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 in^{3}
The volume of 1 lb of plaster is
V_{1} = V(1/4.5) = 35 in^{3} = (35 in^{3})(1 gal / 231 in^{3})(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
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 (in^{3})
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 in^{3}. 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 in^{3}) = 7325.21 in^{3}. Dividing this by the volume of the measured books, we get (7325.21 in^{3}) / (598.78 in^{3}) = 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 in^{3}, cm^{3}, and ounces.
SOLUTION
(a) The volume of the main section is obtained by dividing it into infinitesimal disks and integrating. Each disk has volume
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 in^{3}?
(b) What is the volume of the grout in cm^{3}?
(c) What is the mass of the grout in g?
(d) What is the density of the grout in g/cm^{3}?
(e) What volume of water should be mixed with each cm^{3} of grout?
This amount of water should be mixed with 1790.287 cm^{3} of grout. So for each cm^{3} of grout, (473.1765/1790.287)(1 cm^{3}) = 0.2643 cm^{3} of water is needed.
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, thirteenpart television series first broadcast by the Public Broadcasting Service (PBS) in 1980.
The position x(t) of an object moving with constant acceleration a is given by
x(t) = x_{i} + v_{i}t + (1/2)at^{2}
where x_{i} is the initial position, v_{i} is the initial velocity, and t is the elapsed time. The final velocity v_{f} is related to the initial velocity according to
v_{f} = v_{i} + at
and by
v_{f}^{2} = v_{i}^{2} + 2a(x  x_{i})
If an object moves with constant acceleration, its average velocity is
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 twostory building." Explain.
Consumer Reports 2010, "Pack Your Car Safely," Consumer Reports, August 2010, 1011.
SOLUTION
The initial and final velocities of an object undergoing uniformly accelerated motion in the vertical direction are related by
v_{f}^{2} = v_{i}^{2} + 2a(y_{f}  y_{i})
where v_{f} is the final velocity, v_{i} is the initial velocity, a is the acceleration, y_{f} is the final y coordinate, and y_{i} 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/s^{2}, y_{f}  y_{i} =  h, and v_{i} = 0. Thus,
v_{f}^{2} = 2gh => h = v_{f}^{2} / 2g
If the final velocity of the falling object is v_{f} = 30 mph, which is the assumed initial velocity of the car, then
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 firststory 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 thirdstory floor and ceiling, or about 5 ft above the roof of a twostory building.
++ 35 ft
 3rd 
 story 
++ 25 ft
 2nd 
 story 
++ 15 ft
 1st 
 story 
++ 5 ft
 
++ ground level
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 = v_{i}t + (1/2)at^{2} = v_{i}t  (1/2)gt^{2}
Here y = 5.0 ft is the displacement, t = 0.5 s, v_{i} 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/s^{2} = 9.8 m/s^{2} 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 v_{i}. We solve for v_{i}.
v_{i} = [y + (1/2)gt^{2}] / t = [5 ft + (1/2)(32 ft/s^{2})(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
v_{f}^{2} = v_{i}^{2} + 2ay = v_{i}^{2}  2gh = 0
where v_{f} is the velocity at the top of its motion, which is zero, v_{i} 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 = v_{i}^{2} / 2g = (18 ft/s)^{2} / [(2)(32 ft/s^{2})] = 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
v_{i} = [y + (1/2)gt^{2}] / t = [1.524 m + (1/2)(9.8 m/s^{2})(0.5 s)^{2}] / (0.5 s) = 5.498 m/s
h = v_{i}^{2} / 2g = (5.498 m/s)^{2} / [(2)(9.8 m/s^{2})] = 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/s^{2} and 9.8 m/s^{2}, are not exactly equal, and because there is some roundoff in the above results. In fact, 9.8 m/s^{2} = (9.8 m/s^{2})(100 cm/m)(1 in / 2.54 cm)(1 ft / 12 in) = 32.15223 ft/s^{2}. If this value is used in the above calculation, and we retain all digits that our calculator shows, we get, in English units,
v_{i} = [y + (1/2)gt^{2}] / t = [5 ft + (1/2)(32.15223 ft/s^{2})(0.5 s)^{2}] / (0.5 s) = 18.03806 ft/s
h = v_{i}^{2} / 2g = (18.03806 ft/s)^{2} / [(2)(32.15223 ft/s^{2})] = 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
v_{i} = [y + (1/2)gt^{2}] / t = [1.524 m + (1/2)(9.8 m/s^{2})(0.5 s)^{2}] / (0.5 s) = 5.498 m/s
h = v_{i}^{2} / 2g = (5.498 m/s)^{2} / [(2)(9.8 m/s^{2})] = 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.
An elevator ascends with an upward acceleration of 4.0 ft/s^{2}. 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
y_{bot}(t) = (1/2)at^{2}
where a = 4.0 ft/s^{2} is the acceleration of the elevator. The y coordinate of the top of the elevator at time t is
y_{top}(t) = h + (1/2)at^{2}
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/s^{2}) = 2 s
when the top of the elevator is at
y_{top}(2) = 9.0 ft + (1/2)(4.0 ft/s^{2})(2 s)^{2} = 17 ft
An elevator ascends from the ground with uniform speed. At time T_{1} a boy drops a marble through the floor. The marble falls with uniform acceleration g = 9.8 m/s^{2}, and hits the ground T_{2} seconds later. Find the height of the elevator at time T_{1}.
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 1.13]
SOLUTION
Let v equal the elevator's speed and h equal its height when the marble is dropped. Then
h = vT_{1}
The displacement of the marble is
 h = vT_{2}  (1/2)gT_{2}^{2} =  vT_{1} => v = (1/2)gT_{2}^{2} / (T_{1} + T_{2}) => h = (1/2)gT_{1}T_{2}^{2} / (T_{1} + T_{2})
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 v_{b} = 5 m/s. It is D = 1600 m down range. Shells fired from the gun have an initial velocity of v_{0} = 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/s^{2}. 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 v_{x} = v_{0} cos q.
D = v_{x}t => t = D/v_{x} = D/(v_{0} 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 = v_{y0}t + (1/2)at^{2} = v_{y0}t  (1/2)gt^{2}
v_{y0} is the initial velocity of the shell in the vertical (y) direction, which is
v_{y0} = v_{0} 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/s^{2})(5 s)^{2} = 1200 m  125 m = 1075 m
(c) Let t_{0} 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 + v_{b}(t_{0} + t)
Solving for t_{0}, we get
t_{0} = (y  H)/v_{b}  t = (1075 m  800 m)/(5 m/s)  5 s = 50 s
A baseball player hits a large number of baseballs at various initial speeds v_{0} 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 v_{0}.
(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 v_{0} 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.
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 = v_{x}t = v_{0}t cos q = (2v_{0}^{2}/g) sin q cos q
The maximum height is obtained using
v_{yf}^{2} = v_{yi}^{2} + 2ah = (v_{0} sin q)^{2}  2gh = 0 => h = (v_{0} sin q)^{2} / 2g (II)
From the expression for t (I) above, we see that
v_{0} sin q = gt/2
which we can substitute into (II) to get
h = (g^{2}t^{2}/4) / 2g = gt^{2}/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.
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 v_{x} = 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)gt^{2} => t = sqrt(2h/g)
In this amount of time, the plane travels a horizontal distance
x = v_{x}t = v_{x} sqrt(2h/g) = (69.4 m/s) sqrt[(2)(235 m)/(9.8 m/s^{2})] = 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/v_{x} = (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 v_{0} is the initial velocity in the y direction, we have
 h = v_{0}t  (1/2)gt^{2} => v_{0} = [(1/2)gt^{2}  h] / t
= [(1/2)(9.8 m/s^{2})(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
v_{y} = v_{0}  gt =  8.367 m/s  (9.8 m/s^{2})(6.124 s) =  68.38 m/s
and the horizontal component of the velocity is the same as it was initially, v_{x} = 69.4 m/s. The speed of the goods when they land is
v = sqrt(v_{x}^{2} + v_{y}^{2}) = sqrt[(69.4 m/s)^{2} + ( 68.38 m/s)^{2}] = 97.46 m/s
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 v_{0} 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
v_{0} / 
/ 
/ q v
+
<L>
SOLUTION
The horizontal speed of the projectile, which remains constant, is
v_{x} = v_{0} cos q
The time it takes for the projectile to travel the horizontal distance L is
t = L/v_{x} = L / (v_{0} cos q)
The initial vertical speed of the projectile is
v_{y0} = v_{0} sin q
At time t, the vertical position of the projectile is
y_{1} = v_{y0}t  (1/2)gt^{2} = (v_{0} sin q)[L / (v_{0} cos q)]  (1/2)gt^{2} = L tan q  (1/2)gt^{2} = h  (1/2)gt^{2}
since
tan q = h/L
Also at time t, the vertical position of the object is
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 = v_{x}t = v_{0}t cos q (1)
where
v_{x} = v_{0} cos q
is the horizontal component of the velocity, which is constant, v_{0} 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 = v_{y0}t  (1/2)gt^{2} = v_{0}t sin q  (1/2)gt^{2} => v_{0}t sin q = H + (1/2)gt^{2} (2)
where
v_{y0} = v_{0} sin q
is the initial vertical component of the velocity. Dividing (2) by (1), we get
tan q = [H + (1/2)gt^{2}] / L => q = tan^{1}{[H + (1/2)gt^{2}] / L}
= tan^{1}{[155 m + (1/2)(9.8 m/s^{2})(7.6 s)^{2}] / (195 m)} = 66.00°
From (1), we get
v_{0} = L / (t cos q) = (195 m) / [(7.6 s) cos 66.00°] = 63.09 m/s
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.
v_{0} = 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:
t_{1} = time needed for bullet to travel from window to wall
d_{x} = 6.9 m = horizontal distance between window and wall
d_{y} = 0.50 m = vertical distance between bullet hole in window and point of impact in wall
v_{0} = 340 m/s = initial velocity of bullet
v_{x} = v_{0} = 340 m/s = x component of bullet velocity (constant)
v_{1y} = vertical component of bullet velocity when it passes through window
v_{2y} = vertical component of bullet velocity when it hits wall
The horizontal distance the bullet travels between the window and the wall is
d_{x} = v_{x}t_{1} => t_{1} = d_{x}/v_{x} = (6.9 m) / (340 m/s) = 2.029 x 10^{2} s
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: McGrawHill), problem 1.21]
SOLUTION
^ y
 _
 / v_{0}  /
/ q
+> x
\ f
 \
 \
 \
 \
If we choose the origin of our coordinate system to be at the launch point, the equation of the slope is
y_{slope} =  x tan f (1)
The x coordinate of the rock as a function of time is
x_{rock} = v_{x}t = v_{0x}t = v_{0}t cos q (2)
where v_{x} = v_{0x} = v_{0} cos q is the x velocity of the rock, which is constant. The y coordinate of the rock as a function of time is
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
+
_ / 
/ v_{0} = 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 = v_{x}t = v_{0}t cos q => t = (d + L cos f) / v_{0} cos q (1)
where
v_{x} = v_{0} 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 = v_{y0}t  (1/2)gt^{2} = v_{0}t sin q  (1/2)gt^{2} (2)
where
v_{y0} = v_{0} sin q
is the initial velocity in the vertical direction and g = 9.8 m/s^{2} is the acceleration due to gravity.
Combining (1) and (2), we get
L sin f = v_{0}t sin q  (1/2)gt^{2} => L sin f = (v_{0} sin q)(d + L cos f) / v_{0} cos q
 (1/2)g[(d + L cos f) / v_{0} cos q]^{2} => L sin f = d tan q + L cos f tan q
 (1/2)g[(d^{2} + 2dL cos f + L^{2} cos^{2} f) / (v_{0}^{2} cos^{2} q)]
=> L sin f = d tan q + L cos f tan q  (1/2)gd^{2} / (v_{0}^{2} cos^{2} q)
 (gdL cos f) / (v_{0}^{2} cos^{2} q)  (1/2)(gL^{2} cos^{2} f) / (v_{0}^{2} cos^{2} q)
=>(1/2)(gL^{2} cos^{2} f) / (v_{0}^{2} cos^{2} q) + L sin f  L cos f tan q
+ (gdL cos f) / (v_{0}^{2} cos^{2} q) + (1/2)gd^{2} / (v_{0}^{2} cos^{2} q)  d tan q = 0
=> aL^{2} + bL + c = 0 (3)
where
a = (1/2)(g cos^{2} f) / (v_{0}^{2} cos^{2} q)
b = sin f  cos f tan q + (gd cos f) / (v_{0}^{2} cos^{2} q)
c = (1/2)gd^{2} / (v_{0}^{2} cos^{2} q)  d tan q
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 v_{x} = 88.2 ft/s. The vertical and horizontal components of the raindrop velocity are related to the angle q = 21.6° by
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 W_{A} = 44 N and W_{B} = 22 N.
++
 C 
 
++ T
 A _____{>}_____{o}   /
<+++/ 
f_{s}  
 ^ 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 m_{s} = 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 m_{k} = 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 f_{s} between the table and block A is at least as large as the tension T in the string.
f_{s}> T (I)
The normal force between the table and block A is equal to the combined weight of blocks A and C.
N = W_{A} + W_{C}
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 f_{s}.
f_{s} = m_{s}N = m_{s}(W_{A} + W_{C})
The tension in the string is equal to the weight of block B.
T = W_{B}
Substituting into (I), we get
m_{s}(W_{A} + W_{C}) > W_{B}
Solving for W_{C},
W_{C}> W_{B}/m_{s}  W_{A} = (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 = W_{A}
The force of kinetic friction between the table and block A is
f_{k} = m_{k}N = m_{k}W_{A}
Newton's second law (F = ma) for block A is
T  f_{k} = (W_{A}/g)a
or
T  m_{k}W_{A}
= (W_{A}/g)a (II)
Newton's second law for block B is
W_{B}  T = (W_{B})/g)a (III)
Adding (II) and (III), we get
W_{B}  m_{k}W_{A} = [(W_{A}/g) + (W_{B}/g)]a
Solve for a.
a = (W_{B}  m_{k}W_{A})g / (W_{A} + W_{B}) = [22 N  (0.15)(44 N)](9.8 m/s^{2}) / (44 N + 22 N) = 2.287 m/s^{2}
(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.
[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
T_{3}  T_{2}  T_{1}  F = 0 (1)
For the lower pulley, we have
T_{2} + T_{1}  T_{4} = 0 (2)
For the piano, we have
T_{4}  Mg = 0 => T_{4} = Mg
From (2), we get
T_{1} + T_{2} = T_{4} = Mg
Since there is no friction,
F = T_{2} = T_{1} => T_{1} + T_{2} = 2F = Mg => F = Mg/2
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]
The system shown below uses massless pulleys and rope. The coefficient of friction between the masses and horizontal surfaces is m. Assume that M_{1} and M_{2} are sliding. Gravity is directed downward.
(a) Draw force diagrams for each mass.
(b) How are the accelerations related?
(c) Find the tension in the rope connecting M_{1} and M_{2}
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 2.15]
SOLUTION
(a) Let T_{1} be the tension in the rope connecting M_{1} and M_{2} and let T_{3} be the tension in the rope supporting M_{3}. Then the forces on each mass are as shown in the diagram below.
(b) The movements of M_{1} to the right and of M_{2} to the left release rope allowing M_{3} to move down. M_{3} moves down by a distance which is half of the total length of rope made available by M_{1} and M_{2}. Let (x_{1}, x_{2}) be the displacement of (M_{1}, M_{2}) to the (right, left). Let y_{3} be the displacement of M_{3} downward. Then
(x_{1} + x_{2}) / 2 = y_{3}
If we take the second time derivatives of both sides, we get the relationship between the accelerations,
(a_{1} + a_{2}) / 2 = a_{3}
(c) M_{1} 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 M_{1}, we get
T_{1}  f_{1} = M_{1}a_{1} N_{1}  M_{1}g = 0
where
f_{1} = mN_{1}
is the frictional force and N_{1} is the normal force on M_{1}. If we write Newton's second law for both x and y directions for M_{2}, we get
T_{1}  f_{2} = M_{2}a_{2} N_{2}  M_{2}g = 0
where
f_{2} = mN_{2}
is the frictional force and N_{2} is the normal force on M_{2}. There are only vertical forces on M_{3}, so we write Newton's second law for M_{3} in the vertical direction only,
M_{3}g  T_{3} = M_{3}a_{3}
Newton's second law for the pulley above M_{3} results in
2T_{1}  T_{3} = 0
since the pulley is massless.
If we replace f_{1} with mN_{1} and f_{2} with mN_{2} in the above equations, we have the following seven equations in seven unknowns a_{1}, a_{2}, a_{3}, N_{1}, N_{2}, T_{1}, and T_{3}:
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: McGrawHill), 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 a_{x} and a_{y} are considered positive quantities) is
a_{x} = A + a cos q
a_{y} = a sin q
where q = 45°. Write Newton's second law (F = ma) for the +x and y directions.
N sin q = ma_{x} = m(A + a cos q) (1)
mg  N cos q = ma_{y} = 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 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
a_{c} = v^{2}/R => R = v^{2}/a_{c}> v^{2}/6g = (416.7 m/s)^{2} / [(6)(9.8 m/s^{2})] = 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 = ma_{c} = mv^{2}/R
The force that the seat exerts on the pilot is
N = mg + mv^{2}/R = m(g + v^{2}/R) = (80 kg)[9.8 m/s^{2} + (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 = ma_{c} = mv^{2}/R
The force that the seat exerts on the pilot is
N = mv^{2}/R  mg = m(v^{2}/R  g) = (80 kg)[(416.7 m/s)^{2}/(2953 m)  9.8 m/s^{2}] = 3920 N
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.
Writing Newton's second law (F = ma) for both the horizontal and vertical directions, we get
N sin q = ma_{c} = mv^{2}/R (1)
where
a_{c} = v^{2}/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
(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 = ma_{c} = mv^{2}/R
for the horizontal direction and
N  mg = 0 => N = mg
for the vertical direction. The friction force is
f < mN = mmg => mv^{2}/R < mmg => m > v^{2}/Rg = (16.67 m/s)^{2}/[(150 m)(9.8 m/s^{2})] = 0.1890
The coefficient of friction between the tires and the road must be at least 0.1890.
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
a_{c} = v^{2}/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 v_{0} = 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.
Applying Newton's second law in the horizontal and vertical directions, we get
N sin q = mv_{0}^{2}/r (1)
N cos q  mg = 0 => N cos q = mg (2)
Dividing (1) by (2),
tan q = v_{0}^{2}/rg
When v differs from v_{0}, a frictional force is needed to keep the car from slipping up or down the slope. When v (>, <) v_{0}, 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 = (v_{max}, v_{min}), 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.
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.
We set F = ma in both the horizontal and vertical directions. In the horizontal direction, we have
N sin q = ma_{c} = mv^{2}/R (1)
where
a_{c} = v^{2}/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
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, (v_{min}, v_{max}), 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, v_{min}, we have the following situation:
We set F = ma in both the horizontal and vertical directions. In the horizontal direction, we have
N sin q  f cos q = ma_{c} = mv_{min}^{2}/R
=> N sin q  mN cos q = mv_{min}^{2}/R
=> N(sin q  m cos q) = mv_{min}^{2}/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) = v_{min}^{2}/Rg
=> v_{min} = 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 = v^{2}/Rg, so
A 1250kg 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
a_{c} = v^{2}/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.
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/s^{2} 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 a_{c}, or
N sin q + f cos q = mv^{2}/r => N sin q =  f cos q + mv^{2}/r (2)
Dividing (2) by (1),
tan q = ( f cos q + mv^{2}/r) / (mg + f sin q)
=> mg tan q + f sin q tan q =  f cos q + mv^{2}/r
=> mg sin q / cos q + f sin^{2} q / cos q =  f cos q + mv^{2}/r
=> mg sin q + f sin^{2} q =  f cos^{2} q + (mv^{2}/r) cos q
=> f sin^{2} q + f cos^{2} q = (mv^{2}/r) cos q  mg sin q
=> f(sin^{2} q + cos^{2} q) = (mv^{2}/r) cos q  mg sin q
=> f = (mv^{2}/r) cos q  mg sin q
= mg[(v^{2}/rg) cos q  sin q]
= (1250 kg)(9.8 m/s^{2}){[(23.61 m/s)^{2} / (72 m)(9.8 m/s^{2})] 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.
Newton's second law of motion in the x direction is
mg sin q + f = (mv^{2}/r) cos q
=> f = (mv^{2}/r) cos q  mg sin q = mg[(v^{2}/rg) cos q  sin q]
= (1250 kg)(9.8 m/s^{2}){[(23.61 m/s)^{2} / (72 m)(9.8 m/s^{2})] 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 = (mv^{2}/r) sin q
=> N = mg cos q + (mv^{2}/r) sin q = mg[cos q + (v^{2}/rg) sin q]
= (1250 kg)(9.8 m/s^{2}){cos 14° + [(23.61 m/s)^{2}/(72 m)(9.8 m/s^{2})] sin 14°}
= 14,228 N
(d) The weight of the car is mg. The normal force will be greater than mg if
cos q + (v^{2}/rg) sin q > 1
With the given values of q, v, and r,
cos q + (v^{2}/rg) sin q = cos 14° + [(23.61 m/s)^{2}/(72 m)(9.8 m/s^{2})] 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.
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, T_{up}, and the lower string, T_{low}.
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 2.11]
SOLUTION
(a) The forces on the mass are as shown below. There are the two tensions, T_{up} and T_{low}, and the gravitational force mg.
 ^
  w
\ 
 \
q \  \ T_{up}  L\
 \
 m o q = 45°
 / mg
 L/ v
 / T_{low} q /
 /
/


(b) Write Newton's second law (F = ma) in the radial and vertical directions.
T_{up} sin q + T_{low} sin q = ma_{c} = mv^{2}/R = mRw^{2} = mLw^{2} sin q => T_{up} + T_{low} = mLw^{2} (1)
T_{up} cos q  T_{low} cos q  mg = 0 => T_{up}  T_{low} = mg / cos q (2)
where a_{c} is the centripetal acceleration and
R = L sin q
is the distance from the axle to the mass. Add (1) and (2).
2T_{up} = mLw^{2} + mg / cos q => T_{up} = (m/2)(Lw^{2} + g / cos q) = (m/2)[Lw^{2} + g sqrt(2)]
Use (1) to solve for T_{low}.
T_{low} = mLw^{2}  T_{up} = (m/2)[Lw^{2}  g sqrt(2)]
Suppose that forces F_{1}(r), F_{2}(r), F_{3}(r),…, F_{N}(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
W_{1} = ∫ F_{1} • dr W_{2} = ∫ F_{2} • dr W_{3} = ∫ F_{3} • dr …
W_{N} = ∫ F_{N} • dr
where each integral is over the path taken by the object. The total or net work done on the object is
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
W_{g} = mgh
and the change in the gravitational potential energy is
DPE_{g} =  W_{g} =  mgh
The workenergy principle states that the net work done on an object is equal to the change in the object’s kinetic energy.
(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 workenergy 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
F_{g} = 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
F_{h} = 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
W_{h} = F_{h}d = (mg cos q)(h / cos q) = mgh = (15.0 kg)(9.8 m/s^{2})(10.0 m) = 1470 J
(b) The work done by gravity on the backpack is
W_{g} =  F_{g}d =  (mg cos q)(h / cos q) =  mgh =  1470 J
(c) The net work done on the backpack is
W_{net} = W_{h} + W_{g} = 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 workenergy 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 workenergy 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.
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)mv^{2} = (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/s^{2})(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 W_{snow} =  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 W_{snow} =  36000 J.
(b) The work done by the snow in bringing the paratrooper to rest is
W_{snow} =  F_{snow}h
where F_{snow} 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.
F_{snow} =  W_{snow}/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/s^{2})(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
W_{air} = 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.
A student whose mass is m = 75 kg runs at a speed of v_{i} = 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 = KE_{i} = (1/2)mv_{i}^{2}
When he releases the rope, his energy is all potential energy,
E = PE_{f} = 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
(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.
F_{c} = T  mg cos q = 0 => T = mg cos q = (75 kg)(9.8 m/s^{2}) 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.
F_{c} = T  mg cos q
The centripetal force is also equal to the mass times the centripetal acceleration.
F_{c} = ma_{c} = mv^{2}/L
Equating the two expressions, we have
T  mg cos q = mv^{2}/L
Solving for T,
T = mg cos q + mv^{2}/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
T_{max} = (75 kg)(9.8 m/s^{2}) cos 0 + (75 kg)(5.0 m/s)^{2}/(10.0 m) = 922.5 N
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: McGrawHill), problem 4.11]
SOLUTION
The force exerted on the scale by the chain is of the form
F = F_{1} + F_{2}
where
F_{1} = 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,
F_{2} = dp/dt = v dm/dt = vl dx/dt = v^{2}l
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 onedimensional kinematics,
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
_{_} F_{up} / /
/ /
/ \ /
/ / truck moving up the slope
/ \ /
f/ / 
/ V mg
/ q
/_________________
_ f
/ /
/ /
/ \ /
F_{down} / / 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 F_{up} 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:
F_{up}  f  mg sin q = 0 => F_{up} = 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 F_{down} 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:
F_{down} + mg sin q  f = 0 => F_{down} = 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:
A ball of mass m traveling at speed v_{0} (call this ball 1) collides headon 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 v_{1} and ball 2 moves in the +x direction with speed v_{2}. The initial momentum and kinetic energy are
p_{i} = mv_{0}
and
KE_{i} = (1/2)mv_{0}^{2}
The final total momentum and total kinetic energy are
(b) (i) When m << M, we can neglect m/M relative to 1 and 1 relative to M/m, resulting in v_{1} ≈ v_{0} and
v_{2} ≈ 2v_{0}(m/M) << v_{0}. 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 v_{1} ≈  v_{0} and v_{2} ≈ 2v_{0}. 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 v_{0} (call this the moving reference frame), ball 1 is initially at rest and ball 2 is moving in the x direction at speed v_{0}. 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 v_{0} after the collision, and in the laboratory frame, ball 2 moves in the +x direction at speed v_{0} + v_{0} = 2v_{0}.
(iii) When m = M, v_{1} = 0 and v_{2} = v_{0}. Ball 1 stops and ball 2 moves in the +x direction with the same speed ball 1 had before the collision.
In the centerofmass frame, which is moving in the +x direction at speed v_{0}/2, before the collision ball 1 moves in the +x direction at speed v_{0}/2 and ball 2 moves in the x direction at speed v_{0}/2. After the collision, ball 1 moves in the x direction at speed v_{0}/2 and ball 2 moves in the +x direction at speed v_{0}/2. Thus, in the lab frame, after the collision ball 1 stops and ball 2 moves in the +x direction at speed v_{0}.
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/m^{3}.
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
F_{air} = v dm/dt = v rpR^{2} dy/dt = v rpR^{2}v = v^{2}rpR^{2}
where
dm/dt = rpR^{2} dy/dt = rpR^{2}v
is the rate at which the air hits the parachute. We have
mg = F_{air} = v^{2}rpR^{2} => v = sqrt(mg/rpR^{2}) = sqrt{(70 kg)(9.8 m/s^{2}) / [(1.21 kg/m^{3})p(4 m)^{2}]} = 3.358 m/s
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: McGrawHill), problem 3.11]
SOLUTION
The total momentum of the system at time t is
p(t) = M_{A}(t)v(t) + M_{B}(t)u
where M_{A}(t) is the mass of car A, M_{B}(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
A sandspraying 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.
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), 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) = m_{0} + bt = mass of freight car at time t
m_{0} = 2000 kg = mass of freight car at time t = 0
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 v_{0}, at a constant rate dm/dt.
(a) Find a general formula for the maximum height h_{max} at which the garbage can rides. What assumption must be fulfilled for the maximum height to be reached?
(b) Find h_{max} if v_{0} = 20 m/s, M = 10 kg, and dm/dt = 6 kg/s.
(c) Find h_{max} if v_{0} = 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: McGrawHill), 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
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 = v_{r}r + v_{q}q = v_{r}r + rwq
where v_{r} is a radial component of the linear velocity, r is the radial coordinate,
v_{q} = 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, v_{r} = 1.6 cm/s and
is the normal force between the turntable and the bug, m is the mass of the bug, and m_{s} 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.
Motorized Car Accelerating. A motorized car is a selfpropelled 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 frontwheel drive, the front wheels are being made to turn clockwise. If the car has fourwheel 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 twowheel 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 twowheel 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
If z = 0 (the particle is in the xy plane), and F_{z} = 0 (the force has no z component),
t = k(xF_{y}  yF_{x})
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 xy 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
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,
t_{z} = xF_{y}  yF_{x}
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 xy 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 xy plane, and the system behaves like a twodimensional problem, at least as far as the resulting rotational motion is concerned.
Two masses, m_{1} = 18.0 kg and m_{2} = 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, m_{1} is on the ground and m_{2} rests h = 3.00 m above the ground. If the system is now released, use conservation of energy to determine the speed of m_{2} 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
KE_{i} = 0
and the potential energy is
PE_{i} = m_{2}gh
Let v be the velocity of m_{2} just before it hits the ground. Since m_{1} and m_{2} are connected by a rope, v is also the velocity of m_{1}. Since the rope and pulley move together without slipping, v is also the linear velocity at the outer edge of the pulley.
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
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 q_{F}  RMg sin q_{Mg} = 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
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 centerofmass speed?
[from Physics E1a, 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 centerofmass velocity and is independent of how fast the dumbbell is rotating. Therefore, the centerofmass velocity is the same in both cases.
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 centerofmass speed?
[from Physics E1a, 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 centerofmass velocity is greater in case (ii).
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 m_{s}.
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
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 v_{0} 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?
_{_} v_{0}_{___} / / \ M
/ / \
+\ / q  R 
 >O +>
+/ m  
 \ /
 \___/
<5R>
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 6.4]
SOLUTION
Angular momentum is conserved. The initial angular momentum of the package, measured relative to the center of the planet, is
L_{i} = r_{i} x p_{i} = r_{i}p_{i} sin q k = 5Rmv_{0} 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
L_{f} = r_{f} x p_{f} = Rmv_{f}k
where v_{f} is the velocity of the package, which is entirely in the angular direction. Equating the initial and final angular momenta, we get
L_{i} = L_{f} => 5Rmv_{0} sin q = Rmv_{f} => v_{f} = 5v_{0} sin q
Total energy is conserved. The initial energy of the package is
E_{i} = (1/2)mv_{0}^{2}  GMm/5R
The energy of the package when it reaches its closest approach to the planet is
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.
(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: McGrawHill), 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)MR^{2}a => T = (1/2)MRa (4)
where
I = (1/2)MR^{2}
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)
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.
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: McGrawHill), problem 6.24]
SOLUTION
+

_{/}>
/\
B  + 
\_/
v
T 
^ _
/ \ \
A  +  
\/ v

v Mg
The torques about the centers of drums A and B are
is the moment of inertia of each drum about its axis, and a_{A} and a_{B} are the angular accelerations of each drum about its axis. Newton's second law for drum A is
Mg  T = Ma_{A} => T = Mg  Ma_{A} (3)
where a_{A} is the linear acceleration of drum A. a_{A}, a_{A}, and a_{B} are related by
A marble of mass M and radius R is rolled up a plane of angle q. If the initial velocity of the marble is v_{0}, 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: McGrawHill), problem 6.25]
SOLUTION
The initial energy of the marble is all kinetic and consists of translational and rotational kinetic energy.
A yoyo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR^{2}/2. The yoyo is placed upright on a table and the string is pulled with a horizontal force F as shown. The coefficient of friction between the yoyo and the table is m. What is the maximum value of F for which the yoyo will roll without slipping?
is the moment of inertia of the yoyo about an axis through C and a_{C} is the angular acceleration about the same axis. The yoyo will roll to the right. The linear acceleration of the yoyo is
a = Ra_{C} (2)
Newton's second law in the horizontal direction is
F  f = Ma (3)
where f is the frictional force between the yoyo and the table. Using (1), (2), and (3), we get
F  f = MR(2/3)(R  b)F/MR^{2} = (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 yoyo rolls without slipping is
A bowling ball is thrown down the alley with speed v_{0}. 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)v_{0}.
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 6.30]
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 v_{f} and the angular velocity w_{f} are related by
v_{f} = Rw_{f}
The torque about the center of mass is
t = Rf = Ia = (2/5)MR^{2}a => a = 5f/2MR
where
I = (2/5)MR^{2}
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
w_{f} = at = 5ft/2MR => ft = (2/5)MRw_{f} = (2/5)Mv_{f} (1)
A cylinder of radius R spins with angular velocity w_{0}. 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, w_{f}?
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 6.31]
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 v_{f} and the angular velocity w_{f} are related by
v_{f} = Rw_{f}
The torque about the center of mass is
t =  Rf = Ia = (1/2)MR^{2}a => a =  2f/MR
where
I = (1/2)MR^{2}
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
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: McGrawHill), 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,
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 d^{2}f/dt^{2} (3)
where
a = d^{2}f/dt^{2}
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 => d^{2}f/dt^{2} = (R/b) d^{2}q/dt^{2} (4)
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: McGrawHill), 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 (x_{c}, y_{c}) = (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 x_{com} = (R + L/2) sin q  Rq cos q. If x_{com} < x_{c}, the block will tilt back towards its original position when released.
x_{com} < x_{c} => (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 q ≈ q and cos q ≈ 1. Thus, as long as L < 2R, the block will be stable.
A boy of mass m runs on ice with velocity v_{0} 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: McGrawHill), problem 6.39]
SOLUTION
(a)
++ ^
  
  
  
  
  
M   L
  
  
  
  
m   
o>++ v
v_{0}
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
y_{cm} = (ML/2) / (M + m) = (L/2) / (1 + m/M)
The initial and final momenta, p_{i} and p_{f}, are both entirely in the x direction and are equal.
where v_{cm} is the centerofmass velocity. After the collision, the center of mass moves to the right at speed v_{cm}. The boyplusplank system rotates counterclockwise, at angular velocity w, about its center of mass. Calculate the initial and final angular momenta, L_{i} and L_{f} about the center of mass.
L_{i} = mv_{0}(L/2) / (1 + m/M)
L_{f} = Iw
where
I = I_{boy} + I_{plank}
is the moment of inertia of the boyplusplank about an axis through the center of mass,
I_{boy} = m(L/2)^{2} / (1 + m/M)^{2}
is the moment of inertia of the boy about the axis,
is the moment of inertia of the plank about the same axis, and I_{plank}^{0} 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
I_{stick} = ∫ r^{2} dm = 2 ∫_{0}^{L/2} r^{2} l dr = 2l(r^{3}/3)_{0}^{L/2} = 2(M/L)(L^{3}/24) = ML^{2}/12
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.
(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: McGrawHill), problem 6.40]
SOLUTION
(a) The initial energy of the system is all potential energy and given by
E_{1} = PE_{1} = (1/2)kb^{2}
The wheel moves without rolling until its bottom reaches x = l. When it reaches this point, the energy of the system is all kinetic,
E_{2} = KE_{2} = (1/2)mv_{2}^{2}
where m is the mass of the wheel and v_{2} is its velocity when it reaches x = l. By conservation of energy,
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 w_{3} and rolling with velocity
v_{3} = Rw_{3}
where R is the wheel radius. Conservation of linear momentum requires that
mv_{3} = mv_{2}  Dp (1)
Conservation of angular momentum about the wheel axis requires that
In its first approach to the wall, when the wheel comes to rest all the energy of the system is potential energy,
E_{4} = PE_{4} = (1/2)k(l  x)^{2}
By energy conservation,
E_{3} = E_{4} => kb^{2}/4 = (1/2)k(l  x)^{2} => b^{2}/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,
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
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 w_{3} and its center of mass is moving in the x direction with velocity v_{3}. 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 w_{7} and rolling with velocity
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 twothirds of its initial height.
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), problem 6.41]
The torque about the point of contact with the floor is
t = 2LN_{w} sin q  Lmg cos q = I_{P}a = I_{P} d^{2}q/dt^{2} = (4mL^{2}/3) d^{2}q/dt^{2} (1)
where N_{w} is the normal force that the wall exerts on the plank,
I_{P} = ∫ r^{2} dm = ∫_{0}^{2L} r^{2} l dr = l(r^{3}/3)_{0}^{2L} = (m/2L)(8L^{3}/3) = 4mL^{2}/3
is the moment of inertia of the plank about an axis through the point of contact with the floor,
a = d^{2}q/dt^{2}
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 centerofmass velocity are
dx/dt =  L sin q dq/dt
dy/dt = L cos q dq/dt
The x component of the centerofmass acceleration is
d^{2}x/dt^{2} =  L cos q (dq/dt)^{2}  L sin q d^{2}q/dt^{2} (2)
Newton's second law (F = ma) in the horizontal direction is
N_{w} = ma_{x} = m d^{2}x/dt^{2} =  mL cos q (dq/dt)^{2}  mL sin q d^{2}q/dt^{2}
The initial energy of the plank is all potential and is equal to mg times the initial height of the center of mass,
E_{i} = mgL sin q_{0}
where q_{0} is the initial angle the plank makes with the horizontal. When the plank has slipped so that it makes an angle q < q_{0} with the horizontal, the energy is the sum of the potential energy and the translational and rotational kinetic energies,
E_{f} = mgL sin q + (1/2)m[(dx/dt)^{2} + (dy/dt)^{2}] + (1/2)I_{0}(dq/dt)^{2} = mgL sin q + (1/2)mL^{2}(dq/dt)^{2} + (1/2)I_{0}(dq/dt)^{2}
= mgL sin q + (1/2)(mL^{2} + I_{0})(dq/dt)^{2}
where
I_{0} = ∫ r^{2} dm = 2 ∫_{0}^{L} r^{2} l dr =2l(r^{3}/3)_{0}^{L} = (2)(m/2L)L^{3}/3 = mL^{2}/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,
E_{f} = mgL sin q + (1/2)(mL^{2} + mL^{2}/3)(dq/dt)^{2} = mgL sin q + (2mL^{2}/3)(dq/dt)^{2}
Equating the initial and final energies,
E_{i} = E_{f} => mgL sin q_{0} = mgL sin q + (2mL^{2}/3)(dq/dt)^{2}
=> L sin q_{0} = L sin q + (2L^{2}/3g)(dq/dt)^{2} (3)
At the instant when the plank loses contact with the wall, we have
N_{w} = 0 => d^{2}x/dt^{2} = 0
From (1) and (2) we get
d^{2}q/dt^{2} =  (3g/4L) cos q
(dq/dt)^{2} =  tan q d^{2}q/dt^{2} = (3g/4L) sin q
Substituting into (3),
L sin q_{0} = L sin q + (2L^{2}/3g)(3g/4L) sin q
= (3/2)L sin q => L sin q = (2/3)L sin q_{0} => 2L sin q = (2/3)(2L sin q_{0})
But 2L sin q is the y coordinate of the top of the plank when it loses contact with the wall and 2L sin q_{0} is the initial y coordinate of the top of the plank. Thus, the plank loses contact with the wall when it is at twothirds of its initial height.
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)MR^{2}.]
z ^ w_{z} = W

 ++
 
 
 R 
 
 R 
+======= hoop instantaneously rolling into page
 
 
 
 
 
 ++ ////////////
[from Kleppner, Daniel, and Kolenkow, Robert J. 1973, An Introduction to Mechanics (New York: McGrawHill), 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 ^ w_{z} = W

 ++
 
 
 R 
 
 R 
+=======> y, x out of page
 
 
 
 
 
 ++ ////////////
The z component of the angular velocity is
w_{z} = 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
w_{y} =  v/R =  W
The instantaneous angular velocity vector is
w = w_{x}x + w_{y}y + w_{z}z =  Wy + Wz
It has magnitude
w = sqrt(w_{x}^{2} + w_{y}^{2} + w_{z}^{2}) = sqrt[0^{2} + (W)^{2} + W^{2}] = W sqrt(2)
and makes an angle of
tan^{1}(w_{y} / w_{z}) = tan^{1}(1) = p/4 rad = 45°
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: McGrawHill), 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
Because of the way we defined our coordinate system,
I_{xy} =  ∫ xy dm = 0
I_{zy} =  ∫ yz dm = 0
so the spin angular momentum is
L = I_{yy}w_{y}y
Using the parallel axis theorem, the moment of inertia of the coin about the y axis is
I_{yy} = (1/2)mb^{2} + mb^{2} = (3/2)mb^{2}
Thus,
L = (3/2)mb^{2}(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
L_{h} = 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 = 2pL_{h}/(2pR/v) = vL_{h}/R = (3mbv^{2}/2R) cos f => tan f = 3v^{2}/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.
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.
[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 = t_{1} + t_{2} + t_{3} = r_{1} x F_{1} + r_{2} x F_{2} + r_{3} x F_{3}
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
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 345 right triangle, so
sin q = 4/5 = 0.8
Thus,
T = (50 kg)(9.8 m/s^{2})[1  (2 m) / (2)(3 m)] / (0.8) = 408.3 N
(b) Let F_{h} and F_{v} 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,
F_{h}  T cos q = 0 => F_{h} = T cos q = (408.3 N)(3/5) = 245 N (to the right)
F_{v} + T sin q  Mg/2  Mg/2 = 0 => F_{v} = Mg  T sin q = (50 kg)(9.8 m/s^{2})  (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
dt_{sign} = r x dF_{sign} = xx x [ (Mg/a) dx y] =  (Mg/a) x dx z => t_{sign} =  (Mg/a)z ∫_{ba}^{b} x dx =  Mgz [(1/a) ∫_{ba}^{a} x dx]
Now the average value of x between x = b  a and x = a is
x_{avg} = (1/a) ∫_{ba}^{a} x dx = (1/2)[(b  a) + b] = (1/2)(2b  a)
=> t_{sign} =  (Mg/2)(2b  a)z
This is the same as the torque, due to the sign, that was calculated in part (a), so T, F_{h}, and F_{v} 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
t_{sign} = r_{sign} x F_{sign} = {(1/2)[(b  a) + b]x  (a/2)y} x ( Mgy) =  (Mg/2)(2b  a)z
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.
[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,
a_{1} = 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  a_{2}) and the x coordinate of the center of mass of brick 1 relative to the upper right corner of brick 3 is a_{2}. The x coordinate of the center of mass of bricks 1 and 2 relative to the upper right corner of brick 3 is
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  a_{3}) and the x coordinate of the center of mass of bricks 1 and 2 relative to the upper right corner of brick 4 is a_{3}. The x coordinate of the center of mass of bricks 1, 2, and 3 relative to the upper right corner of brick 4 is
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  a_{4}) and the x coordinate of the center of mass of bricks 1, 2, and 3 relative to the corner of the table is a_{4}. The x coordinate of the center of mass of bricks 1, 2, 3, and 4 relative to the corner of the table is
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  a_{n}) and the x coordinate of the center of mass of bricks 1 through n  1 relative to the pivot point is a_{n}. The x coordinate of the center of mass of bricks 1 through n relative to the pivot point is
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.
[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 = t_{1} + t_{2}
= r_{1} x F_{1} + r_{2} x F_{2} (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
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,
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 (F_{v}, F_{h}) 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
F_{v}  Mg + F = 0 => F_{v} = 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 F_{h}, F_{h} 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 = (ML^{2}/4)a => a = 2g/L
where
I = ML^{2}/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
F_{v}  Mg =  Ma =  Mg => F_{v} = 0
The only horizontal force is F_{h}, 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
E_{i} = 0
After the plank has rotated through an angle q, its kinetic energy is
E_{f} = KE_{f} + PE_{f} = (ML^{2}/8)w^{2}  (MgL/2) sin q = E_{i} = 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 (F_{v}, F_{h}) 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
F_{v}  (M+m)g + F = 0 => F_{v} = (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 F_{h}, F_{h} 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 = [(ML^{2}/4) + (mL^{2}/3)]a = (M/4 + m/3)L^{2}a
=> a = [(M+m) / (M/4 + m/3)]g/2L
where
I = ML^{2}/4 + mL^{2}/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
The only horizontal force is F_{h}, 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
E_{i} = 0
After the plank has rotated through an angle q, its kinetic energy is
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
GM_{e}m/r^{2} = mv^{2}/r => GM_{e}/r = v^{2}
where G = 6.67 x 10^{11} N m^{2}/kg^{2} is the gravitational constant, M_{e} = 5.98 x 10^{24} 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,
GM_{e}/r = 4p^{2}r^{2}/T^{2} => r = (GM_{e}T^{2}/4p^{2})^{1/3} = [(6.67 x 10^{11} N m^{2}/kg^{2})(5.98 x 10^{24} kg)(86164.1 s)^{2}/4p^{2}]^{1/3} = 4.217 x 10^{7} m = 4.217 x 10^{4} m = 42,170 km
This is about 1.05 times the circumference of the Earth.
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 F_{G} on the HST must be equal to its mass m times its centripetal acceleration a_{c}.
F_{G} = ma_{c}
The gravitational force is
F_{G} = GM_{e}m / r^{2}
where G = 6.67 x 10^{11} N m^{2} kg^{2} is the gravitational constant, M_{e} = 5.98 x 10^{24} kg is the mass of the Earth, and r is the orbital radius. The centripetal acceleration is
a_{c} = v^{2} / r
where v is the orbital velocity. Thus,
GM_{e}m / r^{2} = mv^{2} / r => GM_{e} / r = v^{2} => v = sqrt(GM_{e} / r) = 2pr / T
where R_{e} = 6.37 x 10^{6} is the radius of the Earth and h = 600 km = 0.6 x 10^{6} m is the altitude of the HST above the ground. So
r = 6.37 x 10^{6} m + 0.6 x 10^{6} m = 6.97 x 10^{6} m
Substituting into (1), we get
T = 2p{(6.97 x 10^{6})^{3} / [(6.67 x 10^{11} N m^{2} kg^{2})(5.98 x 10^{24} 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
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.
(a) How much work would be required to move a satellite of mass m from a circular orbit of radius r_{1} = 2r_{E} about the earth to another circular orbit of radius r_{2} = 3r_{E}, where r_{E} 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 workenergy 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 F_{g} must equal the mass times the centripetal acceleration a_{c}, or
F_{g} = ma_{c}
The gravitational force is
F_{g} = GMm/r^{2}
where G is the gravitational constant, and M is the mass of the earth. The centripetal acceleration is
a_{c} = v^{2}/r
where v is the orbital velocity of the satellite. Thus,
GMm/r^{2} = mv^{2}/r => GM/r = v^{2}
The kinetic energy of the satellite is
KE = (1/2)mv^{2} = (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
E_{total} = KE + PE = GMm/2r – GMm/r =  GMm/2r
Moving the satellite from a circular orbit of radius 2r_{E} to a circular orbit of radius 3r_{E} increases the energy from
The work required to move the satellite is equal to the change in energy,
W = DE = GMm/12r_{E}
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
W_{g} = ∫ F_{g} • dr
where
F_{g} =  (GMm/r^{2}) r
is the force of gravity and dr is an infinitesimal element of the path taken by the satellite in moving from r = 2r_{E} to r = 3r_{E}.
Since the force of gravity is a radial force,
W_{g} = ∫ F_{g} • dr = ∫  (GMm/r^{2}) r • dr = ∫ – (GMm/r^{2}) r • r dr
where the integral is from r = 2r_{E} to r = 3r_{E}.
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.
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
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,
(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 F_{buoy} on the iceberg is equal to the weight of the water that the iceberg displaces.
F_{buoy} = r_{water}V_{sub}g (I)
where r_{water} is the density of the water, V_{sub} is the volume of the submerged part of the object, and g = 9.8 m/s^{2} is the acceleration of gravity. Let f be the fraction of the iceberg's volume V which is submerged. Then
V_{sub} = fV
The buoyant force supports the weight of the iceberg.
(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.
F_{buoy} = r_{water}Vg = r_{water}(M/r_{ice})g
= (r_{water}/r_{ice})Mg = Mg + W
=> W = [(r_{water}/r_{ice})  1]Mg = [(1.00 g/cm^{3}) / (0.917 g/cm^{3})  1](1000 kg)(9.8 m/s^{2}) = 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.
F_{buoy} = r_{water}V_{above}g = r_{water}(M_{above}/r_{ice})g = (r_{water}/r_{ice})M_{above}g = W
=> W = [(1.00 g/cm^{3}) / (0.917 g/cm^{3})](83 kg)(9.8 m/s^{2}) = 887.0 N
You place a glass beaker, partially filled with water, in a sink. The beaker has a mass of m_{b} = 390 g and an interior volume of V_{in} = 500 cm^{3}. 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 V_{b} be the volume of the beaker itself. Then the volume of the displaced water is V_{in} + V_{b} and the buoyant force is
F_{buoy} = r_{w}(V_{in} + V_{b})g
where r_{w} = 1 g/cm^{3} is the density of water and g = 9.8 m/s^{2} is the acceleration of gravity.
The volume of the water inside the beaker is
V_{w} = (1/2)V_{in}
The weight of the beaker plus the water it contains is
W = m_{b}g + r_{w}V_{w}g = m_{b}g + (1/2)r_{w}V_{in}g
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.
V_{0} = volume of ice cube
r_{0} = density of ice
r_{w} = density of water
V_{d} = volume of water displaced by ice cube.
The buoyant force which supports the ice cube is equal to its weight, so
F_{B} = m_{ice}g = r_{0}V_{0}g (1)
where m_{ice} = r_{0}V_{0} 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
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.
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 cm^{3} of iron
(b) 10 cm^{3} of air
(c) 10 cm^{3} of water
[from Physics E1a, Principles of Physics I: Mechanics, Final Exam, Fall 2003, Harvard Extension School]
SOLUTION
(a) Let m_{ice} be the mass of the ice in the ice block and m_{Fe} be the mass of the iron. The weight of the ice block is
W = (m_{ice} + m_{Fe})g
The buoyant force is equal to the weight of the ice block and is also equal to the weight of the displaced water,
F_{buoy} = r_{w}Vg
where r_{w} is the density of water and V is the volume of the displaced water. Thus,
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/m^{3}) 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?
[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.
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 m^{3}, 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 m_{bear} be the bear’s mass and m_{ice} 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
F_{buoy} = m_{bear}g + m_{ice}g (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 V_{bear} is the total volume of the bear, V_{ice} is the total volume of the ice slab, and f_{2} 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
Intravenous infusions are often made under gravity. Assuming the fluid has a density of 1.00 g/cm^{3}, 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 H_{2}O? (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, P_{0} = 760 mm Hg = 1.013 x 10^{5} N/m^{2} = 1.013 x 10^{6} dyne/cm^{2}. Then the pressure at the needle is
P = P_{0} + rgh => h = (P – P_{0})/rg
The stated pressure of 65 mm Hg is a gauge pressure, specified relative to atmospheric pressure. Thus, P – P_{0} = 65 mm Hg, which should be converted to dyne/cm^{2}, and
h = (P – P_{0})/rg
= (65 mm Hg)[(1.013 x 10^{6} dyne/cm^{2}) / (760 mm Hg)] / [(1.00 g/cm^{3})(980 cm/s^{2})] = 88.41 cm
(b) Since the density of mercury is 13.6 g/cm^{3}, while that of water is 1.0 g/cm^{3}, a pressure of 550 mm H_{2}O is equivalent to (550/13.6) mm Hg = 40.44 mm Hg.
The required value of h is
h = (P – P_{0})/rg
= (40.44 mm Hg)[(1.013 x 10^{6} dyne/cm^{2}) / (760 mm Hg)] / [(1.00 g/cm^{3})(980 cm/s^{2})] = 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 – P_{0})/rg = (18 mm Hg)[(1.013 x 10^{6} dyne/cm^{2}) / (760 mm Hg)] / [(1.00 g/cm^{3})(980 cm/s^{2})] = 24.48 cm
Suppose a person can reduce the pressure in the lungs to – 80 mmHg 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 mmHg 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 P_{0} = 760 mmHg, and the pressure at the end of the straw inside the person’s mouth is P = 760 mmHg – 80 mmHg = 680 mmHg, P_{0} and P are related by
P = P_{0}  rgh
where r = 1.0 g/cm^{3} is the density of water, g = 980 cm/s^{2} is the acceleration of gravity, and h is the height of the water inside the straw. Thus,
h = (P_{0}  P) / rg
= (760 mmHg – 680 mmHg)(1 atm / 760 mmHg)(1.013 x 10^{6} dyne/cm^{2} atm) / [(1.0 g/cm^{3})(980 cm/s^{2})]
= 108.81 cm = 1.0881 m
A patient is being fed through a nasogastric tube. A flow rate of 100 cm^{3}/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/cm^{2}.
[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 P_{0} = 760 mm Hg = 1.013 x 10^{5} N/m^{2} = 1.013 x 10^{6} dyne/cm^{2}. Then the pressure at the end of the tube at the level of the stomach is
P_{s}  P_{0} = (10 mm Hg)[(1.013 x 10^{6} dyne/cm^{2}) / (760 mm Hg)]
= 1.333 x 10^{4} dyne/cm^{2}
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 F_{2} is related to the original flow rate F_{1} according to
(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 = rw^{2}r
(b) Take p = p_{c} at the axis of rotation (r = 0) and show that the pressure p at any point r is
p = p_{c} + (1/2)rw^{2}r^{2}
(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 = w^{2}r^{2}/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
a_{c} =  (v^{2}/r) r =  rw^{2}r
Writing Newton's second law, in the radial direction, for the fluid element, we get
dF = dm a_{c} => [p(r)  p(r + dr)] A(r) r =  r A(r) dr rw^{2}r => p(r)  p(r + dr) =  r dr rw^{2} => [p(r + dr)  p(r)] / dr = rrw^{2} => lim(dr => 0) {[p(r + dr)  p(r)] / dr} = lim(dr => 0) rrw^{2} => dp/dr = rw^{2}r (1)
(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)
where p_{0} 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
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 = (P_{1}  P_{2}) / R
where P_{1} is the pressure where the fluid enters the tube, P_{2} is the pressure where the fluid exits the tube, and R is the resistance.
For the two individual tubes, we have the flow rates
F_{1} = (P_{1}  P_{2}) / R_{1} = 2.0 L/min
and
F_{2} = (P_{1}  P_{2}) / R_{2} = 3.0 L/min
where R_{1} and R_{2} are the resistances of the tubes,
R_{1} = (P_{1}  P_{2}) / F_{1}
and
R_{2} = (P_{1}  P_{2}) / F_{2}
Now suppose that we connect the tubes in series, with the fluid flowing through tube 1 first and then tube 2. Assume that P_{1} is the pressure at the entrance of tube 1, P_{2} 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 = (P_{1}  P’) / R_{1} = (P’ – P_{2}) / R_{2}
because the flow rate must be the same through both tubes. The flow rate through the connected tubes is also
F = (P_{1}  P_{2}) / R
where R is the resistance of the combination. Thus,
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 Utube. The other arm of the Utube 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 v_{A} = 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/r_{air})
where r is the density of the liquid in the Utube, g is the acceleration of gravity, h is the difference in the height of the liquid in the two arms of the Utube, and r_{air} is the density of the air.
(b) If r_{air} = 1.03 kg/m^{3}, the Utube contains alcohol with density r = 0.81 x 10^{3} kg/m^{3}, 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]
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)mv^{2} ≥ mgh => v ≥ sqrt(2gh) = sqrt[(2)(9.8 m/s^{2})(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
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 v_{0} when it emerges in the upward direction from the nozzle is related to its final velocity v_{f} = 0 when it reaches the maximum height y = h = 12.0 m according to
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
If we assume that h_{1} = h_{2} (i.e., neglect any difference in height between the water main and the nozzle), this becomes
P_{1} + (1/2)rv_{1}^{2} = P_{2} + (1/2)rv_{2}^{2}
The rate of flow through the water main and the nozzle must be the same. Thus, if A_{1} is the cross sectional area of the water main, and A_{2} is the cross sectional area of the nozzle, the rate at which water flows is
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,
You need to siphon water from a clogged sink. The sink has an area of 0.375 m^{2} 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]
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.
Identical charges q_{1} = q and q_{2} = 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 q_{3} = 2q be placed so that the electric field at the third vertex of the triangle, at (x, y) = (0, d), is zero?
^ y

q_{3} = 2q o y = ?


/\
/  \
/  \
/  \
/  \
/ d \
/  \
/  \
/ q  q \
o+o>x
q_{1} = q q_{2} = 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
r_{1} = a i r_{2} = a i r_{3} = 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) = S_{i=1}^{3} kq_{i}(r  r_{i}) / r  r_{i}^{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 r_{1} and r_{2} are the same,
[a/sqrt(3)] j + a i = [a/sqrt(3)] j  a i = sqrt(a^{2}/3 + a^{2}) = 2a/sqrt(3)
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, p_{1} and p_{2}. 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 p_{1} = qL pointing from the oxygen to the upper hydrogen, and the other with electric dipole moment p_{2} = qL pointing from the oxygen to the lower hydrogen. The total electric dipole moment points to the right and has magnitude
p = p_{1} cos q + p_{2} 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.
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
e_{0} ∫ E · dS = q_{enc}
where e_{0} = 8.85 x 10^{12} C^{2}/N m^{2} is the permittivity of free space, the integral is over a surface enclosing charge q_{enc}, 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.
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/e_{0} 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,
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
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
(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
E_{1} =  (s/2e_{0})x (region I)
E_{1} = + (s/2e_{0})x (regions II and III)
The electric field due to the plate on the right is
E_{2} =  (s/2e_{0})x (regions I and II)
E_{2} = + (s/2e_{0})x (region III)
A capacitor C_{1} carries a charge Q_{0}. It is then connected directly to a second, uncharged, capacitor C_{2}, as shown. What charge will each carry now? What will be the potential difference across each?
(a) Determine the current through each of the resistors in the diagram if R_{1} = 22 W, R_{2} = 25 W, R_{3} = 15 W, R_{4} = 12 W, R_{5} = 14 W, and E = 6.0 V.
(b) What is the net resistance of the circuit connected to the battery?
[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 I_{i} be the current through resistor R_{i}. Assume I_{1}, I_{2}, I_{4}, and I_{5} flow towards the right, and I_{3} flows downward. Let I be the current through the battery.
Write the loop theorem for the loop which passes through the battery, R_{2}, and R_{4}, in that order. The sum of the potential differences encountered in going around a complete loop must be zero. Thus,
6 – 25I_{2}  14I_{5} = 0 (1)
Write the loop theorem for the loop which passes through R_{1}, R_{3}, and R_{2}.
 22I_{1}  15I_{3} + 25I_{2} = 0 (2)
Write the loop theorem for the loop which passes through R_{3}, R_{4}, and R_{5}.
15I_{3}  12I_{4} + 14I_{5} = 0 (3)
Applying current conservation at each of the four junctions, we get
Factoring the coefficients of I_{4} in (18) and (19), we get
308 = 2^{2} x 7 x 11
168 = 2^{3} 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 2^{3} x 3 x 7 x 11 = 1848. Thus, multiply (18) by 2 x 3 = 6 and multiply (19) by 11.
So we have I_{1} = 3705/21043 A, I_{2} = 3246/21043 A, I_{3} =  24/21043 A, I_{4} = 339/1913 A, and I_{5} = 3222/21043 A. Note that the magnitude of I_{3} 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 = I_{1} + I_{2} = 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
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 p_{1} to another value p_{2} such that
p_{2} = p_{1} + dp
where dp is the change in the momentum during the time interval dt. p_{1} and p_{2} have the same magnitudes but point in slightly different directions due to the curvature of the circular path. The angle between the directions of p_{1} and p_{2} 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  b^{2}), 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.
According to the BiotSavart 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) = (m_{0}/4p) I dl x (r  r') / r  r'^{3}
where dB(r), r, r', and dl are all vectors, m_{0} = 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 BiotSavart law may be rewritten as
dB(r) = (m_{0}/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) = (m_{0}/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 q_{0} > 0 is located a distance d above particle 2, with charge  q_{0} < 0. The direction from particle 2 to particle 1 is the +y direction. Both are moving in the +x direction with velocity v_{0}.
^ 1 o>
 q_{0} v_{0}  y^
d 
 +>
 x
v 2 o>
q_{0} v_{0}
(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 BiotSavart law for the magnetic field due to a moving point charge.
B(r) = (m_{0}/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 =  q_{0} is the charge on particle 2, and v = v_{0}i is the velocity of particle 2. We have
r  r' = d j
and
r  r'^{3} = d^{3}
The magnetic field at the location of particle 1 due to particle 2 is
B(r) = (m_{0}/4p)(q_{0})(v_{0}i) x (d j) / d^{3}
= (m_{0}/4p)(q_{0}v_{0}/d^{2}) k
= (m_{0}/4p)(q_{0}v_{0}/d^{2}) (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 = q_{0}(v_{0}i) x (m_{0}/4p)(q_{0}v_{0}/d^{2}) (k)
= q_{0}v_{0}(m_{0}/4p)(q_{0}v_{0}/d^{2}) j
= (m_{0}q_{0}^{2}v_{0}^{2}/4pd^{2}) (+j)
The magnetic force on particle 1 has magnitude m_{0}q_{0}^{2}v_{0}^{2}/4pd^{2} 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(r_{1/2}). 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,
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 B_{0} 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 B_{0} 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 h_{0}, it will fall and eventually reach a terminal speed v_{max}. 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 =  dF_{B}/dt
where F_{B} is the magnetic flux enclosed by the loop.
F_{B} = ∫ B · dS = B_{0}Dh
where the integral is done over the area enclosed by the loop and h is the height of the rod.
E =  (d/dt) B_{0}Dh =  B_{0}D dh/dt = B_{0}Dv
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 = B_{0}Dv/R
The power dissipated in the resistor is
P = I^{2}R = B_{0}^{2}D^{2}v^{2}/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 v_{max} 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
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
F_{B} = IL x B => F_{B} = IDB_{0}
= B_{0}^{2}D^{2}v/R
The rate at which the magnetic field does work on the rod is
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 10^{6} N/m^{2}. Thus,
5F = Spr^{2} = Spd^{2}/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 10^{6} N/m^{2})] = 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 BiotSavart 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) = (m_{0}I/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) = (m_{0}I/4p)(dx i) x ( x i  a j) /  x i  a j^{3}
=  (m_{0}I/4p)(dx a k) / (x^{2} + a^{2})^{3/2}
Integrating from x =  a to x = + a, we get
B(0) = (m_{0}Ia/4p) k ∫_{a}^{a} dx / (x^{2} + a^{2})^{3/2}
= (m_{0}Ia/2p) k ∫_{0}^{a} dx / (x^{2} + a^{2})^{3/2}
Define x = a tan q => dx = a sec^{2} q dq, x^{2} + a^{2} = a^{2}(tan^{2} q + 1) = a^{2} sec^{2} q
B(0) =  (m_{0}Ia/2p) k ∫_{0}^{p/4} a sec^{2} q dq / (a^{2} sec^{2} q)^{3/2}
=  (m_{0}I/2pa) k ∫_{0}^{p/4} cos q dq
=  (m_{0}I/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 =  dF_{B}/dt =  (d/dt) B · A =  (d/dt) BL^{2} cos wt = BL^{2}w sin wt = E_{0} 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
E_{0} = BL^{2}w
is the amplitude of the induced EMF, which is related to the peak value of the induced current according to
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 stepdown transformer because the output voltage is lower than the input voltage.
V_{S}/V_{P} = N_{S}/N_{P} = 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.
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 dF_{B}/dt
where dF_{B}/dt is the rate at which the magnetic flux through the armature is changing. The magnetic flux through the armature is
F_{B} = AB cos wt
where
A = L^{2}
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 (L^{2}B cos wt) = NL^{2}Bw sin wt = E_{0} sin wt
where
E_{0} = NL^{2}Bw = (125)(6.60 x 10^{2} m)^{2}(0.200 T)(120 rev/s)(2p rad/rev) = 82.11 V
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
V_{R} = IR
If the current is increasing, the inductance causes a back emf of magnitude
V_{L} = L dI/dt
The potential drop across the coil is
V = V_{R} + V_{L} = IR + L dI/dt = (3.00 A)(2.25 W) + (0.440 H)(3.50 A/s) = 8.29 V
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).
In the circuit shown below, resistors 1 and 2 of resistance R_{1} and R_{2}, 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.
(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 R_{1} and R_{2}. Applying the loop theorem around the lefthand loop,
E  IR_{1}  IR_{2} = 0 => I = E/(R_{1} + R_{2})
(b) Applying the loop theorem around the righthand loop, starting at the bottom of R_{2}, we have
(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 R_{1} and the inductor, and none through R_{2}. The total resistance seen by the battery is R_{1}, and the current through the battery is
I = E/R_{1}
(d) The current through the battery starts at E/(R_{1} + R_{2}) and increases asymptotically to E/R_{1}.
(e) Just before the switch is opened, the current through the inductor is E/R_{1}, flowing downward. Just after the switch is opened, the current through the inductor is still E/R_{1}, flowing downward, but now the current flows upward through R_{2}. The voltage across R_{2} is ER_{2}/R_{1} with the bottom of R_{2} at the higher potential.
The potential difference across a given coil is V_{1} = 22.5 V at an instant when the current is I_{1} = 860 mA and is increasing at a rate of dI/dt = 3.40 A/s. At a later instant, the potential difference is V_{2} = 16.2 V whereas the current is I_{2} = 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 = V_{1} = 22.5 V, I = I_{1} = 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 => V_{1} = RI_{1} + L dI/dt => 22.5 = 0.860R + 3.40L (I)
At the second instant, we have V = V_{2} = 16.2 V, I = I_{2} = 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 => V_{2} = RI_{2} + L dI/dt => 16.2 = 0.700R  1.80L (II)
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 V_{rms} = 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[R^{2} + (wL  1/wC)^{2}]
= sqrt[R^{2} + (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
I_{rms} = V_{rms} / 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
P_{avg} = I_{rms}V_{rms} cos f = (230 V)(2.628 A)(0.9254) = 559.3 W
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
I_{0} = V_{0}/Z
where
Z = sqrt[R^{2} + (wL  1/wC)^{2}]
is the impedance. The voltage is of the form
V(t) = V_{0} 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[R^{2} + (wL)^{2}] = V_{0}/I_{0}
and
tan f = wL/R = tan 50 => wL = R tan 50 => sqrt[R^{2} + (R tan 50)^{2}] = V_{0}/I_{0} => R = V_{0}/[I_{0} sqrt(1 + tan^{2} 50)] = (120 V) / [(5.6 A) sqrt(1 + tan^{2} 50)] = 13.77 W
=> L = (R tan 50) / w = (R tan 50) / 2pf = (13.77 W)(tan 50) / [2p(60 Hz)] = 43.54 mH
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 P_{avg} can be expressed in terms of the rms current I_{rms} and the rms voltage V_{rms} as
P_{avg} = I_{rms}V_{rms} cos f
where
cos f = R/Z
is the power factor,
Z = sqrt[R^{2} + (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 = w_{0} = 1/sqrt(LC) = 1/sqrt[(20 x 10^{3} H)(50 x 10^{9} F)] = 31623 rad/s => f = f_{0} = w_{0}/2p = 5033 Hz
On either side of the resonant frequency f_{0}, the power factor decreases. We require that cos f = 0.17. We have
cos f = R / sqrt[R^{2} + (wL  1/wC)^{2}]
= 1 / sqrt[1 + (wL  1/wC)^{2}/R^{2}]
Standard electricity available in the United States is in the form of sinusoidal alternating current (AC) with a rootmeansquare voltage of 120 V and a frequency of 60 Hz (http://www.electricityforum.com/basicelectricity.html). The current drawn from this power source by an electrical device is also sinusoidal. The current rating of a device is its rootmeansquare current. The power rating of a device refers to the product of the rootmeansquare voltage and the rootmeansquare 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 rootmeansquare 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 kilowatthours. 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.
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]
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, m_{melt} is the mass which melts, and L_{f} = 0.25 x 10^{5} J/kg is the heat of fusion,
A house has wellinsulated walls 17.5 cm thick (assume conductivity of air) and area 410 m^{2}, a roof of wood 6.5 cm thick and area 280 m^{2}, and uncovered windows 0.65 cm thick and total area 33 m^{2}.
(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 m^{3}.
(c) If natural gas costs $0.080 per kilogram and its heat of combustion is 5.4 x 10^{7} 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 m^{2})(33 C°)/(0.175 m) = 0.4252 kcal/s
roof: (dQ/dt)_{roof} = (0.3 x 10^{4} kcal/s m C°)(280 m^{2})(33 C°)/(0.065 m) = 4.2646 kcal/s
windows: (dQ/dt)_{windows} = (2.0 x 10^{4} kcal/s m C°)(33 m^{2})(33 C°)/(0.0065 m) = 33.5077 kcal/s
(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 10^{4} kJ)($0.080/kg)/0.9 = $682.22/month
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
pV^{g} = constant
where g = C_{p}/C_{v}, C_{p} is the molar heat capacity at constant pressure, and C_{v} is the molar heat capacity at constant volume.
where p_{1} and V_{1} are the initial pressure and volume, and p_{2} and V_{2} 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 => p_{2}V_{2} = p_{3}V_{3}
where p_{3} and V_{3} 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 p_{3} = p_{1}, and we have
We know that V_{2} > V_{1} because the first process is an adiabatic expansion. Since 1/g = C_{v}/C_{p} is a positive number, we can deduce, from (1), that
p_{1}/p_{2} > 1 => p_{2}/p_{1} < 1
Since C_{p} is larger than C_{v}, regardless of whether the gas is monatomic, diatomic, or polyatomic,
g = C_{p}/C_{v} > 1 => 1/g < 1 => 1 – 1/g > 0
It follows from (2) that
V_{3}/V_{1} < 1 => V_{3} < V_{1}
so the final volume is lower than the initial volume.
A twostory 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 T_{i} = 70 °F to T_{1f} = 65 °F in two hours when the temperature outside is T_{out} = 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 T_{i} = 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 m^{3}
(b) From the ideal gas law, the number of moles of air molecules on the first floor at the average temperature is
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:
is the temperature change, and C_{p} 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
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 T_{2f} is the final temperature on the second floor at the end of the twohour period,
T_{2avg} = (T_{1} + T_{2f}) / 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
(dT_{2}/dx)_{wall} = (T_{2avg}  T_{out}) / a (4)
(dT_{2}/dx)_{ceil} = (T_{2avg}  T_{out}) / b (5)
The average rate at which heat is lost by the second floor is
Equations (3) through (11) can be used iteratively to obtain a value for T_{2f}. A trial value has to be assumed for T_{2f}. If we assume T_{2f} = 65 °F, and use the above equations, we get
One mole of an ideal monatomic gas, initially at point A at a pressure of 1.0 x 10^{5} N/m^{2} and a volume of 25 x 10^{3} m^{3}, is taken through a 3process cycle A => B => C => A (GasCycle060219.01.xls). Each process is done slowly and reversibly. At point B, the pressure is 2.0 x 10^{5} N/m^{2} and the volume is 25 x 10^{3} m^{3}. At point C, the pressure is 1.0 x 10^{5} N/m^{2} and the volume is 50 x 10^{3} m^{3}. For a monatomic gas, the heat capacities for constant volume and constant pressure are, respectively, C_{v} = (3/2)R and C_{p} = (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.
T_{A} = p_{A}V_{A}/nR = (1.0 x 10^{5} N/m^{2})(25 x 10^{3} m^{3})/[(1 mole)(8.32 J/mole K)] = 300.5 K
T_{B} = p_{B}V_{B}/nR = (2.0 x 10^{5} N/m^{2})(25 x 10^{3} m^{3})/[(1 mole)(8.32 J/mole K)] = 601.0 K
T_{C} = p_{C}V_{C}/nR = (1.0 x 10^{5} N/m^{2})(50 x 10^{3} m^{3})/[(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 10^{5} N/m^{2})(25 x 10^{3} m^{3}) + (0.5)(1.0 x 10^{5} N/m^{2})(25 x 10^{3} m^{3}) = 3750 J
W(C=>A) = ∫_{C=>A} p dV = (1.0 x 10^{5} N/m^{2})(25 x 10^{3} m^{3}  50 x 10^{3} m^{3}) =  2500 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 T_{B} = T_{C} = 601.0 K, the change in internal energy for process B=>C is zero. Thus,
(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.
A heat engine consists of an oilfired 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 10^{7} 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 10^{3} kg/m^{3}; specific heat of water is c = 4.2 x 10^{3} 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 10^{6} J/s.
(b) The rate at which oil is burned is (300 x 10^{6} J/s)(1 kg / 4.4 x 10^{7} 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 10^{6} 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 10^{6} J/s)(3600 s/hr)/[(1.0 x 10^{3} kg/m^{3})(200 m)(50 m)(10 m)(4.2 x 10^{3} J/kg °C)] = 1.543 °C
One mole of an ideal monatomic gas is taken through the cycle abca (GasCycle060219.02.xls. State a has volume V_{a} = 17 x 10^{3} m^{3} and pressure P_{a} = 1.2 x 10^{5} N/m^{2}. State b has volume V_{b} = 51 x 10^{3} m^{3} and pressure P_{b} = 1.2 x 10^{5} N/m^{2}. State c has volume V_{c} = 51 x 10^{3} m^{3}. Process ca lies along the 250 K isotherm. The molar heat capacities for the gas are C_{p} = 20.8 J/mole K and C_{v} = 12.5 J/mole K. Determine each of the following:
(a) the temperature T_{b} of state b
(b) the heat Q_{ab} added to the gas during process ab
(c) the change in internal energy U_{b}  U_{a} (d) the work W_{bc} 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) P_{b}V_{b} = nRT_{b} => T_{b} = P_{b}V_{b}/nR = (1.2 x 10^{5} N/m^{2})(51 x 10^{3} m^{3})/[(1 mole)(8.32 J/mole K)] = 735.6 K
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
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 E1a, Principles of Physics I: Mechanics, Final Exam, Fall 2001, Harvard Extension School]
SOLUTION
(a) The first harmonic (i.e., the fundamental) is f_{1} = 50 Hz. The second harmonic (i.e., the first overtone) is f_{2} = 150 Hz = 3f_{1}. The third harmonic (i.e., the second overtone) is f_{3} = 250 Hz = 5f_{1}. 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
f_{1} = v/4L => L = v/4f_{1} = (343 m/s)/[(4)(50 Hz)] = 1.715 m
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 v_{s} = 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 = v_{s}/v => v = v_{s} / sin q = v_{s} / sin[tan^{1}(3/4)] = (330 m/s) / (3/5) = 550 m/s
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 / R_{1} + 1 / R_{2}) (1)
where f is the focal length of the lens, n is the index of refraction, and R_{1} and R_{2} 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, R_{1} =  31.2 cm and R_{2} =  23.8 cm. Solving (1) for f,
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 / R_{1} + 1 / R_{2}) (1)
where f is the focal length of the lens, n is the index of refraction, and R_{1} and R_{2} are the radii of curvature of the surfaces of the lens. A (positive, negative) radius of curvature corresponds to a (convex, concave) surface.
Let R_{1} = 20.0 cm = 0.200 m be the radius of curvature of one of the surfaces. Solving (1) for R_{2},
R_{2} = {[1 / f(n  1)]  1 / R_{1}}^{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} => R_{2} = [(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.
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 q_{1} = 90°  14° = 76°. By Snell's law, at the airwater interface,
n_{1} sin q_{1} = n_{2} sin q_{2}
where (n_{1}, n_{2}) = (1.00, 1.33) is the index of refraction of (air, water). Solving for q_{2}, we get
q_{2} = sin^{1}[(n_{1}/n_{2}) sin q_{1}]
But the width x and depth d of the pool are related by
tan q_{2} = x/d => d = x / tan q_{2} = x / tan{sin^{1}[(n_{1}/n_{2}) sin q_{1}]}
= (5.50 m) / tan{sin^{1}[(1.00/1.33) sin 76°]} = 5.156 m
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 q ≈ q.
[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 d_{o} = d = 3.25 m be the object distance. When the person stands in front of a plane mirror, the image distance is d_{i} =  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 / (d_{o}  d_{i}) = h / (d + d) = h / 2d
If the person stands the same distance in front of a convex mirror, the object distance is d_{o}' = d, the same as in front of the plane mirror, but the image distance and image height are different and can be denoted by d_{i}' =  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
where d_{i}' =  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 q ≈ q (i.e., if sin q ≈ q, then tan q ≈ q 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 =  d_{i}' / d_{o}' =  d_{i}' / d => h' =  d_{i}'h / d_{o}' =  d_{i}'h / d
Substituting into (1), we get
tan q' = h' / (d  d_{i}') =  d_{i}'h / [d(d  d_{i}')] = a =>  d_{i}'h = ad(d  d_{i}') = ad^{2}  add_{i}'
=> d_{i}' = ad^{2} / (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 / d_{o}' + 1 / d_{i}' = 1 / f ' => 1 / d + 1 / d_{i}' = 1 / f ' => 1 / f ' = 1 / (3.25 m)  12 / (13 m) =  8 / (13 m) => f ' =  13 m / 8 =  1.625 m
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
n_{1} sin q_{1} = n_{2} sin q_{2} => 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.
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
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.
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.
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 400nm and 700nm light.
[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 24.17]
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 q_{i} be the angle of incidence on the convex surface. Then
sin q_{i} = h/R => q_{i} = sin^{1}(h/R) (1)
According to Snell's law, the angle of refraction q_{r} on the convex surface is related to q_{i} by
sin q_{i} = n sin q_{r} => q_{r} = sin^{1}[(1/n) sin q_{i}] = 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 q_{i}' at the flat surface is related to q_{i} and q_{r} according to
The angle of refraction q_{r}' at the flat surface is obtained from
n sin q_{i}' = sin q_{r}' => q_{r}' = sin^{1}(n sin q_{i}') (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 q_{r}'. The length of the side of the triangle opposite A can be shown to be equal to R sin q_{i}  R cos q_{i} tan q_{i}'. Thus,
tan q_{r}' = (R sin q_{i}  R cos q_{i} tan q_{i}') / d
=> d = (R sin q_{i}  R cos q_{i} tan q_{i}') / tan q_{r}'
= R(sin q_{i}  cos q_{i} tan q_{i}') / tan q_{r}' (5)
Using equations (1), (2), (3), (4), and (5), we get
q_{i} = sin^{1}(h/R) = sin^{1}(1.0 cm / 12.0 cm) = sin^{1}(1/12) = 4.780192°
q_{r} = sin^{1}(h/Rn) = sin^{1}{(1.0 cm) / [(12.0 cm)(1.50)]} = sin^{1}(1/18) = 3.184739°
q_{i}' = q_{i}  q_{r} = 4.780192°  3.184739° = 1.595453°
q_{r}' = sin^{1}(n sin q_{i}') = sin^{1}[(1.50) sin 1.595453°] = 2.392871°
d = R(sin q_{i}  cos q_{i} tan q_{i}') / tan q_{r}'
= (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
q_{i} = sin^{1}(h/R) = sin^{1}(6.0 cm / 12.0 cm) = sin^{1}(1/2) = 30°
q_{r} = sin^{1}(h/Rn) = sin^{1}{(6.0 cm) / [(12.0 cm)(1.50)]} = sin^{1}(1/3) = 19.47122°
q_{i}' = q_{i}  q_{r} = 30°  19.47122° = 10.52878°
q_{r}' = sin^{1}(n sin q_{i}') = sin^{1}[(1.50) sin 10.52878°] = 15.70443°
d = R(sin q_{i}  cos q_{i} tan q_{i}') / tan q_{r}'
= (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 q_{r}'. 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 q_{r}' = Dy / Dd => Dy = (Dd) tan q_{r}' = (1.489966 cm) tan 15.70443°
= 0.418935 cm
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 convexconvex thin lens and a ray incident on the left side of the lens, parallel to the lens axis. Define
q_{1} = angle of incidence at first surface of lens
q_{2} = angle of refraction at first surface of lens
q_{3} = angle of incidence at second surface of lens
q_{4} = angle of refraction at second surface of lens
At the first surface, by Snell's law,
n' sin q_{1} = n sin q_{2}
Using the small angle approximation,
n'q_{1} = nq_{2} => q_{2} = (n'/n)q_{1} (1)
Similarly, at the second surface,
nq_{3} = n'q_{4} => q_{3} = (n'/n)q_{4} (2)
Define
h_{1} = vertical distance from axis to point on lens at which ray enters lens
h_{2} = 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
R_{1} = radius of curvature of first surface of lens
R_{2} = radius of curvature of second surface of lens
f = focal length of lens (in the general medium with index of refraction n' ≠ 1)
Then
q_{1} = sin q_{1} = h_{1}/R_{1} (3)
a = sin a = h_{2}/R_{2} (4)
b = sin b = h_{2}/f (5)
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/d_{o} + 1/d_{i} = 1/f
where d_{o} is the object distance, d_{i} is the image distance, and 1/f is given by the generalized lensmaker's equation (9).
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]
The image height h_{i} is negative because the image is inverted relative to the object. The magnification is related to the object distance d_{o} and the image distance d_{i} by
M =  d_{i} / d_{o} => d_{o} =  d_{i} / M =  (9.00 m) / ( 75) = (900 cm) / 75 = 12 cm
The focal length f is related to d_{o} and d_{i} by
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 (d_{o} = ∞) 2.0 cm beyond the eye lens. The corresponding power is
P = 1/f = 1/d_{o} + 1/d_{i} = 1/d_{i} = 1 / (0.02 m) = 50 D
The accommodated power is
P_{acc} = 1.1P = (1.1)(50 D) = 55 D
(b) The closest object that she can see clearly corresponds to the accommodated power.
1/d_{o} + 1/d_{i} = 1/f => d_{o} = (1/f – 1/d_{i})^{1} = [55 D – 1 / (0.02 m)]^{1} = 0.2 m = 20 cm
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
P_{acc} = 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/d_{o} + 1/d_{i} = 1/f => d_{o} = (1/f – 1/d_{i})^{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/d_{o} + 1/d_{i} = 1/f => d_{o} = (1/f – 1/d_{i})^{1} = [56.16 D – 1 / (0.02 m)]^{1} = 0.1623 m
= 16.23 cm
(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 d_{o} = 1.50 m and the image distance is d_{i} = 2.0 cm = 0.02 m.
P = 1/f = 1/d_{o} + 1/d_{i} = 1 / (1.50 m) + 1 / (0.02 m) = 50.67 D
(b) Neglecting the distance between the spectacle lens and the eye, d_{o} = 25 cm = 0.25 m and d_{i} =  1.50 m. We get
P = 1/f = 1/d_{o} + 1/d_{i} = 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, d_{o} = 25 cm – 2 cm = 23 cm and d_{i} =  (1.50 m – 0.02 m) =  1.48 m.
P = 1/f = 1/d_{o} + 1/d_{i} = 1 / (0.23 m) – 1 / (1.48 m) = 3.672 D
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.
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,
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/d_{o} + 1/d_{i} => f = 1/(1/d_{o} + 1/d_{i}) = 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/d_{o} + 1/d_{i} => f = 1/(1/d_{o} + 1/d_{i}) = 1/(1/23  1/38) = 58.27 cm = 0.5827 m
=> P = 1/f = 1/0.5827 = +1.716 D
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
A 50yearold 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 d_{o} = 25 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 d_{o} = 25 cm appear at d_{i} =  280 cm.
1/f ' = 1/d_{o} + 1/d_{i} => f ' = 1/(1/d_{o} + 1/d_{i}) = 1/(1/25  1/280) = 27.45 cm = 0.2745 m
=> P' = 1/f ' = +3.643 D
A small reflecting telescope has a focal length of 1.0 m. (a) What magnification does it produce when photographing the sun, 1.50 x10^{8} 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/d_{o} + 1/d_{i} = 1/f => d_{i} = (1/f – 1/d_{o})^{1} = [1 / (1.0 m) – 1 / (1.50 x 10^{11} m)]^{1} = 1.0 m
The magnification is
m =  d_{i}/d_{o} =  (1.0 m) / (1.50 x 10^{11} m) =  6.667 x 10^{12}
(b) h_{i} = mh_{o} = ( 6.667 x 10^{12})(25,000 x 10^{3} m) =  1.667 x 10^{4} =  0.1667 mm
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 = f_{o}/f_{e} => f_{e} = f_{o}/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
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
f_{e} = 1/P_{e} = 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 f_{o} = 85 cm  5 cm = 80 cm. The total magnification of the telescope is
In a doubleslit experiment, it is found that blue light of wavelength 460 nm gives a secondorder 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 l_{1} = 460 nm. The interference maxima for wavelength l_{1} occur when
d sin q = ml_{1}
where d is the separation between the slits and m is an integer. For a secondorder maximum, m = 2 and
d sin q = 2l_{1} (1)
Interference minima occur at the same value of q for wavelength l_{2} if
Two firstorder spectrum lines are measured by an 8500line/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 firstorder 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
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.
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
l_{n} = 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 / l_{n} = 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 l_{a} = 512 nm and l_{b} = 640 nm. Then
N = 2tn / l_{a} (1)
where N is an odd half integer, because there is a minimum for l_{a} = 512 nm, and
N’ = 2tn / l_{b}
where N’ is an integer, because there is a maximum for l_{b} = 640 nm. As the wavelength is increased from l_{a} to l_{b}, the number of wavelengths along the path through the alcohol must decrease by 1/2. Thus,
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
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.
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 = r^{2} 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 = ur^{2} 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 / 4pr^{2})(dA cos q) = (ur^{2} dr sin q dq df / 4pr^{2})(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 / 4pr^{2})(dA cos q)
= (u/4p) dr dA ∫_{0}^{p/2} sin q cos q dq ∫_{0}^{2p} df
= (u/2) dr dA ∫_{0}^{p/2} sin q cos q dq
= (u/2) dr dA (1/2) sin^{2} q _{0}^{p/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,
How much of the Sun's energy is radiated in the range of wavelengths 40007000 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/c^{2})n^{3}/[exp(hn/kT)  1]
where h = 6.63 x 10^{34} J s is Planck's constant, c = 3 x 10^{8} 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
s = (2p^{5}k^{4} / 15c^{2}h^{3}) = 5.642 x 10^{8} W/m^{2}K^{4}
is the StefanBoltzmann constant. The accepted value is 5.67 x 10^{8} W/m^{2}K^{4}.
Now we determine the energy emitted per unit area per unit time in the 40007000 Å wavelength range, which is in the visible part of the spectrum. Let l_{min} = 4000 Å and l_{max} = 7000 Å. The corresponding frequencies are
n_{min} = c / l_{max} = (3 x 10^{8} m/s) / (7 x 10^{7} m) = 4.286 x 10^{14} Hz
and
n_{max} = c / l_{min} = (3 x 10^{8} m/s) / (4 x 10^{7} m) = 7.500 x 10^{14} Hz
The corresponding values of the variable
x = hn/kT
are
x_{min} = hn_{min}/kT = (6.63 x 10^{34} J s)(4.286 x 10^{14} Hz) / [(1.38 x 10^{23} J/K)(5800 K)] = 3.550
and
x_{max} = hn_{max}/kT = (6.63 x 10^{34} J s)(7.500 x 10^{14} 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 x_{min} and x_{max}.
Calculate the fraction of the Sun's energy which is emitted in the following wavelength ranges:
(a) 740 nm  0.1 cm (infrared)
(b) 380740 nm (visible)
(c) 10380 nm (ultraviolet)
(d) 0.0110 nm (Xray)
Assume that the Sun is a blackbody with a surface temperature of T = 5800 K.
SOLUTION
For each of the four wavelength ranges l_{min} < l < l_{max}, the minimum and maximum frequencies n are
n_{min} = c / l_{max}
and
n_{max} = c / l_{min}
where c = 3 x 10^{8} m/s is the speed of light. The minimum and maximum values of the parameter
x = hn/kT
are
x_{min} = hn_{min}/kT
and
x_{max} = hn_{max}/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
E_{min} = hn_{min}
and
E_{max} = hn_{max}
These are summarized in the following table.
Part of Spectrum
Wavelength Range
Frequency Range (Hz)
hn/kT Range
Energy Range
(a) infrared
740 nm  0.1 cm
3 x 10^{11}  4.054 x 10^{14}
2.485 x 10^{3}  3.358
1.243 x 10^{3}  1.680 eV
(b) visible
380740 nm
(4.0547.895) x 10^{14}
3.3586.539
1.6803.271 eV
(c) ultraviolet
10380 nm
7.895 x 10^{14}  3 x 10^{16}
6.539248.5
3.271124.3 eV
(d) Xray
0.0110 nm
3 x 10^{16}  3 x 10^{19}
248.5248501
0.1243124.3 keV
The total energy emitted per unit area per unit time can be shown to be
A beam of 10MeV protons is incident on a copper (Z = 29, A = 65) foil. For protons scattered 90°:
(a) Find the impact parameter.
(b) Using angularmomentum 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 Q_{1} incident on a nucleus of charge Q_{2}, the impact parameter is
b = (k/2)(Q_{1}Q_{2}/E_{k}) cot q/2
where k = 1/4pe_{0} = 8.98755 x 10^{9} N m^{2}/C^{2}, E_{k} is the kinetic energy of the incident particle, and q is the scattering angle.
We have Q_{1} = e, Q_{2} = Ze = 29e, E_{k} = 10 MeV, and q = 90°. Thus, the impact parameter is
b = (k/2)(29e^{2}/E_{k}) cot q/2
= [(8.98755 x 10^{9} N m^{2}/C^{2}) / 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 Q_{1} = e 
o>T_{1} 
b Q_{2} = 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 r_{min}.
The initial energy of the proton, at position 1, is
E_{1} = T_{1} + V_{1}
where
T_{1} = 10 MeV = (10 MeV)(1.6 x 10^{13} J/MeV) = 1.6 x 10^{12} J
is the initial kinetic energy, and V_{1} 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
E_{2} = T_{2} + V_{2}
where T_{2} and V_{2} 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, v_{1} and v_{2},
where r_{1} and p_{1} are the position and momentum vectors of the proton at position 1. Using the right hand rule, L_{1} points into the screen. The magnitude of the initial angular momentum is
L_{1} = rm_{p}v_{1} sin f = bm_{p}v_{1}
where f is the angle between r_{1} and p_{1}. The final angular momentum, when the proton is closest to the nucleus, is
L_{2} = r_{2} x p_{2}
also pointing into the screen, the magnitude of which is
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 incidentparticle 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 centerofmass frame. The minimum incidentparticle energy occurs when the final particles are at rest.
Consider the case where the incident particle is a photon of energy E_{g}.
E_{g} v_{H} o> <O m_{H} >
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,
E_{1} = E_{g} + (1/2)m_{H}v_{H}^{2} + m_{H}c^{2}
where m_{H} = 1.007825 u is the hydrogen mass, v_{H} is the speed of the hydrogen atom, and c = 3 x 10^{8} 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,
E_{2} = m_{p}c^{2} + m_{e}c^{2}
where m_{p} = 1.0072765 u is the proton mass, and m_{e} = 0.000548624 u is the electron mass.
In the centerofmass 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
The velocity of the hydrogen atom before the collision is
v_{H} = E_{g}/m_{H}c
= (115.5 eV)(1.6e19 J/eV) / (1.007825 u)(931.502 MeV/c^{2}u)(1.6 x 10^{13} J/MeV)c
= 1.230 x 10^{7}c = (1.230 x 10^{7})(3 x 10^{8} m/s) = 36.91 m/s
The hydrogen atom was initially moving in the x direction in the centerofmass frame. Since the hydrogen atom is at rest in the laboratory frame, the centerofmass 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,
b = v_{H}/c = (36.91 m/s) / (3 x 10^{8} m/s) = 1.230 x 10^{7} g = 1 / sqrt(1  b^{2}) ≈ 1
Thus,
E_{g}' ≈ E_{g} = 115.5 eV
Consider the case, in the centerofmass frame, where the incident particle is an electron with kinetic energy T_{e} and velocity v_{e}.
v_{e} v_{H} m_{e} o> <O m_{H} >
T_{e} 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,
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,
In the centerofmass 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
The velocity of the hydrogen atom before the collision is
v_{H} = (m_{e}/m_{H})v_{e} = (0.000548624 u / 1.007825 u)(6.377 x 10^{6} m/s)
= 3471 m/s
The hydrogen atom was initially moving in the x direction in the centerofmass frame. Since the hydrogen atom is at rest in the laboratory frame, the centerofmass 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
v_{e}’ = v_{e} + v_{H} = 6.377 x 10^{6} m/s + 3471 m/s = 6.380 x 10^{6} m/s
The speed of the electron is marginally relativistic. Using the nonrelativistic formula for the kinetic energy,
T_{e}’ = (1/2)m_{e}v_{e}’^{2} = (1/2)(9.11 x 10^{31} kg)(6.380 x 10^{6} 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 > E_{g}’
Using the relativistic formula,
T_{e}’ = (g  1)m_{e}c^{2} = {1 / sqrt[1 – (v_{e}’/c)^{2}]}m_{e}c^{2} = {1 / sqrt[1 – (6.380 x 10^{6} m/s / 3 x 10^{8} m/s)^{2}]}(9.11 x 10^{31} kg)(3 x 10^{8} 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 > E_{g}’
So assuming that the electron is nonrelativistic is a good approximation.
(b) Consider the case, in the centerofmass frame, where the incident particle is a proton with kinetic energy T_{p} and velocity v_{p}
v_{p} v_{H} m_{p} o> <O m_{H} >
T_{p} 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,
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,
In the centerofmass 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
(1/2)m_{p}v_{p}^{2} + m_{p}c^{2} + (1/2)m_{H}(m_{p}/m_{H})^{2}v_{p}^{2} + m_{H}c^{2} = 2m_{p}c^{2} + m_{e}c^{2} => (1/2)[m_{p} + m_{H}(m_{p}/m_{H})^{2}]v_{p}^{2} + m_{H}c^{2} = m_{p}c^{2} + m_{e}c^{2} => (1/2)m_{p}(1 + m_{p}/m_{H})v_{p}^{2} + m_{H}c^{2} = m_{p}c^{2} + m_{e}c^{2} => v_{p} = sqrt[2(m_{p} + m_{e}  m_{H})c^{2} / m_{p}(1 + m_{p}/m_{H})]
= c sqrt[2(m_{p} + m_{e}  m_{H}) / m_{p}(1 + m_{p}/m_{H})]
= 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 10^{8} m/s) = 1.053 x 10^{5} m/s
The velocity of the hydrogen atom before the collision is
v_{H} = (m_{p}/m_{H})v_{p} = (1.0072765 u / 1.007825 u)(1.053 x 10^{5} m/s)
= 1.052 x 10^{5} m/s
The hydrogen atom was initially moving in the x direction in the centerofmass frame. Since the hydrogen atom is at rest in the laboratory frame, the centerofmass frame is moving in the +x direction relative to the laboratory frame at velocity 1.052 x 10^{5} m/s. The velocity of the proton in the laboratory frame is
v_{p}’ = v_{p} + v_{H} = 1.053 x 10^{5} m/s + 1.052 x 10^{5} m/s = 2.105 x 10^{5} m/s
The kinetic energy of the proton in the laboratory frame is
T_{p}’ = (1/2)m_{p}v_{p}’^{2} = (1/2)(1.67 x 10^{27} kg)(2.105 x 10^{5} 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 > T_{e}’
A 5.0 MeV electron traveling east passes near a copper nucleus (rest energy, 6.0 x 10^{4} 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 / p_{g} E_{g} ^ north
 / 
e^{} / q = 45° 
o> O +> east
T_{1} e^{} o T_{2} 

v
(a) The (initial, final) kinetic energy of the electron is (T_{1}, T_{2}) = (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
(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
E_{1} = T_{1} + m_{e}c^{2}
where m_{e} is the electron rest mass. The total initial energy of the electron is related to its total initial momentum, p_{1}, and rest mass according to
By conservation of momentum, the x (eastwest) component of the momentum of the recoil nucleus, p_{Nx}, is related to p_{1} and the momentum of the photon according to
p_{1} = p_{Nx} + p_{g} cos q => p_{Nx} = p_{1}  p_{g} cos q
where q = 45° because the photon is traveling in the northeast direction. So
p_{Nx} = p_{1}  p_{g} cos q = 5.487 MeV/c  (0.1 MeV/c) cos 45° = 5.417 MeV/c
Also by conservation of momentum, the y (northsouth) component of the momentum of the recoil nucleus, p_{Ny}, is related to p_{2} and p_{g} according to
0 =  p_{2} + p_{Ny} + p_{g} sin q => p_{Ny} = p_{2}  p_{g} sin q
= 5.387 MeV/c  (0.1 MeV/c) sin 45° = 5.316 MeV/c
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/m_{p}c)(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, m_{p} = 938.2592 MeV/c^{2} 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/m_{p}c)(1  cos q) => 1/E'  1/E = (1/m_{p}c^{2})(1  cos q)
=> 1/E' = 1/E + (1/m_{p}c^{2})(1  cos q)
=> E' = 1/[1/E + (1/m_{p}c^{2})(1  cos q)] = E/[1 + (E/m_{p}c^{2})(1  cos q)]
=> E'_{min} = E/[1 + (2E/m_{p}c^{2})] = (100 MeV)/[1 + (2)(100 MeV)/(938.2592 MeV)] = 82.43 MeV
The maximum possible energy loss for the proton is
A photon collides with a free electron at rest. Show that the minimum photon energy necessary for this interaction to create an electronpositron 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
E_{g}, p_{g} m_{e}
The minimum photon energy E_{g} in the laboratory frame occurs when, in the centerofmass frame, the three final particles (original electron, new positron, and new electron) are at rest.
The energymomentum fourvector has components (ip_{x}, ip_{y}, ip_{z}, E/c), where p_{x} is the x component of the total linear momentum, p_{y} is the y component, p_{z} 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}  p_{x}^{2}  p_{y}^{2}  p_{z}^{2} is invariant and has the same value in the laboratory frame and in the centerofmass frame. In this problem, we can define the +x direction to be the direction in which the incident photon is traveling and set both p_{y} and p_{z} equal to zero, so (E/c)^{2}  p_{x}^{2} is invariant, or
(E/c)^{2}  p_{x}^{2} = constant (1)
In the laboratory frame, the energy is
E_{1} = E_{g} + m_{e}c^{2}
where E_{g} is the photon energy and m_{e} is the electron mass, and the momentum is
A beam of 2.04MeV 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 electronpositron pair at a distance of 1.0 x 10^{3} Å from the nucleus. The created particles have the same kinetic energy E_{k} 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 E_{k}^{+} and E_{k}^{} 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 E_{k}^{+}  E_{k}^{} = 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 E_{k}. Their initial potential energies are
V_{±} = ± (1/4pe_{0})(29e^{2}/r_{min})
where r_{min} = 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
E_{k}^{+}  E_{k}^{} = (2)(1/4pe_{0})(29e^{2}/r_{min})
= (2)(9 x 10^{9} N m^{2} 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  b^{2}), b = v/c, and c = 3 x 10^{8} 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
E_{k}^{±} = E_{k} + V_{±} (4)
When the photon is converted into an electronpositron pair, all of its energy is converted into the total energy of the electronpositron pair, with includes rest energy, kinetic energy, and potential energy. Thus,
v^{+} = b^{+}c = (0.9347)(3 x 10^{8} m/s) = 2.804 x 10^{8} m/s v^{} = b^{}c = (0.5296)(3 x 10^{8} m/s) = 1.589 x 10^{8} m/s
Substituting into (3), and setting q = e = 1.6 x 10^{19} C for both, we get
R^{+} = g^{+}m_{e}v^{+}/qB = (2.813)(9.11 x 10^{31} kg)(2.804 x 10^{8} m/s) / [(1.6 x 10^{19} C)(0.15 T)] = 0.02994 m = 2.994 cm
R^{} = g^{}m_{e}v^{+}/qB = (1.179)(9.11 x 10^{31} kg)(1.589 x 10^{8} m/s) / [(1.6 x 10^{19} C)(0.15 T)] = 0.007109 m = 0.7109 cm
The ionization (binding) energy of the outermost electron in boron is 8.26 eV. (a) Use the Bohr model to estimate the "effective charge," Z_{eff}, 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 = L_{n} = 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 r_{n} is
L = L_{n} = mvr_{n} (2)
where v is the orbital velocity. Combining (1) and (2), we get
mvr_{n} = nħ => v = nħ / mr_{n} (3)
A boron atom contains five electrons. Thus, its ground state electronic configuration is 1s^{2}2s^{2}2p^{1}. The outermost electron is the 2p electron, which orbits a nucleus of effective charge Z_{eff}e, 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
F_{C} = mv^{2} / r_{n} (4)
This force is supplied by the electrical attraction between the electron and the nucleus, given by Coulomb's law, so
F_{C} = kZ_{eff}e^{2} / r_{n}^{2} (5)
where k = 9 x 10^{9} N m^{2} / C^{2}. Equating (4) and (5) gives us
mv^{2} / r_{n} = kZ_{eff}e^{2} / r_{n}^{2}
If we multiply both sides by r_{n} / 2, we get
(1/2)mv^{2} = kZ_{eff}e^{2} / 2r_{n} (6)
The total energy of the 2p electron is the sum of its kinetic and potential energies,
We are told that the ionization energy of the electron is 8.26 eV, so its energy when bound is E_{2} =  8.26 eV. We can now calculate what Z_{eff} is.
Z_{eff} = {[ (8)(6.63 x 10^{34} J s)^{2}( 8.26 eV)(1.6 x 10^{19} J/eV) / 4p^{2}] /
[(9.11 x 10^{31} kg)(9 x 10^{9} N m^{2} C^{2})^{2}(1.6 x 10^{19} C)^{4}]}^{1/2} = 1.5602
(b) Solve (7) for r_{n}.
r_{n} =  kZ_{eff}e^{2} / 2E_{n} => r_{2} =  kZ_{eff}e^{2} / 2E_{2} =  (9 x 10^{9} N m^{2} 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
Estimate the binding energy of the H_{2} 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(p^{2}/2m) = p^{2}/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
KE_{atoms} = {(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 H_{2} molecule is
KE_{molecule} = {(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 = KE_{molecule}  KE_{atoms} = 13.95 eV  27.19 eV =  13.24 eV
Therefore, the binding energy of the H_{2} molecule is 13.24 eV.
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 =  (1/4pe_{0})e^{2}/r
=  (9 x 10^{9} N m^{2}/C^{2})(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.
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 = Ke_{0} 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 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}. 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 = (1/4pe_{0})e/r^{2}
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 = (1/4pKe_{0})e/r^{2}
The Coulomb force experienced by the extra electron is
F = eE = (1/4pKe_{0})e^{2}/r^{2}
and is equal to the electron mass m times the centripetal acceleration
a_{c} = v^{2}/r
where v is the velocity of the electron in its orbit, so
r_{1} = h^{2}Ke_{0}/pme^{2}
= (6.63 x 10^{34}J s)^{2}(12)(8.85 x 10^{12}C^{2}/N m^{2})/[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,
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 R_{load} = 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?
[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 29.45]
SOLUTION
(a) Let V_{min} and V_{max} be the minimum and maximum values of V_{supply} for which V_{out} is regulated at 130 V.
When V_{supply} = V_{min}, the current through R_{load} is (130 V) / (15.0 x 10^{3} 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, V_{min} = V_{supply} = 130 V + 15.6 V = 145.6 V.
When V_{supply} = V_{max}, the current through R_{load} 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, V_{max} = V_{supply} = 130 V + 195.6 V = 325.6 V.
So 195.6 V < V_{supply}< 325.6 V.
(b) Let (R_{load})_{min} and (R_{load})_{max} be the minimum and maximum values of R_{load} for which V_{out} is regulated at 130 V.
If R_{load} = 0, all of the current goes through R_{load} and none goes through the diode. The potential difference across R_{load} and across the diode is zero. If R_{load} is increased, the potential difference across R_{load} 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 R_{load}, Kirchoff's second rule, which states that the sum of the potential differences around a loop is zero, yields
V_{supply}  IR  IR_{load} = 0 => I = V_{supply}/(R + R_{load})
At this value of R_{load}, the diode switches on in the reverse bias direction and current starts flowing through the diode. As we continue to increase R_{load}, the diode continues to maintain its voltage at 130 V no matter how large R_{load} gets because the current through the diode never reaches its maximum current of 100 mA. The potential difference across R is
V_{R} = V_{supply}  V_{load} = 200 V  130 V = 70 V
The current through R is
I = V_{R}/R = (70 V) / (1800 W) = 38.89 mA
So when R_{load} gets very large, the maximum current through the diode is equal to 38.89 mA, the current through R. We have (R_{load})_{max} = ∞ and 3.343 kW < R_{load} < ∞.
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/m^{3}.
^

 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
E_{H} = v_{d}Bw => v_{d} = E_{H}/Bw (1)
where v_{d} is the electron drift velocity. The current density is
j = nev_{d} = I/A => v_{d} = 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 crosssectional area of the conductor. Equating the expressions for v_{d} in (1) and (2), we get
E_{H}/Bw = I/wtne
Solving for the free electron density, we get
n = IB/teE_{H} = (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 10^{20} electrons/m^{3} (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/m^{3})/(28.21 x 10^{3} kg/mole) = 8.260 x 10^{4} mole/m^{3}
The atomic density of silicon is
N_{A}r/M = (6.022 x 10^{3}atoms/mole)(8.260 x 10^{4} mole/m^{3}) = 4.974 x 10^{28} atoms/m^{3} (4)
where N_{A} = 6.022 x 10^{3}/mole is Avogadro's number. Dividing (3) by (4), the number of conduction band electrons per silicon atom is
n/(N_{A}r/M) = (1.042 x 10^{20}) / (4.974 x 10^{28}) electrons/atom = 2.094 x 10^{9} electrons/atom
(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 Qm_{pr} / (m_{pr}  m_{b}), where Q is the energy corresponding to the difference in total mass between products and reactants, m_{b} is the rest mass of the bombarding particle, and m_{pr} 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 ^{6}C^{13}(p, n)^{7}N^{13} 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 centerofmass 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 centerofmass velocity) in the direction in which the bombarding particle was moving.
Then the situations before and after the collision are as follows:
before collision:
v_{i} o> o target
m_{1} m_{2}
Since the kinetic energy of the incident particle is
K = (1/2)m_{1}v_{i}^{2}
we have
K = m_{pr}Q / (m_{pr}  m_{b})
(b) In the reaction ^{6}C^{13}(p, n)^{7}N^{13}, a proton collides with a ^{6}C^{13} nucleus and produces a ^{7}N^{13} nucleus and a neutron. Equivalently, a ^{1}H^{1} nucleus, which is a proton, collides with a ^{6}C^{13} nucleus and produces a ^{7}N^{13} nucleus and a neutron, or
^{1}H^{1} + ^{6}C^{13} => ^{7}N^{13} + ^{0}n^{1}
Using the definitions 1 = bombarding particle, 2 = target particle, 3 = heavier product particle, and 4 = lighter product particle, we have
where m_{n} is the neutron mass and m_{e} is the electron mass. Plugging in values for m(^{6}C^{13}), m(^{7}N^{13}), and m(^{1}H^{1}), which are atomic masses, m_{n}, and m_{e},
Q = (13.005738 u + 1.008665 u)(931.5 MeV/c^{2}u)c^{2}  (1.007825 u + 13.003355 u)(931.5 MeV/c^{2}u)c^{2} = 3.002 MeV
m_{b} = m_{1} = m(^{1}H^{1})  m_{e} = 1.007825 u – 0.000549 u = 1.007276 u
m_{pr} = m_{3} + m_{4} = m(^{7}N^{13}) – 7m_{e} + m_{n} = 13.005738 u – (7)(0.000549 u) + 1.008665 u = 14.01056 u
Thus the minimum kinetic energy is
K = m_{pr}Q / (m_{pr}  m_{b}) = (14.01056 u)(3.002 MeV) / (14.01056 u – 1.007276 u) = 3.235 MeV
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 10^{8} m/s is the speed of light in a vacuum, and g = 1 / sqrt(1  b^{2}). 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
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: McGrawHill), problem 2.8]
SOLUTION
The velocity transformation equation is
v_{x}' = (v_{x}  u) / (1  uv_{x}/c^{2})
where v_{x} is the velocity in the S frame, v_{x}' 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 v_{x} =  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
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  b^{2}). Then
b = 0.935
g = 1 / sqrt(1  0.935^{2}) = 2.8197
The momentum of each proton in the laboratory is
p = gm_{p}v = gm_{p}bc = (2.8197)(938.3 MeV/c^{2})(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 v_{x} =  0.935c. The velocity measured in S' is
A 2000kg spaceship, initially at rest in an inertial system, is accelerated along a straight line for 1.0 yr (3.1 x 10^{7} s) by means of a photon rocket of 300MW 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 = Dmc^{2}
But the energy expended is equal to the product of the power P and the time t or
E = Pt
Thus,
Pt = Dmc^{2} => Dm = Pt / c^{2} = (300 x 10^{6} W)(3.1 x 10^{7} s) / (3 x 10^{8} 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 10^{6} W)(3.1 x 10^{7} s) / [(2000 kg  0.1033 kg)(3 x 10^{8} m/s)] = 15,500 m/s
Note that the rocket is still nonrelativistic since
b = v/c = (15,500 m/s) / (3 x 10^{8} 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 = v_{avg}t = vt / 2 = (15,500 m/s)(3.1 x 10^{7} t) / 2 = 2.403 x 10^{11} m
(d) The kinetic energy of the spaceship is
T = (1/2)(m  Dm)v^{2} = (1/2)(2000 kg  0.1033 kg)(15,500 m/s)^{2} = 2.403 x 10^{11} J
The fraction of the energy released which goes into the kinetic energy of the spaceship is
f = T / Pt = (2.403 x 10^{11} J) / [(300 x 10^{6} W)(3.1 x 10^{7} s)] = 2.583 x 10^{5}
An unknown particle is measured to have a negative charge and a speed of 2.24 x 10^{8} 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  b^{2}), and b = v/c. We have
b = v/c = (2.24 x 10^{8} m/s) / (3 x 10^{8} m/s) = 0.7467
g = 1 / sqrt(1  0.7467^{2}) = 1.503
The particle's rest mass is
m = p / gv = (3.07 x 10^{22} kg m/s) / [(1.503)(2.24 x 10^{8} m/s)] = 9.117 x 10^{31} kg
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: McGrawHill Book Company), problem 3.8]
SOLUTION
Let M_{1} be the mass of the compound nucleus and v' be its speed. Define
b = v/c
g = 1 / sqrt(1  b^{2})
b' = v'/c
g' = 1 / sqrt(1  b'^{2})
A positive pion can be produced through the reaction p + p => p + n + p^{+} by bombarding protons at rest with highenergy 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 zeromomentum frame.
[from Richtmyer, Kennard, and Cooper 1969, Introduction to Modern Physics, Fifth Edition (New York: McGrawHill), problem 3.18]
SOLUTION
In the centerofmass 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' = m_{p}c^{2} + m_{n}c^{2} + m_{p}c^{2}
In the centerofmass 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 = gm_{p}c^{2} + m_{p}c^{2} = (g + 1)m_{p}c^{2}
The total momentum is the momentum of the proton that was initially moving.
p = gm_{p}bc
Here,
g = 1 / sqrt(1  b^{2})
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, E^{2}  (cp)^{2} is invariant.
Antiprotons are produced by bombarding protons at rest with highenergy protons through the reaction p + p => p + p + (p + p*^{}).
(a) In the centerofmass 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: McGrawHill), problem 3.19]
SOLUTION
(a) In the centerofmass frame, the total momentum is zero, and the total energy E_{tot}' must be sufficient to produce the products at rest. The two protons have equal velocities in opposite directions, so the kinetic energies are equal.
E_{tot}' = 2m_{p}c^{2} + 2T' = 4m_{p}c^{2}
where m_{p} is the rest mass of each proton and antiproton, and T' is the kinetic energy of each initial proton in the centerofmass frame. Solving for T', we get
T' = m_{p}c^{2}
(b) The total energy of each proton in the centerofmass frame is
(c) We make use of the fact that since (cp_{tot}, E_{tot}) is a four vector, whose components are related by the Lorentz transformation, the quantity E_{tot}^{2}  (cp_{tot})^{2} is invariant. Thus,
where the unprimed quantities are measured in the lab frame and the primed quantities are measured in the centerofmass 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.
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.
p_{tot} = gm_{p}v = gm_{p}bc
In the centerofmass frame, the total energy, from above, is
E_{tot}' = 4m_{p}c^{2}
and the total momentum is zero. Substituting into (I), we get
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 Z^{2}, 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 Z^{2}. DE is related to the wavelength of the corresponding photon according to
DE = hc/l => l = hc/DE ~ 1/Z^{2}
where h = 6.63 x 10^{34} J s is Planck's constant, and c = 3 x 10^{8} m/s is the speed of light in a vacuum. Thus, the wavelength is proportional to 1/Z^{2}. 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
+>
l_{0}
^
  wavelength of light emitted by singly ionized helium
  in distant star as observed in star's reference frame
+>
l_{0}/4
^
  wavelength of light emitted by singly ionized
  helium in distant star as observed on Earth
+>
l_{0}
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 10^{8} m/s) = 2.647 x 10^{8} m/s
Sputnik I had a perigee (point of closest approach to the Earth) h_{p} = 227 km above the Earth's surface, at which point its speed was v_{p} = 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: AddisonWesley), problem 3.52]
SOLUTION
The distance from the center of the Earth at perigee is
r_{p} = h_{p} + R_{e} = 227 km + 6370 km = 6597 km
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: AddisonWesley), problem 3.53]
SOLUTION
The Earth's radius is R_{e} = 6370 km. At (perigee, apogee), the distance from the center of the Earth to Explorer I is
(r_{p}, r_{a}) = (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
r_{p} = a(1  e) (I)
and
r_{a} = a(1 + e)
If we add these two equations together, we get
r_{p} + r_{a} = 2a => a = (r_{p} + r_{a})/2 = (6730 km + 8919 km)/2 = 7824.50 km
From the above equation (I), we can solve for the eccentricity e.
r_{p} = a(1  e) => e = 1  r_{p}/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  e^{2}) / (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  e^{2}) / [1 + e cos (p/2)] = a(1  e^{2}) = (7824.50 km)(1  0.139881^{2}) = 7671.40 km
The distance above the Earth's surface when q = p/2 is thus 7671.40 km  6370 km = 1301 km.
A comet is observed a distance of r_{i} = 1.00 x 10^{8} km from the Sun, traveling toward the Sun with a velocity of v_{i} = 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 xaxis through the observed position of the comet. (The mass of the Sun is M_{s} = 2.00 x 10^{30} kg.)
[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: AddisonWesley), problem 3.54]
SOLUTION
The equation of the orbit is of the form
1/r = B + A cos (q  q_{0})
where
A = e / [a(1  e^{2})]
and
B = 1 / [a(1  e^{2})]
To calculate the eccentricity e and the semimajor axis a, we first calculate the energy E, angular momentum L, and constant K = GM_{s}m. The energy is
E = (1/2)mv_{i}^{2}  GM_{s}m/r_{i} => E/m = (1/2)v_{i}^{2}  GM_{s}/r_{i} = (1/2)(51.6 x 10^{3} m/s)^{2}  (6.67 x 10^{11} N m^{2} / kg^{2})(2.00 x 10^{30} kg) / (1.00 x 10^{11} m)
=  2.720 x 10^{6} J/kg
The angular momentum is
L = r x p => L = r_{i}mv_{i} sin 45 => L/m = r_{i}v_{i} sin 45 = (1.00 x 10^{11} m)(51.6 x 10^{3} m/s) / sqrt(2) = 3.649 x 10^{15} m^{2}/s
is
K =  GM_{s}m => K/m =  GM_{s} =  (6.67 x 10^{11} N m^{2}/kg^{2})(2.00 x 10^{30} kg) =  1.334 x 10^{20} N m^{2}/kg
The eccentricity is
e = sqrt[1 + 2EL^{2}/mK^{2}] = sqrt[1 + 2(E/m)(L/m)^{2}/(K/m)^{2}]
= sqrt[1 + (2)( 2.720 x 10^{6} J/kg)(3.649 x 10^{15} m^{2}/s)^{2}/(1.34 x 10^{20} N m^{2}/kg)^{2}] = 0.997963102
The semimajor axis is
a = K/2E = (K/m)/2(E/m) = (1.34 x 10^{20} N m^{2}/kg) / [(2)(2.720 x 10^{6})] = 2.452 x 10^{13} m = (2.452 x 10^{13} m)[(1 AU) / (1.49 x 10^{11} m)] = 164.6 AU
The constants A and B are
A = e / [a(1  e^{2})] = 1 x 10^{11} m^{1}
and
B = 1 / [a(1  e^{2})] = 1.00204 x 10^{11} m^{1}
The reference angle q_{0} = 135 deg.
From the equation for r(q) used in this problem, we can find the perihelion and aphelion distances, r_{p} and r_{a}.
1/r_{p} = B + A = 1.00204 x 10^{11} m^{1} + 1 x 10^{11} m^{1} = 2.00205 x 10^{11} m^{1} => r_{p} = 4.995 x 10^{10} m
1/r_{a} = B  A = 1.00204 x 10^{11} m^{1}  1 x 10^{11} m^{1} = 2.041 x 10^{14} m^{1} => r_{a} = 4.899 x 10^{13} m
The perihelion and aphelion distances can also be calculated from the formulas in terms of a and e.
r_{p} = a(1  e) = (2.452 x 10^{13} m)(1  0.997963102) = 4.995 x 10^{10} m
r_{a} = a(1 + e) = (2.452 x 10^{13} m)(1 + 0.997963102) = 4.899 x 10^{13} m
Mariner 4 left the Earth on an orbit whose perihelion distance from the Sun was approximately the distance of the Earth (1.49 x 10^{8} km), and whose aphelion distance was approximately the distance of Mars from the Sun (2.2 x 10^{8} 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: AddisonWesley), problem 3.59]
SOLUTION
The (perihelion, aphelion) distance of Mariner 4 from the Sun is (r_{p}, r_{a}) = (1.49, 2.2) x 10^{8} km. The semimajor axis is
a = (1/2)(r_{p} + r_{a}) = (1/2)(1.49 x 10^{8} km + 2.2 x 10^{8} km) = 1.845 x 10^{8} 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 =  GM_{s}m (II)
where G is the gravitational constant, M_{s} 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)mv^{2}  GM_{s}m/r_{e} (III)
Kepler's third law for an object orbiting the Sun is of the form
T^{2} = 4p^{2}a^{3}/GM_{s}
where T is the period of the orbit and a is the semimajor axis. From this equation, we see that
GM_{s} = 4p^{2}a^{3}/T^{2}
For the planet Earth, T = (365.25 days)(24 hr/day)(3600 s/hr) = 3.156 x 10^{7} s and a = 1.49 x 10^{11} m. It follows that
GM_{s} = 4p^{2}(1.49 x 10^{11} m)^{3}/(3.156 x 10^{7} s)^{2} = 1.311 x 10^{20} m^{3}/s^{2}
A lunar landing craft approaches the Moon's surface. Assume that onethird 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 onesixth 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: AddisonWesley), problem 4.7]
SOLUTION
Let m_{0} = initial mass of ship, including fuel => m_{0}/3 = initial mass of fuel, 2m_{0}/3 = mass of ship without fuel.
The force of gravity is balanced by the force imparted to the ship by the mass being ejected.
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 D_{max} = 1 inch. The minimum diameter (middle) is D_{min} = 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.
[from Symon, Keith R. 1971, Mechanics, Third Edition (Reading, Massachusetts: AddisonWesley), 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 R_{min} = D_{min}/2 = 0.25" to the outer radius R_{max} = D_{max}/2 = 0.5". The volume element is
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
A_{triangle} = (2R)(2R) sin 60 = 4R^{2}sqrt(3)/2 = [4 sqrt(3)/2]R^{2}
The area of the circles enclosed by the equilateral triangle is
A_{circle} = (3)(pR^{2}/6) = pR^{2}/2
The fraction of the area of the triangle that contains parts of the circles is
f = A_{circle}/A_{triangle} = (pR^{2}/2) / {[4 sqrt(3)/2]R^{2}} = p / [4 sqrt(3)]
The volume of space actually occupied by thread is
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 10^{5} lb/in^{2}, 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: AddisonWesley), 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)
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
b_{1} = b + q = 32.09 deg + 30 deg = 62.09 deg
Let (T_{1}, T_{2}) 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 T_{1} and T_{2} are equal.
T_{1} cos a_{1} = T_{2} cos b_{1}
=> T_{1} = T_{2} (cos b_{1}) / (cos a_{1})
The vertical components of T_{1} and T_{2} balance T.
T_{1} sin a_{1} + T_{2} sin b_{1} = T = W
Substitute for T_{1}.
T_{2} cos b_{1} sin a_{1} / cos a_{1} + T_{2} sin b_{1} = W
T_{2} cos b_{1} tan a_{1} + T_{2} sin b_{1} = W
Solve for T_{2}.
T_{2} = W / (cos b_{1} tan a_{1} + sin b_{1}) = (2000 lb) / (cos 62.09 tan 22.83 + sin 62.09) = 1851 lb
(b)In part (a), we obtained a first approximation. To obtain a second approximation, we assume the values of T_{1} and T_{2} 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
stress_{1} = T_{1}/pr^{2}
and
stress_{2} = T_{2}/pr^{2}
The strains in the cable segments AB and BC are
strain_{1} = Db/b
and
strain_{2} = Dc/c
The ratio between stress and strain is the Young's modulus Y, so
stress_{1}/strain_{1} = stress_{2}/strain_{2} = Y
From the above, we can get estimates for the stretched lengths of cable segments AB and BC.
b' = b(1 + strain_{1}) = b(1 + stress_{1} / Y) = b(1 + T_{1}/pr^{2}Y)
= (8 ft)[1 + (1851 lb)/p(0.25 in)^{2}(5 x 10^{5} lb/in^{2})] = 8.077 ft
c' = c(1 + strain_{2}) = c(1 + stress_{2} / Y) = c(1 + T_{2}/pr^{2}Y)
= (12 ft)[1 + (940 lb)/p(0.25 in)^{2}(5 x 10^{5} lb/in^{2})] = 12.226 ft
Using these values for b' and c', we solve for new values of a_{1}, b_{1}, T_{1}, and T_{2}, using the above formulas. The results are
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) twobody problem: Use centerofmass coordinates and relative coordinates.
(3) manybody problem: Use centerofmass 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, centerofmass, and other types of coordinates. JosephLouis Lagrange (17361813), 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 Nparticle system in threedimensional space, we may define generalized coordinates q_{1}, q_{2}, q_{3},..., q_{3N}. 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,
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.
where T is the kinetic energy of the system, q_{i}' = ∂q_{i}/∂t, and Q_{i} is the generalized force corresponding to generalized coordinate q_{i}, defined by
dW_{i} = Q_{i} dq_{i}
where dW_{i} is the work done on the system if q_{i} is increased by an infinitesimal amount dq_{i}, with all other generalized coordinates held constant.
If the forces Q_{i} can be derived from a potential energy function V that depends on the q_{i} but not on the q_{i}', the Lagrangian function
L = T  V
can be defined, and Lagrange's equations can be written in the form
Consider an Atwood's machine, consisting of two masses, m_{1} and m_{2}, connected by a massless string which passes over a massless and frictionless pulley of radius R.
(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.
x_{1} = vertical distance from pulley axis to m_{1} x_{2} = vertical distance from pulley axis to m_{2}
x_{1} and x_{2} are related by the constraint equation
x_{1} + x_{2} + pR = L (1)
where L is the length of the string.
The kinetic energy of the system is
T = (1/2)m_{1}x_{1}'^{2} + (1/2)m_{2}x_{2}'^{2}
where x_{1}' = dx_{1}/dt and x_{2}' = dx_{2}/dt.
The generalized force Q_{1} associated with x_{1} is obtained by considering the work dW_{1} done on m_{1} if x_{1} is increased by an infinitesimal amount dx_{1} while keeping x_{2} constant. The work is done by the force of gravity m_{1}g and the force of the tension t. Thus,
dW_{1} = (m_{1}g  t) dx_{1} = Q_{1} dx_{1}
where
Q_{1} = m_{1}g  t
Similarly, the generalized force Q_{2} associated with x_{2} is
(c) To solve the problem using Newtonian mechanics, we write Newton's first law, F = ma, for each of the two masses. Assuming that m_{1} accelerates downward and m_{2} accelerates upward with the same acceleration a, we have
What are the amplitudes of the various frequencies present in halfwave 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) = B_{0}/2 + S_{n=1}^{∞} (A_{n} sin nwt + B_{n} cos nwt)
where
w = 2p/T
A_{n} = (w/p) ∫_{0}^{T} f(t) sin nwt dt (1)
B_{n} = (w/p) ∫_{0}^{T} 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),
A_{n} = (w/p) ∫_{0}^{T/2} sin wt sin nwt dt
To calculate the A_{n}, 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)
The onedimensional, timedependent Schrodinger wave equation, for a particle of mass m moving in a region where the potential energy is given by V(x), is
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
(h^{2}/2m)(d^{2}y/dx^{2})/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
(h^{2}/2m)d^{2}y/dx^{2} + Vy = Ey
and
(h/i)df/dt = Ef
The first of these is the onedimensional, timeindependent Schrodinger equation. The second can be solved for f.
Y(x,t) = A exp[i(kx  Et/h)] + B exp[i(kx + Et/h)] (I)
The phase of the first term is
q_{1} = kx  Et/h
The time derivative of this is
dq_{1}/dt = k dx/dt  E/h
In order for the phase to remain constant,
dq_{1}/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
The reaction ^{1}H^{1} + ^{3}Li^{7} => ^{4}Be^{7} + ^{0}n^{1} 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 ^{3}Li^{7} 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 M_{1} whose velocity makes an angle f with the velocity of the incident particle, and a second particle of mass M_{2} whose velocity makes an angle q with the velocity of the incident particle.
^{v}1
O ^{M}1
/
/
m M / f
o>O
v \ q
\
\
o M_{2} ^{v}2
If we assume the particle speeds are all nonrelativistic, then we can define
as the kinetic energies of m, M_{1}, and M_{2}. The Q value of the reaction is the change in the total kinetic energy of the particles. Thus,
Q = K_{1} + K_{2}  K_{m} (1)
By conservation of momentum, we have
mv = M_{1}v_{1} cos f + M_{2}v_{2} cos q
=> M_{1}v_{1} cos f = mv  M_{2}v_{2} cos q (2)
0 = M_{1}v_{1} sin f  M_{2}v_{2} sin q
=> M_{1}v_{1} sin f = M_{2}v_{2} sin q (3)
Squaring (2) and (3), and adding the results, we get
(M_{1}v_{1})^{2} = (mv)^{2} + (M_{2}v_{2})^{2}  2mvM_{2}v_{2} cos q (4)