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

Answer

C

Explanation

( c) From law of conservation of energy can neither be created nor be destroyed but it remains conserved In the given case the sum of kinetic energy of rotation and translation is converted to potential energy.
Also moment of inertia of disc is
`I = (2)/(5) MR62`
`(1)/(2) mv_("Translatonal kinetic energy")^2 + (1)/(2) I omega_(Rotational energy)^2 = mgh_("Potential energy")`
`rArr (1)/(2) mv^2 + (1)/(2) ((2)/(5) MR^2) (v^2)/(R^2) = mgh`
where `v = R omega, omega` = angular velocity
`rArr (7)/(10) mv^2 = mgh`
`rArr v = sqrt((10)/(7)) gh`
Hence, to climb the inclined surface the velocity should be greater than `sqrt((10)/(7)) gh`.