Question

A solid sphere of mass $0.50\text{kg}$ is kept at rest on horizontal surface. The coefficient of static friction between the solid sphere and the surface in contact is $\frac{2}{7}$. The maximum force that can be applied at the highest point of sphere in the horizontal direction so that the sphere dosen't slip on the surface is given by $\frac{x}{3}\text{N}$, then value of $x$ is
(Assume $g=10{\text{ms}}^{-2}$)

Answer 1

Let the maximum force applied at the highest point of solid sphere be $F$. It will cause sphere to rotate (angular acceleration $\alpha$) about its centre of mass as well as translate with acceleration $a$.
$\therefore$ The point on sphere in contact with ground will have tendency to slip backwards, hence maximum static frictional force$(f=\mu N)$ will act in forward direction to prevent slipping.


on applying newton's second law along horizontal direction,
$F+f=ma$
$\Rightarrow F+\mu N=ma$
$\Rightarrow F+\mu mg=ma...\left(1\right)$\)
on applying equation of torque on sphere about instantaneous axis of rotation taken at bottom-most point (point of contact),
$\because$In case of pure rolling, the point of contact will remain at rest always hence $\text{I.A.R}$ can be taken passing through it.
$\Rightarrow {\tau }_{\text{net}}=I\alpha$
${\tau }_N=0,{\tau }_{mg}=0,{\tau }_F=F\left(2r\right),{\tau }_f=0$
$2r\times F=(\frac{2}{5}m{r}^2+m{r}^2)\alpha ...\left(2\right)$
$\because [\tau =r_{\perp }F,I=I_0+m{d}^2\text{and}{I}_0=\frac{2}{5}m{r}^2]$
$\Rightarrow 2rF=\frac{7}{5}m{r}^2\times \frac{a}{r}$
$\text{Since},a=\alpha r\text{condition for pure rolling}$
$\Rightarrow 10F=7ma$
Substituting value of $ma$ from Eq.$\left(1\right)$,
$\Rightarrow 10F=7(F+\mu mg)$
$\Rightarrow 3F=7\mu mg$
$\Rightarrow F=\frac{7}{3}\mu mg$
$\Rightarrow F=\frac{7}{3}\times \frac{2}{7}\times \frac{1}{2}\times 10$
$[\mu =\frac{2}{7},m=\frac{1}{2}\text{kg}]$
$\Rightarrow F=\frac{10}{3}\text{N}$
Comparing with given value, $F=\frac{x}{3}\text{N}$
$\therefore x=10$