Angular momentum problem giving me a headache

[karokr94]

Homework Statement

A heavy stick of length L = 3.3 m and mass M = 20 kg hangs from a low-friction axle. A bullet of mass m = 0.014 kg traveling at v = 117 m/s strikes near the bottom of the stick and quickly buries itself in the stick. Just after the impact, what is the angular speed ω of the stick (with the bullet embedded in it)? (Note that the center of mass of the stick has a speed ωL/2. The moment of inertia of a uniform rod about one end is (1/3)ML^2.)

http://www.webassign.net/userimages/bullet_stick.jpg?db=v4net&id=175469

2. Some background information

Okay so this was a homework problem we got two weeks ago (with different numerical values) and we were made to re-do it because the website we use to do homework on (webassign) was apparently marking this question wrong. I never did it the first time. I have a dilemma because I worked out this question and I got an answer of 0.0743rad/s BUT when I use this method for the homework problem from two weeks ago, I get the same answer as in the answer key. The answer key even gives the same formula I got when I worked out this question. The problem here is that, apparently this is wrong! I am not sure what's going on here, if anyone can confirm I've used the right method to solve this problem, that would be greatly appreciated!

3. Attempt at solution

Im sure I'm correct but to confirm it to myself I used three different methods to solve this problem, all which gave the same answer.

M=mass of rod
m=mass of bullet
(1/12)ML² = Moment of inertia for rod about center
(1/3)ML² = Moment of inertia for rod about one end
A=point about which rod rotates

There is no doubt in my mind that, initially, only the bullet has angular momentum about A
[Li]=initial angular momentum=Lmv
[Li]=5.405kgm²/s

Now I've modeled the final state in three ways, first as the bullet and and center of mass of the rod having translational angular momentum about A, and the rod having rotational angular momentum about its center of mass
[Lf]=Lm(ωL)+(L/2)M(ωL/2)+(1/12)ML²ω <---This was the equation in the answer key

Next, I modeled it as the bullet and center of mass of rod connected to A by massless rods so that I can say the bullet and center of mass of rod have rotational angular momentum about A. Of course the rod must also have rotational angular momentum about its own center of mass as well. This equation turns out to be the same as the above equation.
[Lf]=(mL²+M(L/2)²)ω+(1/12)ML²ω

Finally, I modeled it as the bullet having translational angular momentum about A and the rod having rotational angular momentum about A.
[Lf]=Lm(ωL)+(1/3)ML²ω

Solving for ω in any of these equations gives the EXACT same answer (to the last digit in my calculator) of 0.0743rad/s. I don't get how this can be wrong even though it apparently is, someone please help!

 

[Simon Bridge]

The final angular velocity of the center of mass of the stick is the same as the final angular velocity of the tip of the stick which is $\omega$. That $\omega L/2$ thing is the tangential speed. Remember $v=r\omega$?

Rework the equation with that in mind.

 

[karokr94]

The final angular velocity of the center of mass of the stick is the same as the final angular velocity of the tip of the stick which is $\omega$. That $\omega L/2$ thing is the tangential speed. Remember $v=r\omega$?

Rework the equation with that in mind.

I'm not sure what you mean, haven't I already done this?

 

[haruspex]

Hi karokr94,
I see nothing wrong in your method, and I get the same answer.

 

[Simon Bridge]

You appear to have put the angular velocity of the center of mass of the stick as $\omega L/2$ in each equation. That gives you a bunch of $L/2$'s all over the place... course, I have been known to be wrong... but that is what occurs to me right away.

if the final angular momentum of bullet is:
$L_b =mL^2\omega$

then the final angular momentum of the stick is:
$L_s=\frac{1}{3}ML^2\omega$

Given Haruspex's answer I guess I'd better crunch the numbers...
I get: 0.074299 rad/s ... hmmm... looks like our approaches are equivalent.
Maybe the model answer wasn't wrong after all?

 

[karokr94]

if the final angular momentum of bullet is:
$L_b =mL^2\omega$

then the final angular momentum of the stick is:
$L_s=\frac{1}{3}ML^2\omega$

That's the same as this, no?

[Lf]=Lm(ωL)+(1/3)ML²ω

 

[Simon Bridge]

Yah - I just saw a lot of L/2's and figured that may be the confusion... should have been more alert. Is it just me or has there been a run on bad model answers lately?