Chapter 20
Q. 20.4
Find the moment of inertia of a thin rod of length 2a and mass M about an axis through its centre and perpendicular to the rod.
Step-by-Step
Verified Solution
Imagine that the rod is subdivided into small elements of width δx. Assume that the area, A, of a cross-section of the rod is so small that every point on it can be regarded as being the same distance from the axis. Then the only variable is the distance x of the elementary portion of the rod from the axis.
The mass of the element is ρAδx, where ρ is the density, and its moment of inertia about the axis is approximately (ρAδx)x² = ρAx²δx.
The moment of inertia of the rod about the axis is therefore approximately ∑ρAx²δx. In the limit as δx → 0 this gives
I = \int_{-a}^{a}{ρAx²dx}
= ρA \int_{-a}^{a}{x^{2}dx}
⇒ I = ρA \left[ \frac{x^{3}}{3} \right]^{a}_{-a}
= \frac{ρA}{3} [a^{3} – (-a)^{3}]
= \frac{2}{3}ρAa^{3}.
The mass of the rod is M = 2aAρ = 2ρAa. Hence the moment of inertia of a rod about a perpendicular axis through its centre is
\begin{matrix} I = \frac{1}{3}(2\rho Aa)a^{2} & \longleftarrow \boxed{\text{ or write } \frac{I}{M} = \frac{2\rho Aa^{3}}{3 \times 2\rho Aa} = \frac{1}{3}a^{2}} \end{matrix}
= \frac{1}{3}Ma^{2}.
