A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m².If he pulls in his arms so his moment of inertia decreases to 1.80 kg ∙ m²,what will be his new angular speed? A)2.25 rad/s B)4.60 rad/s C)6.25 rad/s D)0.81 rad/s

Question

Quizplus · Accepted Answer

The conservation of angular momentum states that the product of the moment of inertia and the angular velocity remains constant in the absence of external torques. Therefore, we can write:
I1ω1 = I2ω2
where I1 and ω1 are the initial moment of inertia and angular velocity, and I2 and ω2 are the final moment of inertia and angular velocity.
Using the values given in the problem, we get:
(2.25 kg∙m2) (5.00 rad/s) = (1.80 kg∙m2)ω2
Solving for ω2, we get:
ω2 = (2.25 kg∙m2) (5.00 rad/s) / (1.80 kg∙m2) = 6.25 rad/s
Therefore, the answer is choice C.

A Figure Skater Rotating at 5