# A Collision of Rolling Ball and Cart

1. Nov 18, 2016

### FallenApple

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?

Last edited: Nov 18, 2016
2. Nov 18, 2016

### Staff: Mentor

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.

3. Nov 18, 2016

### FallenApple

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. Nov 19, 2016

### Staff: Mentor

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.

5. Nov 19, 2016

### FallenApple

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. Nov 19, 2016

### Staff: Mentor

That's right. But the ball is rolling.

7. Nov 19, 2016

### FallenApple

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 whats 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.

Last edited: Nov 19, 2016
8. Nov 19, 2016

### FallenApple

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.