Question 1

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the weights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. Then:
A) his angular velocity doubles
B) his angular velocity remains about the same
C) his angular velocity is halved
D) the direction of his angular momentum vector changes
E) his rotational kinetic energy increases

Accepted Answer

The answer of A man, holding a weight in each...

Question 2

A rod rests on frictionless ice. Forces that are equal in magnitude and opposite in direction are simultaneously applied to its ends as shown. The quantity that has a magnitude of zero is its: 
A) angular momentum
B) angular acceleration
C) total linear momentum
D) kinetic energy
E) rotational inertia

Accepted Answer

The net force acting on the rod is zero, so the total linear momentum of the system is conserved and stays at zero. However, the other quantities listed (angular momentum, angular acceleration, kinetic energy, and rotational inertia) are all non-zero.

Question 3

An ice skater with rotational inertia I₀ is spinning with angular speed $ \omega $₀. She pulls her arms in, thereby increasing her angular speed to 4 $ \omega $₀. Her rotational inertia is then: A) I₀ B) I₀/2 C) 2 I₀ D) I₀/4 E) 4 I₀

Accepted Answer

According to the law of conservation of angular momentum, the initial angular momentum of the skater (I₀ multiplied by F1F1F1S1F1F1F10) must equal the final angular momentum of the skater (I₀/4 multiplied by 4F1F1F1S1F1F1F10). Solving for the final rotational inertia yields I₀/4, which is choice D.

Question 4

A single force acts on a particle P. Rank each of the orientations of the force shown below according to the magnitude of the time rate of change of the particle's angular momentum about the point O, least to greatest. 
A) 1, 2, 3, 4
B) 1 and 2 tie, then 3, then 4
C) 1 and 2 tie, then 4, then 3
D) 1 and 2 tie, then 3 and 4 tie
E) All are the same

Accepted Answer

The answer of A single force acts on a particle...

Question 5

A phonograph record is dropped onto a freely spinning turntable. Then:
A) neither angular momentum nor mechanical energy is conserved because of the frictional forces between record and turntable
B) the frictional force between record and turntable increases the total angular momentum
C) the frictional force between record and turntable decreases the total angular momentum
D) the total angular momentum remains constant
E) the sum of the angular momentum and rotational kinetic energy remains constant

Accepted Answer

The total angular momentum remains constant because there are no external torques acting on the system. The frictional forces between the record and turntable only transfer angular momentum within the system, but do not change the total amount of angular momentum. Similarly, the mechanical energy of the system is conserved. Therefore, the correct choice is D.

Question 6

A uniform sphere of radius R rotates about a diameter with angular momentum of magnitude L. Under the action of internal forces the sphere collapses to a uniform sphere of radius R/2. The magnitude of its new angular momentum is:
A) L/4
B) L/2
C) L
D) 2L
E) 4L

Accepted Answer

The answer of A uniform sphere of radius R rotates...

Question 7

A 2.0-kg stone is tied to a 0.50 m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:
A) 0 N∙m
B) 6.0 N∙m
C) 14 N∙m
D) 72 N∙m
E) 108 N∙m

Accepted Answer

The torque ($\tau$) about the origin can be calculated using the formula $\tau = rF\sin(\theta)$, where $r$ is the radius of the circle, $F$ is the force acting on the object, and $\theta$ is the angle between the force vector and the radius vector. Here, the radius $r$ is the distance from the origin to the circle, which is 0.75 m. The force $F$ can be found from the centripetal force formula $F = m\omega^2r$, where $m$ is the mass of the stone, $\omega$ is the angular velocity, and $r$ is the length of the string (0.50 m in this case). Substituting the given values, $F = 2.0 \times 12^2 \times 0.50 = 144 N$. Since the circle is parallel to the xy plane and centered on the z axis, the angle $\theta$ between the force vector (which points radially inward) and the radius vector (from the origin to the circle) is 90 degrees, making $\sin(\theta) = 1$. Therefore, the torque is $\tau = 0.75 \times 144 = 108 N\cdot m$.

Question 8

A wheel, with rotational inertia I, mounted on a vertical shaft with negligible rotational inertia, is rotating with angular speed $ \omega $_0.A nonrotating wheel with rotational inertia 2I is suddenly dropped onto the same shaft as shown. The resultant combination of the two wheels and shaft will rotate at: A) $ \omega $₀/2 B) 2 $ \omega $₀ C) $ \omega $₀ /3 D) 3 $ \omega $₀ E) $ \omega $₀/4

Accepted Answer

The answer of A wheel, with rotational inertia I, mounted...

Question 9

A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms:
A) his angular velocity increases
B) his angular velocity remains the same
C) his rotational inertia decreases
D) his rotational kinetic energy increases
E) his angular momentum remains the same

Accepted Answer

When the man extends his arms, his rotational inertia increases, which causes his angular velocity to decrease in order to conserve angular momentum. Angular momentum is conserved because there are no external torques acting on the system.

Question 10

A pulley with radius R is free to rotate on a horizontal fixed axis through its center. A string passes over the pulley. Mass m₁ is attached to one end and mass m₂ is attached to the other. The portion of the string attached to m₁ has tension T₁ and the portion attached to m₂ has tension T₂. The magnitude of the total external torque, about the pulley center, acting on the masses and pulley, considered as a system, is given by: A) $\mid$m₁ - m₂$\mid$gR B) (m₁ + m₂)gR C) $\mid$m₁ - m₂$\mid$gR + (T₁ + T₂)R D) (m₁ + m₂)gR + (T₁ - T₂)R E)$\mid$m₁ - m₂$\mid$gR + (T₁ - T₂)R

Accepted Answer

The answer of A pulley with radius R is free...

Question 11

A 2.0-kg stone is tied to a 0.50-m long string and swung around a circle at a constant angular velocity of 12 rad/s. The net torque on the stone about the center of the circle is:
A) 0 N∙m
B) 6.0 N∙m
C) 12 N∙m
D) 72 N∙m
E) 140 N∙m

Accepted Answer

The net torque on the stone is equal to the product of its moment of inertia and angular acceleration. Since the angular velocity is constant, the angular acceleration is zero and therefore the net torque is also zero.

Question 12

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s². The torque, relative to the origin, acting on it at the end of 2.0 s is: A) 0 N∙m B) (-18 N∙m) C) (+24 N∙m) D) (-144 N∙m) E) (+144 N∙m)

Accepted Answer

The answer of A 2.0-kg block starts from rest on...

Question 13

A uniform disk has radius R and mass M. When it is spinning with angular velocity $ \omega $about an axis through its center and perpendicular to its face its angular momentum is I $ \omega $. When it is spinning with the same angle velocity about a parallel axis a distance h away its angular momentum is: A) I $ \omega $ B) (I + Mh²) $ \omega $ C) (I - Mh²) $ \omega $ D) (I + MR²) $ \omega $ E) (I - MR²) $ \omega $

Accepted Answer

When spinning about a parallel axis a distance h away, the total moment of inertia of the system becomes (I + Mh2) because the parallel axis theorem states that the moment of inertia about a parallel axis a distance h away is equal to the moment of inertia about the center of mass plus the product of the total mass and the square of the distance between the two axes. Therefore, the new angular momentum can be calculated as (I + Mh2)ω, which is equivalent to choice B.

Question 14

A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then: A) mv/I B) v/R C) mRv/I D) 2mRv/I E) mRv/(mR² + I)

Accepted Answer

The correct answer is derived from the conservation of angular momentum. Initially, the angular momentum of the system (merry-go-round plus child) is solely due to the child, since the merry-go-round is at rest. The child's angular momentum before jumping on is given by $L = mvr$, where $r = R$ is the radius of the merry-go-round. After the child jumps on, the total angular momentum of the system must remain constant, so $L = I_{\text{total}}\omega$, where $I_{\text{total}} = I + mR^2$ (the sum of the merry-go-round's rotational inertia and the rotational inertia of the child treated as a point mass at radius $R$), and $\omega$ is the angular velocity of the system. Solving for $\omega$ gives $\omega = \frac{mRv}{I + mR^2}$, which matches the formula in choice E.

Question 15

When a woman on a frictionless rotating turntable extends her arms out horizontally, her angular momentum:
A) must increase
B) must decrease
C) must remain the same
D) may increase or decrease depending on her initial angular velocity
E) tilts away from the vertical

Accepted Answer

Angular momentum is conserved in a closed system without external torques, so when the woman extends her arms, her moment of inertia increases, but her angular momentum remains constant.

Question 16

A pulley with radius R and rotational inertia I is free to rotate on a horizontal fixed axis through its center. A string passes over the pulley. A block of mass m₁is attached to one end and a block of mass m₂_, is attached to the other. At one time the block with mass m₁ is moving downward with speed v. If the string does not slip on the pulley, the magnitude of the total angular momentum, about the pulley center, of the blocks and pulley, considered as a system, is given by: A) (m₁ - m₂)vR + Iv/R B) (m₁ + m₂)vR + Iv/R C) (m₁ - m₂)vR - Iv/R D) (m₁ + m₂)vR - Iv/R E) none of the above

Accepted Answer

The angular momentum of the system is given by the sum of the angular momenta of the two blocks and the pulley:
L = L1 + L2 + Lp
where L1 is the angular momentum of block 1, L2 is the angular momentum of block 2, and Lp is the angular momentum of the pulley. Since the pulley rotates about its center with no external torque, its angular momentum is conserved and is equal to Iv/R, where v is the speed of the blocks and I is the rotational inertia of the pulley. 
The angular momentum of each block is given by the product of its moment of inertia about the pulley axis and its angular velocity:
Li = Iiωi
where i = 1,2. Since the string does not slip on the pulley, the angular velocity of each block is related to the angular velocity of the pulley by:
ωi = v/(2R)
Therefore, the angular momentum of each block is:
Li = (miR^2/2)(v/2R) = (mi/4)vR
Finally, the total angular momentum of the system is:
L = L1 + L2 + Lp = (m1/4 + m2/4 + I/R)v
= [(m1 + m2)/4 + I/R]vR + Iv/R 
= (m1R/2 + m2R/2 + I/R)vR + Iv/R 
= (m1vR + m2vR + Iv)/R
= (m1vR + m2vR) + Iv/R
= (m1vR + m2vR) +- Iv/R
Therefore, the correct choice is B, as it corresponds to the expression above.

Question 17

When a man on a frictionless rotating stool extends his arms horizontally, his rotational kinetic energy:
A) must increase
B) must decrease
C) must remain the same
D) may increase or decrease depending on his initial angular velocity
E) may increase or decrease depending on his angular acceleration

Accepted Answer

This is due to conservation of angular momentum. When the man extends his arms, his moment of inertia increases (since the mass is farther from the axis of rotation), causing his angular velocity to decrease in order to conserve angular momentum. Therefore, his rotational kinetic energy must also decrease.

Question 18

A playground merry-go-round has a radius of 3.0 m and a rotational inertia of 600 kg∙m². It is initially spinning at 0.80 rad/s when a 20-kg child crawls from the center to the rim. When the child reaches the rim the angular velocity of the merry-go-round is: A) 0.62 rad/s B) 0.73 rad/s C) 0.77 rad/s D) 0.91 rad/s E) 1.1 rad/s

Accepted Answer

The answer of A playground merry-go-round has a radius of...

Question 19

A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a fixed axis through its center. Assume the hoop is connected to the rotation axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their angular momenta after a given time t, least to greatest. 
A) all tie
B) disk, hoop, sphere
C) sphere, disk, hoop
D) hoop, sphere, disk
E) hoop, disk, sphere

Accepted Answer

The answer of A uniform disk, a thin hoop, and...

Question 20

Two objects are moving in the x, y plane as shown. If a net torque of 44 N∙m acts on them for 5.0 seconds, what is the change in their angular momentum? A) 0 kg∙m²/s B) 6 kg∙m²/s C) 30 kg∙m²/s D) 150 kg∙m²/s E) 220 kg∙m²/s

Accepted Answer

The answer of Two objects are moving in the x,...

Quiz 11: Rolling, Torque, and Angular Momentum

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the weights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. Then:

A rod rests on frictionless ice. Forces that are equal in magnitude and opposite in direction are simultaneously applied to its ends as shown. The quantity that has a magnitude of zero is its:

An ice skater with rotational inertia I₀ is spinning with angular speed $\omega$ ₀. She pulls her arms in, thereby increasing her angular speed to 4 $\omega$ ₀. Her rotational inertia is then:

A single force acts on a particle P. Rank each of the orientations of the force shown below according to the magnitude of the time rate of change of the particle's angular momentum about the point O, least to greatest.

A phonograph record is dropped onto a freely spinning turntable. Then:

A uniform sphere of radius R rotates about a diameter with angular momentum of magnitude L. Under the action of internal forces the sphere collapses to a uniform sphere of radius R/2. The magnitude of its new angular momentum is:

A 2.0-kg stone is tied to a 0.50 m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:

A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms:

A 2.0-kg stone is tied to a 0.50-m long string and swung around a circle at a constant angular velocity of 12 rad/s. The net torque on the stone about the center of the circle is:

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s². The torque, relative to the origin, acting on it at the end of 2.0 s is:

A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then:

When a woman on a frictionless rotating turntable extends her arms out horizontally, her angular momentum:

When a man on a frictionless rotating stool extends his arms horizontally, his rotational kinetic energy:

A playground merry-go-round has a radius of 3.0 m and a rotational inertia of 600 kg∙m². It is initially spinning at 0.80 rad/s when a 20-kg child crawls from the center to the rim. When the child reaches the rim the angular velocity of the merry-go-round is:

Two objects are moving in the x, y plane as shown. If a net torque of 44 N∙m acts on them for 5.0 seconds, what is the change in their angular momentum?