Effect on rolling motion after an inelastic collision.

[nerdvana101]

Hi,

I have a problem which I can't figure out.

Let us take a sphere which is rolling purely at a constant velocity v_o.
Now, if the sphere were to collide inelastically with a wall, with coeff. of restitution = e.
Then what is the time after which the sphere starts pure rolling again?
Given coeff, friction = μ

I went about by considering that after the collision, the particle will have ev_o velocity, but the same angular velocity ω=v_o/r.

Now, since the sphere is translating, the angular velocity is in the opp. direction. So the frictional force will cause a torque to change ω.

So, if the body was initially moving rightwards, with clockwise ω, it's new velocity will be leftwards, but ω will remain clockwise. So the frictional force must act towards right to apply a counter-clockwise torque.

After this, I couldn't figure out which way to go.

Any help appreciated.!

 

[A.T.]

I went about by considering that after the collision, the particle will have ev_o velocity, but the same angular velocity ω=v_o/r.

Usually the collision will change the angular velocity as well. If the ball rolls the force of the wall on the ball is not horizontal. It is slightly upwards generating a torque.

But can you assume zero friction with the wall to avoid that. Then the ball will slide on the ground until the friction makes it roll. Use the torque from friction and moment of inertia to find the angular acceleration. Keep in mind that the friction also reduces the velocity, while it changes the angular velocity.

 

[nerdvana101]

So, taking the torque about the COM and then finding the angular acceleration, what happens to the linear velocity? I think the friction force will act in it's opposite direction and stop it from sliding.
so that a = f/m.

Then, I cud equate it as -

v-at = r(ω - αt)
and get the time?

 

[nerdvana101]

thx, just worked it out. got the right answer.

Anyway, I was thinking that if the wall were shorter than the sphere itself? let's say that the wall has a height h, where h < r, i.e., the wall has a height less than that of the height of the center of the sphere. SO, only a corner of that wall will touch that sphere.
Let the friction of the wall also be high enough that it resists slipping of the sphere upon contact. Then what?

 

[PJC01]

I can provide some insight into the effect of an inelastic collision on rolling motion. In an inelastic collision, the colliding objects stick together and move with a common velocity after the collision. In the case of a sphere rolling at a constant velocity before the collision, the collision will affect both its translational and rotational motion.

After the collision, the sphere will continue to move with a velocity of evo, but its angular velocity will change due to the frictional force acting on it. This frictional force will cause a torque on the sphere, which will change its angular velocity and eventually bring it back to pure rolling motion.

The exact time it takes for the sphere to start pure rolling again will depend on factors such as the coefficient of restitution, the coefficient of friction, and the moment of inertia of the sphere. A higher coefficient of restitution will result in a shorter time for the sphere to start pure rolling again, while a higher coefficient of friction will result in a longer time. The moment of inertia will also play a role, as a larger moment of inertia will require more torque to change the angular velocity.

In order to determine the exact time after which the sphere starts pure rolling again, a detailed analysis of the forces and torques acting on the sphere is necessary. This can be done using mathematical equations and principles of mechanics. I suggest consulting a textbook or a colleague for further assistance in solving this problem.