Question

(a) What is the angular momentum of a figure skater spinning (with arms in close to her body) at 2.0 rev/s, assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 16 cm, and a mass of 55 kg.

_______ kg·m²/s

(b) How much torque is required to slow her to a stop in 5.0 s, assuming she does not move her arms?
_______ m·N

Answer 1

part (A) according to L = I * w

where L = angular momentum I = moment of inertia w = angular velocity

where we have 2 rev/s , therefore w = 2*pi*2 rad/s = 4*pi rad/s.

now I = (mr²)/2 , about the axis of rotation (assuming axis to be about head leg axis)

therefore I = [ 55 kg * (16 cm )^2 ]/2 = 0.704 kg/m²

therefore, L = I*w = 0.704 * (4*pi) = 8.846 kg.m2/s

part (B) torque needed to stop skater in 5 sec.

using equation w_final = w_initial + a*t where a is the rotational acceleration and wfinal is final rotational angular velocity and winitial is initial rotational angular velocity.

wfinal = 0 and winitial = 4*pi rad/s t= 5 sec

threfore we get a = [ - 0.8*pi] = -2.513 rad/s2

therefore torque is equal to T = I*a where I = moment of inertia about axis of rotation

T = 0.704 *( 2.513 ) = 1.769 m.N