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

Here, `I_(1) = 7.6 xx 2 xx 5 (0.9)^(2) = 15.7 kg m^(2), omega_(1) = 30` rpm
`I_(2) = 7.6 + 2 xx 5(0.2)^(2) = 8.0 kg m^(2), omega_(2) = ?`
According to the principle of conservation of angular momentum,
`I_(2) omega_(2) = I_(1)omega_(1)`
`omega_(2) = (I_(1))/(I_(2))omega_(1) = (15.7 xx 30)/(8.0) = 58.88`rpm
No, kinetic energy is not conserved in the process. Infact, as moment of inertia decreases. K.E. of rotation increases. This change comes about as work is done by the man in bringing his arms closer to his body.