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

Here, `theta 30^(@), v = 5 m//s`
Let the cylinder go up the plane upto a height h.
From `(1)/(2)m v^(2) + (1)/(2)I omega^(2) = mgh`
`(1)/(2)m v^(2) + (1)/(2) ((1)/(2)mr^(2)) omega^(2) = mgh`
`(3)/(4)m v^(2) = mgh`
`h = (3v^(2))/(4g) = (3 xx 5^(2))/(4 xx 9.8) = 1.913 m`
If `s` is the distance up the inclined plane, then as `sin theta = (h)/(s), s = (h)/(sin theta) = 3.826 m`
Time taken to return to the bottom
`t = sqrt((2s(1 + K^(2)//r^(2)))/(g sin theta)) = sqrt((2 xx 3.826 (1 + 1//2))/(9.8 sin 30^(@))) = 1.53 s`