A man stands in the centre of a Zhukovskii turntable (a rotating platform with frictionless bearings) and rotates with it at 30 r.p.m. The moment of inertia of the man's body with respect to the axis of rotation is about `1.2 kg.m^2` . The man holds in his outstretched hands two weights of mass 3 kg each. The distance between the weights is 160 cm. What will be the change in the speed of rotation of the system, if the man lets his hands fall so that the distance between the weights becomes 40 cm? The moment of inertia of the turntable is 0.6 kg. `m^2` the change in the moment of inertia of the man's hands and the friction are to be neglected,

Answer

83 r.p.m

Explanation

From the law of conservation of momentum
`(I_(man)+I_(b)+2mr_1^2)omega_1=(I_(man)+I_b+2mr_2^2)omega_2`