Question

A sphere of mall $5\text{kg}$ rolls without slipping on an inclined plane of inclination ${30}^{\circ }$. Choose the correct statement(s).

• A. Force of friction acting on sphere is $\frac{5g}{7}$.
• B. Linear acceleration of sphere down the plane is $\frac{5g}{14}$.
• C. For pure rolling, $\mu >0.165$.
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A Force of friction acting on sphere is $\frac{5g}{7}$.
B Linear acceleration of sphere down the plane is $\frac{5g}{14}$.
C For pure rolling, $\mu >0.165$.


$N=$ Normal force
$f=$ friction force
$mg=$ weight of sphere
Let the linear acceleration of the sphere down the plane be $a$.
So, equation of linear motion for the centre of mass
$mg\sin \theta -f=ma.....\left(1\right)$
$\because$ Sphere rolls without slipping
So $\alpha =\frac{a}{r}$
where $\alpha =\text{Angular acceleration},$
$a=\text{Linear acceleration}$

For Rotational motion,
${\tau }_0=I\alpha$
where ${\tau }_0$= Torque about centre of mass
$\Rightarrow fr=\frac{2}{5}m{r}^2\left(\frac{a}{r}\right)$
$\Rightarrow f=\frac{2}{5}ma......\left(2\right)$
From eq (1) & (2)
$mg\sin \theta =\frac{7}{5}ma$
$\Rightarrow a=\frac{5}{7}g\sin \theta =\frac{5}{7}\times g\times \sin {30}^{\circ }=\frac{5}{14}g$
$\Rightarrow a=\frac{5g}{14}$
and $f=\frac{2}{7}mg\sin \theta =\frac{2}{7}\times 5\times g\times \frac{1}{2}=\frac{5g}{7}$
$f=\frac{5g}{7}$
For pure rolling,
$f_{max}\ge f_{kinetic}$
$\Rightarrow \mu mg\cos \theta \ge \frac{2}{7}mg\sin \theta$
$\Rightarrow \mu \ge \frac{2}{7}\text{tan}\theta =0.165$
i.e $\mu >0.165$
Options A, B & C are correct.