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
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
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