A solid cylinder of mass `M` and radius `R` rolls down an inclined plane of height `h` without slipping. The speed of its centre when it reaches the bottom is.

Options

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

Answer

B

Explanation

For a solid cylinder, `I=(1)/(2)MR^(2) and v=(omega)/(R)`
and Rolling K.E. `=(1)/(2)Iomega^(2)+(1)/(2)Mv^(2)`
`K_(R)=(1)/(2)(MR^(2))/(2).(v^(2))/(R^(2))+(1)/(2)Mv^(2)=(3)/(4)Mv^(2)`
When it rolls from A to B.
Loss in P.E. = Gain in K.E.
`Mgh=(3)/(4)Mv^(2)`
`therefore v^(2)=(4gh)/(3) therefore v=sqrt((4)/(3)gh)`
`therefore` The speed of its centre of mass, when it reaches the bottom `sqrt((4)/(3)gh)`.