Rod elastically impacting a ball - impact location

[davidwinth]

TL;DR

If a cylindrical rod travels at a steady speed perpendicular to its axis, does the impact force felt by an initially stationary ball matter?

Say we have a cylindrical rod of length L and mass M traveling at constant speed S in a direction that is perpendicular to the axis. Further, say that the rod impacts an initially stationary ball of mass m, and the two collide elastically. My question is: does the force experienced by the ball depend on where on the rod the impact takes place?

My intuition says that if the ball and rod impact at, say, L/2 (in the middle), the ball would experience a greater force than if the impact was right at the end of the rod at L. This is because when the ball hits the rod in the end, some of the impact energy goes into imparting a rotation to the rod. When the ball hits in the middle, there is no imparted rotation and the rod only slows down somewhat.

I tried to work this out with math but couldn't quite figure out how to get the right variable relations. As I see it, there are three equations we can use: conservation of energy, conservation of angular momentum, and conservation of linear momentum. On the other hand, there are three unknowns: The final speed of the rod, the final speed of the ball, and the final angular velocity of the rod.

So by the COE and COLM, we can relate the initial (known) velocities to the final (unknown) velocities. That part is ok to me. But I don't understand how to relate the final angular velocity of the rod to anything. It seems to me that the COE and COLM are two equations in two unknowns and therefore determine both final velocities no matter how the rod rotates after the collision. But if the final velocity of the ball is the same no matter where the impact takes place, then the force on the ball is the same no matter where the impact takes place. If the force is the same no matter where the impact takes place, then the torque on the rod varies only by impact location since torque depends on force and distance from axis of rotation, which always passes through the rod centroid. So we have a situation where the ball always feels the same force and yet the rod will not spin at the same rate after the impact independent of impact location! That seems contradictory to me.

I must be missing something. I'd appreciate any input.

 

[haruspex]

First, let’s tidy the question up. The impact is given as elastic. The impact takes some time. The force the two bodies exert on each other varies over that time, reaching a maximum at the half way point. So maybe you are asking how the maximum force depends on impact location, but it would be cleaner to stop referring to force and consider impulse - the integral of the force over the duration of the impact - instead.
The impulse transfers momentum. For a given impact location, you have three unknowns: the speed of the ball afterwards (its direction is obvious), the speed of the mass centre of the rod afterwards (ditto) and the rate of angular rotation of the rod afterwards.
You have three equations available: conservation energy, linear momentum and angular momentum. To be safe, take some fixed point in space (I suggest either the initial position of the ball or the position of the rod's mass centre at the moment of impact). Before impact, neither object has angular momentum about that axis; after impact, it depends which axis you choose.
If you choose the rod's mass centre, both have angular momentum about it (for the ball, it's the linear momentum multiplied by the perpendicular distance to the axis), and these two angular momenta must be equal and opposite.
If you choose the ball's initial position, it will not acquire angular momentum about that point, but the rod will have angular momentum in one direction due its spin and equal and opposite angular momentum due to its linear momentum multiplied by the perpendicular distance from the momentum to the axis.

I suggest trying both to see how the two produce the same answer.

 

[wrobel]

TL;DR: If a cylindrical rod travels at a steady speed perpendicular to its axis, does the impact force felt by an initially stationary ball matter?

My question is: does the force experienced by the ball depend on where on the rod the impact takes place?

Why don't you write down formulas?

 

[davidwinth]

Thank you for the responses. I figured out that I forgot the KE of rotation. I wrote out the equations and was able to solve them numerically. As I suspected, the larger the distance from the rod end, the larger the final velocity of the ball.

 

[erobz]

First, let’s tidy the question up. The impact is given as elastic. The impact takes some time. The force the two bodies exert on each other varies over that time, reaching a maximum at the half way point. So maybe you are asking how the maximum force depends on impact location, but it would be cleaner to stop referring to force and consider impulse - the integral of the force over the duration of the impact - instead.
The impulse transfers momentum. For a given impact location, you have three unknowns: the speed of the ball afterwards (its direction is obvious), the speed of the mass centre of the rod afterwards (ditto) and the rate of angular rotation of the rod afterwards.
You have three equations available: conservation energy, linear momentum and angular momentum. To be safe, take some fixed point in space (I suggest either the initial position of the ball or the position of the rod's mass centre at the moment of impact). Before impact, neither object has angular momentum about that axis; after impact, it depends which axis you choose.
If you choose the rod's mass centre, both have angular momentum about it (for the ball, it's the linear momentum multiplied by the perpendicular distance to the axis), and these two angular momenta must be equal and opposite.
If you choose the ball's initial position, it will not acquire angular momentum about that point, but the rod will have angular momentum in one direction due its spin and equal and opposite angular momentum due to its linear momentum multiplied by the perpendicular distance from the momentum to the axis.

I suggest trying both to see how the two produce the same answer.

The impulse is delivered over a duration of time, so on an off center hit, do we really know the balls direction?

I'm imagining something like this is going to happen:

The normal force will change magnitude and direction from $N \rightarrow N'$, and the ball doesn't have on obvious direction after (nor will the rod for that matter)?

 

[haruspex]

The impulse is delivered over a duration of time, so on an off center hit, do we really know the balls direction?

True, but then we have to start worrying about friction and rotation of the ball, so we have to take the materials to be elastic but arbitrarily rigid.

 

[pbuk]

The impulse is delivered over a duration of time

Clearly we have to consider the limiting case as the duration tends to zero.

 

[erobz]

Should maybe the author just restrict the domain of the analysis to rods significantly more massive than the ball so the angle of rotation is negligible over the duration of the impact?

 

[erobz]

Clearly we have to consider the limiting case as the duration tends to zero.

I wouldn't have asked if it was "clear".

 

[pbuk]

Should maybe the author just restrict the domain of the analysis to rods significantly more massive than the ball so the angle of rotation is negligible over the duration of the impact?

Do you think that that would lead to an interesting solution? How would it be different from two balls colliding?

Like many problems, the first step here is to come up with a model that is neither so complicated that the solution is intractable nor so simple that that the solution is trivial and the problem becomes uninteresting. @haruspex detailed such a model yielding three equations and three unknowns. It should be clear that introducing another variable (the duration of impact) will not yield a unique solution. It should also be clear that if you eliminate one of the unknowns (by forcing the angular velocity of the rod to zero) the solution becomes trivial.

 

[davidwinth]

Do you think that that would lead to an interesting solution? How would it be different from two balls colliding?

Like many problems, the first step here is to come up with a model that is neither so complicated that the solution is intractable nor so simple that that the solution is trivial and the problem becomes uninteresting. @haruspex detailed such a model yielding three equations and three unknowns. It should be clear that introducing another variable (the duration of impact) will not yield a unique solution. It should also be clear that if you eliminate one of the unknowns (by forcing the angular velocity of the rod to zero) the solution becomes trivial.

Also, just in case it was missed, the author presented such a system of equations.

 

[erobz]

Do you think that that would lead to an interesting solution? How would it be different from two balls colliding?

Like many problems, the first step here is to come up with a model that is neither so complicated that the solution is intractable nor so simple that that the solution is trivial and the problem becomes uninteresting. @haruspex detailed such a model yielding three equations and three unknowns. It should be clear that introducing another variable (the duration of impact) will not yield a unique solution. It should also be clear that if you eliminate one of the unknowns (by forcing the angular velocity of the rod to zero) the solution becomes trivial.

I would assume that if they are of comparable mass then the solution one gets from simplified analysis is horsecrap (far from observerable) anyhow... I also think its damaging to the learning process not to explore the simplifications we need to make in detail. So the best model must lie somewhere in between. If some complexity needs to be added to describe the reality we would see in an experiment sometimes there aren't great alternatives that avoid it? So I ask, what conditions make this observable in reality?

 

[haruspex]

I would assume that if they are of comparable mass then the solution one gets from simplified analysis is horsecrap

No, as I posted, you just need to assume the objects are arbitrarily rigid. That does not conflict with assuming the collision to be perfectly elastic.

 

[A.T.]

I would assume that if they are of comparable mass then the solution one gets from simplified analysis is horsecrap (far from observerable) anyhow...

Assuming negligible contact duration is pretty standard in such collision problems, and often the only way to get a solution, without modeling deformation based on material properties.

So I ask, what conditions make this observable in reality?

No idealized solution is ever exactly observed in reality. But for rigid materials like steel, ceramics, glass etc. you should get a pretty good prediction this way.

 

[erobz]

No, as I posted, you just need to assume the objects are arbitrarily rigid. That does not conflict with assuming the collision to be perfectly elastic.

Ok, basically like a glass marble impacting a steel rod and the duration of impact tends to zero. (I.e normal force doesn’t have time to change the marbles direction).