A man stands on a rotating platform with his arms outstretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30rpm. The man then brings his arms back to his body with the distance of each weight from the axis changing from 0.90m to 0.20m. The moment of inertia of the man together with the platformmay be taken to be constant and equal to 7.6kgm^2 Is K.E conserved in the process ? If not, from where does the change come about ?

A man stands on a rotating platform with his arms outstretched horizon

1.	A man stands on a rotating platform with his arms outstretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30rpm. The man then brings his arms back to his body with the distance of each weight from the axis changing from 0.90m to 0.20m. The moment of inertia of the man together with the platformmay be taken to be constant and equal to 7.6kgm^2 Is K.E conserved in the process ? If not, from where does the change come about ?
Answer» Solution :From the LAW of conservation of angular momentum `Iomega=I'OMEGA'` i.e `(2xx5xx0.9+7.6)30=(2xx5xx0.2xx0.2+7.6)w'` `:.omega'=(15.7xx30)/(8)=58.875` i.e `omega'~~59rpm`