JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A diver having a moment of inertia of

6.0 \mathrm{~kg}-\mathrm{m}^2

about an axis through its centre of mass rotates at angular speed of

2 \mathrm{~rad} / \mathrm{s}

about this axis. If he folds his hands and feet to decrease the moment of inertia to

5.0 \mathrm{~kg}-\mathrm{m}^2,

what will be the new angular speed?

View answer and explanation

Correct option: A

2.4 \mathrm{~rad} / \mathrm{s}

Option A (

2.4 \mathrm{~rad} / \mathrm{s}

) is correct because the angular speed is calculated using the conservation of angular momentum.

More Questions from SYSTEM OF PARTICLES

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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)