Exercises — 6DOF equations — translational (Newton), rotational (Euler's equations)

• $\dot{v}$ = the rate of change of the sideways velocity component as measured in the body frame — this is the $y$-component of $\dot{\vec{v}}{|}_B$ (the "seen in the body" derivative defined above).
• $ru$ and $-pw$ = the $y$-component of $\vec{\omega }\times \vec{v}$. Explicitly $\vec{\omega }\times \vec{v}$ has $y$-component $ru-pw$ (from $(\vec{\omega }\times \vec{v}{)}_y=ru-pw$). What it looks like: even if $v$ never changes in the body ($\dot{v}=0$), a rolling ($p$) or yawing ($r$) rocket still accelerates sideways in the inertial frame — that is what $ru-pw$ encodes.

Every term containing a product of two $\omega$-components, or one $\omega$ times a velocity/momentum, vanishes.

• Newton → $F_x=m\dot{u},F_y=m\dot{v},F_z=m\dot{w}$ (plain $\vec{F}=m\dot{\vec{v}}$).
• Euler → $M_x=I_x\dot{p},M_y=I_y\dot{q},M_z=I_z\dot{r}$ (plain torque = inertia × angular acceleration). This is the "textbook" non-spinning limit.

$F_y=m(\dot{v}+ru-pw)=0$. With $u=200,r=0.5,p=0$: $0=\dot{v}+(0.5)(200)-0\Rightarrow \dot{v}=-100{\text{m/s}}^2.$ What it means: even with no side force, the sideways body-velocity component grows negative because the whole velocity vector is being swept around by the yaw. It is a bookkeeping acceleration, not a real inertial-frame one.

The lever arm $\ell$ turns a force into a moment: $M_y=F\ell =800\times 3=2400\text{N⋅m}$ (force times perpendicular distance). Euler-$y$: $M_y=I_y\dot{q}+(I_x-I_z)rp$. With $r=p=0$ the coupling term is $0$: $\dot{q}=\frac{M_y}{I_y}=\frac{2400}{1200}=2{\text{rad/s}}^2.$ What it looks like: a clean, decoupled angular acceleration — the pitch axis is not fighting any roll/yaw yet.

Step 0 — read the figure's axes. In the figure the horizontal axis is $q$ (pitch rate) and the vertical axis is $r$ (yaw rate). A single point $(q,r)$ is the current transverse part of the angular velocity — the part perpendicular to the spin axis. Our whole job is to find how this point moves. It starts at the yellow dot $(q_0,0)$ (given: $q(0)=q_0$, $r(0)=0$).

Step 1 — spin rate is constant. Euler-$x$: $M_x=I_x\dot{p}=0\Rightarrow \dot{p}=0\Rightarrow p=p_0$. ✓ So the roll rate never changes; only the transverse point $(q,r)$ can move.

Step 2 — write the two transverse Euler equations, sign carefully. Take the parent's boxed forms exactly as written and substitute the symmetric values $I_y=I_z=I_t$, $p=p_0$, and $M_y=M_z=0$: $\text{Euler-}y:0=I_y\dot{q}+(I_x-I_z)rp\Rightarrow 0=I_t\dot{q}+(I_x-I_t)r{p}_0.$ Solve for $\dot{q}$ by moving the coupling term to the other side (this is the only algebra — no cyclic re-indexing needed): $\boxed{I_t\dot{q}=-(I_x-I_t)r{p}_0}.$ Do the identical substitution into Euler-$z$, whose parent coefficient is $(I_y-I_x)$ with $I_y=I_t$: $\text{Euler-}z:0=I_z\dot{r}+(I_y-I_x)pq\Rightarrow 0=I_t\dot{r}+(I_t-I_x)p_{0}q,$ $\boxed{I_t\dot{r}=-(I_t-I_x)p_{0}q=(I_x-I_t)p_{0}q}.$ Why the two boxed equations carry opposite signs on $(I_x-I_t)$ — the real "why": the $\dot{q}$ equation inherits the coefficient $(I_x-I_z)=(I_x-I_t)$ from Euler-$y$, while the $\dot{r}$ equation inherits $(I_y-I_x)=(I_t-I_x)$ from Euler-$z$. These two coefficients are negatives of each other — that is baked into the parent's cyclic pattern ($x:I_z-I_y$, $y:I_x-I_z$, $z:I_y-I_x$). It is precisely this sign flip between the $q$- and $r$-equations that turns straight-line growth into a rotation of the point $(q,r)$, as the next step shows.

Step 3 — name the frequency and get simple-harmonic motion. Divide both boxed equations by $I_t$ and define $\lambda =\frac{I_x-I_t}{I_t}{p}_0.$ Then, tidying signs, $\dot{q}=-\lambda r$ ... let us instead pick the convention that matches the initial data cleanly. Rewrite the boxed pair as $\dot{q}=-\frac{(I_x-I_t)}{I_t}r{p}_0=-\lambda r,\dot{r}=\frac{(I_x-I_t)}{I_t}q{p}_0=\lambda q.$ Differentiate the first and substitute the second: $\ddot{q}=-\lambda \dot{r}=-\lambda (\lambda q)=-{\lambda }^{2}q.$ This is simple harmonic motion: $\ddot{q}=-{\lambda }^{2}q$. With $q(0)=q_0$, $r(0)=0$ (so $\dot{q}(0)=-\lambda r(0)=0$), $q(t)=q_0\cos (\lambda t),r(t)=q_0\sin (\lambda t),$ which you can verify satisfies $\dot{q}=-\lambda r$ and $\dot{r}=\lambda q$ and both initial conditions.

Step 4 — read the answer back off the figure. Because $q^2+r^2=q_0^2{\cos }^2(\lambda t)+q_0^2{\sin }^2(\lambda t)=q_0^2$, the point $(q,r)$ stays exactly $q_0$ from the origin — it rides the pink circle of radius $q_0$ in the figure. The blue arrow shows its travel direction; it completes the circle at angular rate $|\lambda |$. That closed, non-growing circle is steady coning: the spin axis nods around in a cone but never tumbles. Why SHM and not blow-up? The sign flip from Step 2 means $q$ and $r$ feed into each other like $\dot{q}=-\lambda r$, $\dot{r}=+\lambda q$ — a pure rotation of the vector $(q,r)$, so its length is conserved and energy merely sloshes between pitch and yaw. That is the whole reason spinning stabilizes: no transverse rate can grow.

• Long thin rocket (prolate): $I_x<I_t\Rightarrow \lambda <0$. The coning circulates the other way but is still SHM, still bounded: $|\lambda |$ real.
• Flat disk (oblate): $I_x>I_t\Rightarrow \lambda >0$, opposite circulation direction, again bounded. Either way $\lambda$ is real, so $\ddot{q}=-{\lambda }^{2}q$ gives oscillation, never exponential growth. A symmetric body is coning-stable about its spin axis for both signs; only the sense (direction) of the cone flips. The dangerous case is $I_x\to I_t$ ($\lambda \to 0$): coning becomes infinitely slow — the axis wanders freely, no restoring "stiffness." Full instability needs three distinct moments (see the L4 trap and Dzhanibekov effect / intermediate axis theorem).

Euler-$y$ and Euler-$z$ with $M=0$, keeping $p=p_0$ constant and treating $q,r$ small: $I_y\dot{q}=(I_z-I_x)r{p}_0,I_z\dot{r}=(I_x-I_y)q{p}_0.$ Solve for $\ddot{q}$: differentiate the first, substitute $\dot{r}$ from the second: $\ddot{q}=\frac{(I_z-I_x)p_0}{I_y}\dot{r}=\frac{(I_z-I_x)(I_x-I_y)}{I_y{I}_z}{p}_0^{2}q.$ Call the bracketed coefficient $C=\frac{(I_z-I_x)(I_x-I_y)}{I_y{I}_z}{p}_0^2$. Numbers: $(I_z-I_x)=420-90=330$, $(I_x-I_y)=90-400=-310$, so numerator $=330\times (-310)=-102300$, denominator $=400\times 420=168000$. $C=\frac{-102300}{168000}\times 10^2=-60.89{\text{(rad/s)}}^2.$ $C<0\Rightarrow \ddot{q}=Cq=-|C|q$: oscillation at frequency $\sqrt{|C|}\approx 7.80\text{ rad/s}$. Spinning about the smallest axis is stable (bounded coning), consistent with the major/minor-axis stability rule. Why the sign test works: stability $\iff C<0\iff (I_z-I_x)$ and $(I_x-I_y)$ have opposite signs $\iff I_x$ is the largest OR the smallest of the three. Intermediate axis $\Rightarrow$ same signs $\Rightarrow C>0\Rightarrow$ blow-up.

Let $T$ denote the thrust — the forward force the exhaust exerts on the rocket. Its magnitude is $T=-\dot{m}{u}_e$ (the minus makes $T>0$ since $\dot{m}<0$): $T=-\dot{m}{u}_e=-(-4)(2500)=10000\text{N},\text{along }+x.$ Total body-$x$ external force: $F_x=T+F_x^{\text{grav}}=10000-4000=6000\text{N}$. Newton-$x$: $F_x=m(\dot{u}+qw-rv)$. With $q=0.2,w=10,r=0$: $qw-rv=0.2\times 10=2$. $6000=500(\dot{u}+2)\Rightarrow \dot{u}+2=12\Rightarrow \dot{u}=10{\text{m/s}}^2.$ Why this is the correct route: we never wrote $\frac{d(mv)}{dt}$; the momentum carried off by exhaust is already baked into the external thrust force $T$, so the clean $\vec{F}=m\dot{\vec{v}}$ form is valid. Mass loss $\dot{m}$ is tracked separately for the next time-step. See Meshchersky / Tsiolkovsky variable-mass dynamics.

(a) Coning frequency. Use $\lambda =\frac{I_x-I_t}{I_t}{p}_0$ from L3-1: $\lambda =\frac{120-500}{500}\times 8=\frac{-380}{500}\times 8=-6.08\text{rad/s}\Rightarrow |\lambda |=6.08\text{rad/s}.$ The sign is negative (prolate-like here, $I_x<I_t$), meaning the cone circulates one particular way; only the magnitude $|\lambda |$ sets the rate.

(b) Time for one full circle. The transverse point $(q,r)$ from L3-1 goes as $\cos (\lambda t),\sin (\lambda t)$, so it returns to its start after the angle $\lambda t$ advances by $2\pi$: $T_{\text{cone}}=\frac{2\pi }{|\lambda |}=\frac{2\pi }{6.08}\approx 1.0334\text{s}.$

(c) Constant transverse magnitude. From L3-1, $q(t)=q_0\cos (\lambda t)$ and $r(t)=q_0\sin (\lambda t)$, so $\sqrt{q^2+r^2}=\sqrt{q_0^2{\cos }^2(\lambda t)+q_0^2{\sin }^2(\lambda t)}=|q_0|=0.05\text{rad/s},$ a constant — the tip rides a circle of radius $0.05$ rad/s, confirming steady coning with no growth. This is exactly the pink circle picture of the figure in L3-1, now with concrete numbers.

Stability of spin about an axis requires that axis be the largest or the smallest moment of inertia (the two transverse differences must have opposite sign). Here sorted: $I_x=90$ (smallest), $I_y=95$ (intermediate), $I_z=400$ (largest).

• Spin about $x$ (smallest): safe — coefficient $\propto (I_z-I_x)(I_x-I_y)=(310)(-5)<0\Rightarrow$ oscillatory (stable).
• Spin about $z$ (largest): safe — $\propto (I_x-I_z)(I_z-I_y)=(-310)(305)<0\Rightarrow$ stable.
• Spin about $y$ (intermediate): forbidden — $\propto (I_x-I_y)(I_y-I_z)=(-5)(-305)>0\Rightarrow$ exponential growth, the Dzhanibekov flip. Design answer: spin about $x$ or $z$; never about $y$. The near-equal $I_x\approx I_y$ (90 vs 95) makes the $x$-spin only weakly stable, so $z$ (large margin) is the robust choice.

L1: read the terms. ::: The $\omega \times$ pieces cycle $x\to y\to z\to x$ and vanish only instantaneously at $\vec{\omega }=0$. L2: plug in. ::: Keep signs from the cyclic template; a decoupled axis ($p=r=0$) gives clean $\dot{q}=M_y/I_y$. L3–L4: analyze coupling. ::: Symmetric body cones (real $\lambda$); stability sign is $(I_a-I_b)(I_b-I_c)$. L5: design. ::: Spin only about the largest or smallest principal axis; never the intermediate one.