Question

(m = 4.0 kg) the cockroach rides on the rim (at radius R) of a disk (M = 6.0 kg) that turns about its center like a merry-go-round at the rate of 2.0 rad/s. She then crawls from the rim to a point that is 0.50R from the center. At what rate (rad/s) does she and the disk then turn? [Chapter 11, Conservation of angular momentum.]

Answer 1

Answer:

ω' = 2.5 rad/s

Explanation:

mass of cockroach, m = 4 kg

mass of disk, M = 6 kg

Radius of disc= R

initial angular velocity, ω = 2 rad/s

Let the final angular velocity is ω'

As no external torque is applied, so the angular momentum is constant.

Angular momentum = Moment of inertia x angular velocity

I ω = I' ω'

$\frac{1}{2}\left ( M+m \right )R^{2}\times {2} = \left (\frac{1}{2}MR^{2}+m(0.5R)^{2})  \right )\omega '$

$10R^{2} = 4R^{2}\omega '$

ω' = 2.5 rad/s