A man stands on a rotating platform, with his arms stretched horizontal holding a `5 kg` weight in each hand. The angular speed of the platform is `30` revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from `90 cm` to `20 cm`. moment of inertia of the man together with the platform may be taken to be constant and equal t `7.6 kg m^(2)`. (a) What is the his new angular speed ? (Neglect friction.)
(b) Is kinetic energy conserved in the process ? If not, from where does the change come about ?

Explanation

Let `I_(1) and I_(2)` be initial and final moment of inertia of man `omega_(2) and omega_(2)` be initial and final angular velocities.
Given
`I_(1)=M.I.` of man and platform + M.I. of two
5 kg weights
`7.6+2xx2xx(0.2)^(2)=15.7 kg m^(2)`
`omega_(1)30 rpm`
similarly, `I_(1)=7.6+2xx5(0.2)^(2)=8.0 kg m^(2)`
From law of conservation of angular momentum
`I_(1) omega_(1)=I_(2) omega_(2)`
so, `omega_(2)=(I_(1)omega_(1))/(I_(2))`
`=(8xx(59)^(2))/(15.7xx(30)^(2))=58.88 rpm=59 rpm`
`("Initial kinetic energy")/("Final kinetic energy")=((1)/(2)I_(1) omega_(1)^(2))/((1)/(2)I_(2)omega_(2_)^(2))=(8xx(59)^(2))/(15.7xx(30)^(2))=1.97`
Thus, final kinetic energy is approximately double than the initial energy. So, we find that kinetic energy is not conserved. Here, the man uses his internal energy to increase his rotational kinetic energy.