Question

A solid sphere is rolling down an inclined plane without slipping. If the inclined plane has inclination $\theta$ with the horizontal, then the coefficient of friction $\mu$ between the sphere and the inclined plane should be

• A. $\mu \ge 2/7\cot \theta$
• B. $\mu \ge 2/7\tan \theta$
• C. $\mu \ge 2/7\cos \theta$
• D. $\mu \ge 4/7\sin \theta$

Answer 1

The correct option is B $\mu \ge 2/7\tan \theta$

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

$\alpha$ is the angular acceleration of the sphere.

Moment of inertia of the sphere about centre of mass is $I=\frac{2}{5}m{R}^2$.N is the normal reactional force on the sphere &N $=mgcos\theta$

$f=\mu N=\mu mgcos\theta$

Torque about centre of mass sphere is

$I\alpha =f.R$

For downward motion let the acceleration along inclined plane $\alpha$

So, $mgsin\theta -f=ma$

for pure rotation $\alpha =\frac{a}{R}$

So, $mgsin\theta -f=m\alpha R$

$⟹mgsin\theta =f+\frac{5}{2}f$

$⟹mgsin\theta =\frac{7}{2}f$

$⟹\frac{7}{2}\mu mgcos\theta =mgsin\theta$

$⟹\mu =\frac{2}{7}tan\theta$

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

$\mu \ge \frac{2}{7}tan\theta$