Question

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.410 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 if his angular velocity drops to 1.70 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 was exerted if this takes 14.0 seconds?

Answer 1

Angular momentum
\[L=I*\omega\]\[=0.410*（ 6*2\pi）\approx 15.45\]
conservation of angular momentum
\[15.45=I _1*1.7*2\pi\]\[I _1 \approx 1.446\]
\[\tau=I*（V_1-V_2）/T=0.415*（6*2\pi-3*2\pi）/14 \approx 0.558\]