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?

1 Answer
Mar 15, 2018

(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)