# Revolutions frequency after collision

1. Feb 20, 2012

### bznm

1. The problem statement, all variables and given/known data
A bar, whose mass is 10kg, is 3 cm long and is free to rotate about a vertical axle through its centre.
At the beginning, the bar is vertically quiet. Then, the two extremities are hit at the same time by two bullets, one by the left and the other by the right. Their masses are 5 kg each and their velocity is 62.8 m/s.
In the inelastic collision, the bullets become ... (I don't know the exact word for "loyal", "several", "they are moving exactly like the bar", "they are embedded into the bar") each respectly to an extremity.

2. What I need:
I'd like to know the revolutions frequency of the bar after the collision.

3. Attempts to the solution
My result is 500 revolutions per second. I have some doubts, so I ask you for a confirm.

Thank you very very much
Sorry for my bad english and for my horrible diagram :grumpy:
http://img571.imageshack.us/img571/6713/diagramo.png [Broken]

Last edited by a moderator: May 5, 2017
2. Feb 20, 2012

### Staff: Mentor

Your result looks fine. I can't comment on your method of finding that result, since you didn't provide it.

3. Feb 20, 2012

### bznm

I have tried to solve it. Can you confirm? Is it right?

$Lp=$angular momentum of a bullet $= rmv= 0.015 *5*62.8=4.71*10^-2$

$A=$angular momentum of 2 bullets $=2*Lp=9.42 \frac{m^2 * kg} {s}$

Inertia bar = $\frac {1}{12}*M*length^2=\frac {1}{12}*10*(3*10^-2)^2=7.5*10^-4$

Inertia 1 bullet = $mr^2=5*(1.5*10^-2)^2=1.125*10^-3$

inertia 2 bullets = 2 *( Inertia 1 bullet) = $2.25*10^3$

$I$=Sum of inertia = Inertia Bar + Inertia bullets = $7.5+10^-4+2.25*10^-3=3*10^-3$

$A=Iw$

$9,42 = 3*10^-3 w$ --> $w=3410$

frequency=$\frac {w}{2\pi}=500 hz$

4. Feb 20, 2012

### Staff: Mentor

Yes, that is correct. Nicely done.

5. Feb 20, 2012

### bznm

Thank you very very much. It was my university test today. I am so grateful :D :D