Question

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?

Answer 1

`(3u)/(7mug)`

Explanation:

Well,while taking an attempt to solve this,we can say that initially pure rolling was occurring just because of `u=omegar` (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 `1/2` so after the collision the sphere will have a velocity of `u/2` in the opposite direction.

So,new angular velocity becomes `omega=-u/r` (taking the clockwise direction to be positive)

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

So,`r*mumg =2/5 mr^2 alpha`

so,`alpha =(5mug)/(2r)`

And,considering linear force,we get, `ma=mumg`

so,`a=mug`

Now,let after time `t` angular velocity will be `omega'` so `omega'=omega +alphat`

and,after time `t` linear velocity will be `v`,so `v=(u/2) -at`

For pure rolling motion,

`v=omega'r`

Putting the values of `alpha,omega` and `a` we get, `t=(3u)/(7mug)`