A person is spinning with his hands outstretched at the rate of 4 rads s−1. When he brings his hands close to the body, he spins at the rate of 16 rad s−1 . The ratio of M.1 . in the two cases successively is
A
4:1
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B
1:14
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C
16:1
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D
1:2
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Solution
The correct option is C4:1
Let, the initial moment of inertia of person (incase(1))=I0. and, given initial rate of rotation(ω1)=4 rad/s. Let, the final moment of inertia (case(2))=If and, given final angular velocity =16 rad/s. (ω2)
By angular momentum conservation Principal, I0ω1=Ifω2⇒I0×4=IF×16. Thus, I0If=41