JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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For a body rolling along a level surface, the translational and rotational K.E are equal. The body is

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Correct option: C

Disc

Option C (Disc) is correct because for a disc, the moment of inertia is

I = \frac{1}{2}mr^2

, making the translational and rotational kinetic energies equal when rolling.

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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)