A body of mass $M$ and radius $r$ , rolling with velocity $v$ on a smooth horizontal floor, rolls up a rough irregular inclined plane up to a vertical height $(\frac{3 v^{2}}{4 g})$ . Compute the moment of inertia of the body and comment on its shape.

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$I = \frac{1}{2} M v^{2}$ This result clearly indicates that the body is not a cylinder.

$I = (\frac{1}{2}) M r^{2}$ This result clearly indicates that the body is either a disc or a cylinder.

$I = \frac{M R^{2}}{2}$ This result does not clearly indicates that the body is either a disc or a cylinder.

$I = M R^{2}$ This result clearly indicates that the body is not a disc

answer is A.

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Detailed Solution

The total Kinetic energy of the body $E = E_{1} + E_{1} = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}$

$\Rightarrow E = \frac{1}{2} M v^{2} [1 + (\frac{I}{M r^{2}})] [a s v = r ω]$

When it rolls up on an irregular inclined plane of height $h = (\frac{3 v^{2}}{4 g})$ , its $K E$ is fully converted into $P E$ , so by conversation of mechanical energy $\frac{1}{2} M v^{2} [1 + \frac{I}{M r^{2}}] = M g [\frac{3 v^{2}}{4 g}]$ which on simplification gives $I = (\frac{1}{2}) M r^{2}$ . This result clearly indicates that the body is either a disc or a cylinder.