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

(i) Here, `I_(1) = 7.6 kg m^(2) + 5 kg xx (0.9 m)^(2) + 5 kg xx (0.9 m)^(2) = 15.7 kg m^(2)`
`I_(f) = 7.6 kg m^(2) + 5 kg xx (0.2 m)^(2) + 5kg xx (0.2 m)^(2) = 8.0 kg m^(2)`
`omega_(i) = 30 rpm`.
Let `omega_(f)` be the new angualr speed.
Applying law of conservationof angular momentum,
`I_(1)omega_(i) = I_(f)omega_(f)` or `omega_(f) =(I_(i)/I_(f))omega_(i)` or `omega_(f) =(15.7)/(8.0)(30) = 59 rpm`
(ii) `("Final KE of rotation")/("Initial KE of rotation") =((1//2) I_(f)omega_(f)^(2))/((1//2)I_(1)omega_(i)^(2)) = (I_(f))/(I_(i)) ((omega_(f))/(omega_(i))^(2))`
`=((8.0)/(15.7))(59/30)^(2) ~~ (1/2) (2)^(2)=2`
Final kinetic energy increases because of energy lost by man.