Why Do Angular Momentum and Energy Calculations Differ in a Collision Problem?

[uq_civediv]

the problem is the following:

we have a vertical wooden bar pivoted from the top end, length $$2 l$$, mass $$M$$

a bullet with mass $$m$$ hits it in the middle at velocity $$v$$ and gets stuck

i am asked to find the angular velocity $$\omega$$ of the system bar+bullet immediately after the hit

i do know this calls for applying the conservation of energy or angular momentum, for some reason however i get different results

Both if them involve the moment of inertia of the combined system, $$I_{\Sigma}=\frac{1}{3} M (2l)^2 + m l^2 = \frac{4}{3} M l^2 + m l^2$$

Conservation of angular momentum gives me $$m v l = \omega I_{\Sigma}$$, from which $$\omega = \frac{m v l}{I_{\Sigma}} = \frac{m v l}{\frac{4}{3}M l^2+m l^2} = \frac{m}{\frac{4}{3}M+m} \cdot \frac{v}{l}$$

Whereas conservation of energy says $$\frac{m v^2}{2} = \frac{\omega^2 I_{\Sigma}}{2}$$, which gives $$\omega = \sqrt{\frac{m}{I_{\Sigma}}} v = \sqrt{\frac{m}{\frac{4}{3}M + m}}\cdot \frac{v}{l}$$

So the big question is where did I mess up this time. I know it's something really basic because I can't see it. Usually i ask a deskmate or someone to have a look if they spot something really simple but since nobody's around I had to come here.

P.S. while you're at it, why do my $$m$$, $$v$$ and $$\omega$$ look superscripted but $$M$$ and $$2 l$$ don't ?

 

[Doc Al]

Your only mistake is in thinking that mechanical energy is conserved--it's not. This is an example of a perfectly inelastic collision.

P.S. while you're at it, why do my $$m$$, $$v$$ and $$\omega$$ look superscripted but $$M$$ and $$2 l$$ don't ?

To use Latex in the middle of a line of text and have it appear even, use "itex" as your delimiter, not "tex". It gives you $m$ instead of $$m$$.

 

[uq_civediv]

so the angular momentum one is correct ? (just clarifying...)

and the loss in energy is the good ol' $$\int F ds$$ over the distance the bullet travels into the rod !

 

[Doc Al]

so the angular momentum one is correct ? (just clarifying...)

Yes.

and the loss in energy is the good ol' $$\int F ds$$ over the distance the bullet travels into the rod !

Good luck calculating that! To find the loss in energy, just calculate the final KE and compare it to the initial.

 

[uq_civediv]

Good luck calculating that! To find the loss in energy, just calculate the final KE and compare it to the initial.

o no wasn't going to do that, good luck indeed
just realising where the energy went.
case closed anyway