Question

A person holding weights sits on a spinning stool. Then the person pulls the weights in,decreasing the rotational inertia of the system of the man, the weights,and the stool by a factor of 2. The stool can rotate about its axle with negligible friction.How does the kinetic energy of the system change?

Answer 1

Ķ rot = 2K rot

When the arms and weight are retracted toward the body, the energy increases by twice the initial kinetic energy because the moment of inertia has decreased coupled with the energy needed to bring the hands towards the body.

Step-by-step explanation:

K rot = kinetic energy of system when hands + weight are stretched out = ½ Iw²

Where w = angular velocity when arms and weight are stretched out

I = inertia of the system = ( Ie + Ip) where Ie= moment of inertia of the extra weight carried and Ip = moment of inertia of the person.

Ķ rot = final rotational energy when arms and external weight are pulled in = ½Ìŵ² where Ì =( Ìe + Ìp)

And Ìe = inertia of weight when hands are retracted and Ìp = inertia of person when hand are retracted

Using conservation of angular momentum which is

Iw = Ìŵ

Substitute ŵ = (I/Ì)*W in Ķ rot to give

Ķ rot =½Ì *[ (I/Ì) * W ]²

Ķ rot = ½ ( I²/Ì) * w²

Since K rot = ½ I w²

So Ķ rot = K rot ( I/Ì )...eq3

From the question above system inertia reduces by a factor of 2

Ì = ½I

Sub expression into equ3

Ķ rot = 2K rot