Question

A sphere of mass M and radius r shown in figure slips on a rough horizontal plane. At some instant it has translational velocity $v_0$ and rotational velocity about the centre $\frac{v_0}{2r}$. Find the translational velocity after the sphere starts pure rolling.

• A. 2v₀
• B.
• C. v₀
• D.

Answer 1

The correct option is D.

Velocity of the centre = $v_0$ and the angular velocity about the centre = $\frac{v_0}{2r}$.

Thus, $v_0>{\omega }_{0}r$.

The sphere slips forward and thus the friction by the plane on the sphere will act backward. As the friction is kinetic, its value is $\mu N=\mu Mg$ and the sphere will be decelerated by $a_{cm}=\frac{f}{M}$.

Hence,

$v\left(t\right)=v_0-\frac{f}{M}t$. .......(i)

This friction will also have a torque $\tau =f.r$ about the centre. This torque is clockwise and in the direction of ${\omega }_0$. Hence the angular acceleration about the centre will be

$\alpha =\frac{\tau }{I}=\frac{fr}{\left(\frac{2}{5}\right)M{r}^2}=\frac{5f}{2Mr}$

and the clockwise angular velocity at time t will be

$\omega \left(t\right)={\omega }_0+\frac{5f}{2Mr}t=\frac{v_0}{2r}+\frac{5f}{2Mr}t$

pure rolling starts when v(t)=$r\omega \left(t\right)$

i.e., $v\left(t\right)=\frac{v_0}{2}+\frac{5f}{2M}t$ ...(ii)

Eliminating f from (i) and (ii),

$\frac{5}{2}v\left(t\right)+v\left(t\right)=\frac{5}{2}{v}_0+\frac{v_0}{2}$

or, $v\left(t\right)=\frac{2}{7}\times 3v_0=\frac{6}{7}{v}_0$.

Thus, the sphere rolls with translational velocity $\frac{6v_0}{7}$ in the forward direction.

Alternative: Let us consider the torque about the initial point of contact A. The force of friction passes through this point and hence its torque is zero. The normal force and the weight balance each other. The net torque about A is zero. Hence the angular momentum about A is conserved.

Initial angular momentum is,

$L=L_{cm}+Mr{v}_0=I_{cm}\omega +MR{v}_0$

$=\left(\frac{2}{5}M{r}^2\right)\left(\frac{v_0}{2r}\right)+Mr{v}_0=\frac{6}{5}Mr{v}_0$.

Suppose the translational velocity of the sphere, after it starts rolling, is v. The angular velocity is v/r. The angular momentum about A is,

$L=L_{cm}+Mrv$

Thus, $\frac{6}{5}Mr{v}_0=\frac{7}{5}Mrv$

or, $v=\frac{6}{7}{v}_0$.