# Velocity of CM of Rod if struck below CM ?

1. Mar 30, 2012

### sweetpete28

Let's say there is an elastic collision b/t a particle and a rod. The rod (initially motionless) is struck by the particle at some distance below its CM and particle comes to rest.

So we have an elastic collision where v of particle is completely transferred to rod. BUT how do I analyze what the velocity of the CM of the rod is if it is struck below the CM????

2. Mar 30, 2012

### ehild

The rod gets both linear momentum and angular momentum from the particle, so it will both translate and rotate about the CM. The CM will move like a point mass with momentum gained from the collision.

ehild

3. Mar 30, 2012

### sweetpete28

Ok...

so:

1) m1v1 (particle) = (center of mass for rod = length/2)v2 and solve for v2...is what you are saying?

2) How is exactly is energy lost in this elastic collision because I put zero for delta E, energy loss in the collision and that is wrong...???

4. Mar 30, 2012

### sweetpete28

and...you're saying axis of rotation for rod is CM even though it is struck at distance = x below CM...?

5. Mar 30, 2012

### ehild

No. Momenum of the particle before collision is the same as the momentum of the rod after collision. The momentum is mass of the rod times velocity of CM.

The energy of the particle will be transferred both to the translation and rotation of the rod.

ehild

6. Mar 30, 2012

### ehild

If the rod is free, its motion is equivalent to the translation of CM and rotation about the CM.

Is the rod free, or is it fixed to an axis?

ehild

7. Mar 30, 2012

### sweetpete28

1) I meant:

m1v1 (particle) = (rod length/2)(mass of rod)(v of rod) and solve for v of rod. Is this what you are saying?

2) So the following is wrong for the collision: (1/2)m1v1^2 = (1/2)m2v2^2 + (1/2)Iω^2 (b/c this means energy is conserved)...right?

8. Mar 30, 2012

### sweetpete28

The rod is free to move in any direction on the table.

9. Mar 30, 2012

### ehild

Momentum is mv. Why do you multiply the momentum of the CM by length?

m1v1=MrodVCM

It is correct. But you need the angular momentum after collision. It is the same as the angular momentum of the particle before collision, with respect to the CM. If its velocity was normal to the rod, the angular momentum of the particle is v1m1L/2 and this is equal to the angular momentum of the rod after collision.

In case the velocity is at some other angle with the rod, you need to take the angle into account.

ehild

10. Mar 30, 2012

### sweetpete28

ok....

The rod is uniform with M = 1.9kg, L = .795m

The particle moves at right angles to the rod at speed v = 5.99 m/s with m = .801kg and strikes rod at distance x = .13m below CM of rod and stops (but does not stick).

So...

1) V center of mass of rod post collision = momentum of particle / mass of rod

2) Length of the particle is not given, so to find ω for which rod rotates after collision, can I use: (1/2)m1v1^2 = (1/2)MVcm^2 + (1/2)(1/12)ML^2(ω^2) [**I for rod rotating through CM = 1/12 ML^2***]....or do I need to factor in that rod was struck .13m below CM?

3) How do I find the energy lost if the initial total kinetic energy of the particle is equal to the final total kinetic energy of the rod??

I'm sorry but I'm really confused here...thanks so much for your help (and patience)...

11. Mar 30, 2012

### ehild

Correct.
The angular momentum conserves. The particle (a point mass with no length) has angular momentum before collision which is its linear momentum multiplied with the distance from the CM (0.13 m) where it struck the rod. The angular momentum of the rod is Iω. Both angular momenta and the moment of inertia are calculated with respect to the CM.

You said the collision was elastic, but the final zero velocity of the particle does not correspond to an elastic collision. Determine the initial and final energies.
You got the angular velocity ω from conservation of the angular momentum, and the linear velocity of the CM from conservation of linear momentum. Knowing Vcm and ω, calculate the final energy and compare to the initial KE of the particle.

ehild