Solving Angular Momentum Problem: Dividing by L/4 Explained

[paki123]

I got a homework problem the other day, and it was a conservation of angular momentum problem. Basically a bullet hits a rod, and a rod starts to spin. I needed to find how fast the rod was rotating.

I didn't get the answer right, but I was looking up the answers, and it says that to convert it to the angular momentum, I had to divide my linear momentum by l/4(because the bullet strikes the rod l/4 over the center of mass.)

So it looked like this:

P = Linear Momentum = M(b)*V(b) + M(r)*V(r)

AM =Angular Momentum = (I(b)+I(r))*ω

P*L/4 = AM and then solve for ω
Why does dividing by L/4 get me to Angular momentum?

Or was initial momentum suppose to be covering linear momentum and angular momentum?

So Initial Momentum should have been (initial Linear momentum + initial Angular momentum )
and Final Momentum should have been (final linear momentum + final angular momentum)?

 

[rcgldr]

The answer you've shown seems that it would only apply if the bullet completely stops after impact with the rod, a somewhat elastic collision. If the bullet continues to move after impact (including getting imbedded into the rod), then part of the angular and linear momentum of the system composed of bullet + rod remains in the bullet.

If the bullet completely stops, then it's linear momentum equals the impulse the bullet imparts to the rod. You can then treat the problem as an impulse applied to the rod at a point 1/4 the length of the rod away from the rod's center of mass.

 

[paki123]

I forgot to mention that. You are right, the bullet does get embedded into the rod. But I'm still not getting the intuition of why L/4 is being multiplied. Is there any connection between linear and angular momentum?

 

[nasu]

P*L/4 = AM and then solve for ω
Why does dividing by L/4 get me to Angular momentum?

It's multiplying and not dividing.
P*L/4 is the angular momentum of the bullet (in respect to the center of mass) right before collision. What you have there is conservation of angular momentum for the collision.
You don't need to "convert" linear momentum to angular momentum. The bullet has both.