# Impulse on a free rod

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1. Feb 7, 2015

### physicsisgreat

1. The problem statement, all variables and given/known data
I have a rod of mass m and length l on a table without any kind of friction. I give it an impulse J in any point of distance d from the center of the rod, parallel to the table and perpendicular to the rod.
Find the angular velocity ω and the velocity of the center of mass v0.

2. Relevant equations
Moment of inertia of the rod rotating around its center: I = m l2 / 12
L = I · ω

3. The attempt at a solution
From the impulse theorem:
J = ΔP = P'
I can calculate ω from the angular momentum relations:
L = d x J = I · ω
ω = d J / I = 12 d J / (m l2),
which is 0 if I hit the rod on its center and max if I hit it on d = l/2.
Now I fail to calculate v0 :P

Last edited: Feb 7, 2015
2. Feb 7, 2015

### BvU

Hello P.i.G., welcome to PF :)

Your relevant equation needs one or two colleagues: currently v0 doesn't feature there, $\omega$ doesn't, etc.

Your relevant equation also needs improvement. There is a factor missing.

3. Feb 7, 2015

### dean barry

Should you not apply a couple of equal torques (one clock and one anti-clock) to stop any motion of the centre of mass?

4. Feb 7, 2015

### BvU

Not a good idea. 1. It doesn't work. 2. There is nothing that can be considered a cause for these $\tau$. 3. The center of mass is not "stopped"

5. Feb 7, 2015

### physicsisgreat

Hi, I edited my post adding the factor missing and an equation about ω (which I actually used later in my attempt at a solution).
I cannot give another equation about v0 though, as it is exactly what I am looking for! :)

Last edited: Feb 7, 2015
6. Feb 7, 2015

### physicsisgreat

Mmh, following the impulse-momentum theorem
J = ΔP
I would say that
v0 = J / m.

At the same time I wonder: can this be true? Can the velocity of the center of mass of the rod be independent from the point in which the impulse is applied?

7. Feb 7, 2015

### haruspex

Yes, it's correct. It feels wrong, until you realise that to impart the same impulse further from the centre of mass of the rod you have to 'work' harder. The rod tends to swing out of the way, so takes less momentum off the impacting object. To achieve the same imparted momentum the impacting object has to start with more momentum.

8. Feb 8, 2015

### physicsisgreat

Thank you very much haruspex! :)