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Let $t=0$ be the start of deposition and $I_0=M{R}^2/2$ be the original moment of inertia of a uniform disc. Assume that all deposited sand sticks and remains at radius $r$, with negligible thickness and no losses. The sand's incoming angular momentum about the vertical axis is zero. Neither the bearing nor gravity supplies an axial torque, so conservation of angular momentum givesConsequently the angular velocity and angular acceleration areThe rotational kinetic energy isThus, for $\alpha r^2>0$,This is mass-loaded rotation: the sticking process dissipates mechanical energy even though angular momentum is conserved. Indeed $\dot{K}=-\alpha r^2{\omega }^2/2$. If $r=0$ or $\alpha =0$, no additional axial moment of inertia is deposited and the nonzero initial rotational kinetic energy never halves.

For an equal-speed elastic impact on a free rod, let $h$ be the signed distance of the impact point from the rod's center of mass, $I=m{l}^2/12$, and $\mathcal{J}$ the horizontal impulse delivered to the rod. For a smooth impact the impulse is perpendicular to the rod. If the ball's signed final velocity is $u$, the rod's center of mass velocity is $V$ and its angular velocity is $\omega$, momentum balance and angular impulse giveBoth objects have positive final translational velocities for a nonzero impact away from the centre. More explicitly, restitution in an elastic collision gives $V+\omega h-u=v_0$, and henceEqual final speeds thus require $u=V=v_0/2$. One can also see that the alternative $u=-V$ would violate $u+V=v_0$.

By conservation of kinetic energy in the elastic collision,Thus $I{\omega }^2=m{v}_0^2/2$. Since $I\omega =mVh=m{v}_{0}h/2$, elimination of $\omega$ givesEither side of the center of mass works; the nearer end is a distance $l(1/2-1/\sqrt{6})$ away. The final angular speed is $|\omega |=\sqrt{6}{v}_0/l$. These values also satisfy the restitution condition because the struck point's velocity is $V+\omega h=3v_0/2$, reversing the initial relative contact velocity.

The free rod has the same mass $m$ and length $l$ as in part (a). There is no pivot. Put the rod's initial center of mass at $y=0$ and the impact end at $y=l/2$. At impact the combined center of mass isConservation of momentum gives the combined center of mass velocity $V_C=v_0/2$. Evaluate angular momentum about this combined center of mass; the incident ball's perpendicular lever arm is $l/4$, so its initial magnitude is $m{v}_{0}l/4$.

After the perfectly inelastic collision, the parallel axis theorem gives the combined moment of inertiaWith negligible external angular impulse, conservation of angular momentum therefore givesIts sense is that of the incident ball's angular momentum, clockwise in the source diagram. As a consistency check, the final kinetic energy isThus $m{v}_0^2/10$, or one fifth of the initial kinetic energy, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect angular velocity.

A circular orbit has constant $r=r_0>0$. Both its initial radial velocity and radial acceleration must vanish, givingEquivalently, the initial tangential speed and angular velocity obeyThe initial polar angle is arbitrary, and the energy is fixed at $E=V_{\mathrm{eff}}(r_0)$. Either orientation is possible. These conditions give an exact circular orbit even if it is unstable; stability requires an additional positive second derivative of the effective potential.

The magnetic dipole moment is $\mathbf{\mu }=IA\hat{\mathbf{n}}$ and its magnetic torque is ${\mathbf{\tau }}_{\mathrm{mag}}=\mathbf{\mu }\times \mathbf{B}$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving torque needed to keep the angular velocity constant isAveraging over a complete rotation, with $⟨{\sin }^2\omega t⟩=1/2$, givesAs an independent sign and dimension check, $\omega ⟨{\tau }_{\mathrm{drive}}⟩=R⟨I^2⟩$. Thus the mechanical input power equals the mean Joule heating; a passive loop cannot supply a sustained driving torque. This average assumes the same negligible-self-inductance approximation as the current expression.

For a steady filamentary electric current $I$, the Biot-Savart law isHere $\mathbf{r}$ is the observation position, ${\mathbf{r}}^{\prime }$ the source position, $d{\mathbf{\ell }}^{\prime }$ points in the direction of positive electric current, and ${\mu }_0$ is the vacuum permeability. A circular loop has centre magnetic field ${\mu }_{0}I/(2r)$: the cross product has magnitude $rd{\ell }^{\prime }$ and a constant axial direction, and $\int d{\ell }^{\prime }=2\pi r$.

Introduce the constant signed angular velocity $\Omega$ about the positive axis. A ring's electric charge $dq=2\pi \sigma rdr$ crosses any radial cut once per rotation, so its electric current is $dI=\Omega dq/(2\pi )=\sigma \Omega rdr$. Adding loop contributions yields the magnetic field of a rotating charged annulus:The direction follows the right-hand rule for positive electric charge; reversing either $Q$ or $\Omega$ reverses the magnetic field. The source supplies no rotation rate, so the answer necessarily retains $\Omega$.

Relative to an origin, the inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-r_i{r}_j)dm$, with $\mathbf{J}=\mathsf{I}\mathbf{\omega }$. Its orthogonal eigenvectors are principal axes, with eigenvalues the principal moments of inertia. Different components of $\mathbf{\omega }$ are multiplied by different eigenvalues, so angular momentum need not point along angular velocity.

For a uniform thin disc, radial integration gives $I_z=\int r^{2}dm=m{a}^2/2$. The perpendicular axis theorem and symmetry then give $I_x=I_y=I_z/2$, henceFor the elastic eccentric impact of a thin disc, neglect gravity's impulse and take the pole's impulse as vertical. Let $z$ point up and let the contact position relative to the centre be $r\hat{x}$. An upward impulse $P$ givesFor an elastic impact with a fixed pole, the contact point reverses its normal velocity, so $-u+P/m+r^{2}P/I_d=+u$. ThereforeHere $v_C>0$ denotes upward motion; the centre can still move down if $r>a/2$. Substitution givesThis also follows by combining conserved angular momentum about the fixed contact with conserved kinetic energy and selecting the separating, nonzero-impulse solution. At $r=0$ the disc simply rebounds without rotation; at $r=a/2$ its centre is instantaneously stationary. The rebound result concerns the contacting material point immediately after this ideal impact, not its subsequent contact history.

For the inertia tensor of a uniform cube, reflection symmetry makes every off-diagonal element of the central inertia tensor vanish. Permutation of the three axes makes its diagonal elements equal. Since $M=8\rho a^3$ and the mean of each squared coordinate in the cube is $a^2/3$,Every central axis is therefore a principal axis, including the body diagonal. The angular momentum is parallel to the angular velocity, and its magnitude is

Let a frame's origin have inertial acceleration ${\mathbf{a}}_0$ and its axes rotate with angular velocity $\mathbf{\Omega }(t)$. For position $\mathbf{r}$ and velocity ${\mathbf{v}}^{\prime }$ measured in that frame, Newton's law becomesThe extra terms are translational, Coriolis, Euler and centrifugal inertial forces. Centrifugal force points away from the rotation axis; Coriolis force deflects moving bodies sideways and does no instantaneous work on their relative motion, since it is perpendicular to ${\mathbf{v}}^{\prime }$. The Euler term appears when the rotation rate or axis changes.

For constant $\mathbf{\Omega }$, centrifugal force has potential $U_{\mathrm{cf}}=-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. It can be combined with a conservative real potential to find rotating-frame equilibria. The Coriolis term remains velocity dependent. These forces arise from describing acceleration in a noninertial frame; they are not additional interactions between bodies. They are essential for trajectories on rotating apparatus and for large-scale terrestrial motions. Frame-dependent terms must be included consistently with the chosen coordinates.

Choose body-fixed principal axes through a fixed point or the centre of mass, with principal moments $I_1,I_2,I_3$. Angular momentum has components $L_i=I_i{\omega }_i$. The inertial derivative is the body derivative plus $\mathbf{\omega }\times \mathbf{L}$, so Euler's equations readThey avoid differentiating the rotating inertia tensor in laboratory coordinates. Torques must refer to the chosen point, and the displayed fixed moments require a rigid body.

For zero torque, laboratory angular momentum is constant, while its components can change in the body frame. Both $L^2$ and rotational kinetic energy $T=\frac{1}{2}\sum I_i{\omega }_i^2$ are conserved. The intersections of their constant-value surfaces constrain the motion. For distinct $I_1<I_2<I_3$, steady rotation about either extreme principal axis is stable against small perturbations, whereas rotation about the intermediate axis is unstable. For example perturbing rotation $\Omega$ about axis 2 giveswhose coefficient is positive. A symmetric body has additional simplifications and can precess. Conservation of laboratory angular momentum does not imply constant body-frame angular velocity.

For torque-free rotation, angular momentum is fixed in inertial space, but angular velocity need not point along it. In principal body axes,and cyclic permutations. A symmetric top with transverse moment $A$ and axial moment $C$ has constant ${\omega }_3$, while ${\omega }_1+i{\omega }_2$ rotates in body coordinates at signed frequency $(C-A){\omega }_3/A$. The symmetry axis precesses around the fixed angular momentum in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.

Poinsot construction expresses the same torque-free motion geometrically. The energy ellipsoid $\sum I_i{\omega }_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf{L}$, and $\mathbf{L}\cdot \mathbf{\omega }=2T$ places that point in a fixed plane perpendicular to $\mathbf{L}$. The ellipsoid rolls on this invariable plane without slipping: the instantaneous velocity of its contact point is proportional to $\mathbf{\omega }\times \mathbf{\omega }=0$. The contact curve is the polhode on the body ellipsoid and the herpolhode on the plane.^[ref]

For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis givesThe extreme-moment axes are stable; the intermediate one is unstable. The major-axis relaxation theorem adds weak internal dissipation while retaining negligible external torque. At fixed $L$,Energy loss favours rotation about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. A perfectly rigid torque-free body does not dissipatively settle; its energy is conserved. Distinguishing stability from relaxation also explains why minimum-inertia rotation can be stable in the ideal rigid model.

Forced precession occurs when an external torque changes $\mathbf{L}$, with $\dot{\mathbf{L}}=\mathbf{\tau }$. For steady precession, $\mathbf{\tau }={\mathbf{\Omega }}_p\times \mathbf{L}$. A rapidly spinning, pivoted heavy gyroscope has $L\simeq C{\omega }_3$ along its symmetry axis, givingThe torque is perpendicular to the nearly axial angular momentum, so it mainly changes its direction. For a steady inclined symmetric top, the more complete condition isIt exhibits slow and fast precession branches when real solutions exist. A generic release excites nutation as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which angular momentum and pivot are being used.

For a rigid body about a chosen origin,The inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity for rotation about a principal axis, but need not be parallel in a general direction. Its eigenvalues are the principal moments of inertia.

If the origin is translated from the centre of mass by $\mathbf{a}$, the parallel-axis theorem becomes $I_{ij}=I_{ij}^{\mathrm{CM}}+M(a^2{\delta }_{ij}-a_i{a}_j)$. In body axes, Euler's equation is $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{body}}+\mathbf{\omega }\times \mathbf{L}$. Thus the full tensor is needed whenever the rotation is not confined to one fixed principal axis.

Take the initial thrower and receiver positions to be $(-a,0)$ and $(a,0)$, with counterclockwise angular velocity $\omega$. Interpret $v$ as the launch speed relative to the thrower. The ball inherits the thrower's tangential velocity, so in an inertial frame its initial horizontal velocity is $(v,-a\omega )$. The flight time across the diameter is $t_f\simeq 2a/v$. During that time the receiver moves approximately $+a\omega t_f$ sideways while the ball moves $-a\omega t_f$. Thus the transverse miss on a rotating platform isThe ball passes behind the receiver in the sense of rotation. Equivalently, in the rotating frame its transverse Coriolis acceleration has magnitude $2\omega v$ to leading order; integrating over $t_f$ gives $d=\frac{1}{2}(2\omega v)t_f^2$. Terms neglected here are higher order in $a\omega /v$. Specifying an initially radial launch velocity in the inertial frame instead would define a different throw and would lose the inherited tangential velocity.

In Lagrangian mechanics, choose independent generalized coordinates $q_i$ compatible with the mechanical constraints, write $L=T-U$, and apply the Euler-Lagrange equationEquivalently, the physical path makes the action stationary under variations with fixed endpoints. For ideal holonomic constraints, their unknown reaction forces need not appear in the reduced equations. Direct use of Newton's laws of motion instead resolves vector forces and reactions and enforces the constraints separately. Both approaches describe the same mechanics; the scalar Lagrangian is often shorter in curvilinear coordinates. Nonconservative generalized forces must be added separately. Here a drive maintains the prescribed azimuthal angular velocity $\Omega$.

The mass has polar speed $a\dot{\theta }$ and azimuthal speed $a\Omega \sin \theta$, so its kinetic energy is $m{a}^2({\dot{\theta }}^2+{\Omega }^2{\sin }^2\theta )/2$. The elastic potential energy is $\mu {\theta }^2/2$, and the height is $a\cos \theta$, giving gravitational potential energy $mga\cos \theta$. Thereforeand the Euler-Lagrange equation givesFor $\mu =2mga$, define $q={\Omega }^{2}a/g$ and $F(\theta )=q\sin \theta \cos \theta -2\theta +\sin \theta$, so $\ddot{\theta }=(g/a)F(\theta )$. Near zero,When $q<1$ the upright equilibrium is linearly stable and the small oscillations have ${\omega }_1^2=g/a-{\Omega }^2$.

For $q>1$, the graph of $F$ initially rises, whereas $F(\pi /2)=1-\pi <0$. There must therefore be a zero inside the interval. More strongly, for $0<\theta <\pi /2$,is strictly decreasing: $\cos \theta$ decreases and $\theta /\sin \theta$ increases because $\sin \theta -\theta \cos \theta >0$. The ratio goes from $q-1>0$ to $1-\pi <0$, giving exactly one zero ${\theta }_0$ in that interval. At the zero, $F^{\prime }({\theta }_0)<0$, so it is a stable equilibrium. The rotating torsion-stabilized inverted pendulum also has a symmetric stable angle $-{\theta }_0$ if both signs of tilt are permitted.

At ${\theta }_0=\pi /4$, setting $F({\theta }_0)=0$ gives $q/2-\pi /2+1/\sqrt{2}=0$, soFor $\theta ={\theta }_0+\xi$, the linearized equation is $\ddot{\xi }=(g/a)F^{\prime }({\theta }_0)\xi$. Since $F^{\prime }=q\cos 2\theta -2+\cos \theta$,At the threshold $q=1$ the linear restoring term vanishes, but the cubic term remains:The upright state remains nonlinearly stable, with no nonzero infinitesimal harmonic frequency. Small finite disturbances produce quartic oscillator motion whose frequency is proportional to its small angular amplitude; its period tends to infinity as that amplitude tends to zero. Zero linear stiffness is not the same as an absent restoring torque.

A torque-free rigid body has constant angular momentum $\mathbf{L}$ in inertial space. Its angular velocity need not be parallel to $\mathbf{L}$, because $\mathbf{L}=\mathbf{I}\mathbf{\omega }$. In body principal axes, Euler equations for a torque-free rigid body areFor a symmetric top $I_1=I_2=I_{\perp }$, $I_3=I_{\parallel }$, the component ${\omega }_3$ is constant andThis is the signed rotation frequency of the transverse angular-velocity components in the body frame. In space the symmetry axis instead obeysIts angle to $\mathbf{L}$ is constant, with $\cos \theta =I_{\parallel }{\omega }_3/L$. The body-frame and space-frame frequencies of free symmetric-top precession describe different frames. A spherical body has $\mathbf{\omega }\parallel \mathbf{L}$ and no corresponding wobble. For a triaxial body, rotations about the largest and smallest principal-inertia axes are stable, whereas rotation about the intermediate axis is unstable.^[ref]

For positions measured from the chosen origin, the inertia tensor isIt relates angular momentum and rotational kinetic energy to angular velocity:The real symmetric tensor has orthogonal principal axes, in which $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$. Its eigenvalues are nonnegative moments of inertia. Rotation about a principal axis has $\mathbf{L}\parallel \mathbf{\omega }$; a general rotation does not. The moment about a unit axis $\mathbf{n}$ is $I_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{In}$.

Changing from the centre of mass to a parallel origin displaced by $\mathbf{a}$ gives the tensor parallel-axis theorem,For example, a uniform solid sphere has $\mathbf{I}=(2M{R}^2/5)1$, while a thin rod along $x$ has principal moments $0,M{l}^2/12,M{l}^2/12$ in the ideal zero-radius limit. The tensor is needed to describe off-axis rotation and torque-free precession; a single scalar inertia only suffices after an axis is specified.^[ref]

For a body-fixed mass element at position $\mathbf{r}$ relative to the fixed tip, $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$. ThereforeThe inertia tensor is real symmetric. Its orthogonal eigenvectors define principal axes and its eigenvalues are principal moments of inertia. angular momentum is parallel to angular velocity when rotation is along a principal axis, but generally not for arbitrary rotation.

Write $\mathbf{n}={\hat{\mathbf{e}}}_3$ and $s=I_3{\omega }_3$. Axial symmetry gives $\mathbf{L}=I_1\mathbf{\omega }+(I_3-I_1){\omega }_3\mathbf{n}$. Since $\dot{\mathbf{n}}=\mathbf{\omega }\times \mathbf{n}$, the vector triple-product identity gives $\mathbf{n}\times \dot{\mathbf{n}}=\mathbf{\omega }-{\omega }_3\mathbf{n}$. HenceDifferentiate in the inertial frame, using constant $s$ and $\dot{\mathbf{n}}\times \dot{\mathbf{n}}=0$. The gravitational torque about the fixed tip is $(\ell \mathbf{n})\times (-mg\hat{\mathbf{z}})$, yielding the symmetry-axis equation of a heavy symmetric top:The tip reaction exerts no moment about that tip.

For steady precession, $\mathbf{n}=(\sin \theta \cos {\Omega }_{P}t,\sin \theta \sin {\Omega }_{P}t,\cos \theta )$. At azimuth zero, put $s_{\theta }=\sin \theta$ and $c_{\theta }=\cos \theta$. Then $\dot{\mathbf{n}}=(0,{\Omega }_P{s}_{\theta },0)$, $\ddot{\mathbf{n}}=(-{\Omega }_P^2{s}_{\theta },0,0)$, and the $y$ component gives $s{\Omega }_P{s}_{\theta }-I_1{\Omega }_P^2{s}_{\theta }{c}_{\theta }=mg\ell s_{\theta }$. For a nonvertical axis this becomesReal steady branches require a nonnegative discriminant. In particular, for a top pointing above the horizontal there is a spin threshold for these fixed-angle solutions. At $\cos \theta =0$ use the quadratic's linear limit ${\Omega }_P=mg\ell /s$, rather than dividing by zero. For a vertical axis, azimuth does not describe a physical conical motion and the division by $\sin \theta$ is inappropriate.

For rapid positive spin $s$ with $s^2\gg 4mg\ell I_1\cos \theta$, the slow and fast steady precession roots areThe approximation assumes a nonsingular fixed inclination. For $g=0$, the quadratic instead gives ${\Omega }_P=0$ or ${\Omega }_P=s/(I_1\cos \theta )$. The zero branch has a stationary symmetry axis and pure axial spin. On the other branch angular momentum is constant and parallel or antiparallel to the chosen $z$ axis; $L_z=I_1{\Omega }_P$ and $s=L_z\cos \theta$. This is torque-free precession in space, not the different rate at which transverse components rotate in body coordinates.

Separate a ball's orbital motion from its spin. For a rolling ball between concentric races, the ball centre lies $2a$ from the axle axis. Let its tangential centre velocity be $v$ in the direction of axle motion and let its signed spin angular velocity be $\Omega$, positive in the same sense as $\omega$. Rolling without slipping at the moving inner contact and stationary outer contact requiresHenceThe spin is opposite to the axle's rotation, while the centre orbits in the same sense. The total angular momentum about the axle is the sum of its orbital angular momentum and its angular momentum about the centre of mass:For a uniform solid spherical ball, $I_{\mathrm{cm}}=2m{a}^2/5$, soThe solid-ball assumption supplies the usual moment of inertia; the preceding expression in terms of $I_{\mathrm{cm}}$ also states the answer for a different internal mass distribution.

Changing basis and acceleration. Attach orthonormal unit vectors ${\mathbf{e}}_i(t)$ to the rotating frame, with a common origin fixed in the inertial frame. An infinitesimal rotation $d\mathbf{\theta }=\mathbf{\omega }dt$ changes a vector fixed in those axes by $d\mathbf{A}=d\mathbf{\theta }\times \mathbf{A}$: its tip moves tangentially to the circle about the rotation axis. More generally, write $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$. Differentiating the components and the rotating basis separately proves the rotating-frame derivative formula,

Apply the formula first to position: ${\dot{\mathbf{r}}}_0=\mathbf{v}+\mathbf{\omega }\times \mathbf{r}$, where $\mathbf{v}=(d\mathbf{r}/dt{)}_S$. Apply it again to both terms. For constant $\mathbf{\omega }$,One cross term comes from differentiating $\mathbf{v}$ in the inertial frame and another from differentiating $\mathbf{r}$ in $\mathbf{\omega }\times \mathbf{r}$. The displayed source formula implicitly requires constant angular velocity and an unaccelerated common origin. For changing rotation rate there is also $\dot{\mathbf{\omega }}\times \mathbf{r}$; an accelerating origin supplies its own translational acceleration. Moving the rotation terms to the force side gives the apparent Coriolis force and centrifugal force with negative signs.

Southward motion on Earth. Use local east, north and up axes, with latitude $\lambda$ positive northward. Then $\mathbf{\Omega }=\Omega (0,\cos \lambda ,\sin \lambda )$ and $\mathbf{v}=(0,-v,0)$. The Coriolis acceleration in local Earth coordinates givesIt is westward in the northern hemisphere, eastward in the southern hemisphere and zero for exactly southward motion at the equator. There is no vertical Coriolis component for this particular velocity.

Correct steering. On the desired meridional track, drag has no eastward component because it is antiparallel to the velocity. If the thrust makes signed angle $\varphi$ east of south, its eastward component is $T\sin \varphi$. The cross-track force balance is thereforesoThe argument must have magnitude at most one for this simple thrust-only correction to be feasible. The signed angle changes sign at the equator: correction is east of south in the north and west of south in the south. Thus the phrase “east of south” cannot describe a positive correction throughout the entire flight.

Wrong-sign correction and the miss distance. A literally constant velocity vector cannot develop a cross-track displacement. The physically useful interpretation of the last task is constant speed, with an erroneous lateral steering correction. In the usual leading-track model, thrust and drag balance along the actual motion, and the applied transverse correction has the correct magnitude but the wrong sign. It therefore reinforces rather than cancels the Coriolis force. Starting at the equator, use the unperturbed southward track $\lambda \simeq -vt/R$, and let $x_E$ be the eastward departure from that track. To first order in the departure,There is a factor two relative to uncompensated Coriolis deflection because the wrong correction adds an equal lateral force. Integration gives the wrong-sign Coriolis steering error,For southern angular distance $\Lambda =16\pi /180=0.27925$, $R=6.4\times 10^6\mathrm{m}$ and $v=800/3.6=222.2\mathrm{m}{\mathrm{s}}^{-1}$, the nominal flight time after the equator isTaking $\Omega \simeq 2\pi /(86400\mathrm{s})$ givesThe small-latitude approximation $\sin (vt/R)\simeq vt/R$ gives ${\ddot{x}}_E\simeq 4\Omega v^{2}t/R$ and $x_E\simeq 2\Omega v^2{t}^3/(3R)$, or $1.95\times 10^5\mathrm{m}$, confirming the scale and direction.

This is an estimate under an explicit steering model, not an exact trajectory forced by the wording. For example, retaining spherical transport of the local axes while still linearizing in the cross-track motion gives, with $s=-\lambda$, $\dot{u}-(v/R)u\tan s=4\Omega v\sin s$. Consequently $u\cos s=2\Omega R{\sin }^{2}s$ and the longitude departure is $\Delta \ell =2\Omega R(\tan \Lambda -\Lambda )/v$. The distance along the destination parallel is then $R\cos \Lambda \Delta \ell \simeq 1.93\times 10^5\mathrm{m}$. Corrections from the changed track and a more detailed control law are not fixed by the problem. If “keeps the velocity constant” instead means the full vector, the pilot would need an additional compensating force and there would be no miss; if thrust is kept at a specified geographical bearing despite drift, its along-track component and lateral drag must also be modelled.

Take axes fixed in a rigid body, chosen along its principal axes, with moments of inertia $I_1,I_2,I_3$. The angular momentum components are $H_i=I_i{\omega }_i$. Newton's rotational equation is $\mathbf{\tau }=(d\mathbf{H}/dt{)}_{\mathrm{inertial}}$, while the rotating-frame transport rule givesThus Euler's equations areThe nonlinear cross terms arise because the principal axes rotate. Even with no external torque, the body-frame components of angular momentum can vary while the angular momentum vector stays fixed in inertial space.

With zero torque, the equations conserveUniform rotation about any principal axis is a torque-free solution. If $I_1<I_2<I_3$, linearization about the intermediate-axis solution ${\omega }_2=\Omega$ givesThe positive coefficient gives exponential growth: intermediate-axis rotation is unstable, whereas rotation about the largest or smallest moment is linearly stable. For a symmetric body, Euler's equations also describe the precession of the transverse angular velocity around its symmetry axis. Torques from gravity or an applied couple enter through the right-hand sides rather than being absorbed into the inertial cross terms.