Question

A solid sphere of mass $5\text{kg}$ and radius $1\text{m}$ is kept on a rough surface. A force $F=30\text{N}$ is acting at the top most point. Friction required for pure rolling is

• A. $70/9\text{N}$
• B. $90/7\text{N}$
• C. $90/5\text{N}$
• D. $50/9\text{N}$

Answer 1

The correct option is B $90/7\text{N}$

Let us suppose force of friction required is $f$ and will act in forward direction as sphere has tendency to slip back when $F=30\text{N}$ is applied.
On taking instantaneous axis of rotation about bottom most point.


On applying $\sum T=I\alpha$
$30\times 2=(\frac{2}{5}\times 5\times 1^2+5\times 1^2)\alpha$ $[T=F{r}_{\perp }=I\alpha ,I=I_o+m{d}^2]$
$\Rightarrow \alpha =\frac{60}{7}\text{rad/s}$ ... $\left(1\right)$
We know, for pure rolling, $a=\alpha R$
$\Rightarrow a=\frac{60}{7}\text{m/s}$... $\left(2\right)$ [from $\left(1\right)$ and $R=1\text{m}$]
On applying $\sum F=ma$ along horizontal,
$\Rightarrow 30+f=5\times \frac{60}{7}$ [from $\left(2\right)$]
$\Rightarrow f=\frac{90}{7}\text{N}$