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Q.

The moment of inertia of solid cylinder about its axis is I. It is allowed to roll down an inclined plane without slipping. If its angular velocity at the bottom be $ω$ then kinetic energy of the cylinder will be

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a

$\frac{1}{2} {Iω}^{2}$

b

${Iω}^{2}$

c

$\frac{3}{2} {Iω}^{2}$

d

$2 {Iω}^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} K.E. = \frac{1}{2} {Iω}^{2} + \frac{1}{2} {Mv}^{2} \\ K.E. = \frac{1}{2} {Iω}^{2} + \frac{1}{2} M (R^{2} ω^{2}) \\ K.E. = \frac{1}{2} {Iω}^{2} + (\frac{1}{2} {MR}^{2}) ω^{2} \end{array}$
For a cyclinder, $I = \frac{1}{2} {MR}^{2}$
$∴ K.E. = \frac{1}{2} {Iω}^{2} + {Iω}^{2} = \frac{3}{2} {Iω}^{2}$

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The moment of inertia of solid cylinder about its axis is I. It is allowed to roll down an inclined plane without slipping. If its angular velocity at the bottom be ω then kinetic energy of the cylinder will be