In the figure two particles

Question

In the figure, two particles, each with mass m=0.88 kg, are fastened to each other and to a rotational axis at O, by two thin rods, each with length d= 5.5 cm and mass M=101 kg. The combination rotates around the rotation axis with angular speed w=0.25 rad/s. measured about O, what is the combination's (a) rotational inertial and (b) kinetic energy?

Answer 1

Moment Of Inertia measures the resistance of a body to Angular Acceleration and

is equal to the product of (the body's mass times the square of the body's distance

from an axis of rotation).

A mass of 101 kilograms comes to 222.66688481 pounds.

One would more realistically consider the problem 

with M at 0.88 kilograms (1.9400679072 pounds)

and each rod at 0.055 meters (2.1654 inches) length

and 1.01 kilograms (2.2266688481 pounds) mass.

Then diagram:

M=0.88kg-----------------0.055m&1.01kg-----------------O-----------------0.055m&1.01kg-----------------M=0.88kg

First treat the rods without the end masses. Moment Of Inertia about O is found by

I_rods = ∫r²dm equal to ∫_{[from -0.055 to 0.055]}(2×1.01)/(2×0.055)x²dx which translates to

18.3636363636 {x³/3|_{[from -0.055 to 0.055]}} or 2.036833334E-3 kg•m².

Then go to the end masses while treating the rods as massless. Write I_ends = ∑m_ir_i²

equal here to (0.88)(0.055²) + (0.88)(0.055²) or 5.324E-3 kg•m².

Then reckon Rotational Kinetic Energy as 

0.5(2.036833334E-3 kg•m²)(0.25 rad/s)² + 0.5(5.324E-3 kg•m²)(0.25 rad/s)² or

0.06381741669 kg•m²/s² equivalent to 0.0638 Joules.