MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 6: ROTATION (PART 1)
WORKSHEET SOLUTIONS
1. A point on a wheel rotating at 6 rev/s and located 0.3 m from the axis has what tangential velocity?
1A.
(1)  = (6 rev/s)(2 rad/rev) = 37.7 rad/s
(2) v = r
(3) v = (0.3 m)(37.7 rad/s)
(4) v = 11.3 m/s
2. A point on a wheel rotating at 5 rev/s and located 0.2 m from the axis experiences what centripetal
acceleration?
2A.
(1)  = (5 rev / s)(2 rad / rev) = 31.42 rad/s
(2) v = r
(3) v = (0.2 m)(31.4 rad/s)
(4) v = 6.28 m/s
(5) ac = v2/r
(6) ac = (6.28 m/s)2 / 0.2 m
(7) ac = 197.4 m/s2
3. A Ferris wheel, rotating initially at an angular velocity of 0.65 rad/s, accelerates over a 7 s interval at
a rate of 0.03 rad/s2. What angular displacement does the Ferris wheel undergo?
3A.
(1)  = ot + ½ t2
(2)  = (0.65 rad/s)(7 s) + ½ (0.03 rad/s2)(7 s)2
(3)  = 5.3 rad
4. A point on the rim of a 0.15 m radius rotating wheel has a centripetal acceleration of 7.0 m/s2. What
is the angular velocity of the wheel?
4A.
(1) ac = v2/r
(2) v2 = rac
(3) v = (rac)1/2
(4) v = [(0.15 m)(7.0 m/s2)]1/2
(5) v = 1.02 m/s
(6) v = r
(7)  = v/r
(8)  = 1.02 m/s / 0.15 m
(9)  = 6.83 rad/s
PHYSICS
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
5. A Ferris wheel, starting at rest, builds up to a final angular velocity of 0.92 rad/s while rotating
through an angular displacement of 3.5 rad. What is its average angular acceleration?
5A.
(1) 2 = o2 + 2
(2) 2 = 0 + 2
(3) 2 = 2
(4)  = 2 / 2
(5)  = (0.92 rad/s)2 / 2(3.5 rad)
(6)  = 0.121 rad/s2
6. What angular velocity (in revs/sec) is needed for a centrifuge to produce an acceleration of 750 g on a
radius arm of 9 cm?
6A.
(1) ac = v2/r
(2) v2 = rac
(3) v = (rac)1/2
(4) v = [(0.09 m)(750)(9.8 m/s2)]1/2
(5) v = 25.7 m/s
(6) v = r
(7)  = v/r
(8)  = (25.7 m/s)/(0.09 m)
(9)  = 285.8 rad/s
(10) (285.8 rad / 1 sec)(1 rev / 2 rad)
(11)  = 45.5 rev/s
7. A 0.15 kg mass, attached to the end of a 0.8 m string, is whirled around in a circular horizontal path.
If the maximum tension that the string can withstand is 315 N, then what maximum velocity can the
mass have if the string is not to break?
7A.
(1) Fc = mv2/r
(2) v2 = rFc/m
(3) v = (rFc/m)1/2
(4) v = [(0.8 m)(315 N)/(0.15 kg)]1/2
(5) v = 41.0 m/s
8. A roller coaster, loaded with passengers, has a mass of 700 kg; the radius of curvature of the track at
the bottom point of the dip is 14 m. If the vehicle has a speed of 17 m/s at this point, what force is
exerted on the vehicle by the track?
8A.
The track must support the weight of the coaster and the centripetal force of its motion:
(1) Track = weight + Fc
(2) Track = mg + mv2/r
(3) (700 kg)(9.8 m/s2) + (700kg)(17m/s)2 / 14 m
(4) Track = 6860 N + 14,450 N
(5) Track = 21,310 N
PHYSICS
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. Consider a point on a bicycle tire that is momentarily in contact with the ground as the bicycle rolls
across the ground with constant speed. The direction for the acceleration for this point at that moment
is:
(A) upward
(B) down toward the ground
(C) forward
(D) backward
(E) the acceleration is zero
9A. (A) ac is always directed towards the axis of rotation. In this case, it is upward.
10. A car rounds a circular turn of radius r = 75 m in a horizontal (unbanked) road. A rear view of the
car is given below. The coefficient of static friction between tires and road is 0.60. The car’s weight mg
is 1.33 x 104 N.
10a. What is the direction of the car’s acceleration?
A. Towards the center of the circular path (left).
10b. What force is available to maintain this acceleration?
A. Static friction from the tires. The vertical forces cancel leaving only static friction which acts
parallel to the surface to the left.
10c. What is the magnitude of the static force of friction when the car’s speed v = 20 m/s?
A.
(1) Fc = mv2/r
(2) w = mg
(3) m = w/g
(4) m = (1.33 x 104 N) / (9.8 m/s2)
(5) m = 1357 kg
(6) Fc = (1357 kg)(20 m/s)2 / (75 m)
(7) Fc = 7238 N
10d. What is the maximum value of the static frictional force?
A.
(1) Fc (max): f = n
(2) f = w
(3) f = (0.6)(1.33 x 104 N)
(4) f = 7980 N
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
10e. Determine the maximum speed for the car to make it around the curve.
A.
(1) Fc = mv2/r
(2) mv2 = rFc
(3) v2 = (rFc) / m
(4) v = (rFc /m)1/2
(5) v = [(75 m)(7980 N) / (1357 kg)]1/2
(6) v = 21 m/s
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 6: ROTATION
QUIZ SOLUTIONS
1. A point on a wheel rotating at 8 rev/s and located 0.3 m from the axis has what tangential velocity?
1A.
(1)  = (8 rev/s)(2 rad/rev) = 50.24 rad/s
(2) v = r
(3) v = (0.3 m)(50.24 rad/s)
(4) v = 15.1 m/s
2. A point on a wheel rotating at 6 rev/s and located 0.4 m from the axis experiences what centripetal
acceleration?
2A.
(1)  = (6 rev/s)(2 rad/rev) = 37.68 rad/s
(2) v = r
(3) v = (0.4 m)(37.68 rad/s) = 15.07 m/s
(4) ac = v2/r
(5) ac = (15.07 m/s)2 / 0.4 m
(6) ac = 567.9 m/s2
3. A Ferris wheel, rotating initially at an angular velocity of 0.35 rad/s, accelerates over a 5 s interval at
a rate of 0.05 rad/s2. What angular displacement does the Ferris wheel undergo in this 5 s interval?
3A.
(1)  = ot + ½ t2
(2)  = (0.35 rad/s)(5 s) + ½ (0.05 rad/s2)(5 s)2
(3)  = 2.4 rad
4. A point on the rim of a 0.27 m radius rotating wheel has a centripetal acceleration of 4.0 m/s2. What
is the angular velocity of the wheel?
4A.
(1) ac = v2/r
(2) v2 = rac
(3) v = (rac)1/2
(4) v = [(0.27 m)(4.0 m/s2)]1/2
(5) v = 1.04 m/s
(6) v = r
(7)  = v / r
(8)  = 1.039 m/s / 0.27 m
(9)  = 3.8 rad/s
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
5. A Ferris wheel, starting at rest, builds up to a final angular velocity of 0.79 rad/s while rotating
through an angular displacement of 3.5 rad. What is its average angular acceleration?
5A.
(1) 2 = o2 + 2
(2) 2 = 0 + 2
(3) 2 = 2
(4)  = 2 / 2
(5)  = (0.79 rad/s)2 / 2(3.5 rad)
(6)  = 8.9 x 10-2 rad/s2
6. What angular velocity (in revolutions / second) is needed for a centrifuge to produce an acceleration
of 950 g (950 times the force of gravity) on a radius arm of 12 cm?
6A.
(1) ac = v2/r
(2) v2 = rac
(3) v = (rac)1/2
(4) v = [(0.12 m)(950)(9.8 m/s2)]1/2
(5) v = 33.42 m/s
(6) v = r
(7)  = v/r
(8)  = (33.42 m/s)/(0.12 m)
(9)  = 278.5 rad/s
(10) (278.5 rad / 1 sec)(1 rev / 2 rad)
(11)  = 44.4 rev/s
7. A 0.34 kg mass, attached to the end of a 1.15 m string, is whirled around in a circular horizontal path.
If the maximum tension that the string can withstand is 250 N, then what maximum velocity can the
mass have if the string is not to break?
7A.
(1) Fc = mv2/r
(2) v2 = rFc/m
(3) v = (rFc/m)1/2
(4) v = [(1.15 m)(250 N)/(0.34 kg)]1/2
(5) v = 29.1 m/s
8. A roller coaster, loaded with passengers, has a mass of 614 kg; the radius of curvature of the track at
the bottom point of the dip is 17 m. If the vehicle has a speed of 21 m/s at this point, what force is
exerted on the vehicle by the track?
8A.
(1) The track must support the weight of the coaster and the centripetal force of its motion:
(2) Track = train weight + train Fc
(3) Track = mg + mv2/r
(4) Track = (614 kg)(9.8 m/s2) +
(614 kg)(21 m/s)2 / 17 m
(5) Track = 6017 N + 15,928 N
(6) Track = 21,945 N
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. A car rounds a circular turn of radius r = 74 m in a horizontal (unbanked) road. A rear view of the
car is given below. The coefficient of static friction between tires and road is 0.61. The car’s weight mg
is 1.28 x 104 N.
9a. What is the magnitude of the static force of friction when the car’s speed v = 20 m/s?
A.
(1) Fc = mv2/r
(2) w = mg
(3) m = w/g
(4) m = (1.28 x 104 N)/(9.8 m/s2)
(5) m = 1306 kg
(6) Fc = (1306 kg)(20 m/s)2 / (74 m)
(7) Fc = 7060 N
9b. What is the maximum value of the static frictional force?
A.
(1) Fc (max): f = n
(2) f = w
(3) f = (0.61)(1.28 x 104 N)
(4) f = 7808 N
9c.
A.
(1)
(2)
(3)
(4)
(5)
(6)
Determine the maximum speed for the car to make it around the curve.
Fc = mv2/r
mv2 = rFc
v2 = (rFc) / m
v = (rFc /m)1/2
v = [(74 m)(7808 N) / (1306.1 kg)]1/2
v = 21.0 m/s
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