# Solid Cone

MathJax TeX Test Page The important thing to remember is that the total inertia is the sum of many moment of inertias of solid discs. $$dI = \dfrac{1}{2}dm*r^2$$ Why is this true? Because the total moment of inertia is the sum of moment of inertias of solid disc, and the formula is $\frac{1}{2}mr^2$. $$dm = \rho \text{ (density) } * \pi{}r^2 \text{ (area) } \mathrm{d}x \text{ (width) }$$ $$\rho = \dfrac{M}{\pi{}R^2h\frac{1}{3}} = \dfrac{3M}{\pi{}R^2h}$$ $$\mathrm{d}m = \dfrac{3M}{hR^2}*r^2\mathrm{d}x$$ $$\int_0^h \frac{1}{2}\left(\dfrac{3M}{hR^2}*r^2\right)* r^2\mathrm{d}x$$ $r$ is equal to $y$, which is equal to $\frac{R}{h}x$ $$\dfrac{3M}{2hR^2}\int_0^h y^4 \,\mathrm{d}x$$ Looking at the picture above, $y = \dfrac{R}{h}x$, so the integral is $$\dfrac{3M}{2hR^2}\int_0^h \frac{R^4}{h^4}x^2 \,\mathrm{d}x$$ $$\dfrac{3MR^2}{2h^5}\int_0^h x^4 \,\mathrm{d}x$$ $$\dfrac{3MR^2}{2h^5} * \frac{h^5}{5} = \boxed{\frac{3MR^2}{10}}$$
David Witten