Question

A solid sphere of mass $m$ and radius $R$ is gently placed on a conveyer belt moving with constant velocity $V_0$. If coefficient of friction between belt and sphere is $2/7$, the distance travelled by the centre of the sphere before it starts pure rolling is

• A. $\frac{V_0^2}{7g}$
  
• B. $\frac{2V_0^2}{7g}$
• C. $\frac{2V_0^2}{5g}$
• D. $\frac{2V_0^2}{49g}$

Answer 1

The correct option is A $\frac{V_0^2}{7g}$


Given,
mass of solid sphere$=m$
radius of solid sphere$=R$
speed of belt$=V_0$
coefficient of friction$=\mu$

If $a$ be the acceleration of sphere. Then from free body diagram,
$ma=\mu mg$

$\Rightarrow a=\frac{\mu mg}{m}$

$\Rightarrow a=\mu g=\frac{2}{7}g$


Now fro the torque equation,
$I\alpha =\mu mgR$

where, $\alpha$ is angular acceleraion of sphere and $I$ is the moment of inertia about centre .i.e.,$I=\frac{2}{5}m{R}^2$

$\Rightarrow \alpha =\frac{\mu mgR}{\frac{2}{5}m{R}^2}$

$\Rightarrow \alpha =\frac{5\mu g}{2R}$

Substituting the value of $\mu$
$\Rightarrow \alpha =\frac{5g}{7R}$

Pure rolling will start when velocity at point $\text{P}$,
$v_P=v+R\omega ....\left(1\right)$

Since the velocity at the point of contact is the velocity of belt, so
$v_P=V_0....\left(2\right)$

From equation (1) and (2) we have,
$v+R\omega =V_0....\left(3\right)$

Here, velocity of the centre of sphere
$v=at$
$t$ is the time to gain pure rolling motion,
$\omega =\alpha t$

Substituting the values in equation (3), we have

$at+R\alpha t=V_o$

$\Rightarrow \frac{2}{7}gt+R\times \frac{5g}{7R}\times t=V_0$

$\therefore t=\frac{V_0}{g}$

Now, distance travelled by the centre of sphere before its starts pure rolling is
$S=\frac{1}{2}a{t}^2$

$\Rightarrow S=\frac{1}{2}\left(\frac{2}{7}g\right){\left(\frac{V_0}{g}\right)}^2$

$\therefore S=\frac{V_0^2}{7g}$

Hence, option (a) is correct answer.