A figure skater is spinning with an initial angular speed of 9.4rad/s. He then pulls his arms in, reducing his moment of inertia to 0.30 times its original value.

Question

What is his angular speed after pulling in his arms?

Answer 1

So long as there are no external forces slowing the system down (like friction or someone trying to rotate the figure skater externally), the angular momentum of the skater will be conserved.

Angular momentum is given by L = I * (omega), where I is the moment of inertia (usually in kg*m^2) and omega is the angular velocity in rad/s.

If angular momentum is conserved, then L1 = L2 before and after he pulls his arms in.

L1 * (omega1) = L2 * (omega2)

If he reduces his moment of inertia to 0.3 times its initial value, then we can say I2 = 0.3*I1 and substitute that into our equation:

L1 * (omega1) = 0.3 * L1 * (omega2)

L1 cancels from both sides and we have:

omega1 = 0.3 * omega2

9.4 rad/s = 0.3 * omega2

so omega2 = 31.33 rad/s

If we consider this intuitively, when the skater brings their hands inward they reduce their moment of inertia. This means that they should rotate more quickly in exchange in order for total angular momentum to stay the same.