Help with angular momentum question

1. Nov 20, 2007

pakmingki2

A. At a particular instant, a particle of mass M = 3 kg is at the position (x,y,z) = (4,4,6) m and has velocity (2,1,-2) m/s.

B. An identical particle is placed at (x,y,z) = (-4,-4,-6) m, with velocity (-2,-1,2) m/s.
Find the angular momentum of the pair of particles about the origin.

C. A thin rod of mass 9 kg is added between the two particles (Irod = M L^2/12). Since the particles are moving perpendicular to the line separating them and in opposite directions (you should convince yourself that this is true), then the particles + rod system rotates about its center of mass. Compute the angular frequency of rotation of the rod + particles system.

D. Compute the magnitude of the total angular momentum of the system (particles plus rod).

ok, so i got parts a,b,c pretty easily.

A. <-42, 60, -12>
B. <-84, 120, -24>
C. .363

Ok, so for part D, the main approach i used was total angular momentum = sum of all individual components in the system.

So, i can easily find the angular momentum of the two particles, but the problem is finding the moment of inertia of the thin rod. The formula for inertia is (ML^2)/12, but i cant seem to get the length L from anywhere in the problem.

Any help is appreciated.
thanks alot!

2. Nov 20, 2007

rl.bhat

Distance between the two particals is the length of the rod.

3. Nov 21, 2007

missfearless017

how do you find the angular momentum of the particles?

4. Nov 21, 2007

rl.bhat

The angular momentum of the particles = Total moment of inertia*angular velocity=(Moment of inertia of rod + moment of inertia of particals) *angular velocity.