## 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}.