Question

A hollow sphere of mass $M$ and radius $R$ is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force $F$ as shown in figure.


The linear acceleration $\left(a\right)$ of the sphere is :

• A. $a=\frac{10F}{7M}$
• B. $a=\frac{7F}{5M}$
• C. $a=\frac{6F}{5M}$
• D. $a=\frac{F}{M}$

Answer 1

The correct option is C $a=\frac{6F}{5M}$

Point of contact will tend to slip in backward direction, hence friction $\left(f\right)$ will act in forward direction.


Let $a$ and $\alpha$ be the linear and angular accelerations of the sphere respectively.
Applying Newton's $2^{nd}$ law for translational motion,
$F+f=Ma...\left(1\right)$
The magnitude of the net torque acting on the sphere $\tau =FR-fR$.
Applying equation of torque for rotational motion,
$\tau =I\alpha$
[$\because a=\alpha R$ for pure rolling]
$\Rightarrow FR-fR=I\alpha =\frac{Ia}{R}...\left(2\right)$

For a hollow sphere, $I=\frac{2}{3}M{R}^2$.
Substituting in Eq $\left(2\right)$,
$FR-fR=\frac{2}{3}M{R}^2\times \frac{a}{R}$
$\Rightarrow FR-fR=\frac{2}{3}MRa$
$\Rightarrow F-f=\frac{2}{3}Ma...\left(3\right)$
From Eq.$\left(1\right)\&\left(3\right)$
$2F=\frac{5Ma}{3}$
$\therefore a=\frac{6F}{5M}$