Angular momentum of two particles and a rod system

[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 can't seem to get the length L from anywhere in the problem.

Any help is appreciated.
thanks a lot!

 

[rl.bhat]

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

 

[missfearless017]

how do you find the angular momentum of the particles?

 

[rl.bhat]

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