Question

A sphere of mass $m$ and radius $r$ is placed on a rough plank of mass $M$. The system is placed on a smooth horizontal surface. A constant force $F$ is applied on the plank such that the sphere rolls purely on the plank. If the acceleration of the sphere is $\frac{2F}{xM+2m}$, find the value of $x$.

Answer 1

As the sphere rolls purely on the rough plank, thus acceleration of point A must be same as that of plank.

Acceleration of plank $a=\frac{F-f}{M}$

Angular acceleration of sphere $\alpha =\frac{\tau }{I}$

$\therefore$ $\alpha =\frac{fr}{\frac{2}{5}m{r}^2}=\frac{5f}{2mr}$

Using $a=r\alpha$ (pure rolling)

$\frac{F-f}{M}=\frac{5f}{2m}$

Solving this we get $f=\frac{2mF}{2m+5M}$

Acceleration of sphere $a_s=\frac{f}{m}$

$\therefore$ $a_s=\frac{2mF}{(2m+5M)m}$

$⟹$ $a_s=\frac{2F}{2m+5M}$

Thus $x=5$