A solid cylinder rolls up an inclined plane of inclination `theta` with an initial velocity `v`. How far does the cylinder go up the plane ?

Options

A. `(3v^(2))/(2g sintheta)`
B. `(v^(2))/(4g sintheta)`
C. `(3v^(2))/(g sintheta)`
D. `(3v^(2))/(4g sintheta)`

Answer

D

Explanation

Let the cylinder go up the plane upto a height h. Let M and R be the mass and radius of the cylinder respectively. According to law of conservation of mechanical energy, we get
`1/2Mv^(2)+1/2Iomega^(2) = Mgh`
`1/2Mv^(2) + 1/2(MR^(2))/(2) omega^(2)= Mgh` `therefore` For a solid cylinder `I=1/2MR^(2)`
`1/2Mv^(2) + 1/4 MR^(2)omega^(2)=Mgh`
`1/2Mv^(2)+1/4Mv^(2)=Mgh (therefore v=Romega)`
`3/4 Mv^(2)=Mgh`
`h=(3v^(2))/(4g)`................(i)
Let s be distance travelled by the cylinder up the plane.
Then `sintheta=h/s` or `s=h/(sintheta) = (3v^(2))/(4gsintheta)` (using (i))