Easy.
MI= rigid rod I + m1r2^2 + m2r2^2
as I read the problem r1=r2=5.55/2
A light, rigid rod l = 5.55 m in length rotates
in the xy plane about a pivot through the
rod’s center. Two particles of masses m1 =
9.8 kg and m2 = 2.4 kg are connected to its
ends
find the moment of inertia
1 answer