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For a rigid body with three distinct positive principal moments of inertia $0<I_1<I_2<I_3$ and zero external torque about its centre of mass, steady rotation about the first or third principal axis is Lyapunov stable in angular velocity, while rotation about the intermediate principal axis is unstable. Here steady rotation means $\mathbf{\omega }=\Omega {\hat{e}}_i$ in the principal body frame, with constant $\Omega \ne 0$. A small sideways perturbation remains small about the extreme principal moments of inertia, but generically grows about the intermediate one. Stable does not mean that the wobble decays: without damping, there is no asymptotic stability. Coincident principal moments of inertia and a stationary body are excluded.^[ref]

No external torque is needed for a body to tumble. Conservation of angular momentum fixes $\mathbf{L}$ in the fixed space frame, but the body can turn relative to $\mathbf{L}$, so its angular velocity components in the principal body frame change. The torque-free rigid-body invariants constrain this motion. A supported gyroscope with gravitational torque has different governing assumptions.

Put $\mathbf{r}=(x,y,z)$ with ${\mathbf{e}}_3$ along $z$, and interpret the symmetry as axisymmetry of the whole mass density. A half-turn about ${\mathbf{e}}_3$ sends $(x,y,z)$ to $(-x,-y,z)$ and leaves the integration domain and mass density invariant. Therefore $\int \rho xz{d}^{3}r=\int \rho yz{d}^{3}r=0$. A quarter-turn sends $(x,y,z)$ to $(-y,x,z)$; it givesAll off-diagonal entries of the inertia tensor consequently vanish. Writing $s^2=x^2+y^2$, the principal moments of inertia satisfyThus every orthonormal basis with third vector along the symmetry axis is a principal body frame. This is axisymmetry of the inertia tensor. In general there is no requirement that $I_3=I_1$.

Let $J=I_1=I_2$, ${\omega }_3=c$ andHere $c$ and $\nu$ are constant. The first two of Euler's equations give ${\dot{\omega }}_1=-\nu {\omega }_2+{\tau }_0/J$ and ${\dot{\omega }}_2=\nu {\omega }_1+{\tau }_0/J$. Combining them producesMultiplication by the integrating factor $e^{-i\nu t}$ gives, for $\nu \ne 0$,For completeness, $\nu =0$ instead gives $z(t)=z(0)+{\tau }_0(1+i)t/J$. These formulas describe a symmetric top under constant transverse body torque.

The stated initial conditions require $b\ne 0$ and $I_3\ne J$. They give $c=b/(I_3-J)$, $\nu =b/J$ and $z(0)=i{\tau }_0/b$. Substitution and simplification yieldTaking the real part and imaginary part, the angular velocity in the principal body frame isIts value at $t=0$ is the required initial condition; differentiating the first two components verifies both forced Euler's equations.

For the symmetric top, set $I_1=I_2=J$. The third of Euler's equations reduces to $I_3{\dot{\omega }}_3={\tau }_3=0$. Thus the axial angular velocity component is a conserved quantity:The transverse angular velocity components are generally variable. A torque constant in the principal body frame generally rotates in the inertial frame; constancy in the two frames is different.

For a vector $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$ in the principal body frame, the derivative of a body-fixed basis vector is $(d{\mathbf{e}}_i/dt{)}_{\hat{S}}=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the expansion gives the rotating-frame derivative formula, also called the Coriolis formula:Apply this to angular momentum $\mathbf{L}={\sum }_i{I}_i{\omega }_i{\mathbf{e}}_i$. The principal moments of inertia are constant in the principal body frame, so external torque balance becomesResolving the cross product gives Euler's equations:All components and dotted components in this display refer to the principal body frame.

Now take zero torque and $\mathbf{\omega }(0)=(1,0,1)$. The axial angular velocity is ${\omega }_3=1$ and the complex transverse component obeys $\dot{z}=i\nu z$ with $z(0)=1$ and $\nu =(I_3-I_1)/I_1$. Hence the angular momentum components in the principal body frame areThe transverse projection therefore traces a circle with signed angular frequency $\nu$. Its magnitude is constant, $L=\sqrt{I_1^2+I_3^2}$.

In the inertial frame, zero torque means that $\mathbf{L}$ is fixed. Usingand the derivative of a body-fixed basis vector, the symmetry axis satisfiesThus there is uniform torque-free precession of a symmetric top about the fixed angular momentum, at a constant cone angle $\beta$ and space angular frequencyThe cone angle stays constant, with no nutation. The body-frame and space-frame frequencies of free symmetric-top precession differ: $\nu$ describes rotating components, whereas ${\Omega }_{\mathrm{space}}$ describes the symmetry axis in space. The formulas also cover $I_3=I_1$, where the angular momentum components in the principal body frame are constant.

A principal body frame is an orthonormal basis ${\mathbf{e}}_a(t)$ fixed in the rotating body and aligned with its principal axes. If $R(t)$ is the rotation matrix whose columns are these vectors, differentiating $R^{T}R=I$ shows that $\dot{R}{R}^T$ is skew-symmetric. Every skew-symmetric linear map in three dimensions is cross product with a unique vector $\mathbf{\omega }$, so

For any vector $\mathbf{A}=A_a{\mathbf{e}}_a$,Applying this to the conserved angular momentumgives the Euler equations for a torque-free rigid body:

Multiplying these equations respectively by ${\omega }_i$ and summing proves conservation of kinetic energy; multiplying by $I_i{\omega }_i$ and summing proves conservation of squared angular momentum. Thus the torque-free rigid-body invariants areSolving these two linear equations for the first two squared components givesUsing $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2$ yields the quartic equation for one component of torque-free angular velocity:

Let $e_1,e_2,e_3$ be a principal body frame, let $I_1,I_2,I_3$ be the corresponding principal moments of inertia, and write the angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{e}_i$. The body-frame components of the angular momentum areNo external torque acts, so $\mathbf{L}$ is constant in the fixed space frame. Since a body-fixed basis vector satisfies ${\dot{e}}_i=\mathbf{\omega }\times e_i$,Taking body-frame components gives the Euler equations for a torque-free rigid body:

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega |$; it precesses in the positive axial sense when $(I_3-I){\omega }_3>0$ and in the opposite sense when this product is negative.

A fixed space frame is an inertial orthonormal frame whose axes remain fixed in space. A principal body frame is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.

If $e_i(t)$ are the body axes and $\omega$ is the angular velocity, then the derivative of a body-fixed basis vector isIn the principal frame,For torque-free motion, the angular momentum has zero space derivative. ThereforeTaking body-frame components gives the Euler equations for a torque-free rigid body:

For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega ={\omega }_1{e}_1+{\omega }_2{e}_2+{\omega }_3{e}_3$. ThenThus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the coplanarity in a torque-free axisymmetric rigid body.

Inertia and principal directions. About the centre of mass, the inertia tensor isIt relates angular momentum and angular velocity by $\mathbf{J}=\mathsf{I}\mathbf{\omega }$ and gives rotational kinetic energy $T=\mathbf{\omega }\cdot \mathsf{I}\mathbf{\omega }/2$. The principal axes are orthogonal eigenvector directions of this real symmetric tensor; in that basis it is diagonal, with entries the principal moments of inertia $I_1,I_2,I_3$.

Derivatives in a rotating frame. Write $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$ in a body-fixed orthonormal basis. Each basis vector rotates according to $(d{\mathbf{e}}_i/dt{)}_0=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the components and basis separately proves the rotating-frame derivative formula,The body derivative differentiates the scalar components in that basis. A fixed pivot can replace the centre of mass when the corresponding angular-momentum balance is valid.

Euler equations. The inertial angular-momentum equation is $d\mathbf{J}/dt=\mathbf{G}$. Since the principal moments are constant in the body frame, substitute $J_i=I_i{\omega }_i$:These are Euler's equations in the principal body frame.

Linear perturbation of free rotation. After the impulse, $\mathbf{G}=0$. Let ${\omega }_3=\Omega$ to leading order and ${\omega }_1,{\omega }_2$ be small. The third Euler equation changes ${\omega }_3$ only at quadratic order, so the linear transverse equations areFor ${\omega }_1=a{e}^{pt}$ and ${\omega }_2=b{e}^{pt}$, eliminate $a,b$ fromThis givesIts units are inverse square seconds. Positive $p^2$ gives a growing real exponential; negative $p^2$ gives bounded oscillatory perturbations.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.

For strictly positive principal moments of inertia and a fixed rotational kinetic energy $K_0>0$, the allowed angular velocity components form an ellipsoid: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.

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.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

Let a symmetric top have principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Apply a torque $({\tau }_1,{\tau }_2,0)$ constant in a principal body frame. Then the axial angular velocity $c={\omega }_3$ is a conserved quantity. Define $z={\omega }_1+i{\omega }_2$ and $\nu =(I_{\parallel }-I_{\perp })c/I_{\perp }$. The first two of Euler's equations combine into the equation in the title. This derivation uses constancy in the principal body frame, which generally does not mean constant torque in an inertial frame.

Multiplying by the integrating factor $e^{-i\nu t}$ proves, for $\nu \ne 0$,Equivalently $z$ rotates about the constant point $z_{\ast }=i({\tau }_1+i{\tau }_2)/(\nu I_{\perp })$ in the complex transverse plane. When $\nu =0$, the solution is instead $z(t)=z(0)+({\tau }_1+i{\tau }_2)t/I_{\perp }$. The angular velocity is $(\mathrm{Re}z,\mathrm{Im}z,c)$.

A useful readout of a tumble is the alignment of the intermediate principal axis ${\hat{e}}_2$ with the conserved angular momentum direction $\hat{L}$. In the principal body frame, write $L_i=I_i{\omega }_i$ and $L=|\mathbf{L}|$. ThenA change from $c_2\simeq +1$ to $c_2\simeq -1$ means that the directed body axis has turned nearly $180^{\circ }$ relative to $\mathbf{L}$. It does not mean that $\mathbf{L}$ reversed in the fixed space frame. The unsigned line through that principal axis hides this distinction, so a marked end matters in an actual demonstration.

For the normalized intermediate-axis model with $\mathbf{\omega }(0)=(0.01,1,0)$, the rotational kinetic energy and squared angular momentum are $E=1.00005$ and $L^2=4.0001$. Hence $L^2-2E{I}_2=-0.0001$: this initial condition lies close to, but off, the intermediate-axis separatrix. An independent calculation evolves ${\dot{\mathbf{L}}}_{\mathrm{body}}={\mathbf{L}}_{\mathrm{body}}\times I^{-1}{\mathbf{L}}_{\mathrm{body}}$ and a rotation matrix taking body components to space components. The intermediate principal axis crosses $c_2=0$ at normalized times approximately $10.3773$ and $31.1319$, alternating long nearly aligned intervals with quicker reversals.

Both calculations use step size $0.005$ without rescaling either invariant. Over $0\le t\le 50$, the tumbling calculation changes rotational kinetic energy and squared angular momentum by less than $4\times 10^{-13}$ relative to their initial values. Reconstructing $\mathbf{L}$ in the fixed space frame gives relative error below $3\times 10^{-12}$. Halving the step changes the body components and rotation matrix entries by less than $3\times 10^{-10}$. These checks distinguish physical tumbling from numerical drift. On the exact intermediate-axis separatrix, approach to an intermediate-axis rotation takes infinite time; the recurrent off-separatrix example here is different.