Question

A ball of mass M and radius R is released on a rough inclined plane of inclination theta. Friction is not sufficient to prevent slipping. The coefficient friction between the ball and the plane is mu. Find (a). The linear acceleration of the ball down the plane. (b). the angular acceleration of the ball about its centre of mass.

Answer 1

Answer:

f is the frictional force is upward direction along inclined plane.

α is the angular acceleration of the sphere.

Moment of inertia of the sphere about centre of mass is I=52mR2.N is the normal reactional force on the sphere &N =mgcosθ

f=μN=μmgcosθ

Torque about centre of mass sphere is

Iα=f.R

For downward motion let the acceleration along inclined plane α 

So, mgsinθ−f=ma

for pure rotation α=Ra

So, mgsinθ−f=mαR

⟹mgsinθ=f+25f

⟹mgsinθ=27f

⟹27μmgcosθ=mgsinθ

⟹μ=72tanθ

This is the minimum coefficient of friction So, it should be 

μ≥72tanθ