A puck moves on a horizontal air table. It is attached to a string that passes through a hole in the center of the table. As the puck rotates about the hole, the string is pulled downward very slowly
And shortens the radius of rotation, so the puck gradually spirals in towards the center. By what
Factor will the puck's angular speed have changed when the string's length has decreased to
One-half of its original length?
A) 1
B) 4
C) $\sqrt { 2 }$
D) 1/2
E) 2

Question

A puck moves on a horizontal air table. It is attached to a string that passes through a hole in the center of the table. As the puck rotates about the hole, the string is pulled downward very slowly
And shortens the radius of rotation, so the puck gradually spirals in towards the center. By what
Factor will the puck's angular speed have changed when the string's length has decreased to
One-half of its original length? 
A) 1
B) 4
C) $\sqrt { 2 }$
D)  1/2 
E) 2

Quizplus · Accepted Answer

The angular momentum of the puck is conserved because the net external torque acting on the puck is zero. Angular momentum $L$ is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular speed. For a point mass moving in a circle, $I = mr^2$, so $L = mr^2\omega$. When the radius $r$ is halved, the new moment of inertia is $I' = m(\frac{r}{2})^2 = \frac{1}{4}mr^2$. To conserve angular momentum, if $I$ is reduced by a factor of 4, $\omega$ must increase by a factor of 4 to keep $L$ constant. Hence, the puck's angular speed increases by a factor of 4.

A Puck Moves on a Horizontal Air Table 2\sqrt { 2 }2​

A Puck Moves on a Horizontal Air Table $\sqrt { 2 }$