A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg/m2.?

If the arms are pulled in so the moment of inertia decreases to 1.80 kg/m2. What is the final angular speed in rad/s?

2 Answers

  • As external torque is absent, angular momentum (L) is conserved

    moment of inertia with arms extended I1 = 2.25 kg/m2.

    angular velocity w1=5.00 rad/s

    Final angular momentum L2=I2w2=2.25*5=11.25 Js

    moment of inertia with arms pulled I2 = 1.80 kg/m2.

    final angular velocity w2= ?

    Final angular momentum L2=I2w2 =1.80*w2 Js

    As angular momentum is conserved,

    Initial angular momentum=Li=I1w1=final angular momentum=L2=I2w2

    1125=180w2

    w2=1125/180=6.25 rad/s

    final angular speed is 6.25 rad /s

  • I1W1=I2W2

    I1=2.25, I2=1.80 W1=5.00 W2=???

    so W2=2.25*5/1.8=6.25rad/sec

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