Moment of Inertia of a Cone

The moment of inertia of a cone tells us how the cone’s mass is spread out in relation to the axis it's rotating around. This concept is important in physics and engineering, especially when studying how objects spin. For a solid cone, the moment of inertia depends on its mass, height, base radius, and the specific axis of rotation. Whether the cone is spinning around its central axis or tilted to spin around its base, knowing its moment of inertia helps us understand how much force—or torque—is needed to make it rotate.

1.0Moment of Inertia of A Hollow Cone

Consider a hollow cone (a thin, conical shell with no thickness) that has a total mass M, a base radius R, and a slant height L. The cone is symmetric about its vertical axis, which passes through its apex and the center of the circular base. Our objective is to determine the moment of inertia I of this hollow cone about its central vertical axis. Since the cone is hollow and thin, the mass is distributed uniformly over its curved surface, and we assume negligible thickness.

Radius of the ring at that point $r=\frac{R}{L}\cdot l$

By using similar Triangle, Width of ring dl

Cone Surface Area $A=\pi RL$

Mass per unit Area, $\sigma =\frac{M}{\pi RL}$

Area of the Ring, $dA=2\pi rdl=2\pi \left(\frac{R}{l}l\right)dl$

Mass of the Ring, $dm=\sigma \cdot dA=\frac{M}{\pi RL}\cdot 2\pi \left(\frac{R}{L}l\right)dl=\frac{2Ml}{L^2}dl$

Moment of Inertia of Ring,

$dI=r^{2}dm={\left(\frac{R}{L}l\right)}^2\cdot \frac{2Ml}{L^2}dl=\frac{2M{R}^2}{L^4}{l}^{3}dl$

$I={\int }_0^L\frac{2M{R}^2}{L^4}{l}^{3}dl=\frac{2M{R}^2}{L^4}{\int }_0^L{l}^{3}dl=\frac{2M{R}^2}{L^4}{\left[\frac{l^4}{4}\right]}_0^L=\frac{2M{R}^2}{L^4}\cdot \frac{L^4}{4}=\frac{1}{2}M{R}^2$

Moment of inertia of a hollow cone about its central (vertical) axis is 12MR2

2.0Moment of Inertia of A Solid Cone

By similarity of Triangles,

$\frac{r}{x}=\frac{R}{h}\Rightarrow r=\left(\frac{R}{H}x\right)$

Mass of Elemental Disc, $dm=\frac{M}{\frac{1}{3}\pi R^{2}H}\left(\pi r^{2}dx\right)$

$dm=\frac{3M}{\pi R^{2}H}\pi \left(\frac{R^2}{H^2}{x}^2\right)dx$

$dm=\frac{3M}{H^3}{x}^{2}dx$

Moment of Inertia of Disc

$dI=\frac{1}{2}dm{r}^2=\frac{1}{2}\left(\frac{3M}{H^3}{x}^{2}dx\right)\left(\frac{R^2}{H^2}{x}^2\right)$

$I=\int dI=\frac{3}{2}\frac{M}{H^5}{R}^2{\int }_0^H{x}^{4}dx=\frac{3}{2}\frac{M}{H^5}\left(\frac{H^5}{5}\right)=\frac{3}{10}M{R}^2$

$I=\frac{3}{10}M{R}^2$

Moment of inertia of a solid cone about its central vertical axis is 310MR2

3.0Moment of Inertia of Solid and Hollow Cones (Various Axes)