Question

A sphere of mass M and radius R is released from the top of an inclined plane of inclination $\theta$. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

• A. $\mu =tan\theta$
• B. $\mu =\frac{2}{3}tan\theta$
• C. $\mu =\frac{2}{5}tan\theta$
• D. $\mu =\frac{2}{7}tan\theta$

Answer 1

The correct option is D $\mu =\frac{2}{7}tan\theta$

When the sphere rolls down the plane, its acceleration is given by
$a=\frac{gsin\theta }{1+\frac{K^2}{R^2}}$
Where K is the radius of gyration of the sphere about its diameter. Now, the moment of inertia of the sphere about its diameter is
$I=\frac{2}{5}M{R}^2$
$\therefore K^2=\frac{I}{M}=\frac{2}{5}{R}^2$
Therefore, $a=\frac{gsin\theta }{1+\frac{2}{5}\frac{R^2}{R^2}}=\frac{5}{7}gsin\theta$
For rolling without sliding, the frictional force f provides the necessary torque $\tau$ which is given by
$\tau =force\times momentarm=fR$
But $\tau =I\alpha$, where $\alpha$ is the angular acceleration of the sphere. Thus, $I\alpha =fR$. Also, linear acceleration $a=\alpha R$.
Therefore,
$f=\frac{I\alpha }{R}=\frac{I\alpha }{R^2}=\frac{2}{5}Ma$ $(\because I=\frac{2}{5}M{R}^2)$
Now Force of friction $=\mu \times normalreaction=\mu Mgcos\theta$
Or $a=\frac{5}{2}\mu gcos\theta$ $\cdots \left(ii\right)$
Equation (i) and (ii) we have
$\frac{7}{5}gsin\theta =\frac{5}{2}\mu gcos\theta$
or $\mu =\frac{2}{7}tan\theta$