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A plank $P$ is placed on a hollow cylinder $C,$ which rolls on a horizontal surface as shown. No slippage is there at any of the surfaces in contact. Both have equal mass say $M$ (each) and if $v$ is the velocity of center of mass of the cylinder $C,$ , then the ratio of the kinetic energy of plank $P$ to the cylinder $C,$ is :

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$1 : 1$

$2 : 1$

$3 : 8$

$8 : 11$

answer is B.

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Detailed Solution

Velocity of plank will be $2 v$ .

Kinetic energy of plank, $K_{p l a n k} = \frac{1}{2} M {(2 v)}^{2} = 2 M v^{2}$

Kinetic energy of hollow cylinder, $K_{c y l i n d e r} = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2} = M v^{2}$

$\frac{K_{p l a n k}}{K_{c y l i n d e r}} = 2$

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