Determine the moment of inertia of a homogeneous solid

Determine the moment of inertia of a homogeneous solid cone with base radius $R$, height $h$, density $\mu$, and mass $M$, with respect to its base plane.

Step 1: Set up

\[I_{\text{base}}=\int_0^h\int_0^{2\pi}\int_0^{Rz/h}(h-z)^2\rho\,d\rho\,d\theta\,dz = \pi\frac{R^2}{h^2}\int_0^h(h-z)^2z^2\,dz\]

Step 2: Expand and integrate

\[(h-z)^2z^2=h^2z^2-2hz^3+z^4\]

\[\int_0^h(h^2z^2-2hz^3+z^4)\,dz = h^5\left(\frac13-\frac12+\frac15\right)=\frac{h^5}{30}\]

Step 3: Combine

\[I_{\text{base}}=\frac{\pi R^2}{h^2}\cdot\frac{h^5}{30}=\frac{\pi R^2h^3}{30}\cdot\mu\]

\[\boxed{I_{\text{base}}=\frac{\mu\pi R^2h^3}{30}}\]

Step 4: In terms of mass

With $M=\mu\cdot\frac13\pi R^2h$: \[I_{\text{base}}=\frac{\mu\pi R^2h^3}{30}=\frac{Mh^2}{10}\]

\[\boxed{I_{\text{base}}=\frac{Mh^2}{10}}\]

Step 5: Interpretation

This depends only on the height and mass, not the base radius — for an axis lying in the base plane, only the axial (height-wise) distribution of mass matters.