# A solid sphere is rolling purely on a rough horizontal surface (coefficient of kinetic friction = mu) with speed of center = u. It collides inelastically with a smooth vertical wall at a certain moment. The coefficient of restitution being 1/2?

## The time when the sphere will start pure rolling is? Proceed with your approach. But I have a question, can this problem be solved using linear impulse-momentum theorem and/or angular impulse-momentum theorem?

Mar 15, 2018

$\frac{3 u}{7 \mu g}$

#### Explanation:

Well,while taking an attempt to solve this,we can say that initially pure rolling was occurring just because of $u = \omega r$ (where,$\omega$ is the angular velocity)

But as the collision took place,its linear velocity decreases but during collision there was no change inhence $\omega$,so if the new velocity is $v$ and angular velocity is $\omega '$ then we need to find after how many times due to the applied external torque by frictional force,it will be in pure rolling,i.e $v = \omega ' r$

Now,given,coefficient of restitution is $\frac{1}{2}$ so after the collision the sphere will have a velocity of $\frac{u}{2}$ in the opposite direction.

So,new angular velocity becomes $\omega = - \frac{u}{r}$ (taking the clockwise direction to be positive)

Now,external torque acting due to frictional force, $\tau = r \cdot f = I \alpha$ where, $f$ is the frictional force acting ,$\alpha$ is angular acceleration and $I$ is the moment of inertia.

So,$r \cdot \mu m g = \frac{2}{5} m {r}^{2} \alpha$

so,$\alpha = \frac{5 \mu g}{2 r}$

And,considering linear force,we get, $m a = \mu m g$

so,$a = \mu g$

Now,let after time $t$ angular velocity will be $\omega '$ so $\omega ' = \omega + \alpha t$

and,after time $t$ linear velocity will be $v$,so $v = \left(\frac{u}{2}\right) - a t$

For pure rolling motion,

$v = \omega ' r$

Putting the values of $\alpha , \omega$ and $a$ we get, $t = \frac{3 u}{7 \mu g}$