A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity `v m//s`. If it is to climb the inclined surface then `v` should be :

Options

A. `2gh`
B. `(10)/(7)gh`
C. `gt sqrt(2gh)`
D. `ge sqrt((10)/(7)gh)`

Answer

D

Explanation

When the solid sphere rolls, its rolling K.E. at A.
`K_(A)=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
`=(1)/(2)mv^(2)+(1)/(2).(2)/(5)mR^(2).(v^(2))/(R^(2))`
`=(1)/(2)mv^(2)+(1)/(5)mv^(2)=(7)/(10)mv^(2)`
Its P.E. at B = mgh
Thus to reach `B, (7)/(10)mv^(2)=mgh`
`therefore v^(2)=(10gh)/(7)" "therefore v=sqrt((10gh)/(7))`
`therefore` This is the minimum velocity
`therefore" v should be " le sqrt((10)/(7)gh)`