A thick walled hollow sphere has outer radius `R`. It rolls down an inclined plane without slipping and its -speed at the bottom is `v`. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom `5v//4`. What is the radius of gyration of the sphere?

Options

A. `3R_(0)//2`
B. `3R_(0)//4`
C. `9R_(0)//16`
D. `3R_(0)`

Answer

B

Explanation

When body rolls down on inclined plane with velocity `V_(0)` at bottom then body has both rotational and translational kinetic energy.
Therefore, by law of conservation of energy,
`P.E.=K.E_("trans")+K.E_("rotational")=(1)/(2)mV_(0)^(2)+(1)/(2)Iomega_(2)`
`=(1)/(2)mV_(0)^(2)+(1)/(2)mk^(2)(V_(0)^(2))/(R_(0)^(2))" "...(i) [because I=mk^(2),omega=(V)/(R_(0))]`
When body is sliding down then body has only translatory motion.
`therefore P.E.=K.E_("trans")=(1)/(2)m((5)/(4)v_(0))^(2)" "...(ii)`
Dividing (i) by (ii) we get
`(P.E.)/(P.E.)=((1)/(2)mv_(0)^(2)[1+(K^(2))/(R_(0)^(2))])/((1)/(2)xx(25)/(16)xxmV_(0)^(2))=(25)/(16)=1+(K^(2))/(R_(0)^(2))`
`implies (K^(2))/(R_(0)^(2))=(9)/(16)or,K=(3)/(4)R_(0)`