Question

A tangential force acting on the top of sphere of mass m kept on a rough horizontal place as showing in figure
If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is

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

Answer 1

Answer: (A)

$\frac{10F}{7m}$

Let the angular acceleration of the sphere be $\alpha$.

Moment of inertia of the sphere, $I=\frac{2}{5}m{r}^2$

Using: $\tau =I\alpha$

$\therefore$ $(F-f)r=\frac{2}{5}m{r}^2\alpha$

$⟹\alpha =\frac{5(F-f)}{2mr}$ ............(1)

Acceleration of its centre, $a=\frac{F+f}{m}$ ......(2)

As the sphere performs pure rolling, thus acceleration of the point O is zero i.e. $a=r\alpha$

$\therefore$ $\frac{F+f}{m}=r\frac{5(F-f)}{2mr}$

Or, $2F+2f=5F-5f$

$⟹$ $f=\frac{3}{7}F$

Thus from (2), we get: $a=\frac{F+\frac{3}{7}F}{m}=\frac{10F}{7m}$