Quiz 8

Physics 6A

3/5/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A solid spherical lump of pizza dough of radius $r_i = 0.19 m$ is thrown in the air. It initially rotates about its center of mass at $\omega = 462$ revolutions per minute. It stretches out, reaching the final shape of a solid circular disk of radius $r_f = 0.76 m$ (with uniform thickness), still rotating about the same axis. Ignore air resistance so that no external forces act on the dough except gravity.

(a) (3 points) What is the net torque acting on the pizza dough about its axis of rotation?

(b) (3 points) What is the change in angular momentum of the dough between its initial and final shape?

(c) (9 points) What is the final angular velocity of the dough in its final shape?

Hint: This is similar to this week's homework problem about star collapse.

Useful information: Moment of inertia of a solid sphere of radius $R$ and mass $m$ is $(2/5) m R^2$. Moment of inertia of a solid circular disk of radius $R$ and mass $m$ is $(1/2) m R^2$.

Solution

(a): Net torque

Gravity is in the same direction as the axis of rotation. So it applies no torque, Since this is the only force acting on the system, the net torque is zero.

(b): Change in angular momentum

If the net torque is zero, the angular momentum is conserved. So the change in angular momentum is zero.

(c): Final angular velocity

The initial angular momentum equals the final angular momentum. For a solid object, $L = I\omega$. So $I_i \omega_i = I_f \omega_f$ or $\omega_f = (I_i/I_f)\omega_i$. $I_i = (2/5)m r_i^2$ and $I_f = (1/2)m r_f^2$. So

$${I_i \over I_f} = {(2/5)m r_i^2 \over (1/2)m r_f^2} = {4\over 5} ({r_i\over r_f})^2 = 0.05$$ (1)

Therefore $\omega_f = (I_i/I_f)\omega_i ~=~ 23.10 rev/min$.

Quiz 8

Physics 6A

3/5/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A solid spherical lump of pizza dough of radius $r_i = 0.18 m$ is thrown in the air. It initially rotates about its center of mass at $\omega = 384$ revolutions per minute. It stretches out, reaching the final shape of a solid circular disk of radius $r_f = 0.54 m$ (with uniform thickness), still rotating about the same axis. Ignore air resistance so that no external forces act on the dough except gravity.

(a) (3 points) What is the net torque acting on the pizza dough about its axis of rotation?

(b) (3 points) What is the change in angular momentum of the dough between its initial and final shape?

(c) (9 points) What is the final angular velocity of the dough in its final shape?

Hint: This is similar to this week's homework problem about star collapse.

Useful information: Moment of inertia of a solid sphere of radius $R$ and mass $m$ is $(2/5) m R^2$. Moment of inertia of a solid circular disk of radius $R$ and mass $m$ is $(1/2) m R^2$.

Solution

(a): Net torque

Gravity is in the same direction as the axis of rotation. So it applies no torque, Since this is the only force acting on the system, the net torque is zero.

(b): Change in angular momentum

If the net torque is zero, the angular momentum is conserved. So the change in angular momentum is zero.

(c): Final angular velocity

The initial angular momentum equals the final angular momentum. For a solid object, $L = I\omega$. So $I_i \omega_i = I_f \omega_f$ or $\omega_f = (I_i/I_f)\omega_i$. $I_i = (2/5)m r_i^2$ and $I_f = (1/2)m r_f^2$. So

$${I_i \over I_f} = {(2/5)m r_i^2 \over (1/2)m r_f^2} = {4\over 5} ({r_i\over r_f})^2 = 0.09$$ (2)

Therefore $\omega_f = (I_i/I_f)\omega_i ~=~ 34.13 rev/min$.

Quiz 8

Physics 6A

3/5/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A solid spherical lump of pizza dough of radius $r_i = 0.17 m$ is thrown in the air. It initially rotates about its center of mass at $\omega = 305$ revolutions per minute. It stretches out, reaching the final shape of a solid circular disk of radius $r_f = 0.85 m$ (with uniform thickness), still rotating about the same axis. Ignore air resistance so that no external forces act on the dough except gravity.

(a) (3 points) What is the net torque acting on the pizza dough about its axis of rotation?

(b) (3 points) What is the change in angular momentum of the dough between its initial and final shape?

(c) (9 points) What is the final angular velocity of the dough in its final shape?

Hint: This is similar to this week's homework problem about star collapse.

Useful information: Moment of inertia of a solid sphere of radius $R$ and mass $m$ is $(2/5) m R^2$. Moment of inertia of a solid circular disk of radius $R$ and mass $m$ is $(1/2) m R^2$.

Solution

(a): Net torque

Gravity is in the same direction as the axis of rotation. So it applies no torque, Since this is the only force acting on the system, the net torque is zero.

(b): Change in angular momentum

If the net torque is zero, the angular momentum is conserved. So the change in angular momentum is zero.

(c): Final angular velocity

The initial angular momentum equals the final angular momentum. For a solid object, $L = I\omega$. So $I_i \omega_i = I_f \omega_f$ or $\omega_f = (I_i/I_f)\omega_i$. $I_i = (2/5)m r_i^2$ and $I_f = (1/2)m r_f^2$. So

$${I_i \over I_f} = {(2/5)m r_i^2 \over (1/2)m r_f^2} = {4\over 5} ({r_i\over r_f})^2 = 0.03$$ (3)

Therefore $\omega_f = (I_i/I_f)\omega_i ~=~ 9.76 rev/min$.

Quiz 8

Physics 6A

3/5/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A solid spherical lump of pizza dough of radius $r_i = 0.15 m$ is thrown in the air. It initially rotates about its center of mass at $\omega = 527$ revolutions per minute. It stretches out, reaching the final shape of a solid circular disk of radius $r_f = 0.6 m$ (with uniform thickness), still rotating about the same axis. Ignore air resistance so that no external forces act on the dough except gravity.

(a) (3 points) What is the net torque acting on the pizza dough about its axis of rotation?

(b) (3 points) What is the change in angular momentum of the dough between its initial and final shape?

(c) (9 points) What is the final angular velocity of the dough in its final shape?

Hint: This is similar to this week's homework problem about star collapse.

Useful information: Moment of inertia of a solid sphere of radius $R$ and mass $m$ is $(2/5) m R^2$. Moment of inertia of a solid circular disk of radius $R$ and mass $m$ is $(1/2) m R^2$.

Solution

(a): Net torque

Gravity is in the same direction as the axis of rotation. So it applies no torque, Since this is the only force acting on the system, the net torque is zero.

(b): Change in angular momentum

If the net torque is zero, the angular momentum is conserved. So the change in angular momentum is zero.

(c): Final angular velocity

The initial angular momentum equals the final angular momentum. For a solid object, $L = I\omega$. So $I_i \omega_i = I_f \omega_f$ or $\omega_f = (I_i/I_f)\omega_i$. $I_i = (2/5)m r_i^2$ and $I_f = (1/2)m r_f^2$. So

$${I_i \over I_f} = {(2/5)m r_i^2 \over (1/2)m r_f^2} = {4\over 5} ({r_i\over r_f})^2 = 0.05$$ (4)

Therefore $\omega_f = (I_i/I_f)\omega_i ~=~ 26.35 rev/min$.

Quiz 8

Physics 6A

3/5/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A solid spherical lump of pizza dough of radius $r_i = 0.14 m$ is thrown in the air. It initially rotates about its center of mass at $\omega = 448$ revolutions per minute. It stretches out, reaching the final shape of a solid circular disk of radius $r_f = 0.42 m$ (with uniform thickness), still rotating about the same axis. Ignore air resistance so that no external forces act on the dough except gravity.

(a) (3 points) What is the net torque acting on the pizza dough about its axis of rotation?

(b) (3 points) What is the change in angular momentum of the dough between its initial and final shape?

(c) (9 points) What is the final angular velocity of the dough in its final shape?

Hint: This is similar to this week's homework problem about star collapse.

Useful information: Moment of inertia of a solid sphere of radius $R$ and mass $m$ is $(2/5) m R^2$. Moment of inertia of a solid circular disk of radius $R$ and mass $m$ is $(1/2) m R^2$.

Solution

(a): Net torque

Gravity is in the same direction as the axis of rotation. So it applies no torque, Since this is the only force acting on the system, the net torque is zero.

(b): Change in angular momentum

If the net torque is zero, the angular momentum is conserved. So the change in angular momentum is zero.

(c): Final angular velocity

The initial angular momentum equals the final angular momentum. For a solid object, $L = I\omega$. So $I_i \omega_i = I_f \omega_f$ or $\omega_f = (I_i/I_f)\omega_i$. $I_i = (2/5)m r_i^2$ and $I_f = (1/2)m r_f^2$. So

$${I_i \over I_f} = {(2/5)m r_i^2 \over (1/2)m r_f^2} = {4\over 5} ({r_i\over r_f})^2 = 0.09$$ (5)

Therefore $\omega_f = (I_i/I_f)\omega_i ~=~ 39.82 rev/min$.

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