A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be :

Options

A. `(1)/(R ) sqrt((gh )/(2))`
B. `(2)/(R ) sqrt((gh )/(3))`
C. `(1)/(R ) sqrt((2gh )/(3))`
D. `(3)/(R ) sqrt((gh )/(2))`

Answer

B

Explanation

`I=(1) /(2) MR^(2)`
when the cylinder rails down the inclined plane from a vertical height h its potential energy Mgh is partly convered into translations kinetic energy `(1)/(2) Mv ^(2)` and partly into rotational kinetic energy `(1)/(2) lomega ^(2)` . therefore
`Mgh =(1)/(2) Mv ^(2) +(1)(2) l omega ^(2)`
`= (1)/(2) mv^(2) + (1)/(2) ((1)/(2) Mr^(2) MR ^(2)) omega ^(2) `
`=(1)/(2) M ( R omega )^(2) + (1)/(2) MR ^(2) omega ^(2) ( :' R omega )`
`Mgh =(3)/(4) MR ^(2) omega ^(2)`
`oemga = sqrt((4gh )/( 3R ^(2)))=(2)/(R)sqrt((gh)/(3))`