The M.I. of a solid cylinder about its axis is I. It is allowed to rool down an incline plane without slipping. If its angular velocity at the bottom be `omega`, then kinetic energy of rolling cylinder will be

Options

A. `I omega^(2)`
B. `(3)/(2)I omega^(2)`
C. `2 I omega^(2)`
D. `(1)/(2)I omega^(2)`

Answer

B

Explanation

`K E_("Roll")=(1)/(2)I omega^(2)+(1)/(2)mv^(2)=(1)/(2)I omega^(2)+(1)/(2) mr^(2)omega^(2)`
`= (1)/(2) I omega^(2)+I Omega^(2)=(3)/(2)I omega^(2)`