Rotational energy ellipsoid

{title2=$\boldsymbol\omega^T\mathbf I_O\boldsymbol\omega=2K_0$}

For strictly positive <principal moments of inertia> and a fixed <rotational kinetic energy> $K_0>0$, the allowed <angular velocity> components form an <ellipsoid>:
$$
I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2=2K_0,
\qquad
s_i=\sqrt{\frac{2K_0}{I_i}}.
$$
Its semiaxis lengths $s_i$ have units of <angular speed>. The longest semiaxis corresponds to the smallest <principal moment of inertia>: that direction permits the greatest <angular speed> for the same <rotational kinetic energy>. This is a surface in <angular velocity> space, not the physical outline of the body. If one <principal moment of inertia> vanishes in an ideal line-mass model, this constant-energy surface is unbounded in that direction and is no longer an <ellipsoid>.

\Image[/inertia-tensor-principal-axis-energy-ellipsoid.png]
{title=<Principal axes> and the constant-<rotational kinetic energy> <ellipsoid>}
{description=Original exact model using the six <point masses> in <inertia tensor of paired point masses>. With $K_0=ma^2\Omega_0^2$, the displayed surface in $\boldsymbol\omega/\Omega_0$ has semiaxes $\sqrt{2/(9-\sqrt{13})}$, $\sqrt{2/(9+\sqrt{13})}$ and $\sqrt{2/14}$. The left view uses the original coordinate directions and the right view the <principal body frame>. Colors identify the same <principal axes> in both views; the surface describes possible <angular velocities>, not a material shell or a measured trajectory.}

The <rotational energy ellipsoid> gives the geometry behind <Poinsot's ellipsoid>, but a constant-energy surface alone does not determine a free-rotation trajectory. A torque-free trajectory must also satisfy the conserved <angular momentum> constraint and the <Euler equations for a torque-free rigid body>.


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 Incoming links: Rotational energy ellipsoid