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 ?

Explanation

Here, `theta=30^(@), v=5 m//s`
Let the cylinder go up the plane up to a height h.
From `1//2 mv^(2)+1//2lW^(2)=mgh`
`(1)/(2)mv^(2)+(1)/(2)((1)/(2)mr^(2))omega^(2)=mgh`
`(3)/(4) mv^(2)=mgh`
`h=(3v^(2))/(4g)=(3xx5^(2))/(4xx9.8)=1.913m`
If s is the distance up the inclined plane, then as
`sin theta =(h)/(s), s=(h)/(sin theta)=(1.913)/(sin 30^(@))3.856 m`
Time taken to return to the bottom
`t=sqrt((2s(1+(k^(2))/(r^(2))))/(g sin theta))=sqrt((2xx3.826(1+(1)/(2)))/(9.8 sin 30^(@)))=1.53s`