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

Answer

D

Explanation

According to conservation of mechanical energy,
`(("Total potential energy"),("of cylinder at the top"),("of inclined plane"))=(("Rotational kinetic"),("energy at the bottom"),("of inclined plane"))+(("The linear kinetic energy"),("of centre of mass of cylinder"),("at the bottom of inclined plane"))`
`therefore mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)`
putting `omega=(v)/(R )` in this formula,
`mgh=(1)/(2)I(v^(2))/(R^(2))+(1)/(2)mv^(2)`
Now Moment of inertia of solid cylinder
`I=(mR^(2))/(2)`
`therefore mgh=(1)/(2).(mR^(2))/(2).(v^(2))/(R^(2))+(1)/(2)mv^(2)`
`therefore gh=(1)/(4)v^(2)+(1)/(2)v^(2)`
`gh=(3)/(4)v^(2)`
`therefore v=sqrt((4)/(3)gh)`