A solid cylinder rolls without slipping down an inclined plane of height h. The velocity of the cylinder when it reaches the bouom is

Options

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

Answer

B

Explanation

Let M and R be the mass and radius of the solid cylinder respectively .
According to conservation of mechanical energy ,
`Mgh = (1)/(2) Mv^(2) + (1)/(2) I omega^(2) = (1)/(2) Mv^(2) + (1)/(2) ((1)/(2) MR^(2)) omega^(2)`
= `(1)/(2) Mv^(2) + (1)/(4) MR^(2) ((v)/(R))^(2) = (1)/(2) Mv^(2) + (1)/(4) Mv^(2) = (3)/(4) Mv^(2)`
or `v^(2) = (4)/(3) gh = v = sqrt((4)/(3) gh)`