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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}.

fig 20.8