Calculate the kinetic energy of a homogeneous right

Calculate the kinetic energy of a homogeneous right circular cone of mass $M$, base radius $R$, and height $H$, rotating with angular velocity $\omega$ about its axis of symmetry.

Step 1: Set up cylindrical coordinates

Place the apex at the origin, axis along $z$, base at $z=H$ with radius $R$; at height $z$ the cone's radius is $r(z)=Rz/H$. With uniform density $\rho_m=M/V=3M/(\pi R^2H)$ (since $V=\tfrac13\pi R^2H$):

\[I=\int_0^H\!\!\int_0^{2\pi}\!\!\int_0^{Rz/H}\rho^2\cdot\rho_m\,\rho\,d\rho\,d\theta\,dz.\]

Step 2: Inner ($\rho$) integral

\[\int_0^{Rz/H}\rho^3\,d\rho=\frac14\left(\frac{Rz}{H}\right)^4=\frac{R^4z^4}{4H^4}.\]

Step 3: Angular integral and combine

\[I=\rho_m\cdot2\pi\cdot\frac{R^4}{4H^4}\int_0^Hz^4\,dz=\frac{3M}{\pi R^2H}\cdot\frac{\pi R^4}{2H^4}\cdot\frac{H^5}{5}=\frac{3MR^2}{10}.\]

\[\boxed{I=\frac3{10}MR^2}\]

Step 4: Kinetic energy

\[E_k=\frac12I\omega^2=\frac{3MR^2\omega^2}{20}.\]

\[\boxed{E_k=\frac{3MR^2\omega^2}{20}}\]

Step 5: Remarkable independence from $H$

The result depends only on $M$ and $R$, not on the height $H$. This is because a taller cone (for fixed $M,R$) spreads more mass toward larger $z$ but proportionally less of it stays out near radius $R$ (the cone narrows faster along its length), and these effects exactly cancel in the $z$-integral.

Step 6: Comparison with other shapes about a central axis

Cone: $E_k=0.15\,MR^2\omega^2$; solid cylinder or disk: $E_k=0.25\,MR^2\omega^2$; thin ring: $E_k=0.5\,MR^2\omega^2$. The cone stores less rotational energy than a cylinder of the same $M,R$ because more of its mass is concentrated close to the axis.