Question

The moment of inertia of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to $m$ and the radius of its base to $R$ is $I=\frac{3m{R}^2}{y}$. Find the value of $y$.

Answer 1

Consider an element disc of radius $r$ and thickness $dx$ at a distance $x$ from the point $O$.
Then $r=x\tan \alpha$ and volume of the disc $=\pi x^2{\tan }^2\alpha dx$
Hence, its mass $dm=\pi x^2\tan \alpha dx\cdot \rho$ (where $\rho =$ density of the cone
$=\frac{m}{\frac{1}{3}\pi R^{2}h}$)
Moment of inertia of this element, about the axis $OA$,
$dI=dm\frac{r^2}{2}$
$=\left(\pi x^2{\tan }^2\alpha dx\right)\frac{x^2{\tan }^{2}x}{2}$
$=\frac{\pi \rho }{2}{x}^4{\tan }^4\alpha dx$
Thus the sought moment of inertia $=\frac{\pi \rho }{2}{\tan }^4\alpha {\int }_0^h{x}^{4}dx$
$=\frac{\pi \rho R^4\cdot h^5}{10h^4}(as\tan \alpha =\frac{R}{h})$
Hence $I=\frac{3m{R}^2}{10}(putting\rho =\frac{3m}{\pi R^{2}h})$