A solid cylinder rolls up an inclined plane of angle of inclination `30^(@)`. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of `5 m//s`.
(a) How far will the cylinder go up the plane ? (B) How long will it take to return to the bottom ?

Options

A. `(10)/(4)m`
B. `(3)/(10)m`
C. `(4)/(13)m`
D. `(15)/(4)m`

Answer

D

Explanation

`sin theta=(h)/(L)" "therefore" L"=(h)/(sin theta)=(h)/(1//2)=2h`
Suppose that the solid cylinder of a radius R goes from A to B, where AB = L and height h =BC .
At A, the rolling energy of te cylinder
`=(1)/(2)m^(2)+(1)/(2)Iomega^(2)`
`"For the solid cylinder, I"=(mR^(2))/(2) and omega=(v)/(R)`
`therefore" "E_("Rolling")=(1)/(2)mv^(2)+(1)/(2).(mR^(2))/(2)xx(v^(2))/(R^(2))=(3)/(4)mv^(2)`
It slops at B and has the P.E. = mgh
By the principle of conservation of energy,
`therefore" mgh "=(3)/(4)mv^(2)" "therefore h=(3v^(2))/(4g)`
`therefore" "L=2h=(3v^(2))/(2g)=(3xx5xx5)/(2xx10)=(15)/(4)m`