Lesson 8 · 40 min

Gyroscopic Motion

A spinning top leans over but does not fall; it swings slowly around instead. A gyroscope held at one end of its axle floats horizontally. Nothing mysterious is going on: gravity supplies a moment, the moment changes the angular momentum, and a large spin makes that change a sideways swing rather than a fall.

Learning objectives

Euler angles

For a body with an axis of symmetry \(z\), three angles fix its orientation: the precession angle \(\phi\) (the axis swings about the fixed vertical \(Z\)), the nutation angle \(\theta\) (the tilt of \(z\) from \(Z\)), and the spin angle \(\psi\) (rotation about \(z\) itself). Use axes \(x, y, z\) that follow the symmetry axis but not the spin: \(x\) horizontal (the line of nodes) and \(y\) in the vertical plane through \(z\). In those axes

\[ \wvec = \dot\theta\,\ihat + \dot\phi\sin\theta\,\jhat + (\dot\phi\cos\theta + \dot\psi)\,\khat, \qquad \Wvec = \dot\theta\,\ihat + \dot\phi\sin\theta\,\jhat + \dot\phi\cos\theta\,\khat \]

The frame lags the body only by the spin \(\dot\psi\khat\): Lesson 7's modified Euler equations apply, with principal moments \(I_x = I_y = I\) and \(I_z\) about the fixed point \(O\).

Steady precession

In steady precession \(\theta\), \(\dot\phi\) and \(\dot\psi\) are constant. Then \((\dot{\Hvec})_{xyz} = \mathbf{0}\) and \(\sum\Mvec_O = \Wvec \times \Hvec\), which has only an \(x\)-component:

Steady precession of an axisymmetric body

\[ \sum M_x = -I\dot\phi^2\sin\theta\cos\theta + I_z\,\dot\phi\sin\theta\,(\dot\phi\cos\theta + \dot\psi), \qquad \sum M_y = \sum M_z = 0 \]

For a top pivoted at \(O\) with its center of mass a distance \(r\) up the axis, gravity gives \(\sum M_x = m g r\sin\theta\).

The gyroscope

With the axis horizontal (\(\theta = 90^\circ\)), the first term vanishes and \(\omega_z = \dot\psi = \omega_s\). The equation becomes exact and simple:

Gyroscope (axis horizontal)

\[ \sum M_x = I_z\,\Omega\,\omega_s \quad\Rightarrow\quad \Omega = \frac{m g r}{I_z\,\omega_s} \]

\(\Omega = \dot\phi\). The same formula is a good approximation at any tilt when the spin is fast (\(I_z\omega_s \gg I\Omega\)). In vector form \(\Mvec = \Wvec \times \Hvec\): the spin axis swings toward the moment vector, not in the direction the force pushes.

Example 8.1 — The disk on the axle, let go

The disk of Lessons 6 and 7 (\(4\ \text{kg}\), \(I_s = 0.08\ \text{kg·m}^2\), center \(0.5\ \text{m}\) from the pivot) spins at \(30\ \text{rad/s}\) with its axle horizontal, supported only at the pivot. At what rate does it precess?

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Gravity's moment about the pivot is \(m g r = 4(9.81)(0.5) = 19.62\ \text{N·m}\), horizontal and perpendicular to the axle. With \(\theta = 90^\circ\):

\[ \Omega = \frac{m g r}{I_s\,\omega_s} = \frac{19.62}{0.08(30)} = 8.175\ \text{rad/s} \]

In Lesson 7 an external moment of \(4.8\ \text{N·m}\) held the precession at \(2\ \text{rad/s}\). Released, gravity's larger moment drives a faster precession, \(8.175\ \text{rad/s}\), so that \(I_s\omega_s\Omega\) matches \(m g r\).

Figure 8.1 A top (a \(0.5\ \text{kg}\) rotor of radius \(35\ \text{mm}\), its center \(50\ \text{mm}\) up the axle from the pivot) simulated with the full equations of motion. Released from rest at a tilt, it drops a little, then precesses while nodding: the axis tip draws cusps (nutation). Launched at the steady-precession rate, it glides round smoothly. Lower the spin and the nutation grows, and the gyroscope approximation drifts from the exact rate.

Example 8.2 — Exact and approximate precession of a top

The top of Figure 8.1 has \(I_z = 3.063 \times 10^{-4}\ \text{kg·m}^2\) and \(I = 1.403 \times 10^{-3}\ \text{kg·m}^2\) about the pivot. It spins with \(\omega_z = 300\ \text{rad/s}\) at \(\theta = 30^\circ\). Find its slow steady precession rate, exactly and with the gyroscope approximation.

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Dividing the steady-precession equation by \(\sin\theta\), with \(\omega_z = \dot\phi\cos\theta + \dot\psi\):

\[ I\cos\theta\,\dot\phi^2 - I_z\omega_z\,\dot\phi + m g r = 0 \quad\Rightarrow\quad 1.2151\times10^{-3}\,\dot\phi^2 - 0.091875\,\dot\phi + 0.24525 = 0 \] \[ \dot\phi = 2.771\ \text{rad/s} \quad (\text{slow root; the fast root is } 72.84\ \text{rad/s}) \]

Approximation: \(\Omega \approx m g r/(I_z\omega_z) = 0.24525/0.091875 = 2.669\ \text{rad/s}\), \(3.7\%\) low. The top usually settles into the slow precession; the fast one needs a hard, precise push.

Example 8.3 — Gyroscopic moment on a jet engine

A turbofan's rotating assembly has \(I_s = 12\ \text{kg·m}^2\) and turns at \(1000\ \text{rad/s}\) (about \(9500\) rev/min). The aircraft pitches up at \(0.2\ \text{rad/s}\). What moment must the engine mounts transmit?

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\[ M = I_s\,\omega_s\,\Omega = 12(1000)(0.2) = 2400\ \text{N·m} \]

The moment acts about the yaw axis (\(\Mvec = \Wvec \times \Hvec\) is perpendicular to both the pitch axis and the engine axis); its sense depends on which way the rotor spins. Pilots of single-engine propeller aircraft feel the same effect as a yaw when they pitch.

Torque-free motion

With no moment about \(G\) (a satellite, a thrown football, a coin in the air), \(\Hvec_G\) is constant in magnitude and direction. For an axisymmetric body the symmetry axis keeps a constant angle \(\theta\) to \(\Hvec_G\) and swings round it steadily:

Torque-free motion of an axisymmetric body

\[ \dot\phi = \frac{H_G}{I}, \qquad \dot\psi = \frac{I - I_z}{I\,I_z}\,H_G\cos\theta \]

\(I\) is the transverse moment and \(I_z\) the moment about the symmetry axis, both about \(G\). If \(I > I_z\) (a long body, like a football or a rocket) the precession is direct: \(\dot\phi\) and \(\dot\psi\) have the same sense. If \(I \lt I_z\) (a flat body, like a coin or the Earth) it is retrograde.

Example 8.4 — A wobbling satellite

A spin-stabilized satellite has \(I_z = 300\ \text{kg·m}^2\) about its symmetry axis and \(I = 500\ \text{kg·m}^2\) transverse. It spins with \(\omega_z = 2\ \text{rad/s}\), but its axis is \(5^\circ\) off its angular momentum vector. Find the precession and spin rates.

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The \(z\)-component of \(\Hvec_G\) is \(I_z\omega_z = H_G\cos\theta\), so \(H_G = 300(2)/\cos 5^\circ = 602.3\ \text{kg·m}^2/\text{s}\).

\[ \dot\phi = \frac{602.3}{500} = 1.205\ \text{rad/s}, \qquad \dot\psi = \frac{500 - 300}{500(300)}(602.3)\cos 5^\circ = 0.800\ \text{rad/s} \]

\(I > I_z\): direct precession. The satellite's axis traces a \(5^\circ\) cone about \(\Hvec_G\) once every \(2\pi/1.205 = 5.2\ \text{s}\). Being a minimum-axis spinner, it is also at risk of the slow drift toward flat spin described in Lesson 7 if it dissipates energy.

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Key takeaways