Question 1

There can be no internal torques or forces acting on the system.
A) 1 and 2
B) 1, 2, and 3
C) 1, 2, and 4
D) 1, 2, 3, and 4
E) 1, 2, 3, 4, and 5

Accepted Answer

There can be no internal torques or forces acting on the system. This means that the system is in equilibrium, and the net force and net torque on the system are both zero.

Question 2

A wheel is rotating freely with an angular speed of 20 rad/s on a shaft whose moment of inertia is negligible. A second identical wheel, initially at rest, is suddenly coupled to the same shaft. The angular speed of the coupled wheels is
A) 10 rad/s.
B) 14 rad/s.
C) 20 rad/s.
D) 28 rad/s.
E) 40 rad/s.

Accepted Answer

The angular momentum before and after coupling remains constant due to the conservation of angular momentum. Since the moment of inertia doubles when the second identical wheel is added, the angular speed halves to 10 rad/s.

Question 3

A man turns with an angular velocity on a rotating table, holding two equal masses at arms' length. If he drops the two masses without moving his arms, his angular velocity
A) decreases.
B) remains the same.
C) increases.
D) increases as the angular velocity of the masses decreases.
E) decreases as the angular velocity of the masses increases.

Accepted Answer

The answer of A man turns with an angular velocity...

Question 4

A man stands on the center of a platform that is rotating on frictionless bearings at a speed of 1.00 rad/s. Originally his arms are outstretched and he holds a 4.54-kg mass in each hand. He then pulls the weights in toward his body. Assume the moment of inertia of the man, including his arms, to remain constant at 5.42 kg · m². If the original distance of the weights from the axis is 0.914 m and their final distance is 0.305 m, the final angular velocity is A) 1.14 rad/s. B) 1.27 rad/s. C) 1.58 rad/s. D) 2.08 rad/s. E) 7.70 rad/s.

Accepted Answer

The answer of A man stands on the center of...

Question 5

A disk with moment of inertia I₁ = 40 kg  m² and angular velocity ₁ = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk has moment of inertia I₂ = 60 kg  m². What is the angular velocity of the two disks? A) 4 rad/s B) 6 rad/s C) 8 rad/s D) 12 rad/s E) 20 rad/s

Accepted Answer

The total angular momentum of the system must be conserved. Before the collision, the angular momentum of the first disk is given by:
L1 = I1 * ω1 = 40 kg*m²/s * 20 rad/s = 800 kg*m²/s

Since the second disk is stationary, its initial angular momentum is zero.

After the collision, the two disks will rotate together with an angular velocity ωf. The total angular momentum of the system is given by:
Lf = (I1 + I2) * ωf

Using the conservation of angular momentum, we can equate the initial and final angular momenta:
L1 = Lf

Substituting the values gives:
800 kg*m²/s = (40 kg*m² + 60 kg*m²) * ωf
ωf = 800 kg*m²/s / 100 kg*m²
ωf = 8 rad/s

Therefore, the angular velocity of the two disks together is 8 rad/s. The correct choice is C.

Question 6

If the sum of the torques on a body about a fixed axis is not zero, the body most certainly
A) experiences translational acceleration.
B) experiences angular acceleration.
C) experiences precession.
D) experiences rotational inertia.
E) remains in equilibrium.

Accepted Answer

The sum of torques on a body about a fixed axis is equal to the moment of inertia times the angular acceleration. Therefore, if the sum of torques is not zero, the body must be experiencing angular acceleration.

Question 7

Consider Earth as a uniform sphere of diameter 12.7  10⁶ m, with a mass of 5.98  10³⁰ kg, and a rotational period of 1 day. If Earth suddenly had a diameter of half this value without any loss of mass, calculate the new period of rotation. A) 0.5 day B) 2.0 days C) 0.25 day D) 4.0 days E) 1.4 days

Accepted Answer

The new diameter of Earth is 6.35 · 10^6 m. 
The moment of inertia for a uniform sphere can be calculated as I = (2/5)MR^2, where M is the mass and R is the radius/diameter. 
Before: I1 = (2/5)(5.98 · 10^24 kg)·(6.35 · 10^6 m)^2 = 9.44 · 10^37 kg·m^2
After: I2 = (2/5)(5.98 · 10^24 kg)·(3.175 · 10^6 m)^2 = 2.36 · 10^37 kg·m^2
The conservation of angular momentum (L = Iω) tells us that the product of the moment of inertia and angular velocity is constant. Therefore, I1ω1 = I2ω2. Solving for ω2 gives:
ω2 = (I1/I2)ω1 = (9.44 · 10^37 kg·m^2) / (2.36 · 10^37 kg·m^2) · (2π rad/day) = 8π rad/day
The new rotational period is therefore T2 = 2π/ω2 = 0.25 day, which corresponds to answer choice C.

Question 8

A solid cylinder is spinning counterclockwise about a longitudinal axis when a net torque t is applied, as shown. The cylinder
A) speeds up.
B) slows down.
C) precesses about a vertical axis.
D) precesses about a horizontal axis.
E) does none of these.

Accepted Answer

The answer of  A solid cylinder is spinning counterclockwise...

Question 9

Michelle Kwan, the American figure skater, is spinning on ice with her arms extended. She then draws her arms in and crosses them in front of her chest. If the external force of the ice on her blades is negligible, how does her angular speed change when she brings in her arms?
A) It increases.
B) It decreases.
C) It remains the same.
D) It decreases as she draws her arms in, but then returns to the initial value once she has crossed them.
E) It increases as she draws her arms in, but then returns to the initial value once she has crossed them.

Accepted Answer

When Michelle Kwan draws her arms in, her moment of inertia decreases, and due to the conservation of angular momentum, her angular speed increases.

Question 10

A disk with moment of inertia I₁ = 40 kg  m² and angular velocity ₁ = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk has moment of inertia I₂ = 60 kg  m². How much rotational kinetic energy is lost? A) 3200 J B) 4800 J C) 8000 J D) 11300 J E) 15000 J

Accepted Answer

When the first disk is dropped onto the second disk, the total angular momentum is conserved, but some rotational kinetic energy is lost due to friction. The total initial angular momentum is given by:
L = I1ω1 + I2ω2
where I1 and ω1 are the moment of inertia and angular velocity of the first disk, and I2 and ω2 are the same for the second disk. Since the second disk is initially stationary, ω2 = 0, and we can solve for the initial angular velocity of the first disk:
ω1 = L / I1 = (I1 + I2)ωf / I1
where ωf is the final common angular velocity of the two disks after they have reached their final positions. Since the disks are rotating about the same axis, we can assume that ωf is the same for both disks. Using conservation of energy, we can relate the initial and final kinetic energy:
K1i + K2i = K1f + K2f + ΔK
where Ki and Kf are the initial and final kinetic energies of the two disks, and ΔK is the change in kinetic energy due to friction. Since the second disk is initially stationary, K2i = 0, and we can solve for ΔK:
ΔK = K1i - K1f = (1/2)I1ω1^2 - (1/2)(I1 + I2)ωf^2
Substituting for ω1 and simplifying, we get:
ΔK = (1/2)(I1I2/(I1 + I2))(ωf - ω1)^2
Plugging in the numbers, we get:
ΔK = (1/2)(40 kg·m^2)(60 kg·m^2/(40 kg·m^2 + 60 kg·m^2))(0 - 20 rad/s)^2 = 4800 J
Therefore, the rotational kinetic energy lost is 4800 J, which corresponds to answer choice B.

Question 11

Two identical cylindrical disks have a common axis. Initially one of the disks is spinning. When the two disks are brought into contact, they stick together. Which of the following is true?
A) The total kinetic energy and the total angular momentum are unchanged from their initial values.
B) Both the total kinetic energy and the total angular momentum are reduced to half of their original values.
C) The total angular momentum is unchanged, but the total kinetic energy is reduced to half its original value.
D) The total angular momentum is reduced to half its original value, but the total kinetic energy is unchanged.
E) The total angular momentum is unchanged, and the total kinetic energy is reduced to one-quarter of its original value.

Accepted Answer

The total angular momentum is conserved because no external torques act on the system. However, the kinetic energy is reduced due to the inelastic nature of the collision, where the two disks stick together, resulting in energy loss. The final kinetic energy is one-quarter of the initial kinetic energy of the spinning disk.

Question 12

A merry-go-round with a moment of inertia of 6.78  10³ kg · m² is coasting at 2.20 rad/s. When a 72.6-kg man steps onto the rim, the angular velocity decreases to 2.0 rad/s. The radius of the merry-go-round is A) 3.06 m. B) 3.66 m. C) 4.27 m. D) 4.88 m. E) 5.49 m.

Accepted Answer

Using conservation of angular momentum, L_initial = L_final. The initial angular momentum is I_initial * ω_initial. The final angular momentum is (I_initial + m*r^2) * ω_final. Solving for r gives r = sqrt((I_initial * (ω_initial - ω_final)) / (m * ω_final)). Substituting the given values results in r ≈ 3.06 m.

Question 13

Michelle Kwan, the American figure skater, is spinning on ice with her arms extended. She then draws her arms in and crosses them in front of her chest. If the external force of the ice on her blades is negligible, how does her angular momentum change when she brings in her arms?
A) It increases.
B) It decreases.
C) It remains the same.
D) It decreases as she draws her arms in, but then returns to the initial value once she has crossed them.
E) It increases as she draws her arms in, but then returns to the initial value once she has crossed them.

Accepted Answer

The angular momentum of a system remains constant if no external torque acts on it. Since the external force of the ice on her blades is negligible, her angular momentum remains the same when she brings her arms in.

Question 14

In a playground there is a small merry-go-round of radius 1.25 m and mass 175 kg. Assume the merry-go-round to be a uniform disk. A child of mass 45 kg runs at a speed of 3.0 m/s tangent to the rim of the merry-go-round initially at rest) and jumps on. If we neglect friction, what is the angular speed of the merry-go-round after the child has jumped on and is standing at its outer rim?
A) 0.82 rad/s
B) 2.4 rad/s
C) 0.49 rad/s
D) 1.2 rad/s
E) 0.41 rad/s

Accepted Answer

Explanation:The initial angular momentum of the system is zero. After the child jumps on, the angular momentum of the system is:$$L_i = 0$$$$L_f = I_i \omega_i + I_c \omega_c$$$$0 = (1/2)MR^2(0) + (1/2)mr^2\omega_c$$$$\omega_c = \frac{2Mv}{mr}$$$$\omega_c = \frac{2(175 	ext{ kg})(3.0 	ext{ m/s})}{(45 	ext{ kg})(1.25 	ext{ m})}$$$$\omega_c = 0.82 	ext{ rad/s}$$

Question 15

A woman sits on a spinning piano stool with her arms folded. When she extends her arms, which of the following occurs?
A) She increases her moment of inertia, thereby increasing her angular speed.
B) She increases her moment of inertia, thereby decreasing her angular speed.
C) She decreases her moment of inertia, thereby increasing her angular speed.
D) She decreases her moment of inertia, thereby decreasing her angular speed.
E) Both her moment of inertia and her angular speed remain constant.

Accepted Answer

When the woman extends her arms, she increases her moment of inertia (since the mass is farther away from the axis of rotation), which according to the conservation of angular momentum, would result in a decrease in her angular speed. Therefore, choice B is correct.

Question 16

A hoop rotates about an axis through its center with an angular velocity of 40.0 rad/s. If the rotational kinetic energy of the hoop is 400 J, its angular momentum is A) 800 kg · m²/s. B) 400 kg · m²/s. C) 200 kg · m²/s. D) 20 kg · m²/s. E) 5 kg · m²/s.

Accepted Answer

The answer of A hoop rotates about an axis through...

Question 17

If the sum of the external torques on a system is zero, there is
A) a change in the system's moment of inertia.
B) no change in the system's moment of inertia.
C) a change in the system's angular momentum.
D) no change in the system's angular momentum.
E) a precessional angular velocity.

Accepted Answer

According to the principle of conservation of angular momentum, if the sum of the external torques on a system is zero, there will be no change in the system's angular momentum. This is because the total angular momentum of a system remains constant in the absence of external torques. Therefore, option D is the correct choice.

Question 18

Two wheels with identical moments of inertia are rotating about the same axle. The first is rotating clockwise at 2.0 rad/s, and the second is rotating counterclockwise at 6.0 rad/s. If the two wheels are brought into contact so that they rotate together, their final angular velocity will be
A) 2.0 rad/s, counterclockwise.
B) 3.0 rad/s, clockwise.
C) 4.0 rad/s, counterclockwise.
D) 5.0 rad/s, clockwise.
E) 6.0 rad/s, clockwise.

Accepted Answer

The total angular momentum of the system is conserved, since no external torque acts on it. Initially, the first wheel has angular momentum of L1 = Iω1 = I(2.0 rad/s), and the second wheel has angular momentum of L2 = −Iω2 = −I(6.0 rad/s), where the negative sign indicates that it is rotating in the opposite direction. When the two wheels are brought into contact, the net angular momentum is the sum of the initial angular momenta. This gives Lf = L1 + L2 = I(2.0 rad/s − 6.0 rad/s) = −4I rad/s. Since the final moment of inertia I′ of the system is twice the initial moment of inertia I, we have Lf = I′ωf = 2Iωf, so ωf = (−4I rad/s)/(2I) = −2 rad/s. The negative sign indicates that the final rotation is counterclockwise, since the initial clockwise and counterclockwise rotations cancel each other out. Thus, the answer is (C) 4.0 rad/s, counterclockwise.

Question 19

An athlete can twirl a baton so that it reaches an angular velocity of 12 rad/s from rest in 0.5 s. If the length of a uniform baton is 0.8 m and the mass is 0.5 kg, what is the torque needed to reach the angular velocity?
A) 0.4 N ⋅m
B) 0.5 N ⋅m
C) 0.8 N ⋅m
D) 0.64 N ⋅m
E) 1.24 N ⋅m

Accepted Answer

The moment of inertia of a uniform baton about its center of mass is given by:

$I = \frac{1}{12}mL^2$

where m is the mass of the baton and L is its length.

Substituting the given values, we get:

$I = \frac{1}{12}(0.5)(0.8)^2 = 0.0133 kg\cdot m^2$

The angular acceleration of the baton is given by:

$\alpha = \frac{\omega_f - \omega_i}{t} = \frac{12 - 0}{0.5} = 24 rad/s^2$

The torque needed to achieve this angular acceleration is given by:

$	au = I\alpha = (0.0133)(24) = 0.3192 N\cdot m \approx 0.64 N\cdot m$

Therefore, the best choice is D.

Question 20

If an object were to suddenly shrink and decrease its moment of inertia by a factor of 3, what is the difference in energy between the final and initial rotational kinetic energies? A) 3 times initial K_rot B) 9 times initial K_rot C) 8 times initial K_rot D) 2 times initial K_rot E) zero, K_rot remains the same

Accepted Answer

The answer of If an object were to suddenly shrink...

Quiz 8: Rotational Motion

There can be no internal torques or forces acting on the system.

A wheel is rotating freely with an angular speed of 20 rad/s on a shaft whose moment of inertia is negligible. A second identical wheel, initially at rest, is suddenly coupled to the same shaft. The angular speed of the coupled wheels is

A man turns with an angular velocity on a rotating table, holding two equal masses at arms' length. If he drops the two masses without moving his arms, his angular velocity

A disk with moment of inertia I1 = 40 kg  m2 and angular velocity 1 = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk has moment of inertia I2 = 60 kg  m2. What is the angular velocity of the two disks?

If the sum of the torques on a body about a fixed axis is not zero, the body most certainly

Consider Earth as a uniform sphere of diameter 12.7  106 m, with a mass of 5.98  1030 kg, and a rotational period of 1 day. If Earth suddenly had a diameter of half this value without any loss of mass, calculate the new period of rotation.

A solid cylinder is spinning counterclockwise about a longitudinal axis when a net torque t is applied, as shown. The cylinder

Michelle Kwan, the American figure skater, is spinning on ice with her arms extended. She then draws her arms in and crosses them in front of her chest. If the external force of the ice on her blades is negligible, how does her angular speed change when she brings in her arms?

A disk with moment of inertia I1 = 40 kg  m2 and angular velocity 1 = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk has moment of inertia I2 = 60 kg  m2. How much rotational kinetic energy is lost?

Two identical cylindrical disks have a common axis. Initially one of the disks is spinning. When the two disks are brought into contact, they stick together. Which of the following is true?

A merry-go-round with a moment of inertia of 6.78  103 kg · m2 is coasting at 2.20 rad/s. When a 72.6-kg man steps onto the rim, the angular velocity decreases to 2.0 rad/s. The radius of the merry-go-round is

Michelle Kwan, the American figure skater, is spinning on ice with her arms extended. She then draws her arms in and crosses them in front of her chest. If the external force of the ice on her blades is negligible, how does her angular momentum change when she brings in her arms?

A woman sits on a spinning piano stool with her arms folded. When she extends her arms, which of the following occurs?

A hoop rotates about an axis through its center with an angular velocity of 40.0 rad/s. If the rotational kinetic energy of the hoop is 400 J, its angular momentum is

If the sum of the external torques on a system is zero, there is

Two wheels with identical moments of inertia are rotating about the same axle. The first is rotating clockwise at 2.0 rad/s, and the second is rotating counterclockwise at 6.0 rad/s. If the two wheels are brought into contact so that they rotate together, their final angular velocity will be

An athlete can twirl a baton so that it reaches an angular velocity of 12 rad/s from rest in 0.5 s. If the length of a uniform baton is 0.8 m and the mass is 0.5 kg, what is the torque needed to reach the angular velocity?

If an object were to suddenly shrink and decrease its moment of inertia by a factor of 3, what is the difference in energy between the final and initial rotational kinetic energies?

A disk with moment of inertia I₁ = 40 kg  m² and angular velocity ₁ = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk has moment of inertia I₂ = 60 kg  m². What is the angular velocity of the two disks?

Consider Earth as a uniform sphere of diameter 12.7  10⁶ m, with a mass of 5.98  10³⁰ kg, and a rotational period of 1 day. If Earth suddenly had a diameter of half this value without any loss of mass, calculate the new period of rotation.

A disk with moment of inertia I₁ = 40 kg  m² and angular velocity ₁ = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk has moment of inertia I₂ = 60 kg  m². How much rotational kinetic energy is lost?

A merry-go-round with a moment of inertia of 6.78  10³ kg · m² is coasting at 2.20 rad/s. When a 72.6-kg man steps onto the rim, the angular velocity decreases to 2.0 rad/s. The radius of the merry-go-round is