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. `(3 v^2)/(2 g sin theta)`
B. `(v^2)/(4 g sin theta)`
C. `(3 v^2)/(g sin theta)`
D. `(3 v^2)/(4 g sin theta)`

Answer

D

Explanation

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