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## 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.

## 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

$\begin{matrix} I = \frac{1}{3}(2\rho Aa)a^{2} & \longleftarrow \boxed{\text{ or write } \frac{I}{M} = \frac{2\rho Aa^{3}}{3 \times 2\rho Aa} = \frac{1}{3}a^{2}} \end{matrix}$

$= \frac{1}{3}Ma^{2}.$