The moment of inertia of a uniform rod of len | vedclass

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(A) Let the rod lie along the $x$-axis with its centre at the origin. The mass per unit length of the rod is $\lambda = \frac{m}{2l}$.Consider a small

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$\frac{m l^2}{3} \sin^2 \alpha$

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(A) Let the rod lie along the $x$-axis with its centre at the origin. The mass per unit length of the rod is $\lambda = \frac{m}{2l}$.Consider a small element of length $dx$ at a distance $x$ from the centre. The mass of this element is $dm = \lambda dx = \frac{m}{2l} dx$.The perpendicular distance of this element from the axis $xx'$ is $r = x \sin \alpha$.The moment of inertia of this element about the axis $xx'$ is $dI = (dm) r^2 = \left(\frac{m}{2l} dx\right) (x \sin \alpha)^2$.Integrating this from $x = -l$ to $x = +l$:$I = \int_{-l}^{l} \frac{m}{2l} x^2 \sin^2 \alpha dx = \frac{m \sin^2 \alpha}{2l} \int_{-l}^{l} x^2 dx$.$I = \frac{m \sin^2 \alpha}{2l} \left[ \frac{x^3}{3} \right]_{-l}^{l} = \frac{m \sin^2 \alpha}{2l} \left( \frac{l^3}{3} - \frac{(-l)^3}{3} \right) = \frac{m \sin^2 \alpha}{2l} \left( \frac{2l^3}{3} \right) = \frac{1}{3} m l^2 \sin^2 \alpha$.

The moment of inertia of a uniform rod of length $2l$ and mass $m$ about an axis $xx'$ passing through its centre and inclined at an angle $\alpha$ is

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