Conservation of angular momentum: 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? A) 2.25 rad/s B) 4.60 rad/s C) 6.25 rad/s D) 1.76 rad/s E) 0.810 rad/s

Question

Quizplus · Accepted Answer

According to the conservation of angular momentum, the initial angular momentum of the skater with arms extended is equal to the final angular momentum with arms pulled in. This can be expressed as:

I1ω1 = I2ω2

where I1 and ω1 are the moment of inertia and angular velocity with arms extended, and I2 and ω2 are the moment of inertia and angular velocity with arms pulled in.

Rearranging this equation to solve for ω2, we get:

ω2 = (I1/I2)ω1 = (2.25 kg ∙ m2/s)/(1.8 kg ∙ m2/s) × 5.00 rad/s = 6.25 rad/s

Therefore, the final angular speed is 6.25 rad/s, and the correct choice is C.

Conservation of Angular Momentum: a Figure Skater Rotating at 5