Question

A man stands on a frictionless platform that is rotating with an angular speed of 3.5 rad/s; his arms are outstretched and he holds a weight in each hand. With his hands in this position the rotational inertial of the system of man, weights, and platform is 6.0 kg*m².

1. If by moving the weights the man decreases the rotational inertial of the system to 3.5 kg*m², what is the resulting angular speed of the platform?

2. What is the change of the kinetic energy when the man moves his arms?

Answer 1

here,

the initial angular speed ,w0 = 3.5 rad/s

the initial moment of inertia , I0 = 6 kg.m^2

1)

the final moment of inertia , I = 3.5 kg.m^2

let the new angular speed be w

using conservation of angular momentum

I0 * w0 = I * w

6 * 3.5 = 3.5 * w

w = 6 rad/s

the new angular speed is 6 rad/s

2)

the change of the kinetic energy , KE = KEf - KEi

KE = 0.5 * ( I * w^2 - I0 * w0^2)

KE = 0.5 * ( 6^2 * 3.5 - 3.5^2 * 6) J

KE = 26.25 J