Gyroscope — steady precession derivation

WHAT is steady precession?

The three Euler angles:

• $\phi$ = precession angle (rotation about vertical $\hat z$)
• $\theta$ = nutation angle (tilt of the axis from vertical)
• $\psi$ = spin angle (rotation about the symmetry axis)

In steady precession: $\dot\theta = 0$, $\ddot\phi=0$, $\ddot\psi = 0$.

WHY does it precess instead of fall? (first-principles)

The master equation of rotational dynamics about the fixed pivot $O$:

$\vec\tau = \frac{d\vec L}{dt}$

HOW to derive it — the fast-top (gyroscopic) approximation

Setup. Symmetric top, mass $m$, pivot at $O$, center of mass at distance $\ell$ along the axis. Axis makes angle $\theta$ with vertical. Spin angular momentum magnitude along axis:

$L_s = I_3\,\omega_s$

where $I_3$ is the moment of inertia about the symmetry axis and $\omega_s$ is the spin rate.

Step 1 — Torque from gravity. Weight $m\vec g$ acts at the center of mass, position $\vec r = \ell\,\hat e_3$ (along axis). $\vec\tau = \vec r \times m\vec g$

Why this step? Torque is defined about the pivot; only gravity and the (torque-free) pivot reaction act, and the reaction at $O$ has zero moment arm. Magnitude: $|\vec\tau| = m g \ell \sin\theta$ Why $\sin\theta$? The lever arm perpendicular to gravity is $\ell\sin\theta$ — the horizontal offset of the CM from the vertical line through $O$. The torque is horizontal, perpendicular to the vertical plane containing the axis.

Step 2 — How $\vec L$ moves under precession. If the axis precesses about $\hat z$ at rate $\Omega$, the tip of $\vec L$ traces a horizontal circle. Only the horizontal component of $\vec L$ rotates. That horizontal component has magnitude $L_s\sin\theta$. A vector of horizontal magnitude $L_h$ rotating at $\Omega$ changes at rate: $\left|\frac{d\vec L}{dt}\right| = \Omega\, L_h = \Omega\, L_s \sin\theta$

Why this step? For any vector rigidly rotating about $\hat z$ with rate $\Omega$, $\dfrac{d\vec L}{dt} = \vec\Omega \times \vec L$, and $|\vec\Omega\times\vec L| = \Omega L \sin(\angle) = \Omega\,(L\sin\theta)$ for the part perpendicular to $\hat z$.

Step 3 — Equate. $m g \ell \sin\theta = \Omega\, I_3 \omega_s \sin\theta$

The $\sin\theta$ cancels (remarkable!):

Exact treatment (Lagrangian — keeps all terms)

Kinetic energy with $I_1=I_2$ (transverse) and $I_3$ (axial): $T = \tfrac12 I_1(\dot\theta^2 + \dot\phi^2\sin^2\theta) + \tfrac12 I_3(\dot\psi + \dot\phi\cos\theta)^2$ $V = m g \ell \cos\theta$

For steady precession set $\dot\theta=0,\ \ddot\theta=0$. The $\theta$ Euler–Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial\dot\theta} - \frac{\partial L}{\partial\theta}=0$ gives, after dropping $\ddot\theta,\dot\theta$ terms:

$I_1\dot\phi^2\sin\theta\cos\theta - I_3(\dot\psi+\dot\phi\cos\theta)\dot\phi\sin\theta + mg\ell\sin\theta = 0$

Divide by $\sin\theta$ and let $\omega_s \equiv \dot\psi+\dot\phi\cos\theta$ (the true axial spin), $\Omega\equiv\dot\phi$:

$I_1\Omega^2\cos\theta - I_3\omega_s\Omega + m g \ell = 0$

This is a quadratic in $\Omega$: $\Omega = \frac{I_3\omega_s \pm \sqrt{(I_3\omega_s)^2 - 4 I_1 m g \ell \cos\theta}}{2 I_1\cos\theta}$

Why the $-$ root reduces to the simple formula? Expand the square root for large $\omega_s$: $\sqrt{a^2 - b}\approx a - \frac{b}{2a}$ with $a=I_3\omega_s$, $b=4I_1mg\ell\cos\theta$. Then $\Omega_- \approx \frac{b/2a}{2I_1\cos\theta} = \frac{mg\ell}{I_3\omega_s}$.

Worked examples

Common mistakes (steel-manned)

Flashcards

Connections

• Euler Angles — $\phi,\theta,\psi$ parametrization used in the Lagrangian.
• Angular Momentum — the $\vec\tau = d\vec L/dt$ backbone.
• Lagrangian Mechanics — derives the exact quadratic precession condition.
• Moment of Inertia Tensor — distinction between $I_1$ and $I_3$.
• Nutation — the nodding motion superimposed when precession isn't steady.
• Symmetric Top — the general object; steady precession is a special solution.

Concept Map

Hinglish (regional understanding)