# Effect on rolling motion after an inelastic collision.

• nerdvana101
In summary: If the wall was shorter than the sphere, the sphere would collide with the wall and stop. If the wall was taller than the sphere, the sphere would slide along the wall.
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 vo.
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 evo velocity, but the same angular velocity ω=vo/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.!

nerdvana101 said:
I went about by considering that after the collision, the particle will have evo velocity, but the same angular velocity ω=vo/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.

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?

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?

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.

## What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. This means that after the collision, the objects involved will have less total kinetic energy than before the collision. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

## How does an inelastic collision affect rolling motion?

In an inelastic collision, the objects involved will stick together after colliding. This can affect the rolling motion of the objects, as they will no longer be able to roll freely. The objects will now move together as one unit, rather than independently rolling.

## What factors influence the effect of an inelastic collision on rolling motion?

Some factors that can influence the effect of an inelastic collision on rolling motion include the mass and velocity of the objects involved, the angle and surface of the collision, and the materials of the objects. These factors can determine how much kinetic energy is lost during the collision and how the objects will move after the collision.

## How is momentum conserved in an inelastic collision?

In an inelastic collision, momentum is conserved. This means that the total momentum of the objects involved before the collision will be equal to the total momentum after the collision. However, in an inelastic collision, kinetic energy is not conserved, as some of the momentum is transformed into other forms of energy.

## What are some real-world examples of inelastic collisions and their effects on rolling motion?

One example of an inelastic collision affecting rolling motion is a car accident. When two cars collide, the kinetic energy of the cars is not conserved, as some of the energy is converted into sound and heat. This can cause the cars to stick together and roll as one unit after the collision. Another example is a bowling ball hitting pins. The pins will stick to the ball after the collision, changing the rolling motion of the ball.

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