Bullet lodged in rotating rod: angular momentum conservation

[nchin]

A uniform thin rod of length 0.500 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact?

Solution with diagram on page 4m problem #11.55:
http://www.ifm.liu.se/edu/coursescms/TFYA16/lessons/Le-8-extra.pdf

Moment inertia of system: I = Irod + mr^2

Solution:
conservation of ang momentum:

rmvsinθ = (1/12 ML^2 + mr^2)ω


So conservation of ang momentum is

Linitial = Lfinal

but this problem uses: ang momentum of a particle = ang momentum of a rigid body?

What i don't understand:

Why is the Moment inertia of system I = Irod + mr^2? what i did was before looking up the solution was:

ML^(2)/12 ω = rmvsinθ + MR^(2)ω

because ang momentum inital was just the rod rotating by itself and then comes the bullet so ang momentum of a bullet plus ang momentum of the rod. rmvsinθ = (Irod + mr^2)ω doesn't make sense to me. Also can someone explain to me what a rigid body is? because i know rigid body is L = Iω. Are rigid bodies like a rod, ring, disk or shell?

 

[tiny-tim]

hi nchin!

(try using the X² button just above the Reply box )

Also can someone explain to me what a rigid body is? because i know rigid body is L = Iω. Are rigid bodies like a rod, ring, disk or shell?

a rigid body is anything, of any shape, that doesn't change shape

(ie it has no separately-moving parts …

eg a rod with a ring free to move along it is not a rigid body )

Why is the Moment inertia of system I = Irod + mr^2? what i did was before looking up the solution was:

ML^(2)/12 ω = rmvsinθ + MR^(2)ω

because ang momentum inital was just the rod rotating by itself and then comes the bullet so ang momentum of a bullet plus ang momentum of the rod. rmvsinθ = (Irod + mr^2)ω doesn't make sense to me.

the angular momentum of the bullet after the collision is rmv_f

but v_f = rω, doesn't it?

so rmv_f = mr²ω

(but since the bullet has become part of the rod, it's easier to use the moment of inertia of the bullet-plus-rod about the centre of the rod, which is the sum of the two moments of inertia, I + mr² …

mr² of course comes from the parallel axis theorem, 0 + mr² )

 

[nchin]

the angular momentum of the bullet after the collision is rmv_f

but v_f = rω, doesn't it?

so rmv_f = mr²ω

(but since the bullet has become part of the rod, it's easier to use the moment of inertia of the bullet-plus-rod about the centre of the rod, which is the sum of the two moments of inertia, I + mr² …

mr² of course comes from the parallel axis theorem, 0 + mr² )

so the moment of inertia can only be used when two moments of inertia come together as one?

so what about the ang momentum before? since the rod is at rest it would just be I(0) + rmvsinθ = rmvsinθ, right?

 

[nchin]

also when we calculate the r, that would be half of the length of the rod? i thought radius only applied to circles.

 

[tiny-tim]

hi nchin!

so the moment of inertia can only be used when two moments of inertia come together as one?

the moment of inertia of a large body can only be used when the large body moves as one

if its parts move separately, they will have separate angular velocities, and maybe separate centres of rotation, and so we must use the separate moments of inertia, and calculate each part's angular momentum separately

so what about the ang momentum before? since the rod is at rest it would just be I(0) + rmvsinθ = rmvsinθ, right?

right

also when we calculate the r, that would be half of the length of the rod? i thought radius only applied to circles.

which r are you talking about?

the r in r x mv is the displacement from the point about which you're taking moments

the r in the parallel axis theorem is usually written d (so as not to confuse it with radius!), and is the distance from the centre of mass of the part to the centre of rotation

 

[nchin]

Got it. Thanks!