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 moment of inertia of the system doubles. The kinetic energy of the system now is:

2 K

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\frac{K}{2}

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\frac{K}{4}

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4 K

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Solution

The correct option is B $\frac{K}{2}$
By angular momentum conservation: $I ω_{o} = I_{1} ω_{1}$
So, $I_{1} = 2 I$ , $ω_{1} = \frac{ω_{o}}{2}$
Initial $K E = \frac{I ω_{o}^{2}}{2} = K$
Final $K E = \frac{I_{1} ω_{1}^{2}}{2} = \frac{2 I (\frac{ω_{o}}{2})^{2}}{2} = \frac{K}{2}$

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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 moment of inertia of the system doubles. The kinetic energy of the system now is:

The correct option is B K2By angular momentum conservation: Iωo=I1ω1So, I1=2I , ω1=ωo2Initial KE=Iω2o2=KFinal KE=I1ω212=2I(ωo2)22=K2

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 moment of inertia of the system doubles. The kinetic energy of the system now is:

The correct option is B $\frac{K}{2}$
By angular momentum conservation: $I ω_{o} = I_{1} ω_{1}$
So, $I_{1} = 2 I$ , $ω_{1} = \frac{ω_{o}}{2}$
Initial $K E = \frac{I ω_{o}^{2}}{2} = K$
Final $K E = \frac{I_{1} ω_{1}^{2}}{2} = \frac{2 I (\frac{ω_{o}}{2})^{2}}{2} = \frac{K}{2}$