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 centre of mass when the cylinder reaches its bottom ?

Options

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

Answer

C

Explanation

Potential energy of the solid cylinder at height h = Mgh
K.E. of centre of mass when reached at bottom
`=1/2Mv^(2)+1/2Iomega^(2)=1/2Mv^(2)+1/2Mk^(2)v^(2)//R^(2)`
`=1/2Mv^(2)(1+(k^(2))/(R^(2)))`
For a solid cylinder, `(k^(2))/(R^(2))=1/2therefore` K.E.`=3/4Mv^(2)`
`thereforeMgh=3/4Mv^(2)rArrv=sqrt(4/3gh)`