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

AI Thread Summary
In a collision problem involving a wooden bar and a bullet, the discussion highlights the discrepancy between angular momentum and energy calculations. The conservation of angular momentum yields one formula for angular velocity, while conservation of energy provides a different result, leading to confusion. The key point is that mechanical energy is not conserved in this perfectly inelastic collision, which explains the differing outcomes. Participants clarify that the angular momentum calculation is correct, while energy loss must be assessed by comparing initial and final kinetic energies. The conversation concludes with an acknowledgment of the energy loss during the collision process.
uq_civediv
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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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Your only mistake is in thinking that mechanical energy is conserved--it's not. This is an example of a perfectly inelastic collision.

uq_civediv said:
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.
 
Last edited:
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 !
 
uq_civediv said:
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.
 
Doc Al said:
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
 
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