Question

A 50.0-kg skater begins a spin with an angular speed of 4.0 rad/s. By changing the position of her arms, the skater decreases her moment of inertia to one-half its initial value. What is the skater's final angular speed?

Answer 1

Answer:

The Skater's final angular speed =8rad/s

Explanation:

Let initial momentum of inertia be I1

Let final momentum of inertia be I2

But I2= I1/2

It is decreased by one-half

There is no external torque on the system, therefore

I1w1=I2w2

I1×4=(I1/2)×w2

Dividing through by I1 on both sides leaves you with:

4=w2/2

Cross multiply

W2=4×2=8rad/s