Question

A hamster sits 0.10 m from the center of a lazy Susan of negligible mass. The wheel has an angular velocity of 1.0 rev/ s. How will the angular velocity of the lazy Susan change if the hamster walks to 0.30 m from the center of rotation? Assume zero friction and no external torque.A. It will speed up to 2.0 rev/s.B. It will speed up to 9.0 rev/s.C. It will slow to 0.01 rev/s.D. It will slow to 0.02 rev/s.

Answer 1

C. It will slow to 0.1 rev/s.

Step-by-step explanation:

As we know that there is no external torque on the system

so here angular momentum must be conserved

so we will have

`I_1\omega_1 = I_2\omega_2`

here initially he is at distance of r = 0.10 m from the center

so we have

`I_1 = m_1r_1^2`

`I_1 = m(0.10)^2`

now when he walks to distance of r = 0.30 m from center then

`I_2 = m(0.30)^2`

now we have

`m(0.10)^2 (1 rev/s) = m(0.30)^2\omega`

`\omega = (0.01)/(0.09)(1 rev/s) = 0.11 rev/s`