Two particles, each with mass m=0.0036kg, are fastened to each other and to a rotation axis at P, by two thin rods, each with length L=0.68m and each with a mass of 0.0086kg. The combination rotates around the rotation axis with an angular velocity of 13.0rad/s. Find the moment of inertia of the combination about P.
Icm for thin rod about the end =1/3*M*L2
I=∫(x2+y2 dm and Rod is thin so y≈0 which makes
I=∫x2 dm and dm=M/L dx. After integration, I=1/3*ML2.
The Attempt at a Solution
At first I just tried adding the lengths and masses and put that into I=1/3*ML2 but that didn't work. Then I tried using the center of mass = L/2, after some integration, and using that as the distance, d, in I=Icm+Md2 and that didn't work either. I don't know if I am supposed to multiply that number by two since there are two rod/particle combos or if the other one has a different center of mass or what. And I have no clue how or where to use the angular velocity unless it's in the conservation of mechanical energy equation. Any suggestions?