Answer:
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.
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
