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?

Question

Physics

Gianni Sharp

4 April, 03:28

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?

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Answers (1)

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Howard Cannon · Answer 1

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