A 5.0 kg skater begins a spin with an angular speed of 4 rad/s. By changing the position of her arms, the skater decreases her moment of inertia to one half its initial value. What is the skater's final angular speed? I thank you in advance for your time. JL

1 answer

I omega = constant angular momentum
I/2 (2 omega) = I omega
twice 4 = 8 radians/second