# Particle hits rod in space

1. Jan 7, 2015

### jojotank

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Hello :)
I was doing some problem and i don't quite understand it, so i thought you could help a bit. The problem is:
rod with mass M floats in space and then it is hit at distance b (measuring from the end of the rod) by a small particle with mass m. Collision is elastic.

I know how i should solve this. Energy, linear momentum and angular momentum are conserved. But i do t know how to explain for myself when we wrote for conservation of angular momentum:

bmv=bmv1 + Mv2*l/2+Iω

I understand the first and second parts on the right side of equation. The first one is angular momentum of particle right after the collision. The third is angular momentum of stick which rotates around the center of it's gravity, right? But what is Mv2*l/2. Why? i simply can t figure that out. If someone would explain it i would be really grateful.
Thank you.

2. Jan 7, 2015

### TSny

Hello, and welcome to PF!

Angular momentum is always defined relative to some origin. Note that b is the distance measured from one end of the rod. When you write bmv for the initial angular momentum of the ball, what point are you choosing as the origin? Then think about how you should write the final total angular momentum of the rod relative to this origin.

3. Jan 7, 2015

### jojotank

Hmm. I still don t know why l/2 instead of b...

4. Jan 7, 2015

### TSny

b is associated with the position of the ball. But the term Mv2(L/2) is associated with the rod.

Can you specify the point that you are choosing for the origin for calculating angular momentum? Knowing the location of the origin is essential to understanding this problem.

5. Jan 7, 2015

### jojotank

it should be the same at the end, whatever origin i choose. But let's choose the end of the rod. I don t understand what the middle part does in the problem. To what does it relate?

6. Jan 7, 2015

### TSny

Yes, but if you choose the origin at some random point then it would generally not be correct to write the initial angular momentum of the ball as bmv. When you write the angular momentum as bmv, then that implies that you have some origin in mind. The angular momentum of the rod must then be expressed relative to this same origin.

OK, if you choose the origin at the end of the rod, then it will be correct to write bmv for the initial angular momentum of the ball. Suppose for the moment that the rod does not rotate after the collision, but just translates with velocity v2. (That would be the case if b = L/2 and the ball struck the center of the rod.) Would the rod have any angular momentum about the origin? If so, how much?

7. Jan 7, 2015

### jojotank

yes it would - mvl/2 right ?

8. Jan 7, 2015

Yes.

9. Jan 7, 2015

### haruspex

Yes.
If an object's centre of mass is moving and the object is also rotating then you can express this movement as a sum of a translation and a rotation. For this purpose, you can take the rotation to be about whatever point you choose, but the translation velocity will depend on that choice. Mostly one chooses the mass centre, so the motion is a rotation rate ω about the mass centre plus a velocity v2 of the mass centre. Both of these contribute to the angular momentum about the chosen origin.
Another choice is the 'instantaneous centre of rotation', such as the point of contact of a wheel with the road. In this view the motion is entirely rotation, but you have to use the parallel axis theorem to get the moment of inertia. The same answer results.

A word of warning: you do have to be careful how you choose your axis in angular momentum problems. It is always ok to chose a point that's fixed in the inertial reference frame, and it's always ok to choose the mass centre of the system. It is not OK to use an axis anchored to an object that gets accelerated in the process.

10. Jan 7, 2015

### jojotank

So it was wrong, that they wrote mv2*(l/2). It should be (b-(l/2))mv2 right? If they choose such point of origin - end of the rod.
I can't thank you guys enough. I would totally buy you a beer right now. Thanks.

11. Jan 7, 2015

### haruspex

No, l/2 is the distance from the chosen axis (end of rod) to mass centre of rod.

12. Jan 7, 2015

### haruspex

In case my last post was not clear, let me elaborate. As I cautioned, you should not literally take the axis to be the end of the rod, since that will be accelerated. Rather, you take the axis as that point in space where the end of the rod was before impact.
About that point, only the particle has a.m. beforehand: bmv.
After impact, the particle has speed v1, the rod mass centre has speed v2, and the rod rotates about its centre at rate $\omega$. These contribute a.m. of bmv1, Mv2(l/2), I$\omega$ respectively.

13. Jan 8, 2015

### jojotank

Now it is totaly clear. I can' t thank you enough. I literally love you more than a birthday cake. A chocolate one... With sprinkles on the top... No, seriously - thank you very much!