MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 7: ROTATION (PART 2)
TORQUE WORKSHEET SOLUTIONS
1. A majorette takes two batons and fastens them together in the middle at right angles to make an “x”
shape. Each baton is 0.65 m long and each ball on the end is 0.2 kg. (Ignore the mass of the rods.)
What is the moment of inertia if the arrangement is spun around an axis formed by one of the batons?
1A.
(1) I = mr2
Since there are two batons:
(2) I = 2 [(0.2 kg)(0.325 m)2]
(3) I = 4.23 x 10-2 kg.m2
2. A solid cylinder (I = ½ mr2), a hoop (I = mr2), and a sphere (I = 2/5 mr2) are released and roll down a
slope. The mass of all objects is the same, but the cylinder has twice the radius of the other two. Which
reaches the bottom first?
2A.
The sphere because it has the lowest moment of inertia I.
Questions 3 - 6. A pulley that rotates about a fixed axis O has string wrapped around it and attached to a
mass m that hangs vertically as shown. T is the tension in the string.
3. According to the diagram, what is the net torque on the pulley about O?

 = TR
4. Choose the rotational equation that applies:
(A) TR = I
(B) T = I
(C) T/R = I
(D) T = I
4A.
(1)  = F┴r
(2)  = I
(3) F┴r = I
(4) TR = I(from 1a. )
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
5. According to the diagram, what is the net downward force on mass m?
5A.
(1) FNET = ma
(2) T – mg = - ma
(3) -1 (T – mg) = -1 (- ma)
(4) mg – T = ma
Fy: mg – T
6. Express Fy in terms of torque.
6A.
(1) Fy: mg – T
(2)  = TR
(3) T =  / R
(4) Fy: mg – /R
Questions 7 - 9. Two particles move in opposite directions about their center of mass, a point halfway
between them. The CM is 4 m from each particle.
7. Compute the angular momentum of one 2 kg particle about the axis through the CM in kg.m2/s units.
Take counterclockwise as positive.
7A.
(1) L = I
(2) Find the value of I: I = mr2
(3) I = (2 kg)(4 m)2
(4) I = 32 kg.m2
(5) Find the value of : v = R
(6)  = v / R
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7A. (continued…)
(7)  = (- 15 m / s) / (4 m)
(8)  = - 3.75 rad /s
(9) Substitute I and  into L = I:
L = (32 kg.m2)(- 3.75 rad /s)
(10) L = - 120 kg.m2/s
8. Compute the angular momentum of the other 2 kg particle about the axis through the CM in kg.m2/s
units.
8A.
L = - 120 kg.m2/s (same steps as 2a.)
9. Compute the total angular momentum of both particles about the axis through the CM in kg.m2/s
units.
9A.
(1) LT = L1 + L2
(2) LT = - 120 kg.m2/s – 120 kg.m2/s
(3) LT = - 240 kg.m2/s
Questions 10 - 13. A child of mass 60 kg runs to the left at 5 m/s and jumps on the rim of a merry-goround that is rotating clockwise at 4 rad/s. The radius R is 3 m and the moment of inertia about the
center is 400 kg.m2.
10. Let counterclockwise momentum be positive. What is the running girl’s angular momentum about
O in units of kg.m2/s?
10A.
(1) L = mvr
(2) L = (60 kg)(300 kg.m/s)(3 m) (clockwise direction)
(3) L = - 900 kg.m2/s
11. What is the merry-go-round’s angular momentum about O before the girl jumps on it?
11A.
(1) L = I
(2) L = (400 kg.m2)(- 4 rad/s)
(3) L = - 1600 kg.m2/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
12. What is the total angular momentum of the system of girl plus merry-go-round?
12A.
(1) L = LM + LG
(2) L = - 1600 kg.m2/s – 900 kg.m2/s
(3) L = - 2500 kg.m2/s
13. Compute the moment of inertia of the whole system with the girl on the rim in kg.m2 units.
13A.
(1) IT = IM + IG
(2) IG = mR2
(3) IG = (60 kg)(3 m)2
(4) IG = 540 kg.m2
(5) IT = (400 kg.m2) + (540 kg.m2)
(6) IT = 940 kg.m2
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 7: ROTATION (PART 2)
ROTATIONAL DYNAMICS WORKSHEET SOLUTIONS
1. A phonograph turntable has a moment of inertia of 3.15 x 10-2 kg.m2 and spins freely on a frictionless
bearing at 33.3 rev/min. A 0.17 kg ball of putty is dropped vertically on the turntable and sticks at a
point 0.12 m from the center. What is the new rate of rotation?
1A.
(1) I11 = I1’1’
(2) 1 = (33.3 rev / min)(1 min / 60 sec)(2 rad / 1 rev)
(3) 1 = 3.49 rad/s
(4) I1’ = ITT + IP
(5) I1’ = ITT + mr2
(6) 1’ = (II)(I)/(IF)
(7) 1’ = [(3.15 x 10-2 kg.m2)(3.49 rad/s)] / [(3.15 x 10-2 kg.m2) + (0.17 kg)(0.12 m)2]
(8) 1’ = 3.24 rad/s
(9) 1’ = (3.24 rad/s) (60 s / 1 minute)(1 rev / 2)
(10) 1’ = 30.9 rev/min
2. A solid sphere with mass M and radius R rolls along a level surface without slipping with a linear
velocity, v. What is the ratio of its rotational kinetic energy to its linear kinetic energy? (For a sphere,
I = 0.4 mr2.)
2A.
(1) KR = ½ I2
(2) KR = ½ (2/5 MR2)(v/R)2
(3) KR = 2/10 Mv2
(4) KR = 1/5 Mv2 (rotational kinetic)
(5) K = ½ Mv2 (translational kinetic)
(6) KR ~ K = 1/5 ~ 1/2
(7) 0.2 ~ 0.5
(8) KR ~ K = 2 : 5
3. A bus is designed to draw its power from a rotating flywheel that is brought up to its maximum speed
(4250 rpm) by an electric motor. The flywheel is a solid cylinder of mass 800 kg and diameter 0.9 m
(ICYLINDER = ½ mr2). If the bus requires an average power of 10 kilowatts, how long will the flywheel
rotate?
3A.
(1) KR = ½ I2
(2) KR = ½ (1/2 mr2)2
(3) KR = ¼ mr22
(4)  = (4250 rev/min)(2 rad / rev)(1 min / 60 s)
(5)  = 445 rad / s
(6) KR = ¼ (800 kg)(0.45 m)2(445 rad/s)2
(7) KR = 8.02 x 106 J
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3A. (continued…)
(8) P = W / t
(9) tP = W
(10) t = W / P
(11) t = 8.02 x 106 J / 1.0 x 104 watts
(12) t = 802 s
4. A solid sphere with a mass of 3.0 kg and radius of 0.2 m, starts from rest at the top of a ramp,
inclined at 15o, and rolls to the bottom. The upper end of the ramp is 1.2 m higher than the lower end.
What is the linear velocity of the sphere when it reaches the bottom of the ramp? (I = 0.4 MR2 for a
solid sphere)
4A.
(1) UK = K + KR
(2) UK = ½ Mv2 + ½ I2
(3) UK = ½ Mv2 + ½ (2/5 MR2)(v/R)2
(4) UK = ½ Mv2 + ½ (2/5 Mv2)
(5) UK = ½ Mv2 + 2/10 Mv2
(6) UK = 7/10 Mv2
(7) Ug = Mgh
(8) Ug = UK
(9) Mgh = 7/10 Mv2
(10) gh = 7/10 v2
(11) v2 = 10/7 gh
(12) v = (10/7 gh)1/2
(13) v = [10/7 (9.8 m/s2)(1.2 m)]1/2
(14) v = 4.10 m/s
GRAVITY PROBLEMS
5. Two lead balls of mass 12 kg and 16 kg are 2.6 m apart.
5a. Calculate the magnitude of the gravitational force they exert on one another treating the balls as
point masses.
A.
(1) FG = Gm1m2/r2
(2) FG = (6.67 x 10-11 m3/kg.s2)(12 kg)(16 kg) / (2.6 m)2
(3) FG = 1.89 x 10-9 N
5b. If the two balls are moved closer together, to a separation 1.3 m which is half the original, what
happens to the magnitude of their gravitational attraction?
A.
(1) FG = Gm1m2/r2
(2) FG = (6.67 x 10-11 m3/kg.s2)(12 kg)(16 kg) / (1.3 m)2
(3) FG = 7.58 x 10-9 N
(4) It is 4 times larger
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
6. An artificial satellite of mass m travels at a constant speed in a circular orbit of radius r around the
Earth mass M. What is the speed of the satellite?
6A.
(1) FG = Gm1m2/r2
(2) FG = GmM/r2
(3) Fc = mv2/r
(4) Set Fc to FG: GmM/r2 = mv2/r
(5) (r)GmM/r2 = (r)mv2/r
(6) GmM/r = mv2
(7) GM/r = v2
(8) v2 = GM/r
(9) v = (GM/r)1/2
7. A satellite of mass m is in a circular orbit of radius R around the earth (mass M). What is its total
mechanical energy?
7A.
(1) FC = mv2 / r
(2) FG = - GmM / r2
(3) FC = FG
(4) mv2 / r = - GmM /r2
(5) r [ mv2 / r = - GmM / r2]
(6) mv2 = - GmM / r
(7) K = ½ mv2
(8) ½ [mv2 = - GmM / r]
(9) ½ mv2 = - GmM / 2r
(10) K = | - GmM / r | (kinetic energy is scalar and positive)
(11) K = GmM / 2r
(12) E = K + U
(13) E = GmM /2r – GmM / r
(14) E = - GmM / 2r
Questions 8 - 9. A uniform 10 m long plank weighing 300 N is supported by two blocks, A and B, as
shown:
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
8. What is the upward force supplied by block A?
8A.
Solve in two steps.
Step 1:
(1) F = 0
(2) FUP = FDOWN
(3) FA + FB = 300 N
Step 2:
(4)  = 0
(5) CW = CCW
Arbitrarily choose block B as a pivot:
(6) (FA)(8 m) = (FBOARD)(5 m)
(7) (FA)(8 m) = (300 N)(5 m)
(8) FA = (300 N)(5 m) / (8 m)
(9) FA = 187.5 N
9. A 14 N can of paint is placed on the plank directly over block B. What is the upward force supplied
by block B?
9A.
Before the paint can was added, the upward force supplied by block B was previously FB = 300 N –
187.5 N = 112.5 N. Adding the can over block B supplies no additional torque if we still use B as our
pivot. Therefore, now FB = 112.5 N + 14 N = 126.5 N.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 7: ROTATION (PART 2)
QUIZ SOLUTIONS
1. A majorette takes two batons and fastens them together in the middle at right angles to make an “x”
shape. Each baton is 0.7 m long and each ball on the end is 0.22 kg. (Ignore the mass of the rods.)
What is the moment of inertia if the arrangement is spun around an axis formed by one of the batons?
1A.
(1) I = mr2
(2) I = 2 [(0.22 kg)(0.35 m)2]
(3) I = 5.4 x 10-2 kg.m2
Questions 2 - 5. A child of mass 60 kg runs to the left at 5 m/s and jumps on the rim of a merry-goround that is rotating clockwise at 3.2 rad/s. The radius R is 3.7 m and the moment of inertia about the
center is 500 kg.m2. The figure on the right shows the child hanging on the rim and the whole system
rotating as one.
Question 2.
2. Let counterclockwise momentum be positive. What is the running girl’s angular momentum about
O in units of kg.m2/s?
2A.
(1) L = mvr
(2) L = (60 kg)(5 m/s)(3.7 m) (the girl adds clockwise momentum)
(3) L = - 1110 kg.m2/s
3. What is the merry-go-round’s angular momentum about O before the girl jumps on it (in units of
kg.m2/s)?
3A.
(1) L = I
(2) L = (500 kg.m2)(- 3.2 rad/s)
(3) L = - 1600 kg.m2/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4. What is the total angular momentum of the system of girl plus merry-go-round in units of kg.m2/s?
4A.
(1) LTOTAL = LMGR + LGIRL
(2) L = - 1600 kg.m2/s – 1110 kg.m2/s
(3) L = - 2710 kg.m2/s
5. Compute the moment of inertia of the whole system with the girl on the rim in kg.m2 units (see the
right hand figure above).
5A.
(1) ITOTAL = IMGR + IGIRL
(2) IGIRL = mR2
(3) IGIRL = (60 kg)(3.7 m)2
(4) IGIRL = 821.4 kg.m2
(5) I = (500 kg.m2) + (821.4 kg.m2)
(6) I = 1321.4 kg.m2
6. A bus is designed to draw its power from a rotating flywheel that is brought up to its maximum speed
(3750 rpm) by an electric motor. The flywheel is a solid cylinder of mass 1120 kg and diameter 0.9 m
(ICYLINDER = ½ mr2). If the bus requires an average power of 10 kilowatts, how long will the flywheel
rotate?
6A.
(1) KR = ½ I2
(2) KR = ½ (1/2 mr2)2
(3) KR = ¼ mr22
(4) Solve for :
(3750 rev/min)(2 rad / rev)(1 min / 60 s)
(5)  = 392.5 rad / s
(6) KR = ¼ (1120 kg)(0.45 m)2(392.5 rad/s)2
(7) KR = 8.735 x 106 J
(8) P = KR / t
(9) tP = KR
(10) t = KR / P
(11) t = 8.735 x 106 J / 1.0 x 104 watts
(12) t = 874 s
7. A solid sphere with a mass of 7.2 kg and radius of 0.24 m, starts from rest at the top of a ramp,
inclined at 18o, and rolls to the bottom. The upper end of the ramp is 1.4 m higher than the lower end.
What is the linear velocity of the sphere when it reaches the bottom of the ramp? (I = 0.4 MR2 for a
solid sphere and g = 9.8 m/s2.)
7A.
(1) U = K + KR
(2) U = ½ Mv2 + ½ I2
(3) U = ½ Mv2 + ½ (2/5 MR2)(v/R)2
(4) U = ½ Mv2 + ½ (2/5 Mv2)
(5) U = ½ Mv2 + 2/10 Mv2
(6) U = 7/10 Mv2
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7A. (continued)
(7) Mgh = 7/10 Mv2
(8) gh = 7/10 v2
(9) 7/10 v2 = gh
(10) v2 = 10/7 gh
(11) v = (10/7 gh)1/2
(12) v = [10/7 (9.8 m/s2)(1.4 m)]1/2
(13) v = 4.43 m/s
Questions 8 - 9. Two lead balls of mass 3 kg and 9 kg are 7.42 m apart.
8. Calculate the magnitude of the gravitational force they exert on one another treating the balls as point
masses.
8A.
(1) FG = Gm1m2/r2
(2) FG = (6.67 x 10-11 m3/kg.s2)(3 kg)(9 kg) / (7.42 m)2
(3) FG = 3.27 x 10-11 N
9. By what factor does the gravitational force increase if the two balls are moved closer together to a
3.71 m separation?
9A.
(1) FG = Gm1m2/r2
(2) FG = (6.67 x 10-11 m3/kg.s2)(3 kg)(9 kg) / (3.71 m)2
(3) FG = 1.31 x 10-10 N
(4) It is 4 times larger
10. An artificial satellite of mass m travels at a constant speed in a circular orbit of radius r around the
Earth mass M. What is the speed of the satellite?
10A.
(1) FG = Gm1m2/r2
(2) FG = GmM/r2
(3) Fc = mv2/r
(4) Set Fc to FG: GmM/r2 = mv2/r
(5) Multiply both sides by r: GmM/r = mv2
(6) Divide both sides by m: GM/r = v2
(7) v2 = GM/r
(8) v = (GM/r)1/2
11. A satellite of mass m is in a circular orbit of radius R around the earth (mass M). What is its total
mechanical energy?
11A.
(1) FC = mv2 / r
(2) FG = - GmM / r2
(3) FC = FG
(4) mv2 / r = - GmM /r2
(5) r [ mv2 / r = - GmM / r2]
(6) mv2 = - GmM / r
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
11A. (continued. . .)
(7) K = ½ mv2
(8) ½ [mv2 = - GmM / r]
(9) ½ mv2 = - GmM / 2r
(10) K = | - GmM / r | (kinetic energy is scalar and positive)
(11) K = GmM / 2r
(12) E = K + U
(13) E = GmM /2r – GmM / r
(14) E = - GmM / 2r
Questions 12 - 13. A uniform 10 m long plank weighing 200 N is supported by two blocks, A and B, as
shown:
12. What is the upward force supplied by block A?
12A.
Solve in two steps.
Step 1:
(1) F = 0
(2) FUP = FDOWN
(3) FA + FB = 200 N
Step 2:
(4)  = 0
(5) CW = CCW
Arbitrarily choose block B as a pivot:
(6) (FA)(8 m) = (FBOARD)(5 m)
(7) (FA)(8 m) = (200 N)(5 m)
(8) FA = (200 N)(5 m) / (8 m)
(9) FA = 125 N
13. A 12 N can of paint is placed on the plank directly over block B. What is the upward force supplied
by block B?
13A.
Before the paint can was added, the upward force supplied by block B was previously FB = 200 N – 125
N = 75 N. Adding the can over block B supplies no additional torque if we still use B as our pivot.
Therefore, now FB = 75 N + 12 N = 87 N.
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PHYSICS