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26 July, 09:05

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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  1. 26 July, 11:55
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    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
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