Questions per topic
Course Name:
General Physics
Course Code:
PHY60
Grade:
11
Term:
1
Periods per
week
4
Topic 4
Rotational Motion
Subtopic4.1:
Subtopic4.2:
Subtopic4.3:
Describing Rotational Motion
Rotational Dynamics
Equilibrium
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 1 of 14
1. A bicycle has a wheel of radius 68 𝑐𝑚. Find the distance travelled by the bicycle as the wheel makes
65 revolutions.
𝜃 = 65 × 2𝜋 = 408 𝑟𝑎𝑑
𝑠 = 𝑟𝜃 = 0.68 × 408 = 277 𝑚
2. The wheel turns through angles from 1.2 𝑟𝑎𝑑 to 5.2 𝑟𝑎𝑑 during a time 𝑡 = 1.04 𝑠, calculate the
angular velocity of the wheel.
𝜔𝑎𝑣 =
∆𝜃 5.2 − 1.2 38.6
=
=
= 3.85 𝑟𝑎𝑑⁄𝑠
∆𝑡
1.04
10.4
3. One revolution per minute is equivalent to ______.
√
A.
B.
C.
D.
E.
Making a full rotation in one second
Making a full rotation in one minute
Moving in uniform circular motion for one minute
Moving in uniform circular motion for one second
Rotating such that all the points have the same linear speed
4. Points A, B, C, and D are moving with the same angular speed 𝜔. Which point(s) has/have the
largest linear speed?
A
√
A.
B.
C.
D.
E.
𝝎
B
C
D
Point A
Point B
Point C
Point D
All the points have the same linear speed
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 2 of 14
5. The SI unit of angular acceleration is ____________
√
A.
B.
C.
D.
E.
𝑟𝑎𝑑𝑖𝑎𝑛
𝑟𝑎𝑑𝑖𝑎𝑛/𝑠𝑒𝑐𝑜𝑛𝑑
𝑟𝑎𝑑𝑖𝑎𝑛/𝑠𝑒𝑐𝑜𝑛𝑑 2
𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑
𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑 2
6. A car wheel turns from rest to 4.75 𝑟𝑎𝑑⁄𝑠 in 2.25 𝑠. Calculate the angular acceleration of the wheel.
√
2.11 𝑟𝑎𝑑/ 𝑠 2
3.20 𝑟𝑎𝑑/ 𝑠 2
4.52 𝑟𝑎𝑑/ 𝑠 2
5.25 𝑟𝑎𝑑/𝑠 2
A.
B.
C.
D.
7. The angular acceleration of an object, 𝛼, is given by:
√
A.
B.
C.
D.
E.
𝛼
𝜔
𝑑
𝑣
𝛼
= ∆𝜔/∆𝑡
= ∆𝜃/∆𝑡
= 𝑟𝜃
= 𝑟𝜔
= 𝑣𝑟𝜔
8. The angular acceleration is ____________
√
A.
B.
C.
D.
E.
directly proportional to time
directly proportional to linear speed
directly proportional to linear acceleration
inversely proportional to linear speed
inversely proportional to linear acceleration
9. Fill in the blanks in the sentences below by choosing one of the phrases below. with one of the
following options:
smaller than
equal to
greater than
smaller than
In the figure above: the linear speed of lady bug 1 is ______________________
the linear speed of lady
bug 2. Moreover, the angular speed of lady bug 1 is ________________________the
angular speed of
equal to
ladybug 2.
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 3 of 14
10. The wheels of a car initially rotating with an angular speed of 3.0 𝑟𝑎𝑑/𝑠 increase its speed to
15 𝑟𝑎𝑑/𝑠 in 6.0 𝑠𝑒𝑐𝑜𝑛𝑑𝑠. What is the angular acceleration of the wheels?
𝜔𝑓 − 𝜔𝑖 15 − 3
𝛼=
=
= 2.0 𝑟𝑎𝑑/𝑠 2
𝑡
6
11. A wheel accelerates from rest at a rate of 4.0 𝑟𝑎𝑑/𝑠 2 . How long does the wheel take to reach a final
speed of 185 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠/𝑚𝑖𝑛𝑢𝑡𝑒?
𝜔=
185 × 2𝜋
= 19.4 𝑟𝑎𝑑/𝑠
60 𝑠
𝛼𝑎𝑣 =
𝑡=
∆𝜔
∆𝑡
𝜔𝑓 − 𝜔𝑖 19.4 − 0
=
= 4.85 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝛼
4
12. A chainsaw is started by pulling on a rope that is attached to a solid steel flywheel It starts at rest and
must accelerate to 25.12 rad/s in 2.00 seconds in order to start the motor. What is its angular
acceleration?
∆𝜔 25.12
=
= 12.6 𝑟𝑎𝑑⁄𝑠 2
𝑡
2
13. A grindstone has a constant angular acceleration of 0.27 rad/s 2. If the wheel has an initial
velocity of 3.3 rad/s, at what point in time will it momentarily come to rest?
𝛼=
𝛼𝑎𝑣 =
∆𝜔
∆𝑡
t = ( o)/ 
t = [0 (-3.3)]/ 0.27
t = 12 s
14. TRUE OR FALSE
• Radians indicate the ratio between distance traveled along a circumference and the radius.
T
•
Angular velocity is the angular displacement divided by time.
T
15. The relationship between linear displacement and angular displacement is given by:
a.
v = d/t c.
v = r
b.
a = r d.
d = r
D
16. The relationship between linear velocity and angular velocity is given by:
a.
v = d/t c.
v = r
b.
a = r d.
d = r
C
17. The relationship between linear acceleration and angular acceleration is given by:
a.
v = d/t c.
v = r
b.
a = r d.
d = r
B
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 4 of 14
18. The angular velocity of an object, , is given by:
a.
 = /t
c.
d = r
b.
 = /t
d.
v = r
B
19. The angular acceleration of an object, , is given by:
a.
 = /t
c.
d = r
b.
 = /t
d.
v = r
A
20. As an object rotates, the change in the angle is called the ____________________
angular displacement
21. A motorcycle is towing a travel trailer. If the motorcycle has wheels that are twice the diameter of the
trailer’s wheels, what is the ratio of the angular acceleration of the motorcycle wheels to the angular
acceleration of the trailer wheels?
The motorcycle wheels are twice the radius of the trailer wheels, so the angular acceleration of the
motorcycle wheels would be 1/2 that of the trailer wheels. The ratio would be 1:2 or 1/2.
22. On a test stand a bicycle wheel is being rotated about its axle so that a point on the edge moves
through 0.210 m. The radius of the wheel is 0.350 m, as shown in Figure. Through what angle (in radians)
is the wheel rotated?
𝑠 = 𝑟𝜃
𝜃=
𝑠
0.21
=
= 0.6 𝑟𝑎𝑑
𝑟
0.350
23. A disc of radius 6.80 cm rotates about its axis and a point 3.80 cm from the center of the disc moves
30.3 cm in 12.6 s. Calculate the angular speed of the disc.
0.303
𝑣
𝜔 = = 12.6 = 0.633 𝑟𝑎𝑑⁄𝑠
𝑟 0.038
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 5 of 14
24.
𝑣 = 𝑟𝜔 = (0.26) (
0.373𝑟𝑒𝑣
𝑟𝑎𝑑
) (2
) = 0.609 𝑚/𝑠
𝑠
𝑟𝑒𝑣
25. Rank the torques on the five doors shown in Figure 26 from least to greatest. Note that the magnitude
of all the forces is the same.
Recall that τ=Fr sin θ . Thus, 0 = E < D < C < B < A .
26. ________________ is the measure of the twisting action caused by a force that can cause an object to
rotate about an axis.
A. Moment arm
B. Moment mass
C.
Moment time
D. Rotation
E. Torque
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 6 of 14
27. A 25 𝑁, force 𝐹 is applied to a bar that can pivot around its end as shown below.
The force is parallel to the bar and is 𝑟 = 0.75 𝑚 away from the end. What is the torque on the bar?
𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 = 0.75 × 25 × sin 0 = 0
28. A 25 𝑁, force 𝐹 is applied to a bar that can pivot around its end as shown below.
The force is 𝑟 = 0.75 𝑚 away from the end and at an angle θ=30°. What is the torque on the bar?
𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 = 0.75 × 25 × sin −30 = − 9.4 𝑁. 𝑚 OR 9.4 N.m (clockwise)
29. A 25 𝑁, force 𝐹 is applied to a bar that can pivot around its end as shown below.
The force is perpendicular to the bar and is 𝑟 = 0.75 𝑚 away from the end. What is the torque on the
bar?
𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 = 0.75 × 25 × sin 90 = 18.8 𝑁. 𝑚
30. The lug nuts on your car’s tires have been tightened with a torque of 40.0 𝑁 · 𝑚. The tire iron
that came with the car is only 20.0 𝑐𝑚 long, but you are able to place it exactly perpendicular. How
much force do you have to exert in order to break the lug nuts free?
Visualize
𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 = 𝑟𝐹𝑠𝑖𝑛90 = 𝑟𝐹(Lever arm is exactly perpendicular that means the angle 𝜃 = 90)
𝜏= 𝐹𝑟
𝐹 = 𝜏/𝑟
𝐹 = 40 𝑁 · 𝑚 / (0.025 𝑚)
𝐹 = 1600 𝑁
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 7 of 14
31. Calculate the net torque on the square plate about the point O. Force F1 is 32.0 𝑁, F2 is 21.0 𝑁, F3 is
30.0 𝑁.
√
A.
B.
C.
D.
2.88 𝑁. 𝑚
0.99 𝑁. 𝑚 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
0.99 𝑁. 𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
2.83 𝑁. 𝑚 𝑐𝑜𝑢𝑛𝑡𝑒𝑟 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
32. A beam B that weighs 15 𝑁 is 3.5 𝑚 long. Its pivot at 0.5 𝑚 from one end. An object must be
placed on the other end of the beam to balance it. Find the weight of the object O that balances the
beam.
Visualize the problem
(3.5 m)2=1.75m (3.5 m)2=1.75m
Object O
1.75 - 0.5=
1.25 m
0.5 m
FB=15 N (acw)
FO =?? (cw)
Beam B
Balancing for rotation
∑𝜏 = 0
Solve for Force of object
𝜏𝑎𝑐𝑤 − 𝜏𝑐𝑤 = 0
𝜏𝑎𝑐𝑤 = 𝜏𝑐𝑤
𝐹𝐵 × 𝑟𝐵 = 𝐹𝑂 × 𝑟𝑂
𝑟𝐵
1.25
= 15 ×
𝑟𝑂
0.5
𝐹 = 38 𝑁
𝐹𝑂 = 𝐹𝐵 ×
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 8 of 14
33. A uniform meter stick with equally spaced tick marks has a fulcrum at its center, and a downward
force F is applied to the meter stick at the position shown. An upward ↑ force of 4F is then applied to the
meter stick.
a) 4𝐹
a) At which position should the upward force 4F be applied to keep the meter stick in equilibrium?
Balancing for rotation
∑ 𝜏 = 0; 𝜏𝑎𝑐𝑤 − 𝜏𝑐𝑤 = 0; 𝜏𝑎𝑐𝑤 = 𝜏𝑐𝑤
𝐹𝑢𝑝 × 𝑟𝑢𝑝 = 𝐹𝑑𝑜𝑤𝑛 × 𝑟𝑑𝑜𝑤𝑛
4𝐹 × 𝑟𝑢𝑝 = 𝐹 × 4
𝑟𝑢𝑝 = 1
This means the upward force should be places at x3
b) At which position should the downward force 4F be applied to keep the meter stick in equilibrium?
∑ 𝜏 = 0; 𝜏𝑎𝑐𝑤 − 𝜏𝑐𝑤 = 0; 𝜏𝑎𝑐𝑤 = 𝜏𝑐𝑤
𝐹4𝐹 × 𝑟4𝐹 = 𝐹𝐹 × 𝑟𝐹
4𝐹 × 𝑟4𝐹 = 𝐹 × 4
𝑟4𝐹 = 1
This means the downward force should be places at x 2
34. At which position should the upward force 2F be applied to keep the meter stick in equilibrium?
a. Position x1
b. Position x2
c. Position x3
d. Position x4
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 9 of 14
35. What does the formula: mass x (lever arm)2 calculate?
Net Force
Net Torque
Moment of inertia 
Angular acceleration
36. What do we call an object’s tendency to resist changes in its angular velocity?
Friction
Net Force
Net Torque
Moment of inertia 
37. A toy consisting of two balls, each 0.45 kg, at the ends of a 0.46 m-long, thin, lightweight rod is
shown in Figure. Find the moment of inertia of the toy. The moment of inertia is to be found about the
center of the rod.
𝑟 = 0.5 𝑙 = 0.5 × 0.46 = 0.23 𝑚
𝐼𝑡𝑜𝑦
𝐼𝑚𝑎𝑠𝑠 = 𝑚𝑟 2
= 0.45 × 0.232 = 0.024 𝑘𝑔. 𝑚2
= 𝐼𝑚𝑎𝑠𝑠 + 𝐼𝑚𝑎𝑠𝑠 = 2 × 0.024 = 0.048 𝑘𝑔. 𝑚2
38. A simplified model of a twirling baton is a thin rod with two round objects at each end. The length of
the baton is 0.66 m, and the mass of each object is 0.30 kg. Find the moment of inertia of the baton as it is
rotated about an axis at the midpoint between the round objects and perpendicular to the rod.
𝑟 = 0.5 𝑙 = 0.5 × 0.66 = 0.33 𝑚
𝐼𝑏𝑎𝑡𝑜𝑛
𝐼𝑚𝑎𝑠𝑠 = 𝑚𝑟 2
= 0.30 × 0.332 = 0.033 𝑘𝑔. 𝑚2
= 𝐼𝑚𝑎𝑠𝑠 + 𝐼𝑚𝑎𝑠𝑠 = 2 × 0.033 = 0.066 𝑘𝑔. 𝑚2
39. A rope is wrapped around a pulley and pulled with a force of 13.0 N. The pulley’s radius is 0.150 m.
The pulley’s rotational speed increases from 0.0 to 14.0 rev/min in 4.50 s. What is the moment of inertia
of the pulley?
𝜏𝑛𝑒𝑡
𝐼
𝜏𝑛𝑒𝑡
1.95
𝐼=
=
= 5.99 𝑘𝑔. 𝑚2
𝛼
0.326
𝛼=
In order to put the correct values in the above equation, first calculate the net torque, then find the
angular acceleration (substitute the angular velocity with correct units)
Calculation of net torque
𝜏𝑛𝑒𝑡 = 𝐹𝑑 sin 𝜃 = 13 × 0.150 × 1 = 1.95 𝑁. 𝑚
Converting the angular velocity to rad/s
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 10 of 14
𝜔=
14 𝑟𝑒𝑣 2𝜋 𝑟𝑎𝑑
×
= 1.465 𝑟𝑎𝑑/𝑠
60 𝑠
𝑟𝑒𝑣
Calculating the angular acceleration
𝜔𝑓 − 𝜔𝑖 1.465 − 0
𝛼=
=
= 0.326 𝑟𝑎𝑑⁄𝑠 2
𝑡
4.5
𝜏
Then substitute for 𝐼 = 𝑛𝑒𝑡
𝛼
40. A solid disk is shown in the figure below and subject to three external forces as shown in the
figure. The disk accelerates at the rate of -0.75 rad/s2. Calculate its moment of inertia.
0.127 kg.m2
0.232 kg.m2
0.412 kg.m2 
0.549 kg.m2
41. A solid disk is shown in the figure below and subject to three external forces as shown in the
figure. The disk accelerates at the rate of -0.67 rad/s2. Calculate its moment of inertia.
0.584 kg.m2
0.871 kg.m2
1.94 kg.m2 
2.89 kg.m2
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 11 of 14
42. Choose among the list, all the correct statement for a body in static equilibrium.
 The velocity of the object is zero
 The net torque is different from zero
 The velocity of the object is changing
 The net force is equal to zero
 The angular velocity of the object is constant
 The angular velocity of the object is changing
 The angular velocity of the object is zero
43. A uniform 400 N diving board is supported at two points as shown in the diagram. If a 75 kg diver stands
at the end of the board, what is the force F exerted on the left support by the board?
1303 N
2270 N
2605 N 
3340 N
44. A circus performer on a unicycle of total mass 55 kg rides across a uniform 30 kg beam. The
supports are placed at opposite ends of the beam. What is the magnitude of the force exerted by
support B?
49 N
64 N
484 N 
631 N
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 12 of 14
45. Consider the system of three flat objects of uniform density shown in Figure below
The three objects (2 squares and 1 circle) center of masses are at their centers
The positions of the center of masses are as follows
M1 = 1 kg x1 = 6 y1 = 5
M2 = 2 kg x2 = 6 y2 = 8.5
𝑚1 𝑥1 + 𝑚2 𝑥2
1×6+2×6
𝑥𝑐𝑚 =
=
=6
𝑚1 + 𝑚2
1+2
𝑦𝑐𝑚 =
𝑚1 𝑦1 + 𝑚2 𝑦2
1 × 5 + 2 × 8.5
=
= 7.3
𝑚1 + 𝑚2
1+2
Center of mass (𝑥𝑐𝑚 , 𝑦𝑐𝑚 ) = (6,7.3)
46. Two masses are located as shown in the figure. M1=1.0 kg and M2=3.0 kg. What are the
coordinates of the center of mass for the 2-mass system?
(0.75 cm, 5.5 cm)
(2.25 cm, 6.5 cm) 
(5.6 cm, 0.75 cm)
(6.5 cm, 2.25 cm)
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 13 of 14
47. Two masses are located as shown in the figure. M1=2.0 kg and M2=3.0 kg. What are the
coordinates of the center of mass for the 2-mass system?
(2.4 cm, 2.8 cm)
(2.8 cm, 2.4 cm)
(3.2 cm, 3.6 cm) 
(3.6 cm, 3.2 cm)
T1- PHY60 – t4 – Rotational motion- 2020 – 2021- AK
Page 14 of 14