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In the figure mass of both spherical body and block is m. Moment of inertia of the spherical body about center of mass is 2mR2. The spherical body rolls on the horizontal surface. There is no slipping at any surfaces in contact. The ratio of kinetic energy of the spherical body to that of block is

 

a
34
b
13
c
23
d
12

detailed solution

Correct option is C

Let v be the linear velocity of center of mass of the spherical body and ω its angular velocity about centre of mass. ThenAs there is no slipping at B,  v=ω2R ⇒ω=v2RK.E. of spherical body,K1=12mv2+12Iω2=12mv2+12(12m(2R)2)v24R2=34mv2   ...............(i)Let say speed of the block is v'. There is no slipping at A ,  v+Rω=v' ⇒R2ω+Rω=v'⇒v'=(ω)(3R)=3Rω=(3R)v2R=32v∴ K.E. of block K2=12mv′2=12m32v2=98mv2…........ (ii)  From equations (i) and (ii), K1K2=23

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Similar Questions

In both the figures all other factors are same, except that in Figure (i) AB is rough and BC is smooth while in Figure (ii) AB is smooth and BC is rough. In Figure (i), if a sphere is released from rest it starts rolling. Now consider the Figure (ii), if same sphere is released from top of the inclined plane, what will be the kinetic energy of the sphere on reaching the bottom:


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