Question

The moment of inertia of an ice skater is 0.400 kg·m2 when he is spinning at 6.00 rev/s. (a) He reduces 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 decreases to 1.25 rev/s. (b) Suppose instead he keeps his arms in and allows friction on the ice to slow him to 3.00 rev/s What average torque was exerted if this takes 15.0 s?

Answer 1

here,

the initial moment of inertia , I0 = 0.4 kg.m^2

the initial angular speed , w0 = 6 rev /s = 37.68 rad/s

a)

when the new angular speed , w = 1.25 rev/s = 7.85 rad/s

let the new moment of inertia be I

using conservation of angular momentum

I0 * w0 = I * w

0.4 * 6 = I * 1.25

I = 1.92 kg.m^2

the new moment of inertia is 1.92 kg.m^2

b)

when the new angular speed , w = 3 rev /s

w = 18.84 rad/s

t = 15 s

let the angular acceleration be alpha

w = w0 + alpha * t

18.84 = 37.68 + alpha * 15

alpha = - 1.26 rad/s^2

the average torque required , T = I * alpha

T = 0.4 * (-1.26) N.m

T = - 0.502 N.m