(d/dt)(∂T/∂x1') - ∂T/∂x1 = Q1
(d/dt)(∂T/∂x2') - ∂T/∂x2 = Q2

We have ∂T/∂x1' = m1x1', ∂T/∂x2' = m2x2', ∂T/∂x1 = 0, and ∂T/∂x2 = 0. Thus, Lagrange's equations become

m1x1" = m1g - t (2)
m2x2" = m2g - t (3)

If we take the second time derivative of the constraint equation (1), we get

x1" + x2" = 0 => x2" = - x1"

Substituting into (3),

- m2x1" = m2g - t (4)

Subtracting (4) from (2),

(m1 + m2)x1" = (m1 - m2)g => x1" = (m1 - m2)g / (m1 + m2) => x2" = - (m1 - m2)g / (m1 + m2)

Thus, if m1 > m2, m1 accelerates downward at an acceleration of magnitude

|x1"| = (m1 - m2)g / (m1 + m2)

while m2 accelerates upward with an acceleration of the same magnitude.

(b) Use (2) or (3) to solve for the tension t. Using (2),

t = m1g - m1x1" = m1g - m1(m1 - m2)g / (m1 + m2)
= m1(m1 + m2)g / (m1 + m2) - m1(m1 - m2)g / (m1 + m2) = 2m1m2 / (m1 + m2)

(c) To solve the problem using Newtonian mechanics, we write Newton's first law, F = ma, for each of the two masses. Assuming that m1 accelerates downward and m2 accelerates upward with the same acceleration a, we have

m1g - t = m1a (5)
t - m2g = m2a (6)

Adding (5) and (6), we get

(m1 - m2)g = (m1 + m2)a => a = (m1 - m2)g / (m1 + m2)

Solving (5) for t,

t = m1g - m1a = m1g - m1(m1 - m2)g / (m1 + m2)
= m1(m1 + m2)g / (m1 + m2) - m1(m1 - m2)g / (m1 + m2) = 2m1m2g / (m1 + m2)

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Fourier Analysis

PROBLEM

What are the amplitudes of the various frequencies present in half-wave rectified 60 Hz AC, that is,

f(t) = sin 120pt for 0 < t < 1/120 s
f(t) = 0 for 1/120 s < t < 1/60 s

and then the function repeats?

[from Beard, David B. 1963, Quantum Mechanics (Boston, Massachusetts: Allyn and Bacon, Inc.), problem 2.2]

SOLUTION

The specified function has the appearance of a sine wave with period T = 1/60 s and the negative sections set equal to zero.

In general, a periodic function of time can be expressed as a sum or superposition of sine and cosine functions. In particular, if a periodic function of time f(t) has period T, it can be written as a sum of sine and cosine functions of the form

f(t) = B0/2 + Sn=1 (An sin nwt + Bn cos nwt)

where

w = 2p/T
An = (w/p) ∫0T f(t) sin nwt dt (1)
Bn = (w/p) ∫0T f(t) cos nwt dt (2)

The specific function of interest here is defined as

f(t) = sin 120pt for 0 < t < 1/120 s
f(t) = 0 for 1/120 s < t < 1/60 s

Since f(t) repeats with period T = 1/60 s, the corresponding angular frequency is

w = 2p/T = 120p s-1

and we can write

f(t) = sin wt for 0 < t < T/2
f(t) = 0 for T/2 < t < T

Substituting these expressions for f(t) into (1),

An = (w/p) ∫0T/2 sin wt sin nwt dt

To calculate the An, we make use of the trigonometric identities

cos(A ± B) = cos A cos B -/+ sin A sin B => sin A sin B = (1/2)[cos(A - B) - cos(A + B)] (3)

If A = B, (3) yields

sin2 A = (1/2)(1 - cos 2A)

so

A1 = (w/p) ∫0T/2 sin2 wt dt = (w/p) ∫0T/2 (1/2)(1 - cos 2wt) dt
= (w/2p) ∫0T/2 dt - (w/2p) ∫0T/2 cos 2wt dt = wT/4p - (1/4p) sin 2wt |0T/2
= wT/4p - (1/4p) sin wT = 1/2 - 0 = 1/2

For n > 2,

An = (w/2p) ∫0T/2 cos[(1 - n)wt] dt - (w/2p) ∫0T/2 cos[(1 + n)wt] dt
= [1/2p(1 - n)] ∫0T/2 (1 - n)w cos[(1 - n)wt] dt - [1/2p(1 + n)] ∫0T/2 (1 + n)w cos[(1 + n)wt] dt
= [1/2p(1 - n)] sin[(1 - n)wt] |0T/2 - [1/2p(1 + n)] sin[(1 + n)wt] |0T/2
= [1/2p(1 - n)] sin[(1 - n)wT/2] - [1/2p(1 + n)] sin[(1 + n)wT/2]
= [1/2p(1 - n)] sin[(1 - n)p] - [1/2p(1 + n)] sin[(1 + n)p]
= 0

The formula for Bn, from (2), is

Bn = (w/p) ∫0T/2 sin wt cos nwt dt

For n = 0, we have

B0 = (w/p) ∫0T/2 sin wt dt = - (1/p) cos wt |0T/2 = - (1/p)(cos wT/2 - 1) = - (1/p)(cos p - 1) = 2/p

For n = 1,

B1 = (w/p) ∫0T/2 sin wt cos wt dt = (1/2p) sin2 wt |0T/2 = (1/2p) sin2 wT/2 = (1/2p) sin2 p = 0

To calculate Bn for n > 2, we make use of the trigonometric identity

sin(A ± B) = sin A cos B ± cos A sin B => sin A cos B = (1/2)[sin(A + B) + sin(A - B)]

So for n > 2,

Bn = (w/2p) ∫0T/2 sin[(1+n)wt] dt + (w/2p) ∫0T/2 sin[(1-n)wt] dt
= - [1/2p(1+n)] cos[(1+n)wt] |0T/2 - [1/2p(1-n)] cos[(1-n)wt] |0T/2
= - [1/2p(1+n)]{cos[(1+n)wT/2] - 1} - [1/2p(1-n)]{cos[(1-n)wT/2] - 1}
= - [1/2p(1+n)]{cos[(1+n)p] - 1} - [1/2p(1-n)]{cos[(1-n)p] - 1}
= - [1/2p(1+n)]{(- 1)1+n - 1} - [1/2p(1-n)]{(- 1)1-n - 1}

For n even,

Bn = (1/p)[1/(1+n) + 1/(1-n)]

For n odd,

Bn = 0

The first few nonzero coefficients are

B2 = (1/p)[1/(1+2) + 1/(1-2)] = (1/p)(1/3 - 1) = - 2/3p
B4 = (1/p)[1/(1+4) + 1/(1-4)] = (1/p)(1/5 - 1/3) = (1/p)(3/15 - 5/15) = - 2/15p
B6 = (1/p)[1/(1+6) + 1/(1-6)] = (1/p)(1/7 - 1/5) = (1/p)(5/35 - 7/35) = - 2/35p
B8 = (1/p)[1/(1+8) + 1/(1-8)] = (1/p)(1/9 - 1/7) = (1/p)(7/63 - 9/63) = - 2/63p

The first few nonzero terms of the Fourier series are

f(t) = B0/2 + A1 sin wt + B2 cos 2wt + B4 cos 4wt + B6 cos 6wt + B8 cos 8wt + ...

The file FourierAnalysis080425.xls shows the sum of the first few terms of the Fourier series and the actual function f(t) for 0 < wt < 4p.

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Schrodinger Wave Equation

The one-dimensional, time-dependent Schrodinger wave equation, for a particle of mass m moving in a region where the potential energy is given by V(x), is

-(h2/2m)∂2Y(x,t)/∂x2 + V(x)Y(x,t) = -(h/i)∂Y(x,t)/∂t (I)

where

h = h/2p

and h = 6.6256 x 10-34 J s is Planck's constant. If we use the method of separation of variables and assume that

Y(x,t) = y(x)f(t)

substitute into (I) and divide both sides by Y, we get

-(h2/2m)(d2y/dx2)/y + V = -(h/i)(df/dt)/f

Since both sides of this equation are functions of a different variable, each side must be equal to the same constant, which is the energy E. Thus, y and f satisfy

-(h2/2m)d2y/dx2 + Vy = Ey

and

-(h/i)df/dt = Ef

The first of these is the one-dimensional, time-independent Schrodinger equation. The second can be solved for f.

f(t) = C exp(-iEt/h)

where C is a constant.

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PROBLEM

A beam consists of particles of mass m and total energy E moving in the +x direction and subject to the potential energy

V(x) = V0

where V0 < E is a constant. Solve the Schrodinger equation for this system.

SOLUTION

We assume that the complete wave function is of the form

Y(x,t) = y(x)f(t)

where the time-dependent part is

f(t) = C exp(-iEt/h)

where C is a constant, and the time-independent part satisfies

-(h2/2m)d2y/dx2 + V0y = Ey => d2y/dx2 = -[2m(E - V0)/h2]y => d2y/dx2 = - k2y

where

k = sqrt[2m(E - V0)/h2]

The general solution is

y(x) = A exp(ikx) + B exp(-kdx)

The complete wave function is of the form

Y(x,t) = A exp[i(kx - Et/h)] + B exp[-i(kx + Et/h)] (I)

The phase of the first term is

q1 = kx - Et/h

The time derivative of this is

dq1/dt = k dx/dt - E/h

In order for the phase to remain constant,

dq1/dt = 0 => dx/dt = E/hk

Thus, the phase of the wave would remain constant for an observer moving with velocity

v = E/hk

in the +x direction. This means that the first term in (I) corresponds to a wave moving in the +x direction. Similarly, the second term in (I) corresponds to a wave moving in the -x direction. Since the beam of particles is moving in the +x direction, the constant B must be zero, and the wave function reduces to

Y(x,t) = A exp[i(kx - Et/h)]

The value of A depends on the beam density. If we integrate the absolute square of the spatial part of Y(x,t) over some interval from x to x + L and assume a uniform beam density, we get

xx+L |y(x)|2 dx = A2xx+L dx = A2L

If n is the number of beam particles per length L,

n = A2L => A = sqrt(n/L)

and

Y(x,t) = sqrt(n/L) exp[i(kx - Et/h)]

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Nuclear Reactions

PROBLEM

The reaction 1H1 + 3Li7 => 4Be7 + 0n1 is sometimes used to produce monoenergetic neutrons from a source of monoenergetic protons. The Q value of the reaction is - 1.64 MeV. If a 3Li7 target is bombarded by a beam of 5 MeV protons, at what angle to the beam are 2.5 MeV neutrons emitted?

[from Eisberg, Robert, and Resnick, Robert 1985, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, Second Edition (New York: John Wiley & Sons), problem 16.24]

SOLUTION

Consider the collision of a particle of mass m with an initially stationary nucleus of mass M, which produces a residual nucleus of mass M1 whose velocity makes an angle f with the velocity of the incident particle, and a second particle of mass M2 whose velocity makes an angle q with the velocity of the incident particle.

                 v1
                  O M1
                /|
              /
m         M / f
o-------->O-------
     v      \ q
              \
                \|
                  o M2
                 v2


If we assume the particle speeds are all nonrelativistic, then we can define

Km = (1/2)mv2
K1 = (1/2)M1v12
K2 = (1/2)M2v22

as the kinetic energies of m, M1, and M2. The Q value of the reaction is the change in the total kinetic energy of the particles. Thus,

Q = K1 + K2 - Km (1)

By conservation of momentum, we have

mv = M1v1 cos f + M2v2 cos q => M1v1 cos f = mv - M2v2 cos q (2)
0 = M1v1 sin f - M2v2 sin q => M1v1 sin f = M2v2 sin q (3)

Squaring (2) and (3), and adding the results, we get

(M1v1)2 = (mv)2 + (M2v2)2 - 2mvM2v2 cos q (4)

We can write

Km = (1/2m)(mv)2 => mv = (2mKm)1/2
M1v1 = (2M1K1)1/2
M2v2 = (2M2K2)1/2

Substituting these into (4),

2M1K1 = 2mKm + 2M2K2 - 2(2mKm)1/2(2M2K2)1/2 cos q
=> K1 = (m/M1)Km + (M2/M1)K2 - (2/M1)(mM2KmK2)1/2 cos q

Using (1), we can rewrite K1 in terms of Q, Km, and K2,

K1 = Q - K2 + Km

Thus,

Q - K2 + Km = (m/M1)Km + (M2/M1)K2 - (2/M1)(mM2KmK2)1/2 cos q
=> Q = K2(1 + M2/M1) - Km(1 - m/M1) - (2/M1)(mM2KmK2)1/2 cos q
=> cos q = [K2(1 + M2/M1) - Km(1 - m/M1) - Q] / [(2/M1)(mM2KmK2)1/2] (5)

In this particular problem,

m = mp = 1.007825 u
M1 = m(4Be7) = 7.016928 u
M2 = mn = 1.008665 u
Km = 5 MeV
K2 = 2.5 MeV
Q = - 1.64 MeV

Substituting these values into (5), we get

cos q = [(2.5)(1 + 1.008665 / 7.016928) - (5)(1 - 1.007825 / 7.016928) + 1.64] /
{(2 / 7.016928)[(1.007825)(1.008665)(5)(2.5)]1/2} = 0.214076 => q = 77.64°

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