Question

Calculate the angular momentum (in kg·m2/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.390 kg·m2.
(b)
He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kg·m2) if his angular velocity drops to 2.40 rev/s.
(c)
Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque (in N·m) was exerted if this takes 18.0 seconds?

Answer 1

(a) Convert 6.00 rev/s to rad/s: ω1 = 6.00(2π) = 12π rad/s.
L = Iω = 0.390·(12π) = 14.7 kg·m^2/s (to 3 s.f.).

(b) Angular momentum conserved: I2 = L/ω2. With ω2 = 2.40(2π) = 4.8π rad/s,
I2 = [0.390·(12π)]/(4.8π) = 0.975 kg·m^2.

(c) ωi = 12π rad/s, ωf = 3.00(2π) = 6π rad/s. ΔL = I(ωf − ωi) = 0.390(6π − 12π) = −2.34π ≈ −7.35 kg·m^2/s.
Average torque τ = ΔL/Δt = (−7.35)/18.0 ≈ −0.409 N·m (magnitude 0.409 N·m, opposite the rotation).