A solid cylinder of mass `M` and radius `R` rolls without slipping down an inclined plane of length `L` and height `h`. What is the speed of its center of mass when the cylinder reaches its bottom

Options

A. `sqrt(4gh)`
B. `sqrt(2gh)`
C. `sqrt((3)/(4)gh)`
D. `sqrt((4)/(2)gh)`

Answer

D

Explanation

When the cylinder is at A, it has only P.E. = Mgh ….(1)
and when it is at C, it has rolling K.E.
For the solid cylinder `I=(MR^(2))/(2)`
Rolling K.E. = Translation K.E. + Rotational K.E.
`=(1)/(2)Mv^(2)+(1)/(2)I omega^(2)`
`=(1)/(2)Mv^(2)+(1)/(2).((MR^(2))/(2))xx(v^(2))/(R^(2))`
`=(3)/(4)Mv^(2) " "` ......(2)
By the principle of conservation of energy
`Mgh=(3)/(4)Mv^(2)`
`therefore v^(2)=(4)/(3)gh` or `v = sqrt((4)/(3)gh)`