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

False. Newton's law only holds for acceleration in an inertial frame; the body-frame rate must be corrected with the transport theorem, which adds the $\vec{\omega }\times \vec{v}$ term.

False. A rocket flying at constant speed but turning has a rotating velocity direction, so $\vec{\omega }\times \vec{v}\ne 0$ and it genuinely accelerates in the inertial frame.

False in general. Only $\vec{H}$ (in the inertial frame) and kinetic energy are conserved; if $I_x\ne I_y\ne I_z$ the coupling terms make $p,q,r$ trade energy and the body tumbles (Dzhanibekov effect).

True. Along a principal axis $\vec{\omega }\times (\mathbf{I}\vec{\omega })=0$ and $\vec{M}=0$, so $\dot{\vec{\omega }}=0$ — this is the basis of spin stabilization.

False. Glued to the rigid body, $\mathbf{I}$ is constant in the body frame; that is precisely why Euler's equations are written there. It is $\mathbf{I}$ expressed in the inertial frame that rotates and becomes time-varying.

False. Euler's law $\vec{M}=\mathbf{I}\dot{\vec{\omega }}{|}_B+\vec{\omega }\times (\mathbf{I}\vec{\omega })$ holds for any body axes; choosing principal axes merely diagonalizes $\mathbf{I}$ and gives the clean scalar form.

Half-true. They decouple kinematically into "motion of the CoM" + "rotation about the CoM", but re-couple through physics: thrust misalignment creates torque, and — concretely — when the body pitches, its nose points away from the velocity vector, creating an angle of attack $\alpha$; the aerodynamic lift/drag scale with $\alpha$, so attitude (a rotational state) directly resets the force in Newton's equation.

True. Both are the transport theorem applied to a stored vector — linear momentum $m\vec{v}$ for translation, angular momentum $\mathbf{I}\vec{\omega }$ for rotation.

The $\dot{m}\vec{v}$ term is not the thrust; the ejected gas carries away momentum, so the correct variable-mass form is ${\vec{F}}_{ext}+\dot{m}{\vec{v}}_{exh,rel}=m\dot{\vec{v}}$. Cleanest fix: treat thrust as an external force and use constant-mass Newton.

The coupling term $(I_z-I_y)qr$ is genuinely there, but for an axisymmetric rocket $I_y=I_z=I_t$ makes it vanish exactly, so $\dot{p}=0$ is exact — not an approximation.

Wrong — $\dot{\vec{\omega }}=0$ kills the $\mathbf{I}\dot{\vec{\omega }}$ term, but $\vec{\omega }\times (\mathbf{I}\vec{\omega })$ remains and equals the steady torque needed to hold a constant spin off-axis.

Only when $\vec{\omega }$ is along a principal axis. In general $\mathbf{I}$ is a tensor, so $\vec{H}$ and $\vec{\omega }$ point in different directions — the source of coning and precession.

Not neglected — it is exactly zero because we started from $p=r=0$. Pitch decouples cleanly only because of that initial condition, not by approximation.

Internal forces cancel in pairs by the 3rd law; their torques cancel too only if each pair is central (acts along the line joining the two particles), so the moment arms match. That (usually-satisfied) condition is what lets ${\vec{F}}_{ext}=\dot{\vec{p}}$ and ${\vec{M}}_{ext}=\dot{\vec{H}}$ hold — it is the justification, not something to re-drop.

Because in the body frame the inertia tensor is constant; in the inertial frame $\mathbf{I}$ rotates and its time-derivative would clutter the equation far worse than the single cross-product term.

For translation the inertia is a single scalar $m$; for rotation the resistance is a tensor $\mathbf{I}$ that differs about each axis and couples the three rates $p,q,r$, producing gyroscopic effects with no translational analogue.

With $I_y=I_z=I_t$ the transverse equations become $\dot{q}=\lambda r,\dot{r}=-\lambda q$, a pure rotation of $(q,r)$ — the spin axis traces a clean cone at frequency $|\lambda |$, which is why sounding rockets spin-stabilize.

To keep the clean constant-mass Newton form and avoid the Meshchersky bookkeeping trap; the momentum of ejected exhaust is packaged into a single thrust vector $\vec{T}=-\dot{m}{\vec{u}}_e$.

Because Newton's rotational law is $\vec{M}=d\vec{H}/dt{|}_I$; with $\vec{M}=0$ it is $\vec{H}$ in the inertial frame that is fixed. The body-frame $\vec{\omega }$ can still vary because $\mathbf{I}$ links them non-trivially. See Angular momentum conservation.

Both come from applying one theorem — the transport theorem — to a momentum vector stored in body components; only the stored vector differs ($m\vec{v}$ vs $\mathbf{I}\vec{\omega }$).

Finite rotations don't add as vectors, so $\vec{\omega }$ must feed a proper kinematic equation — Euler angles or quaternions — to track orientation without singularities.

All differences $(I_z-I_y)$ etc. vanish, so $\vec{M}=\mathbf{I}\dot{\vec{\omega }}$ — rotation behaves as simply as translation, with $\vec{\omega }$ constant whenever $\vec{M}=0$.

The cross term drops entirely, leaving $\vec{F}=m\dot{\vec{v}}{|}_B$, which reduces to elementary point-mass Newtonian motion — the sanity-check limit in Ex 1.

The motion is unstable: tiny perturbations grow exponentially and the body flips — the intermediate axis theorem. Spin about the largest or smallest axis is stable.

$p=p_0$ stays constant, so the coning frequency $\lambda =\frac{I_x-I_t}{I_t}{p}_0$ is fixed too — a steady, predictable cone, unlike the tri-axial tumbling case.

Its sign follows $(I_x-I_t)$: for a prolate (pencil-like) body $I_x<I_t$ so $\lambda <0$ and the transverse rates rotate one way; for an oblate (disk-like) body $I_x>I_t$ so $\lambda >0$ and they rotate the opposite way. $|\lambda |$ sets the cone's speed, the sign sets whether the coning is prograde or retrograde relative to the spin.

Not necessarily — as fuel drains the CoM shifts, changing moment arms and the inertia tensor; strictly the body-frame $\mathbf{I}$ and CoM are only quasi-constant and must be updated over the burn.

Both cross terms vanish, leaving $\vec{M}=\mathbf{I}\dot{\vec{\omega }}$ — pure angular acceleration from rest, exactly the clean start-of-maneuver case in Ex 3.