A plank P is placed on a solid cylinder S, which rolls on a horizontal surface. The two are of equal mass. 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
$K E o f P = \frac{1}{2} m (2 v)^{2} = 2 m v^{2}$

$K E o f S = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{3}{4} m v^{2} \Rightarrow \frac{K E o f P}{K E o f S} = \frac{8}{3}$

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A plank P is placed on a solid cylinder S, which rolls on a horizontal surface. The two are of equal mass. 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 : 3KE of P=12m(2v)2=2mv2 KE of S=12mv2+12Iω2=34mv2⇒KE of PKE of S=83

The correct option is C 8 : 3
$K E o f P = \frac{1}{2} m (2 v)^{2} = 2 m v^{2}$

$K E o f S = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{3}{4} m v^{2} \Rightarrow \frac{K E o f P}{K E o f S} = \frac{8}{3}$