Collision of Rolling Ball and Cart

In summary, the conversation discusses a scenario where a ball rolls down an incline and collides with a cart. The ball then rolls on the cart with friction, causing it to eventually roll without slipping. The conversation also addresses the conservation of angular momentum and the role of friction in achieving pure roll. It is concluded that the cart will always move ahead of the ball due to the potential well created by its depth.
  • #1
FallenApple
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Say a ball rolls down to the right with slipping with on an arc incline so that at the bottom it leaves horizontally. The instant it leaves horizontally and is now on a cart with mass with frictional surface.

So the ball will roll on the cart with kinetic friction pushing to the left altering the Vcm and forward spin such that at one point, it rolls without slipping.

By Newton's third law, there would be friction from the ball to the cart pushing the cart to the right.

So it seem like this is an inelastic collision where both ball and cart would move forward with the same linear speed.

So is the cart moving to the right with the ball just spinning while its position is stationary at one point on the cart while the cart moves forward? So they both have the same v forward as observed from someone outside the system. So from the reference point of someone on the cart, the ball is just spinning in one spot. Is this right?

But the weird part about this scenario is that the angular momentum (the final spin is faster)is not conserved even though the system is just mass and cart. How to explain this?
 
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  • #2
FallenApple said:
So it seem like this is an inelastic collision where both ball and cart would move forward with the same linear speed.
Why would you think they'd move with the same linear speed? The ball is rolling and slipping along the surface.

FallenApple said:
But the weird part about this scenario is that the angular momentum (the final spin is faster)is not conserved even though the system is just mass and cart. How to explain this?
You must consider the total angular momentum, not just the spin angular momentum of the ball about its center.
 
  • #3
Doc Al said:
Why would you think they'd move with the same linear speed? The ball is rolling and slipping along the surface.You must consider the total angular momentum, not just the spin angular momentum of the ball about its center.
Well its it not possible that we have some configuration of initial angular velocity and linear velocity such that it would eventually not slip? I mean, the friction slows down the velocity and speeds up the cart, eventually they could match up. As long as Vcm-w*R= V of cart then we are good.Ah I see, the angular momentum due to the linear motions of the two objects relative to some point.
 
  • #4
FallenApple said:
Well its it not possible that we have some configuration of initial angular velocity and linear velocity such that it would eventually not slip?
Can the ball end up rolling without slipping? Sure. But then it's rolling and thus moving with respect to the cart.

The only way that the ball could move with the cart and rotate in place would be if there were no friction, which contradicts your set up.
 
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  • #5
Doc Al said:
Can the ball end up rolling without slipping? Sure. But then it's rolling and thus moving with respect to the cart.

The only way that the ball could move with the cart and rotate in place would be if there were no friction, which contradicts your set up.

Interesting. Well usually problems involve objects first rolling with slipping on a surface, with friction, and then its the friction itself that eventually brings it to pure roll, at which point there would no longer be any friction whatsoever since the relative velocity of the point of contact and the surface is 0. So in those cases, the friction vanishes once pure roll is established.
 
  • #6
FallenApple said:
Well usually problems involve objects first rolling with slipping on a surface, with friction, and then its the friction itself that eventually brings it to pure roll, at which point there would no longer be any friction whatsoever since the relative velocity of the point of contact and the surface is 0. So in those cases, the friction vanishes once pure roll is established.
That's right. But the ball is rolling.
 
  • #7
Doc Al said:
That's right. But the ball is rolling.

Oh right. So on the ground, once a ball gets pure roll, it is still moving relative to the ground, its just that the point of contact isn't.

So in the problem I made up, if pure roll is achieved, the ball is still moving relative to the cart, just not slipping over it.

So the logic dictates that the cart will just move ahead of the ball, making the ball drop, for the x linear momentum to be conserved.

But what's interesting is that once the cart moves ahead of ball, the ball falls. Making the y momentum not conserved. I suppose this makes sense since by Noether's theorem, there is only translational symmetry in the x and not the y. I mean even if the initial configuration was that it didn't roll down an incline and then collided horizontally with the cart: Say that it started out as rolling with slipping over a frictionless surface, and then it end up on the cart, the cart has depth to it in the y. So that is a potential well waiting to be used.
 
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  • #8
edit: I just did the calculation: the ball will always move ahead of the cart, regardless of the configuration, back spin or not, vi greater than vi*wi or not. For a cart of any length, moving from left to right, the ball will always move ahead of the cart.
 

1. What factors affect the collision of a rolling ball and cart?

The collision of a rolling ball and cart is affected by several factors, including the mass of the ball and cart, the velocity of the ball and cart, and the surface materials of the ball and cart. Other factors that may impact the collision include the angle of impact and the presence of any external forces.

2. How does the conservation of momentum apply to the collision of a rolling ball and cart?

The conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision. In the case of a rolling ball and cart, the combined momentum of the ball and cart before the collision must equal the combined momentum after the collision.

3. What is the difference between elastic and inelastic collisions in the context of a rolling ball and cart?

In an elastic collision, both the kinetic energy and momentum are conserved. This means that the objects involved in the collision bounce off each other without any loss of energy. In contrast, in an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound. In the context of a rolling ball and cart, an elastic collision would result in the ball and cart bouncing off each other, while an inelastic collision would result in the ball sticking to the cart or the cart slowing down after the collision.

4. How can the coefficient of restitution affect the collision of a rolling ball and cart?

The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach. In the context of a rolling ball and cart, the coefficient of restitution would affect the amount of energy lost or conserved during the collision. A higher coefficient of restitution would result in a more elastic collision, while a lower coefficient of restitution would result in a more inelastic collision.

5. Can the collision of a rolling ball and cart be accurately predicted?

Predicting the collision of a rolling ball and cart can be challenging due to the many factors that can affect the outcome. However, with a thorough understanding of the physical principles involved and accurate measurements of the variables, it is possible to make reasonably accurate predictions of the collision. Computer simulations and experiments can also help to improve the accuracy of these predictions.

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