# Angular momentum vs energy

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 ?

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Doc Al
Mentor
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 $$m$$.

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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
Mentor
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.

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