Moment of inertia... (10 points!!!)

Two particles with masses 2.5 kg and 5.8 kg are connected by a light rod of length 3.5 m. Find the moment of inertia of the system about an axis perpendicular to the rod and passing through:

a) the midpoint of the rod;
kg*m^2
b) the center of mass of the system of particles:
kg*m^2

First one to clearly explain this and have the right answer gets 10 points!!! :)

The moment of inertia, designated by the letter I, is defined as the sum of all the masses times their distances from the axis of rotation squared:

I = Σmr²

For a):

If the axis of rotation is in the center of the rod, that means that each mass will be 1.75 m from the center (half the rod length).

I = Σmr² = (2.5)(1.75)² + (5.8)(1.75)² = 25.419 kg*m²



For b):

The center of mass (I'll use the letter x) is defined as the sum of the masses times the distance from a given point, divided by the sum of the masses:

x = Σmr / Σm

For this system, we'll use the center of the rod as the given point:

x = Σmr / Σm = ((5.8)(1.75) - (2.5)(1.75)) / (5.8 + 2.5) = 0.696 m from the center toward the heavier mass.

This means that the distance from the heavier mass to the new axis of rotation is 1.75 - 0.696 = 1.054

The distance for the lighter mass is 1.75 + 0.696 = 2.446


Calculating the new moment of inertia:

I = Σmr² = (5.8)(1.054)² + (2.5)(2.446)² = 21.4 kg*m²