Understanding Conservation of Angular Momentum - AP Physics C Electricity & Magnetism

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Question

A mouse sits at the edge of a $\small 40cm$ diameter record, rotating on a turntable at fifteen revolutions per minute. If the mouse walks straight inwards to a point $\small 10cm$ from the center, what will be its new angular velocity? Assume the mass of the record is negligible.

Answer

Relevant equations:

$L = mr^2\omega$

Write an expression for the initial angular momentum.

$L_i = mr_i^2\omega_i=m(20cm)^2(15rpm)$

Write an expression for the final angular momentum.

$L_f = mr_f^2\omega_f=m(10cm)^2\omega_f$

Apply conservation of angular momentum, setting these two expressions equal to one another.

$m(20cm)^2(15rpm)=m(10cm)^2\omega_f$

Solve for the final angular velocity.

$w_f=\frac{(20cm)^2(15rpm)}{(10cm)^2}=60rpm$

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Question

An ice skater begins to spin, starting with his arms spread out as far as possible, parallel to the ice. He pulls his arms into his body, and then raises them completely vertically towards the ceiling. As the skater pulls his arms inward, his angular velocity will , and as he raises his arms vertically his angular velocity will .

Answer

Through conservation of angular momentum, as moment of inertia decreases, the angular velocity increases. Moment of inertia is dependent on the distribution of the spinning body's mass away from the center of mass—as the skater brings his arms in, he lowers his moment of inertia, thus increasing his angular velocity. If he raises his arms completely vertically from this point, it does not change the radius of mass distributions (from the center of his body), thus maintaining his angular velocity.

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Question

Sarah spins a ball of mass attached to a string of length around her head with a velocity . If the ball splits in half, losing exactly one-half of its mass instantaneously, what is its new velocity, ?

Answer

By the conservation of angular momentum, the angular momentum , is equal to the product of the mass, angular velocity, and radius (or length of the rope in this case). The equation relating these terms is:

$L_{initial}=m_{i}v_{i}r$

Here, $m_{i}$ is the initial mass, $v_{i}$ is the initial angular velocity, and is the length of the rope, which remains constant. Angular momentum must be conserved, thus:

$L_{initial}=L_{final}$

Substitute.

$m_{i}v_{i}r=m_{f}v_{f}r$

We are given that $m_{f}=\frac{1}{2}m_{i}$ and $v_{f}$ is the final velocity. Plug in and solve.

$m_{i}v_{i}r=\frac{1}{2}m_{i}v_{f}r$

$v_{f}=2v_{i}$

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Question

A child is standing at the center of a frictionless, rotating platform. Both the child and the platform are rotating with and initial angular velocity, $\small \omega_0$ . The child begins to walk slowly toward the edge of the platform. Which quantity will decrease as the child walks?

Answer

Since the platform is frictionless, as the child walks, angular momentum is conserved. However, since there is an $\small \omega ^{2}$ term in the kinetic energy expression, as $\small \omega$ decreased due to the increase of inertia, it affects the energy more, and it decreases. From the reference frame of the disk, the child feels an outward directed force, and thus is doing negative work. This is the same principle that allows ice skaters to increase their spinning speed.

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Question

A mouse sits at the edge of a $\small 40cm$ diameter record, rotating on a turntable at fifteen revolutions per minute. If the mouse walks straight inwards to a point $\small 10cm$ from the center, what will be its new angular velocity? Assume the mass of the record is negligible.

Answer

Relevant equations:

$L = mr^2\omega$

Write an expression for the initial angular momentum.

$L_i = mr_i^2\omega_i=m(20cm)^2(15rpm)$

Write an expression for the final angular momentum.

$L_f = mr_f^2\omega_f=m(10cm)^2\omega_f$

Apply conservation of angular momentum, setting these two expressions equal to one another.

$m(20cm)^2(15rpm)=m(10cm)^2\omega_f$

Solve for the final angular velocity.

$w_f=\frac{(20cm)^2(15rpm)}{(10cm)^2}=60rpm$

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Question

An ice skater begins to spin, starting with his arms spread out as far as possible, parallel to the ice. He pulls his arms into his body, and then raises them completely vertically towards the ceiling. As the skater pulls his arms inward, his angular velocity will , and as he raises his arms vertically his angular velocity will .

Answer

Through conservation of angular momentum, as moment of inertia decreases, the angular velocity increases. Moment of inertia is dependent on the distribution of the spinning body's mass away from the center of mass—as the skater brings his arms in, he lowers his moment of inertia, thus increasing his angular velocity. If he raises his arms completely vertically from this point, it does not change the radius of mass distributions (from the center of his body), thus maintaining his angular velocity.

Compare your answer with the correct one above

Question

Sarah spins a ball of mass attached to a string of length around her head with a velocity . If the ball splits in half, losing exactly one-half of its mass instantaneously, what is its new velocity, ?

Answer

By the conservation of angular momentum, the angular momentum , is equal to the product of the mass, angular velocity, and radius (or length of the rope in this case). The equation relating these terms is:

$L_{initial}=m_{i}v_{i}r$

Here, $m_{i}$ is the initial mass, $v_{i}$ is the initial angular velocity, and is the length of the rope, which remains constant. Angular momentum must be conserved, thus:

$L_{initial}=L_{final}$

Substitute.

$m_{i}v_{i}r=m_{f}v_{f}r$

We are given that $m_{f}=\frac{1}{2}m_{i}$ and $v_{f}$ is the final velocity. Plug in and solve.

$m_{i}v_{i}r=\frac{1}{2}m_{i}v_{f}r$

$v_{f}=2v_{i}$

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Question

A child is standing at the center of a frictionless, rotating platform. Both the child and the platform are rotating with and initial angular velocity, $\small \omega_0$ . The child begins to walk slowly toward the edge of the platform. Which quantity will decrease as the child walks?

Answer

Since the platform is frictionless, as the child walks, angular momentum is conserved. However, since there is an $\small \omega ^{2}$ term in the kinetic energy expression, as $\small \omega$ decreased due to the increase of inertia, it affects the energy more, and it decreases. From the reference frame of the disk, the child feels an outward directed force, and thus is doing negative work. This is the same principle that allows ice skaters to increase their spinning speed.

Compare your answer with the correct one above

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