What are the combinations rotational inertia and kinetic energy?

Question

Two particles each with mass m = 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.6 cm and mass M = 1.2 kg. The combination rotates around the rotation axis with angular speed = 0.30 rad/s. Measured about O, what are the combination's:
(A) rotational inertia
(B) kinetic energy

The answers are:
(A) 0.023 kg*m^2
(B) 11mJ

but I don't know how to get these answers. I really want to understand this problem, so please help me!

Answer 1

It will rotate about the center of mass O at d/2
calculate the moment or inertia about O, include both mass and both rods. see link for formulas if you don't know how
http://en.wikipedia.org/wiki/Moment_of_inertia
The formula for KE is also there

Answer 2

a.) to locate the entire rotational inertia, you're able to sum up all of the guy inertias. For the rod, use the parallel-axis theorem. I = mr^2 ; (d+d) = total length of rod; M total = 2M Parallel axis theorem for a skinny rod: I = (a million/12)Md^2 + M(d/2)^2 => (a million/3)Md^2 --You divide the gap via 2 because is how far the rod is rotating from the midsection of mass I total = md^2 + m(d+d)^2 + (a million/3)(2*M)(d+d)^2 => => (.80 5)(.056)^2 + (.80 5)(.112)^2 + (a million/3)(2.4)(.112)^2 => .023 kgm^2 <-- answer A b.) Rotational Kinetic ability (ok) = (a million/2)Iw^2 --w = omega = angular velocity ok = (a million/2)(.023)(.3)^2 => .001 J <--answer B