Angular Momentum?

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 12 cm, and a mass of 55 kg.
kg*m^2/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

(Inertia) i = 1/2 m r^2 (for cylinder) = 1/2 * 55 * (.12)^2 = 0.396 Kg m^2

L = w * I

L = 2.0 rev/sec * ( 2 * pi ) rad/rev * 0.396 Kg M^2 = 4.976

The units for angular momentum are: (kg m2)(radians/sec) = kg m2/sec.
(b) The angular acceleration of an object is given by Newton's second law in rotational form

α = τ/I

where τ is torque applied, and I is moment of inertia

Then the angular speed (the rate of rotation) after time t is given by

ω = α*t = τ*t/I (1)

Torque = W / t * i = 2*2*pi (rad/sec) / 5 sec * 0.396 Kg M^2 = 4.95.