# Angular momentum: why can't I answer this problem using linear momentum rules?

1. Nov 11, 2011

### mm2424

1. The problem statement, all variables and given/known data

A 1.0 g bullet is fired into a 0.5 kg block attached to the end of a 0.6 m nonuniform rod of mass 0.5 kg. The block-rod-bullet system then rotates in the plane of the figure about a fixed axis at A (the figure shows a vertical rod labeled A at its top. A block is attached to its bottom end. The bullet flies into the block). The rotational inertia of the rod alone about the axis at A is 0.060 (kg)m^2. Treat the block as a particle. If the angular speed (ω) of the system about A just after impact is 4.5 rad/s, what is the bullet's speed just before impact.

2. Relevant equations

I have the worked out solution and see that we need to use L = Iω and L = (linear momentum) x r.

3. The attempt at a solution

When I first tried this problem, I tried to do it as a linear momentum problem. I set it up as (mass of the bullet)(initial velocity of bullet) = (mass of the block, bullet and rod system)(velocity of the block, bullet, rod system). I then found the velocity of the block, bullet, rod system using V=ωR.

I understand that this is an angular momentum problem and that I should have used angular momentum equations to solve it. However, I tried to turn it into a linear momentum problem. My strategy was to use the w given to find v of the block/rod/bullet system after impact (using V=ωR). Since my strategy didn't yield the correct answer, I now see that my reasoning was flawed :).

One thing in particular that confuses me about all of this is that the actual solution requires that we take the linear momentum of the bullet and multiply it by the length of the rod...which seems like taking linear momentum and turning it into angular momentum. I therefore thought/hoped/prayed there was a way to turn angular momentum into linear momentum (by finding V from the w given)...in part because the L = r x p and L = Iω equations seem like magic whereas linear momentum equations seem to make sense.

Can someone please explain the error in my logic? While I'm sure that a completely adequate explanation requires an explanation of the cross product, I think I'm clinically incapable of understanding what it is...so if there's a conceptual way of explaining why I can't tackle this problem with linear momentum...and why the actual method of multiplying the bullet's linear momentum by R and equating it to Iω works, I'd be super grateful!

Thanks!

2. Nov 11, 2011

### SammyS

Staff Emeritus
Welcome to PF .

One end of the rod is fixed, so the center of mass of the rod has only 1/2 of the velocity of the rest of the system.

3. Nov 12, 2011

### mm2424

Oh, I see. I have a quick follow up, if that's ok.

I think what confuses me about this problem is the fact that angular momentum can be written as both r x p and Iω.

The solution to this problem seems to require that we write the angular momentum of the bullet before it strikes as r x p while writing the angular momentum of the system after the strike as Iω. What's conceptually hard for me to reconcile is that it doesn't seem possible to write the angular momentum of the bullet in the Iω form before the strike, since it has no rotational velocity. It seems like ω for the bullet prior to the strike is 0, which would make the angular momentum before the strike 0...but this conflicts with the fact that r x p for the bullet is not 0.

Right now, the decision of whether to express angular momentum as r x p or Iω seems arbitrary, which makes the concept seem a bit mysterious. Is there some way of thinking about these problems that I'm missing? Thanks a million for any advice!