Centre of mass of the cone

The integral is not only useful for areas and volumes: it also allows us to compute the centre of mass of a solid, with the formula $z_{CM} = \dfrac{\int z\,dm}{\int dm}$, where $dm = \rho\,dV$.

For a homogeneous cone (constant density $\rho$, which cancels out), sliced into discs of radius $r(z)$: $z_{CM} = \frac{\int_0^H z\cdot\pi r^2(z)\,dz}{\int_0^H \pi r^2(z)\,dz} = \frac{\int_0^H z(H-z)^2\,dz}{\int_0^H(H-z)^2\,dz}.$

Expanding the numerator $z(H-z)^2 = zH^2 - 2H z^2 + z^3$ and integrating term by term, one finds $z_{CM} = \frac{H}{4}.$ The centroid of a homogeneous cone therefore lies at a quarter of the height starting from the base.

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Topics: Integrale
Concepts: Centro di massa · Integrale definito
Skills: Integrare · Modellizzare