I have a solid cone base radius r, height h, mass M and I need to find the moment of inertia about a diameter of the base as axis. The book's answer is $\frac{1}{10}Mh^2+\frac{3}{20}Mr^2$ however, I got $\frac{3}{20}Mr^2+\frac{3}{5}Mh^2$. What am I doing wrong? I have tried looking it up but all that I can find is the moment of inertia about its central axis or about the vertex perpendicular to the central axis (which is the answer that I obtained). Please can you help! I've put most of the working below.
moment of inertia of a thin disc about a diameter $$=\frac{My^2}{4}=\frac{1}{4}\pi y^4\delta x\rho$$
moment of inertia of thin disc about a parallel axis, distance $x$ away
$$=\frac{1}{4}\pi y^4\delta x\rho+ \pi y^2\delta x\rho x^2$$
So the moment of inertia for all such thin discs becomes when $\delta x\to0$
$$\frac{1}{4}\pi\rho\int^h_0y^4dx + \pi\rho\int^h_0x^2y^2dx$$
The cone is lying on its side with the vertex at the origin so $y=\frac{r}{h}x$ which gives
$$\frac{1}{4}\pi\rho\frac{r^4}{h^4}\int^h_0x^4dx + \pi\rho\frac{r^2}{h^2}\int^h_0x^4dx$$
Which gives
$$\frac{\pi\rho r^4h}{20}+\frac{\pi\rho r^2h^3}{5}$$
substituting $\rho=\frac{M}{V}$ gives
$$\frac{3}{20}Mr^2+\frac{3}{5}Mh^2$$
Best Answer
You should have $y(x) = r (1-{x \over h})$. The axis is at the base. You have $y(x) = r {x \over h}$. This puts the tip at the axis.