A uniform solid sphere of mass M and radius R rolls down an inclined plane making an angle `theta` with the horizontal without slipping. Show that the acceleration of the sphere is `(5)/(7)gsin theta`. [ Given: moment of inertia of the sphere is `(2)/(5) MR^(2)` ]

Explanation

The force acting on the sphere are frictional force (f), normal force (n) and weight (Mg). The net force on the sphere acting parallel to the inclined plane = Mgsin`theta`-f
If the linear acceleration of the sphere is a,
Mgsin`theta` - f = Ma ...(1)
The moment of inertia of the sphere , I = `(2)/(5)MR^(2)`
If the angular of the sphere about its centre is `alpha`, the torque on the sphere, `tau = l alpha`= fR
`therefore " "f = (I alpha)/(R ) = (I a)/(R^(2)) [ because alpha = (a)/(R )]" "`...(2)
From equation (1) we get,
Mgsin`theta - (I a)/(R^(2)) = `Ma
or, a = `(g sin theta)/(1 + (I)/(MR^(2))) =(g sin theta)/(1 + (2)/(5))= (5)/(7)g sin theta`