Answer to Question #139241 in Mechanics | Relativity for Ansh

Question

Find moment of inertia of solid cylinder through an axis which passes through middle of its width

Answer 1

Explanations & Calculations

• Consider radius = r, height = h, mass = M for the cylinder.
• Moment of inertia of about its central axis is $\small \bold{\frac{1}{2}Mr^2}$

Finding that about an axis at $\large\frac{h}{2}$ : passes through middle of width

• For that consider an elemental disc of mass $\small \delta m$
• The easiest way of deriving the final solution is to consider the moment of inertia of it through the axis in consideration.
• For that one should know MI of a solid plate through its middle perpendicular axis as $\frac{1}{2}\delta m r^2$
• And by the law for MI of perpendicular axes, MI of that plate through an axis lie at any diameter as $\small \frac{1}{4}\delta m r^2$
• Finally MI of that plate about an axis lies a distance of x parallel to that previously considered diameter as $\small \frac{1}{4}\delta m r^2 +\delta mx^2$

Then considering this elemental MI of that plate about the axis considered for the whole cylinder and integrating over the full height, answer could be obtained.

$\qquad\qquad \begin{aligned} \small \delta m &= \small \pi r^2\delta x\times \rho \,\,\,\,\,\,\,\,\,: \rho\text{ = volumetric mass}\\ &= \small \pi r^2\delta x\times \frac{M}{\pi r^2h}\\ &= \small \frac{M}{h}\delta x\\ \\ \small I &= \small \int_{-h/2}^{h/2}\delta I\\ &= \small \small \int_{-h/2}^{h/2} \frac{1}{4}\Big(\frac{M}{H}\delta x\Big)r^2+\Big(\frac{M}{h}\delta x\Big) x^2 \\ &= \small \frac{Mr^2}{4h} \int_{-h/2}^{h/2} \delta x + \frac{M}{h} \int_{-h/2}^{h/2} x^2 \delta x\\ & = \small \frac{Mr^2}{4h} [x]_{-h/2}^{h/2} + \frac{M}{h} [\frac{x^3}{3}]_{-h/2}^{h/2}\\ &= \small \frac{Mr^2}{4h} [h] + \frac{M}{h} [\frac{h^3}{12}]\\ \small I &= \small \bold{M\Big[\frac{r^2}{4}+\frac{h^2}{12}\Big]} \end{aligned}$