Question

A uniform rod of mass 2.20 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass
m1 = 4.90 kg
is attached to one end and a second mass
m2 = 2.50 kg
is attached to the other end of the rod. Treat the two masses as point particles.
What is the moment of inertia of the system in kg · m2?
(b)
If the rod rotates with an angular speed of 2.40 rad/s, how much kinetic energy, in joules, does the system have?
(c)
Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined, in kg · m2?
(d)
If the rod is of negligible mass, what is the kinetic energy, in joules, when the angular speed is 2.40 rad/s?

Answer 1

(a) I = I_rod + I_masses. For the rod about its center I_rod = (1/12) M L^2 = (1/12)(2.20)(2.00^2) = 0.7333 kg·m^2.
The two point masses at ±1.00 m give I_masses = m1(1.00)^2 + m2(1.00)^2 = 4.90 + 2.50 = 7.40 kg·m^2.
Total I = 7.40 + 0.7333 = 8.13 kg·m^2 (≈8.133 kg·m^2).

(b) K = 1/2 I ω^2 = 0.5(8.1333)(2.40^2) = 0.5(8.1333)(5.76) = 23.4 J (≈23.42 J).

(c) If rod mass negligible, I = I_masses = 7.40 kg·m^2.

(d) K = 1/2(7.40)(2.40^2) = 0.5(7.40)(5.76) = 21.3 J (≈21.31 J).