A child is standing with folded hands at the center of a platform rotating about its central axis. The kinetic energy of the system is `K`. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is

Options

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

Answer

b

Explanation

Initial kinetic energy, `K=1/2Iomega^(2)`……….(i)
According to the principle of conservation of angualr momentum, `Iomega` =constant
As Iis doubled, `omega` becomes half.
So final kinetic energy,
`K^(') = 1/2(2I)(omega/2)^(2)=1/4 Iomega^(2) = K/2` (using (i))