JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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A ballet dancer, dancing on a smooth floor is spinning about a vertical axis with her arms folded with an angular velocity of

20\;rad/s

. When she stretches her arms fully, the spinning speed decrease in

10\;rad/s

. If I is the initial moment of inertia of the dancer, the new moment of inertia is

View answer and explanation

Correct option: A

2\,I

Option A (

2\,I

) is correct because angular momentum is conserved, and when the angular velocity halves, the moment of inertia doubles.

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\mathrm{AB}

\mathrm{L}

\mu(x)=a+\frac{b x}{L}

x

\mathrm{A}

\mathrm{CM}

\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)