Find the moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its centre and making an angle $\theta$ with the rod.

Question

Aicharya · Accepted Answer

Divide the rod into a very large number of extremely small elements each of length $d x$. Consider one such element at a distance $x$ from the centre $O$ of the rod (Fig.).
Mass of element is $d m=\frac{M}{L} d x$ Perpendicular distance of the element from the axis of rotation is
$r=O A=x \sin \theta$
Moment of inertia of the rod about the given axis is
$\begin{aligned}
I=\int d m r^{2} &=\int \frac{M}{L} d x \times(x \sin \theta)^{2} \\
&=\frac{M \sin ^{2} \theta}{L} \int_{x=-\frac{L}{2}}^{x=+\frac{L}{2}} x^{2} d x \\
&=\frac{M \sin ^{2} \theta}{L}\left|\frac{x^{3}}{3}\right|_{-L / 2}^{+L / 2} \\
&=\frac{M L^{2}}{12} \sin ^{2} \theta \\
\text { If } \theta=90^{\circ}, I=\frac{M L^{2}}{12}
\end{aligned}$

Find the moment of inertia of a thin uniform rod of mass \(M\) and length \(L\) about an axis passing…

Solution