Question

A force $F$ acts tangentially at the highest point of a disc of mass $m$ kept on a rough horizontal plane. If the disc rolls without slipping, the acceleration of centre of the disc is:

• A. $\frac{2F}{3m}$
• B. $\frac{10F}{7m}$
• C. $zero$
• D. $\frac{4F}{3m}$

Answer 1

The correct option is D $\frac{4F}{3m}$

As force $F$ moves the sphere towards right , the point of contact has a tendency to slip towards right so that static friction( force f shown ) on the sphere will act towards left.
Let
$r$: radius of the sphere
$a$: linear acceleration
Since there is slipping, $\alpha =\frac{a}{r}$
For linear motion, $F-f=ma$ ....(1)
where $f$ is the frictional force
For rotational motion, torque about center, $\tau =Fr+fr=I\alpha =\left(\frac{1}{2}m{r}^2\right)\frac{a}{r}$
$\therefore F+f=\frac{1}{2}ma$ ......(2)
From (1) and (2) adding, we get $2F=\frac{3}{2}ma$
$\therefore a=\frac{4F}{3m}$