Worked example 8.3: Moment of inertia of a rod

Question: A rod of mass $M=3 {\rm kg}$ and length $L=1.2 {\rm m}$ pivots about an axis, perpendicular to its length, which passes through one of its ends. What is the moment of inertia of the rod? Given that the rod's instantaneous angular velocity is $60 {\rm deg./s}$, what is its rotational kinetic energy?

Answer: The moment of inertia of a rod of mass $M$ and length $L$ about an axis, perpendicular to its length, which passes through its midpoint is $I=(1/12) M L^2$. This is a standard result. Using the parallel axis theorem, the moment of inertia about a parallel axis passing through one of the ends of the rod is

$$I' = I + M \left(\frac{L}{2}\right)^2 = \frac{1}{3} M L^2,$$

so

$$I' = \frac{3\times 1.2^2}{3} = 1.44 {\rm kg m^2}.$$

The instantaneous angular velocity of the rod is

$$\omega = 60 \times \frac{\pi}{180} = 1.047 {\rm rad./s}.$$

Hence, the rod's rotational kinetic energy is written

$$K = \frac{1}{2} I' \omega^2 = 0.5\times 1.44\times 1.047^2 = 0.789 {\rm J}.$$