Question

A solid sphere $A$ of mass $m$ rolls without slipping on an inclined plane of inclination ${30}^{\circ }$. Coefficient of friction of the inclined plane is $\mu$. Then, which of the following options is satisfied when the sphere undergoes pure rolling motion?

• A. $\mu \ge \frac{2\sqrt{3}}{7}$
• B. $\mu \ge \frac{2}{7\sqrt{3}}$
• C. $\mu <\frac{2\sqrt{3}}{7}$
• D. $\mu <\frac{2}{7\sqrt{3}}$

Answer 1

The correct option is B $\mu \ge \frac{2}{7\sqrt{3}}$

Le us assume the radius of the sphere is $R$.
F.B.D from the data given in the question is shown below.


From the diagram, we can deduce that
$\sum F_x=m{a}_x$ [Net Force along x - direction]
$\Rightarrow mg\sin \theta -f=ma......\left(1\right)$
$\sum F_y=m{a}_y$ [Net Force along y - direction]
$\Rightarrow N=mg\cos \theta .....\left(2\right)$
Torque experienced by the sphere due to frictional force is given by
$fR=I\alpha$
where $I=$ Moment of inertia of solid sphere $=\frac{2}{5}m{R}^2$
$\Rightarrow fR=\left(\frac{2}{5}m{R}^2\right)\left(\frac{a}{R}\right)[\because \alpha =\frac{a}{R}]$
$\Rightarrow f=\frac{2}{5}ma......\left(3\right)$
Substituting $\left(3\right)$ in $\left(1\right)$,
$mg\sin \theta -\frac{2}{5}ma=ma$
$\Rightarrow g\sin \theta =(a+\frac{2}{5}a)$
$\Rightarrow g\sin \theta =\frac{7}{5}a$
$\Rightarrow a=\frac{5}{7}g\sin \theta$
and from $\left(3\right)$, we can say that
$f=\frac{2}{7}mg\sin \theta .....\left(4\right)$
For pure rolling, $\mu N\ge f$
From $\left(2\right)$ and $\left(4\right)$
$\mu mg\cos \theta \ge \frac{2}{7}mg\sin \theta$
$\Rightarrow \mu \ge \frac{2}{7}\text{tan}\theta$
$\Rightarrow \mu \ge \frac{2}{7}\text{tan}{30}^{\circ }$
$\Rightarrow \mu \ge \frac{2}{7\sqrt{3}}$
Hence, option (b) is the correct answer.