Question

A solid sphere of mass $7\text{kg}$ rolls without slipping on an inclined plane of inclination ${60}^{\circ }$. What should be the minimum coefficient of static friction to support pure rolling?

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

Answer 1

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


Given, the sphere is rolling without sliding. Hence,
$a=\alpha R...\left(i\right)$
From, $\sum {\tau }_{com}=I_{com}\alpha$
$-f\times R=-\frac{2}{5}m{R}^2\times \alpha$
or $f=\frac{2}{5}mR\alpha =\frac{2}{5}ma...\left(ii\right)$
[using (i)]
Applying Newton's 2nd law,
$\sum {\overrightarrow{F}}_{ext}=m{\overrightarrow{a}}_{com}$
$mg\sin \theta -f=ma...\left(iii\right)$

Putting (1) & (ii) in (iii)
$mg\sin \theta -\frac{2}{5}ma=ma$
$a=\frac{5g\sin \theta }{7}\&f=\frac{2mg\sin \theta }{7}$

We know, $(f_s{)}_{max}=\mu N=\mu mg\cos \theta$
and, $f\le \left[\right(f_s{)}_{max}]$

$\Rightarrow \frac{2mg\sin \theta }{7}\le \mu mg\cos \theta$
$\Rightarrow \mu \ge \frac{2}{7}\tan \theta$
$\therefore {\mu }_{min}=\frac{2}{7}\tan {60}^{\circ }=\frac{2\sqrt{3}}{7}$