Question

A uniform solid sphere of radius $R$ and mass $m$ rolls down an inclined plane. The coefficient of friction between the sphere and the inclined plane is $\mu$ then maximum value of $\theta$ for pure rolling is

• A. ${\text{tan}}^{-1}\left(\frac{3\mu }{2}\right)$
• B. ${\text{tan}}^{-1}\left(\frac{7\mu }{2}\right)$
• C. ${\text{tan}}^{-1}\left(\frac{5\mu }{3}\right)$
• D. ${\text{tan}}^{-1}\left(\frac{7\mu }{3}\right)$

Answer 1

The correct option is B ${\text{tan}}^{-1}\left(\frac{7\mu }{2}\right)$

F.B.D of the uniform solid sphere


$N=mg\cos \theta ...........\left(1\right)$
$\text{mg sin}\theta -f=ma.........\left(2\right)$
$fR=I\alpha =\frac{2}{5}m{R}^2\times \frac{a}{R}$
[ for uniform rolling, $a=\alpha r$ ]
$\therefore f=\frac{2}{5}\text{ma}.........\left(3\right)$
On putting $\left(3\right)$ in $\left(2\right)$
$mg\sin \theta =\frac{7}{5}\text{ma}$
$a=\frac{5}{7}\text{g sin}\theta ..........\left(4\right)$
On putting $\left(4\right)$ in $\left(3\right)$
$f=\frac{2}{7}\text{mg sin}\theta$
Now, for critical case, $f\le f_s\text{max}=\mu N$
$\Rightarrow \frac{2}{7}\text{mg sin}\theta \le \mu \text{mg cos}\theta$
[ from $\left(1\right)$ ]
$\Rightarrow \theta \le {\text{tan}}^{-1}\left(\frac{7\mu }{2}\right)$