# Answer in Mechanics | Relativity for Ansh #139241

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 “largefrac{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.

“qquadqquadnbegin{aligned}nsmall delta m &= small pi r^2delta xtimes rho ,,,,,,,,,: rhotext{ = volumetric mass}\n&= small pi r^2delta xtimes frac{M}{pi r^2h}\n&= small frac{M}{h}delta x\n\nsmall I &= small int_{-h/2}^{h/2}delta I\n&= small small int_{-h/2}^{h/2} frac{1}{4}Big(frac{M}{H}delta xBig)r^2+Big(frac{M}{h}delta xBig) x^2 \n&= small frac{Mr^2}{4h} int_{-h/2}^{h/2} delta x + frac{M}{h} int_{-h/2}^{h/2} x^2 delta x\n& = small frac{Mr^2}{4h} [x]_{-h/2}^{h/2} + frac{M}{h} [frac{x^3}{3}]_{-h/2}^{h/2}\n&= small frac{Mr^2}{4h} [h] + frac{M}{h} [frac{h^3}{12}]\nsmall I &= small bold{MBig[frac{r^2}{4}+frac{h^2}{12}Big]}nend{aligned}”

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