Question

Calculate the moment of inertia of a uniform solid cone relative to its axis of symmetry, if the mass of the cone is equal to $m$ and the radius of its base is equal to R. Mass is uniformly distributed.

Answer 1

Mass per unit volume of the cone is,

$\rho =\frac{m}{\frac{1}{3}\pi R^{2}h}=\frac{3M}{\pi R^{2}h}$

we choose an elementary disc of radius r at a distance $x$ from apex and width $dx$.

The mass of the disc,

$dm=\left(\pi r^2\right).dx.\rho$

M.i of the disc about $AO$,

$dI=\frac{1}{2}dm{\left(r\right)}^2=\frac{1}{2}\pi \rho {\left(r\right)}^{4}dx$

From $△A{O}_{1}D$ and $△ADC$

$\frac{r}{R}=\frac{x}{h}\therefore r=\frac{Rx}{h}\therefore dI=\frac{1}{2}\pi \rho {\left(\frac{Rx}{h}\right)}^{4}dx.I=\frac{1}{2}\pi \rho \frac{R^4}{h^4}{\int }_0^h{x}^{4}dxI=\frac{1}{2}\pi \rho \frac{R^4}{h^4}\times \frac{h^5}{5}I=\frac{1}{2}\pi \rho R^4\frac{h}{5}=\frac{1}{2}\pi .\frac{3M}{\pi R^{2}h}.R^4.\frac{h}{5}\therefore I=\frac{3}{10}{MR}^2$