A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity `omega_(0)`. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is `K` initially, its final kinetic energy will be

Options

A. 2K
B. `K//2`
C. K
D. `K//4`

Answer

B

Explanation

The angular momentum (L) is conserved, since `tau_("ext")` is zero.
Let `I_(1) = I_(0)`, therefore `I_(r) = 2I_(0)`.
`omega_(1) = omega_(2)`, therefore `omega_(t) = omega` (say) `therefore I_(0)omega_(0) = 2I_(0)omega rArr omega = (omega_(0))/2`
`K=1/2 Iomega^(2), K_(t)=1/2(2I_(0)) xx (omega_(0)/2)^(2) =1/2 xx 2I_(0) xx omega_(0)^(2)/4 =k/2`