A plank $P$ is placed on a solid cylinder $S$ of same mass, which rolls on a horizontal surface. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of $P$ to the kinetic energy of $S$ is:

1 : 1

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2 : 1

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8 : 3

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11 : 8

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Solution

The correct option is C $8 : 3$

As there is no slipping at A and B i.e. $v = R ω$
Also $v_{P} = v + R ω = 2 v$
$K . E_{P} = \frac{1}{2} m v_{p}^{2} = 2 m v^{2}$
$K . E_{S} = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}$
$K . E_{S} = \frac{1}{2} \times \frac{1}{2} m R^{2} ω^{2} + \frac{1}{2} m v^{2}$
$K . E_{S} = \frac{3}{4} m v^{2} (As v = R ω)$
Thus, $K . E_{P} : K . E_{S} = 8 : 3$

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A plank P is placed on a solid cylinder S of same mass, which rolls on a horizontal surface. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of P to the kinetic energy of S is:

The correct option is C 8:3 As there is no slipping at A and B i.e. v=Rω Also vP=v+Rω=2v K.EP=12mv2p=2mv2 K.ES=12Iω2+12mv2 K.ES=12×12mR2ω2+12mv2 K.ES=34mv2 (As v=Rω) Thus, K.EP:K.ES=8:3

A plank $P$ is placed on a solid cylinder $S$ of same mass, which rolls on a horizontal surface. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of $P$ to the kinetic energy of $S$ is: