Question

A person sits on a rotating stool that is spinning at 10 rpm. They are holding a heavy weight in each hand at a distance of 0.745 m from their axis of rotation. When they symmetrically pull the weights closer in, their angular speed increases to 28.5 rpm. Neglecting the mass of the person and any friction in the stool, how far are the weights from their rotational axis after they pull their hands in? (Note that the moment of inertia of a single object with a mass M a distance R away from a rotational axis is MR2.) O 0.441 m O 0.618 m O 0.745 m O 0.309 m O 0.192 m O 0.882 m

Answer 1

0.441 m

Step-by-step explanation:

Metric unit conversion:

`\omega_1 = 10 rpm = (10 * 2\pi (rad/rotation))/(60 (sec/min)) = (\pi)/(3) rad/s`

`\omega_2 = 28.5 rpm = (28.5 * 2\pi (rad/rotation))/(60 (sec/min)) = (57\pi)/(60) rad/s`

By law of angular momentum conservation:

`I_1\omega_1 = I_2\omega_2`

There I is the moment of inertia of the object and
`I = MR^2`

`MR_1^2\omega_1 = MR_2^2\omega_2`

`R_2^2 = (R_1^2\omega_1)/(\omega_2)`

`R_2 = R_1\sqrt{(\omega_1)/(\omega_2)}`

`R_2 = 0.745\sqrt{(\pi/3)/(57\pi/60)}`

`R_2 = 0.745\sqrt{(20)/(57)} = 0.441 m`