Rigid body dynamics

Rigid body dynamics studies translation and rotation of bodies whose internal distances are fixed.

For an axially symmetric rotational Lagrangian with potential $Mgh\cos \theta$, fixed tilt and conservation of the spin and azimuthal momenta give constant precession. The tilt equation becomes $A{\Omega }^2\cos \alpha -Cn\Omega +Mgh=0$. A real precession rate requires the quadratic discriminant to be nonnegative. The inertia and height parameters must correspond consistently to the chosen fixed point.

Expanding the squared velocity $v_G+\omega \times r$ and integrating over mass removes the mixed term because the centre-of-mass first moment is zero. The remaining rotational integral is half the scalar product of angular velocity and angular momentum about the centre of mass. Principal axes diagonalize that rotational quadratic form.

An ideal circular ring whose mass lies at a single radius $R$ from its symmetry axis. About that axis, the moment of inertia is $\int R^{2}dm=M{R}^2$. The result assumes negligible radial and axial thickness; it is different from the $M{R}^2/2$ of a uniform solid disk.

A thin straight rod of length $l$, mass $m$, central moment of inertia $I=\beta m{l}^2$ and midpoint centre of mass starts beside a pointlike pole at that midpoint. An end impulse perpendicular to the rod gives translation $V=J/m$ and rotation $\Omega =J/(2\beta ml)$. At the first quarter-turn the trailing end is at horizontal distance $\pi \beta l-l/2$ from the pole. Strict clearance requires $\beta >1/(2\pi )$; equality is the grazing limit. The criterion is for free motion before contact, a negligibly thin rod and pole, and an impulse directed away from the pole.

A vehicle with centre of mass midway between level rails separated by d, at height h, follows a turn of radius R. Vertical force balance and cross-sectional moment balance give inner rail reaction $mg/2-m{v}^{2}h/(Rd)$. Its first zero occurs at the displayed speed. This is a toppling criterion under maintained lateral contact, not a guarantee against sliding or suspension instability.

A rigid body undergoes translation when every point has the same displacement at each time, so its orientation stays fixed. For a freely moving system of particles coupled only by internal springs, the common displacement is a zero-frequency normal mode: internal spring lengths and potential energy are unchanged. Its mode coordinate is $A+Bt$, representing constant centre-of-mass velocity, rather than an oscillation at nonzero frequency.

The instantaneous velocity of the material point of a rigid body touching another body. In the displayed formula, ${\mathbf{r}}_P$ is its position relative to the centre of mass, and $\mathbf{\Omega }$ is the body's angular velocity. The rotational term is a cross product. This material-point velocity differs from the motion of the geometrical contact location: in rolling without slipping on a stationary plane it vanishes even while the contact location advances. For a sphere touching a horizontal plane, with rightward centre velocity $v$ and anticlockwise spin $\Omega$, its tangential component is $v+a\Omega$.

An ideal elastic collision between a sphere and a stationary plane reverses normal approach velocity. On the fully rough, tangentially elastic branch, the tangential contact-point velocity reverses too. If $K_t$ is the signed tangential impulse, translation and spin obey $M\Delta v=K_t$ and $I\Delta \Omega =a{K}_t$, with positive spin anticlockwise. Restoring tangential translational plus rotational kinetic energy givesThe nonzero root is the displayed impulse; the other root is the smooth-contact branch. For $I=2M{a}^2/5$, the map is $v^{\prime }=(3v-4a\Omega )/7$ and $a{\Omega }^{\prime }=-(10v+3a\Omega )/7$. It preserves kinetic energy and the angular momentum about the contact point, $I\Omega -Mav$. An unspecified rough surface need not realize this idealization: finite friction or a dissipative tangential contact law requires additional constitutive data. Setting the outgoing contact slip to zero instead loses $M(v+a\Omega {)}^2/7$ of kinetic energy for this uniform sphere.

A short transverse impulse $J$ at lever arm $b$ gives a pivoted rigid body angular velocity $\Omega =Jb/I$ and contact-point velocity $b\Omega =J/m_{\mathrm{eff}}$, where $I$ is its pivot moment of inertia. Thus $m_{\mathrm{eff}}=I/b^2$ plays the role of a translational mass in the impact direction. A point projectile elastically striking an initially stationary body stops immediately after impact precisely when its mass equals this effective value. The pivot can exchange linear momentum while doing no work.

A normal impulse $j$ on a free planar rigid body of mass $M$ and moment of inertia $I$, at perpendicular lever arm $b$, changes the contact point's normal velocity by $j(1/M+b^2/I)$. Its effective mass in that direction is therefore $(1/M+b^2/I{)}^{-1}$. This follows by adding the translational change $j/M$ to the rotational contribution $b(jb/I)$.

For an incoming point mass $m$, the inverse reduced contact mass is $1/m+1/M+b^2/I$. A smooth elastic collision with initial relative approach speed $u$ has impulse $2u/(1/m+1/M+b^2/I)$, since the relative normal velocity reverses. This separates effective inertia at a contact from the body's total mass and explains the lever-arm dependence in an equal-speed elastic impact on a free rod.

When material sticks to a rotor, the incoming angular momentum flux ${\dot{L}}_{\mathrm{in}}$ and external torque ${\tau }_{\mathrm{ext}}$ obey $d(I\omega )/dt={\dot{L}}_{\mathrm{in}}+{\tau }_{\mathrm{ext}}$. With zero incoming axial angular momentum and zero external axial torque, $I\omega$ is constant. For deposition at fixed radius $a$ at mass flow rate $\alpha$, $\dot{I}=\alpha a^2$ and the rotational kinetic energy satisfies $\dot{K}=-\alpha a^2{\omega }^2/2$. This loss is the irreversible cost of bringing the incoming material up to tangential speed; it does not violate conservation of angular momentum.

Let a bell have axial moment of inertia $M{k}^2$ and centre of mass h from its axle. A clapper of point mass m at radius l touches the bell at fixed relative angle $\beta$. While they remain in contact, the clapper angle $\theta$ from the upward vertical obeys $(M{k}^2+m{l}^2)\ddot{\theta }=Mgh\sin (\theta -\beta )+mgl\sin \theta$. The contact force opposing increasing angle is $N=mg\sin \theta -ml\ddot{\theta }$. Eliminating the acceleration givesThe first zero on the physical trajectory marks detachment, because a nonadhesive contact cannot support negative normal force. Selecting the correct trigonometric branch is part of the contact problem.

At fixed angular momentum, rotational energy is minimized by rotation about the largest principal moment of inertia. Indeed, $T={\sum }_i{L}_i^2/(2I_i)\ge {\sum }_i{L}_i^2/(2I_{\max })$. Weak internal dissipation can therefore relax a nearly rigid isolated body towards that axis. A perfectly rigid torque-free body conserves energy and does not settle; both extreme-inertia axes are stable in its ideal dynamics. The maximum-inertia axis need not be the geometrically longest axis.

For a body constrained to rotate about a fixed pivot, a perpendicular impulse $K$ delivered at distance $b$ from the pivot produces $\omega =Kb/I_O$. If the centre of mass is distance $h$ from the pivot, its momentum is $M\omega h$. The pivot impulse is $J_O=M\omega h-K$, and vanishes when $b=I_O/(Mh)$. This impact location is the centre of percussion. The result assumes the stated planar rigid-body geometry and a short impulse; it is not a claim that all forces at the pivot vanish afterward.

For a uniform square plate of side $l$ hinged along an edge, $h=l/2$ and $I_H=M{l}^2/3$. An impulse $K$ normal to the plate, applied a distance $a$ below the centre, gives $v_G=3K(1+2a/l)/(4M)$ and signed hinge impulse $J_H=K(3a/(2l)-1/4)$. Therefore the hinge impulse vanishes at $a=l/6$ below the centre, a distance $2l/3$ from the hinge. The sign of $J_H$ identifies the reaction direction; its magnitude is $|J_H|$.

The net external torque about an inertial origin equals the time derivative of angular momentum. For rotation about a fixed axis with constant moment of inertia, its axial component reduces to $I\dot{\omega }=\tau$.

A rigid body with moment of inertia $I$, constant driving power $P$ and linear resisting torque $-\gamma \omega$ has an exponential approach of ${\omega }^2$ to $P/\gamma$. The ideal drive requires torque $P/\omega$, so starting from rest is a limiting energy model with singular initial torque.

For a rigid vehicle turning on a level road, with centre of mass height $h$, half track width $a$ and turn radius $R_b$, sufficient static friction supplies centripetal acceleration $v^2/R_b$. The inner and outer wheel normal forces obey $N_i+N_o=Mg$ and $a(N_o-N_i)=Mh{v}^2/R_b$. The inner contact force first vanishes at the displayed speed. The approximation neglects suspension movement and extra torques; inadequate static friction can cause sliding before rollover.

Two identical uniform rigid bodies of rod shape, each of length $2l$ and mass $M$, form a symmetric arch with a free common hinge and both outer ends sliding without friction on a horizontal floor. At angle $\theta$, each center has speed $l|\dot{\theta }|$, giving total kinetic energy $4M{l}^2{\dot{\theta }}^2/3$. The total gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at $\alpha$ gives the displayed hinge impact speed. Unlike fixing one outer endpoint, allowing both endpoints to slide keeps the arch center horizontally fixed. The floor reaction on each rod is $Mg[1+9{\sin }^2\theta -6\sin \alpha \sin \theta ]/4$, nonnegative throughout this fall.

Each massless rod has endpoint masses $m$ at distance $\ell$ from its centre, hence total mass $2m$ and moment of inertia $2m{\ell }^2$. In a symmetric opposite-end collision with centres approaching at speed $V$, a horizontal impulse $J$ changes each centre speed by $J/(2m)$ and each contact-point speed by $J/m$. The coefficient of restitution gives $2J/m-2V=2eV$. Both rods rotate in the same sense for the staggered vertical configuration; their centre speeds after impact are $V(3-e)/4$.

// Target: electromagnetism.bigb

Two equal uniform rods of length $2l$ and mass $M$, one pivoted to the floor and the other endpoint sliding on it, have arch coordinates $B=(2l\cos \theta ,2l\sin \theta )$ and $C=(4l\cos \theta ,0)$. The kinetic energy decomposition about the center of mass gives $T=\frac{4}{3}M{l}^2(1+3{\sin }^2\theta ){\dot{\theta }}^2$, while gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at angle $\alpha$ gives the displayed equation and hinge impact speed $\sqrt{6gl\sin \alpha }$, provided contact is maintained. The endpoint's normal reaction becomes negative during this constrained fall if $\sin \alpha >35/54$; a unilateral floor then permits detachment and the constrained formula no longer describes the entire motion.

In solid-body rotation the angular velocity is independent of distance from the rotation axis, so tangential speed is proportional to cylindrical radius. A galactic region with this galaxy rotation curve has radial epicyclic frequency $\kappa =2\Omega$.

A symmetric top is a rigid body with two equal principal moments of inertia, conventionally $I_1=I_2$. Its third principal axis is its symmetry axis.

Consider a symmetric top with fixed vertex $Q$, axial unit vector $\mathbf{e}$, principal moments of inertia $A,A,C$, and centre of mass at $h\mathbf{e}$. Let a force $\kappa \ell (\mathbf{k}-\mathbf{e})$ act at $\ell \mathbf{e}$, where $\mathbf{k}$ is the upward vertical unit vector. Together with gravity, it produces torque $(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ and potential energy $mgh\mathbf{k}\cdot \mathbf{e}+\kappa {\ell }^2(1-\mathbf{k}\cdot \mathbf{e})$.

The axial angular velocity of a symmetric top $n$, vertical angular momentum $\mathbf{k}\cdot \mathbf{L}$ and total mechanical energy are conserved. For steady precession, $\dot{\mathbf{e}}=\Omega \mathbf{k}\times \mathbf{e}$ and $\mathbf{k}\cdot \mathbf{e}=\cos \alpha$. Substituting $\mathbf{e}\times \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\times \mathbf{k}$ into $A\mathbf{e}\times \ddot{\mathbf{e}}+Cn\dot{\mathbf{e}}=(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ gives the condition in the title for nonvertical inclination. If the total angular velocity itself is $\Omega \mathbf{k}$, then $n=\Omega \cos \alpha$, and the condition becomes $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

For a symmetric top with axial unit vector $\mathbf{e}$ and angular velocity $\mathbf{\omega }$, the classical axial spin is the scalar $n=\mathbf{\omega }\cdot \mathbf{e}$. The identity $\dot{\mathbf{e}}=\mathbf{\omega }\times \mathbf{e}$ implies $\mathbf{\omega }=\mathbf{e}\times \dot{\mathbf{e}}+n\mathbf{e}$. If the transverse and axial principal moments of inertia are $A$ and $C$, then the angular momentum about the fixed vertex is $\mathbf{L}=A\mathbf{e}\times \dot{\mathbf{e}}+Cn\mathbf{e}$. A torque perpendicular to $\mathbf{e}$ conserves $n$. During steady precession at rate $\Omega$ and inclination $\alpha$, the general angular velocity is $\Omega \mathbf{k}+(n-\Omega \cos \alpha )\mathbf{e}$; the precession rate by itself does not fix $n$.

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 rotor with two equal transverse moments and a smaller symmetry-axis moment, $I_a<I_b=I_c$. Its rotational constants obey $A>B=C$.

An oblate symmetric top has two equal smaller principal moments of inertia and a larger moment of inertia about its symmetry axis. A planar trigonal molecule, such as boron trifluoride, is an example. Its axial rotational constant is smaller than its transverse rotational constant.

For an axially symmetric rigid body with $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, no external torque means that angular momentum $\mathbf{L}$ is fixed in an inertial frame. The axial angular velocity ${\omega }_3$ is constant by Euler equations for a torque-free rigid body, and the symmetry axis obeysIts angle to $\mathbf{L}$ is constant and it precesses in space at $L/I_{\perp }$. The transverse angular-velocity components instead rotate in body coordinates at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. These are different frame descriptions and must not be substituted for one another.

For principal inertias $I_{\perp },I_{\perp },I_{\parallel }$, the transverse angular-velocity components rotate in body axes at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. The symmetry axis precesses about the inertially fixed angular momentum at rate $L/I_{\perp }$. These are different frame-dependent descriptions of the same motion.

A heavy symmetric top is a symmetric top with one point fixed in a uniform gravitational field. Its vertical and body-axis rotational symmetries give two cyclic Euler angles.

After fixing the two cyclic angular momenta of a heavy symmetric top, its inclination has an effective potential $U(u)$ with $u=\cos \theta$. Write the reduced energy equation in the form $P(u)=2A(E-U(u))(1-u^2)$. A steady nonvertical inclination has $E=U(u_0)$ and $U^{\prime }(u_0)=0$, the latter coming from the reduced second-order Euler-Lagrange equation. Therefore $P(u_0)=P^{\prime }(u_0)=0$.

This proves the double-root condition without dividing by a velocity that vanishes in the very solution under study. A simple turning point has $P(u_0)=0$ but nonzero derivative, and is not a steady inclination. Exactly vertical spinning corresponds to $u=\pm 1$, where the inclination coordinate and its reduced denominator are singular; these endpoint configurations require a separate treatment and do not supply an interior double root.

Writing the unit symmetry axis as $n$ and axial spin momentum as $s=I_3{\omega }_3$, angular momentum is $L=I_{1}n\times \dot{n}+sn$. Taking its derivative and applying gravitational torque gives $I_{1}n\times \ddot{n}+s\dot{n}=-mg\ell n\times \hat{z}$.

Variation in the inclination of a rotating body's axis, distinguished from its azimuthal precession. For a heavy symmetric top, the cyclic angular momenta reduce the motion to ${\dot{\mu }}^2=F(\mu )$ with $\mu =\cos \theta$ and $F$ cubic. Its admissible interval and turning points describe the nutation.

A heavy symmetric top precesses uniformly when its inclination is constant and its azimuthal speed is constant. For a nonvertical inclination, the condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$. The real-root condition depends on whether the body axis points above or below the horizontal.

When $s^2\gg 4mg\ell I_1\cos \theta$, the two nonvertical steady-precession roots are approximately $mg\ell /s$ and $s/(I_1\cos \theta )-mg\ell /s$. These asymptotic descriptions do not apply at a singular denominator without taking the original quadratic limit.

Precession is the rotation of a body's spin axis about another direction, such as the vertical or the conserved angular-momentum vector.

The time for a precessing orientation to turn through $2\pi$. An unlabelled line is unchanged by a $\pi$ turn, so its visible pattern can repeat after half the full precession period.

Angular velocity is the vector whose direction gives the instantaneous rotation axis and whose magnitude gives the angular speed.

Angular acceleration is the time derivative of angular velocity, $\mathbf{\alpha }=d\mathbf{\omega }/dt$. For rotation about a fixed axis, its signed scalar component obeys $\tau =I\alpha$ when the axial moment of inertia is constant and $\tau$ is the total applied torque about that axis.

Angular speed is the nonnegative magnitude of angular velocity: $\omega =|\mathbf{\omega }|$. It measures the rate of rotation about an instantaneous axis without specifying its direction. For uniform circular motion of radius $R$, the ordinary speed is $v=R\omega$.

A fixed space frame is an inertial orthonormal basis whose axes do not rotate with the body.

A principal body frame is an orthonormal basis fixed in the body and aligned with the principal axes of its inertia tensor. In this frame the inertia tensor is diagonal with principal moments $I_1,I_2,I_3$.

A principal axis is an eigenvector direction of the inertia tensor. Choosing three orthogonal principal axes diagonalizes the inertia tensor.

If $e_i$ is fixed in a rigid body with angular velocity $\omega$, then its derivative in the fixed space frame is

If a vector $k$ is fixed in space while a body frame rotates with angular velocity $\omega$, its body-frame components satisfy

In principal body coordinates, Euler's equations express angular momentum balance as $I\dot{\omega }+\omega \times (I\omega )=\tau$, with applied torque $\tau$. Setting the torque to zero gives the following torque-free specialization.

In principal-axis body coordinates, torque-free angular velocity satisfies

The Euler equations for a torque-free rigid body conserve rotational kinetic energy and squared angular momentum,Their common level sets are the intersection of two ellipsoids in angular-velocity space.

Eliminating ${\omega }_1^2$ and ${\omega }_2^2$ between the conserved energy and squared angular momentum givesThe right-hand side is a quartic polynomial in ${\omega }_3$; its real turning points delimit the nutation of the body-frame angular velocity.

For principal moments $I_1=I_2=I$ and symmetry-axis unit vector $e_3$, angular momentum and angular velocity obeyThus $L$, $\omega$, and the symmetry axis are coplanar at every instant.

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.

For torque-free rotation, total angular momentum magnitude $J$ and rotational kinetic energy are conserved. Near rotation about principal axis three, eliminate $J_3^2=J^2-J_1^2-J_2^2$ from $T={\sum }_i{J}_i^2/(2I_i)$ to obtain the displayed energy difference from $J^2/(2I_3)$. If $I_3$ is largest, it is a constrained energy minimum; if smallest, a maximum. Both extrema are stable in the conservative problem. An intermediate moment gives opposite signs in the quadratic form, a saddle and instability. Dissipation changes the fixed-energy argument and must not be silently added to torque-free motion.

Choose an inertia unit $I_{\ast }$ and an angular speed unit ${\Omega }_{\ast }>0$, so physical principal moments of inertia are $I_{\ast }(1,2,3)$, physical angular velocity is ${\Omega }_{\ast }\mathbf{\omega }$, and normalized time is $t={\Omega }_{\ast }{t}_{\mathrm{physical}}$. The Euler equations for a torque-free rigid body becomeThis is a physically admissible ideal planar mass distribution: $I_3=I_1+I_2$ is the perpendicular axis theorem. A book of finite thickness need not have these exact ratios.

For an original comparison, start with angular velocity $(1,\epsilon ,0)$, $(\epsilon ,1,0)$ or $(\epsilon ,0,1)$, respectively. With $\epsilon =0.01$, the transverse Euclidean norm stays small about axes one and three, but becomes order one about axis two. For the intermediate-axis start, its linearization givesThe hyperbolic functions contain a growing exponential function with rate ${\lambda }_2=1/\sqrt{3}$. Reducing $\epsilon$ tenfold delays reaching the same small transverse amplitude by approximately $\sqrt{3}\log 10\simeq 3.99$ normalized time units, once the growing term dominates. That delays the instability. Once the transverse angular velocity is comparable to the main component, the linearization ceases to describe the motion.

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.

Clemson's documented apparatus is a book held tightly closed with several rubber bands. Toss it freely with spin about each principal axis in turn: the stated visual readout is steady orientation for the smallest and largest principal moments of inertia, but a flip before the catch for the intermediate one. The same book and the two extreme-axis throws provide comparison controls; the bands support the rigid body approximation by restraining the pages. This is a documented qualitative demonstration, not a measurement of ${\lambda }_2$.^[ref]

For a rectangular book with length $a$, width $b$ and thickness $c$, where $a>b>c$, a uniform mass model gives principal moments of inertia $M(b^2+c^2)/12$, $M(a^2+c^2)/12$ and $M(a^2+b^2)/12$. The intermediate principal axis runs across the width through the centre of mass; the maximum is normal to the cover. A marked cover or end helps distinguish ordinary spin from reversal of a directed body axis.

Interpret only the interval after release and before the catch: hand contact applies torque, whereas gravity in a uniform gravitational field has zero net torque about the centre of mass because $\int \mathbf{r}dm=0$. Air resistance, page motion, an imperfect release and short flight limit the approximation. A toss ending before the perturbation grows need not reveal instability. Comparable initial angular speeds and flight durations would improve the comparison; these are suggested controls, not quantified controls reported by Clemson.

For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

On the invariant level $L^2=2E{I}_2$, the torque-free rigid-body orbit is a separatrix joining the two intermediate-axis rotations. Withone orientation has ${\omega }_2(t)=\mu \tanh (\lambda (t-t_0))$; the other components are constant multiples of $\mathrm{sech}(\lambda (t-t_0))$.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

A uniform disk of radius $R$ and area mass density $\rho$ has a circular hole of radius $r_o$, whose centre H is displaced by $r_o$ from the original centre C. Set $D=\pi \rho R^2$, $h=\pi \rho r_o^2$ and $W=D-h$. The parallel axis theorem gives $I_H=D{R}^2/2+D{r}_o^2-h{r}_o^2/2$. Removing the material shifts the remaining centre of mass G away from H by $CG=h{r}_o/W$, so $HG=D{r}_o/W$ and $I_G=I_H-D^2{r}_o^2/W$. Thus the rotational component of free rigid-body kinetic energy is smaller than fixed-axis kinetic energy about H at the same angular speed, although total rolling kinetic energy also includes translation.