A sphere rolls up an inclined plane whose inclination is `30^(@)`. At the bottom of the inclined plane, the center of mass of the sphere has a translational speed of `5ms^(-1)` (a) How far does the sphere travel up the plane? (b) How long does it take to return to the bottom?

Explanation

`v=5m//s , theta=30^(@)`
a) Let h be the vertical height covered above the bottom.
From law of conservation of energy we can write
`(1)/(2) mv^(2)+(1)/(2) I omega^(2)= mgh`
`(1)/(2) mv^(2)+(1)/(2) ((1)/(2) mr^(2)) omega^(2)=mgh`
`(3)/(4) m r^(2) omega^(2) mgh`
`:. (3)/(4) mv^(2)= mgh`

`rArr h=(3v^(2))/(4g)=(3xx(5)^(2))/(4xx9.8)=1.913m`
Let is the distance covered up the inclined plane.
`sin theta=(h)/(s)`
`s=(h)/(sin theta)=(1.913)/(sin30^(@))=(1.913)/(1)xx2=3.826m`
(b) t= 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`
disclaimer, [ The answer of (b) does not match with NCERT]