A man stands on a rotating platform, with his arms stretched horizontally 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 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to `7.6 kg m^(2)`.
(a) What is 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.