# Calculating angular frequency and velocity after a collision

1. Jun 19, 2017

### Fibo112

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

The Problem is the following: We have a uniform disk of radius r laying still with its center at the origin. Two bullets, with equal mass m and negligible size are approaching the disk, both with trajectories paralell to the x axis and at distance h, -h from the y axis respectively where h<r. The velocities are v1 and v2 whre v2>v1. Both bullets remain stuck in the disk immediately without penetrating it. The question is now to solve for v1 and v2 in dependance of the angular velocity and the velocity of the disk after the collision.
2. Relevant equations

mh(v1+v2)=(1/2 Mr^2 + (2mr^2))w
3. The attempt at a solution
I can solve the question without problems using this relation, which is also used in the solution. mh(v1+v2)=(1/2 Mr^2 + (2mr^2))w where w is the angular velocity. What bothers me is 1. why can we just assume that the disk will rotate about its center( at least I think this has to be assumed based on the moment of inertia we are using), 2. If the disk does rotate about its center it seems to me that it will have to continuously slow down unless the connecting line between the impact points of the bullets goes through the center.

2. Jun 19, 2017

### collinsmark

I think the solution requires the assumption: The bullets strike the disk simultaneously.

Also, I suspect there is a misprint in the problem statement. The problem statement says, "...both with trajectories paralell to the x axis and at distance h, -h from the y axis ..." I think that is a mistake. I'm guessing h and -h should be defined as the distances from the x-axis, not the y-axis. In other words, I think the problem statement is saying that one bullet is traveling at a distance h above the x-axis, and the other is traveling in the opposite direction, and a distance h below the x-axis.

3. Jun 19, 2017

### Fibo112

Thanks for the reply. You are right, I meant distance h from the x axis. Could you ellaborate further on how the assumption of simultanious impact is sufficient.

4. Jun 19, 2017

### collinsmark

Well, that means that the bullets strike the disk on opposite sides of the disk. That means that line that connects the impact locations goes through the disk's center.

Also, keep in mind the part of the problem statement that says, "Both bullets remain stuck in the disk immediately without penetrating it." I take that as meaning you should treat the bullets as point particles that remain on the surface of the disk; i.e., they don't penetrate deeper into the disk.

Given all that, what can you conclude about the system's (disk + bullets) final center of mass?

5. Jun 19, 2017

### Fibo112

That it is the center of the disk. But wouldnt it also be conceivable that the disk ends up rotating about a different point?

6. Jun 19, 2017

### collinsmark

It might if its motion is restricted somehow, such as if the disk was pinned to a hinge or affixed to rails, etc.

But if the final system (disk + attached bullets combo) is allowed to freely move and freely rotate in space, without restriction, it will rotate around its center of mass.

7. Jun 19, 2017

### scottdave

So are they coming in from opposite sides (+x and -x) or from the same side? I cannot find where it specifies that.

8. Jun 19, 2017

### Fibo112

would it be correct to say that the reason for this is that any rotation that is not about the center of mass requires addition forces to sustain it and since there are no forces acting after the collision it must rotate about its center of mass?

9. Jun 19, 2017

### Fibo112

Yeah I forgot to mention that. Opposite sides

10. Jun 19, 2017

### collinsmark

That sounds reasonable to me.

11. Jun 19, 2017

### Fibo112

great, thanks for your help