Collision of a ball with a rod on a track

[f todd baker]

The rod is attached and can frictionlessly slide on the purple track. The questioner specified that the collision is elastic. The first obvious error in this problem is that the direction and magnitude of the final velocity of the ball is impossible. If the ball carries off all the energy it came in with, the rod must end up with no energy. And, conservation of linear momentum in the y direction (not conserved in the x direction) demands that the speed of the center of mass (COM) of the rod along the rail would be 15 m/s meaning it would not be rotating. And, the angular momentum relative to the COM of the incoming ball is equal and opposite that of the outgoing ball, so the rod would have to be rotating to conserve angular momentum. So I thought to simply redraw the picture but with the final ball velocity having unknown components (2 unknowns). Two other unknowns are the final speed of the COM and the angular velocity of the rod about the COM. However, I only can see three equations: conservation of energy, linear momentum, and angular momentum. I would welcome suggestions on how to solve this problem which does not seem like it should be unsolvable. I'm a little rusty on my Goldstein-level classical mechanics when constraints (rail) are present; maybe that's the key? (Incidentally, the questioner would like a more general solution for any initial angle of the rod, but that should be straightforward once I understand a specific example.)

 

[haruspex]

You have, as you say, four unknowns. Add another: the impulse between the rod and the ball. (What is the direction of that?)
Equations:

• Two linear momentum for the ball;
• One linear momentum for the rod (there is an unknown sideways impulse from the track, so don't bother with that direction);
• One angular momentum for the rod;
• Energy.

Seems enough.

 

[f todd baker]

Sorry, doesn't seem enough to me. y-linear momentum is conserved; angular momentum (not for the rod but for the whole system) is conserved, energy is conserved. If you add the equation Δp_x=J (impulse) you have added another variable. Now we have 4 equations and 5 unknowns. The impulse to the system comes only from the track which can only exert forces in the x direction because it is stipulated to be frictionless. What am I missing? Does the impulse make energy conservation impossible? I cannot see how the force from the track does any work.

 

[haruspex]

What am I missing?

The fact that you can answer this question:

What is the direction of [the impulse between the rod and the ball]?

To put it another way, your list of system-wide equations nowhere features the known initial angle of the rod.

 

[f todd baker]

The impulse between the rod and the ball is irrelevant because it is not an external impulse; Newton's third law. Only the track can exert an external force and only in the x-direction. I guess the reason you have found the rod angle missing from my equations is that I have not written any equations. Maybe I should do that. I will call the initial velocity of the ball v₁ and the final velocity v₂ with components v_2x and v_2y, the masses equal are denoted m, and the angle between the rod and the rail (y) θ. The length of the rod is 2L, the moment of inertia about the center of mass of the rod is I=m(2L)²/12=mL²/3, the distance from the COM where the ball hits is denoted as s. The speed of the COM after the collision is u₂.

y linear momentum: mv₁=mv_2y+mu₂
energy: ½mv₁²=½mv₂²+½mu₂²+½Iω²
angular momentum: mv₁ssinθ=Iω-mv₂scosθ

 

[haruspex]

The impulse between the rod and the ball is irrelevant because it is not an external impulse;

The question asks you to find values which are internal to that system. If it only asked for susbsequent angular and linear momenta, plus the KE, of the rod+ball system then you would obviously be able to do that without considering the direction of impulse. Utilising the direction of the impulse is what allows you to find the individual consequences for rod and ball.

you have found the rod angle missing from my equations

True, with the data as given the angle does feature, but that is an artefact of the way the collision point is specified. Suppose instead of being told the impact is 0.5m along the rod from the axis we had been told the ball's trajectory was in a line parallel to the track and ¼√2m from it. Then the sin(θ) would not appear.
Your cos(θ) term is wrong, by the way. It will not bounce off like a reflected light beam, as you noted in post #1.

Can you get the answer to my question about the direction of the impulse? Can you see that then gives you five equations and five unknowns?

 

[f todd baker]

Right you are about error in angular momentum:
angular momentum: mv₁ssinθ=Iω-mv_2xscosθ+mv_2yssinθ
I wrote the equations I have for you (correct now, I think). Why don't you write the other two you have in mind for me?

 

[haruspex]

Why don't you write the other two you have in mind for me?

Forum rules. We point out errors, correct misunderstandings and provide hints.
I have gone somewhat further already by telling you how to proceed:
Invent an unknown, J say, for the impulse between rod and ball.
Figure out the direction of that impulse.
Using J, write separate conservation equations for linear momentum for for the ball (two equations), linear momentum in the y direction for the rod, and angular momentum for the rod about its (original) axis.
Energy gives the fifth equation.

Edit: just realized this is not on a homework forum, though it looks like homework. If you assure me it is not homework I can be more prescriptive.

 

[f todd baker]

I just saw your edit. The person interested in this is doing industrial design of some sort and says: "Glad I found your site, my name is Ben, and this question is something I'm working on professionally, not homework. I use Autodesk Inventor's Dynamic Simulation to model collisions and some of the results for a specific
simulation don't seem consistent. So, a colleague suggested we simplify the problem and work the numbers out by hand, but we can't figure it out. "

I have been following your suggestion of introducing the impulse on rod/ball but have been finding that the end result is one of the equations I already have (angular momentum conservation). Kinda embarrassing since I am usually pretty good at these CM problems! You can look at the current status of my work at this link.

 

[haruspex]

I have been following your suggestion of introducing the impulse on rod/ball

perhaps I am missing it but I do not see where you have answered my question about the direction of the impulse between rod and ball. This is something you should be immediately able to state without any algebra. Until you have that you do not have the extra equation you need.

 

[f todd baker]

Doesn't knowing the two components of the impulse specify its direction? E.g. the angle the impulse on the ball makes below the positive x-axis is tan^-1[(v₁-v_2y)/v_2x]

 

[haruspex]

Doesn't knowing the two components of the impulse specify its direction? E.g. the angle the impulse on the ball makes below the positive x-axis is tan^-1[(v₁-v_2y)/v_2x]

There is a much more direct way to see what the direction is, and it does not use those unknowns. Consequently it provides an additional equation.

 

[f todd baker]

All that comes to mind is that the force on the ball must be normal to the surface. If that is the case, and that would be pretty sweet, I am having trouble convincing myself why. Are we getting close to wrapping this thing up?!

 

[haruspex]

All that comes to mind is that the force on the ball must be normal to the surface.

Bingo.
The normal force between two surfaces is the force of minimum magnitude necessary to prevent interpenetration of the bodies. That it is indeed normal to the surfaces is not hard to prove.
Now, in the real world, contact is not instantaneous - the bounce takes some time. During that time, if there is any frictional coefficient, the ball will start to rotate, taking some of the energy. Since we do not know anything about friction here, and we do not know how long the contact will last, we must ignore that. The result is that the impulse is normal to the surfaces.