# A rod at rest on ice is struck by a piece of clay...

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1. Mar 28, 2017

### Truman I

1. The problem statement, all variables and given/known data

A rod (m=6kg, L=3m) at rest on ice (μ=0) is struck by a piece of clay (m=1kg, V=5m/s). The clay sticks.
1) What is the Velocity of the center of mass after the collision?
2) What is the Angular Velocity of the rod following the collision?

If the clay did not stick but its rebound was V=0m/s...
3) Now what is the velocity of the center of mass?
4) And what is the Angular Velocity of the rod?

2. Relevant equations

I am just getting back from Spring break, so I'm pretty much brain dead right now. All I know is that I have a test soon and the teacher said a problem like this may be on it. So, this is my attempt to recreate the problem he described. I realize that there are three other threads with similar problems, but I am still having a hard time understanding this concept. Any help would be much appreciated.

What I do understand is that obviously this problem has a lot to do with conservation of angular momentum.

Therefore Li=Lf.

I also understand the L=Iω.

3. The attempt at a solution

So, taking this, I plugged in what I know.

L=Iω
L=(Irod+Iparticle
L=(1/3 mL2+mr2

→→→→→→→→→→→→→→→→→→→→

From here I am totally stuck; I have no idea how to solve this problem. So, I turned to the AP Physics C: Mechanics test from 2005. One of the Free Response Questions follows this same concept.

It's Question 3

Solutions Link here: http://apcentral.collegeboard.com/apc/public/repository/_ap05_sg_physics_c_me_46691.pdf

Questions Link here: http://apcentral.collegeboard.com/apc/public/repository/_ap05_frq_physics_c_m_45648.pdf

Once again, it's not the same question, so please don't try and help me with this AP one. It's just a similar problem.

All help is much appreciated.

2. Mar 29, 2017

### ehild

Where does the piece of clay hit the rod, at the end?
You said that the angular momentum conserves during the collision. What else?
Specify the symbols you use. What is r?
You used the moment or inertia of the rod with respect to one end. Is it fixed?

3. Mar 29, 2017

### Truman I

Yeah, there really is not much to go on for this problem. The rod is not fixed. The clay hits at one of the ends of the rod. r is meaningless, it was just me assigning a variable out of desperation because I have no idea what I am doing. To be honest, my attempt at the probably is probably fundamentally wrong. I think what I am failing to understand is something more central to the core of the problem.

Is there anything else I need to clear up?

4. Mar 29, 2017

### Truman I

Is the velocity of the center of mass the same before and after the collision? Because then I could just find it with the center of mass formula and use the 5m/s ν0 to solve that part of the problem.

5. Mar 29, 2017

### haruspex

If you mean the velocity of the centre of mass of the whole system, rod plus clay, yes.

But there is no need to calculate the system's centre of mass. Just apply conservation of linear momentum and conservation of angular momentum.
For the latter, you need to pick a reference axis and use it consistently. Choose some point that remains fixed. I suggest the point on the ice where the centre of the rod is initially.
What is the initial angular momentum of the system about that point?

I would assume the clay hits the rod at one end, perpendicularly to it.

6. Mar 29, 2017

### Truman I

Thanks for your response, I'm starting to understand this a bit more.

Yes, I do mean the center of mass of the whole system.

What I don't understand right now is how linear momentum comes into play for this problem. Could you explain that?

7. Mar 29, 2017

### Truman I

This part makes sense. I can work my way through that. Thanks again.

8. Mar 29, 2017

### haruspex

There are no horizontal external forces, so it will be conserved. That gives a useful equation. Which part don't you understand?

9. Mar 29, 2017

### Truman I

Are you saying like this?

Initial:
Rod: mass=6kg, velocity=0m/s, ρ=0kgm/s
Clay: mass=1kg, velocity=5m/s, ρ=5kgm/s

Final:
Rod and Clay together: mass=7kg, ρ=5kgm/s, so therefore velocity=.714m/s

If that's all I need to do with linear momentum, then I understand that. If there are more steps, or if there is something else I can use linear momentum to solve for in this problem, please let me know.

10. Mar 29, 2017

### haruspex

Yes, that's it for the linear momentum (but the usual symbol is p, not ρ).
The angular momentum equation, which you also need, is a little tougher. And you are likely to need a third equation relating linear and angular velocities.

11. Mar 29, 2017

### Truman I

Thanks for your replies. They have been very insightful. We went over a few more things in class today that I can now connect everything together. I can solve the problem now.

Thanks again.