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A

Anonymous

Oct 11, 2024

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

A figure skater is rotating with an initial angular speed of 5.00 rad/s and a moment of inertia of 2.25 kg/m². If the skater pulls in their arms, reducing the moment of inertia to 1.80 kg/m², what will be the final angular speed in rad/s?

2 Answers

A
Anonymous

Jan 29, 2025

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

A
Anonymous

Jan 04, 2025

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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