Small sphere of mass m and radius r is kept on fixed inclined rough surface. Surface makes an angle with the horizontal. Initially centre of sphere is at a height h above the horizontal floor. Coefficient of friction between the sphere and surface is u. Sphere is released from the state of rest.

Assume that friction is sufficient engough for pure rolling. Accleeration of the shpere is

Options

A. `g sin theta`
B. `(5 g sin theta)/(7)`
C. `(7 g sin theta)/(5)`
D. `(2 g sin theta)/(5)`

Answer

B

Explanation

Force acting on the sphere are shown in figure.

As the sphere tries to move down the plane due to mg sin component of its weight then friction acts up the plane in order to oppose this motion and torque applied by the friction about the centre provides it necessary angular acceleration. In case of pure rolling angular acceleration can be written in terms of linear acceleration as follows:
`alpha=(a)/(r)`
Applying F - ma along the inclined plane we get.
`mg sin theta -f= ma " "...(i)`
Applying `tau = I alpha` about the centre of sphere we get
`fr=I alpha rArr fr(2)/(5) mr^92)((a)/(r))`
`rArr f=(2)/(5) ma`
Substituting equation (ii) in (i) we get
`mg sin theta-(2)/(5) ma= ma`
`a=(5 g sin theta)/(7)`
Hence option (6) is correct.