physics question (idk how to do them :( )

Question

An ice skater (m=76.4kg) is rotating in at 1.00 rad/s while moving linearly 1.50m/s. She pulls in her arms and reduces her moment of inertia from 3.60 kgm^2 to 0.300 kgm^2.

a. Neglecting friction, what is her rotational velocity after this? [ans: 12.0 rad/s]

b. What is her initial and final kinetic energy? Hint: she is not just rotating! [ans: K = 88J, K’ = 108J]

Answer 1

For the first part, you use the principle of conservation of angular momentum.

Angular momentum = moment of inertial x angular velocity.

I₁ω₁ = I₂ω₂ angular momentum 1 = angular momentum 2 solve for ω₂

ω₂ = I₁ω₁ / I₂ = 3.60 kgm^2 x 1.00 rad/s / 0.300 kgm² = 12 rad/s

KE = translational KE + rotational KE = 1/2 mv² + 1/2 I ω²

Initial KE = 1/2 x 76.4 kg x 1.50² m/²s² + 1/2 x 3.6 kgm² x 1.0 ² rad²/s² = 88 joules

You can try the second part yourself using the same approach as above.