Question

A force $10\text{N}$ acts tangentially at the highest point of a solid sphere of mass $10\text{kg}$ kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of the centre of the sphere (in $\text{m}/{\text{s}}^2$).

• A. $\frac{7}{10}$
• B. $\frac{10}{7}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is B $\frac{10}{7}$


The sphere will have a tendency to slip towards left at the point of contact, thus friction will act rightwards as shown.
Since the sphere is rolling without slipping, $a=\alpha R$ . . . $\left(1\right)$

From $\sum {\overrightarrow{F}}_{ext}=m{\overrightarrow{a}}_{com}$
$F+f=ma...\left(2\right)$
and from ${\overrightarrow{\tau }}_{ex{t}_{com}}=I_{com}\overrightarrow{\alpha }$
$-F\times R+f\times R=-\frac{2}{5}m{R}^2\times \alpha$
$\Rightarrow F-f=\frac{2}{5}mR\alpha ...\left(3\right)$

From (1), (2) & (3)
$2F=ma+\frac{2}{5}ma=\frac{7}{5}ma$
$\Rightarrow a=\frac{10F}{7m}=\frac{10\times 10}{7\times 10}=\frac{10}{7}{\text{m/s}}^2$