Rotational Motion Question 221

Question: The spherical body and block is m. Moment of inertia of the spherical body about centre 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

Options:

A) 3/4

B) 1/3

C) 2/3

D) 1/2

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Answer:

Correct Answer: C

Solution:

[c] Let v be the linear velocity of centre of mass of the spherical body and w its angular velocity about centre of mass.

Then ω=v2R

KE of spherical body

K1=12mv2+12Iω2

K1=12mv2+12(2mR)2(v24R2)=34mv2?.

(i) Speed of the block will be

v=(ω)(3R)=3Rω=(3R)(v2R)=32v

KE of block K2=12mv2

=12m(32v)2=98mv2

From equations (i) and (ii), $

K1K2=23



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