Physics · Mechanics · Year 1

Gyroscope and precession

Why doesn't a spinning top fall over, but slowly begins to turn its axis instead?

The situation

At the holiday your little brother spun a wooden top on the table. The top tilted but did not fall — its axis began to trace a slow circle. In the physics room the teacher hung a spinning bicycle wheel by a rope from just one end of its axle: the wheel did not fall but began to circle around the rope. The gyrocompass on a ship and a nut tumbling in a space station obey the same laws. Why does a spinning body not “run” in the direction of the force of gravity, but at right angles to it?

Topic

Gyroscope and precession. A spinning top, a bicycle wheel and an aeroplane's gyroscope

Key ideas

Angular momentum and precession
Euler's equations
Intermediate-axis instability

Formulas

\vec{τ} = \frac{d \vec{L}}{d t}, \vec{L} = I \vec{ω}

Torque changes angular momentum. With no external torque, $\vec{L}$ is conserved.

Ω = \frac{m g r}{I_{3} ω_{3}}, T_{p} = \frac{2 π}{Ω}

Angular speed and period of precession of a fast-spinning gyroscope.

I_{1} {\dot{ω}}_{1} + (I_{3} - I_{2}) ω_{2} ω_{3} = M_{1}

Euler's equations (in the principal axes of the body); the other two are obtained by cycling the indices 1→2→3→1.

λ = ω \sqrt{\frac{(I_{2} - I_{1}) (I_{3} - I_{2})}{I_{1} I_{3}}}

In rotation about the middle axis a small deviation grows like $e^{λ t}$ ( $I_{1} < I_{2} < I_{3}$ ) — the Dzhanibekov effect.

T = 2 π \sqrt{\frac{H}{K Ω_{\oplus︎} \cos φ}}

The undamped oscillation period of a pendulum gyrocompass: $H$ — angular momentum of the rotor, $K$ — pendulum torque coefficient, $φ$ — latitude.

Theory

Angular momentum and the gyroscope

The angular momentum $\vec{L}$ of a spinning body is a vector; if a symmetric body spins about its own axis, it points along the axis and $L = I_{3} ω_{3}$ . Torque $\vec{τ}$ changes it: $d \vec{L} = \vec{τ} d t$ . Because the torque of gravity is perpendicular to $\vec{L}$ , the magnitude of $L$ does not change, only its direction turns — the axis does not fall, it rotates horizontally. This is precession. With no external torque, $\vec{L}$ is conserved: the axis of a free gyroscope keeps its direction in space — this is the basis of the orientation instruments of aircraft and spacecraft. (The relation $L = I ω$ itself is in the “Dynamics of a rigid body” experiment.)

Precession and nutation (the Lagrange top)

For a heavy symmetric top with a fixed tip, at fast spin $Ω \approx m g r / (I_{3} ω_{3})$ . The exact steady precession is the smaller root of the equation $I_{1} \cos θ Ω^{2} - I_{3} ω_{3} Ω + m g r = 0$ ( $I_{1}$ — moment of inertia about a transverse axis through the tip). If the top is released without precession, the axis “dips” slightly and oscillates — this is nutation: the trace of the axis tip draws a line with cusps; if the top is given a push backwards — loops, forwards — waves. An upright (“sleeping”) top is stable only if $I_{3}^{2} ω_{3}^{2} > 4 I_{1} m g r$ — in our model ≈ 12 rev/s. Conserved quantities: energy, $L_{y}$ (the vertical component) and $L_{3}$ .

Euler angles and Euler's equations

The orientation of a rigid body is given by three Euler angles: $φ$ — precession (from the fixed $x$ axis to the line of nodes), $θ$ — nutation (between the vertical and the symmetry axis), $ψ$ — proper rotation (from the line of nodes to the white dot on the body). The motion is described by Euler's equations; they are written in principal axes that rotate with the body. In the simulation, $\vec{L}$ in the body axes and the orientation quaternion are integrated by the RK4 method, and the quaternion is normalised at every step. Check: for the top, the relative error of $E$ and $L_{y}$ is less than $10^{- 7}$ .

The inertia tensor and a free body: Euler's case, the middle axis

The “resistance” of a rigid body to rotation is not a single number but the inertia tensor $I_{i j} = \sum m (r^{2} δ_{i j} - x_{i} x_{j})$ — a symmetric 3×3 table. It connects angular velocity with angular momentum: $L_{i} = I_{i j} ω_{j}$ , and the kinetic energy is $T = \frac{1}{2} I_{i j} ω_{i} ω_{j}$ ; in general $\vec{L}$ and $\vec{ω}$ are not parallel. Every body has three mutually perpendicular principal axes of inertia: along them the tensor is diagonal ( $I_{1}, I_{2}, I_{3}$ — the values in the “Free rotation” section), $\vec{L} = (I_{1} ω_{1}, I_{2} ω_{2}, I_{3} ω_{3})$ and $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2})$ . An axis of symmetry, or an axis perpendicular to a plane of symmetry, is always a principal axis: in the T-handle these are the two axes through the centre of mass parallel to the handles, and the normal to the plane of the handles. With no torque, $\vec{L}$ is fixed in space and the energy is conserved too. In the body axes, however, the tip of $\vec{L}$ moves along the polhode — the line where the sphere ( $| L |$ = const) meets the energy ellipsoid (the small sphere in the scene). Around the axes of smallest and largest $I$ these lines are small loops — the rotation is stable. The separatrix passes through the middle axis: a small deviation grows quickly and the body flips over from time to time (the Dzhanibekov effect, the “tennis racket theorem”). For a symmetric disc ( $I_{1} = I_{2}$ ) the symmetry axis rotates about $\vec{L}$ at a rate $L / I_{1}$ .

The gyrocompass and gyroscopic forces

The Earth rotates at $Ω_{\oplus︎} = 7.29 \cdot 10^{- 5}$ rad/s; at latitude $φ$ its horizontal component $Ω_{\oplus︎} \cos φ$ points north. If the axis of the gyrocompass deviates from the meridian, the Earth's rotation tilts it, while the weight of the pendulum makes the axis precess towards the meridian — the axis oscillates about the meridian and, thanks to damping, settles on true north. If the period is tuned to $2 π \sqrt{R / g} \approx 84.4$ min (the Schuler period), the ship's accelerations do not disturb the reading. Near the pole $\cos φ \to 0$ — the compass stops working. To turn the axis of a spinning rotor at rate $Ω$ , the bearings must supply a torque $M = I ω Ω$ — these are gyroscopic forces (turbines, propellers). The gyroscopic effect of a bicycle wheel helps with balance but is not essential: in 2011 Kooijman and colleagues showed that even a bicycle with extra counter-rotating wheels (total $L \approx 0$ ) keeps itself upright.

Lab work and model simplifications

Lab with the wheel: choose $r$ and the frequency, measure the precession period over $N$ full revolutions as $T_{p} = t / N$ , calculate $Ω = 2 π / T_{p}$ and press “Add to table”. Repeat at several frequencies: the graph of $Ω (1 / ω)$ is a straight line through the origin with slope $m g r / I_{3}$ — from it $I_{3}$ is found. Simplifications: the rope is an ideal fixed support, there is no friction or air resistance (the spin does not decay), and the tip of the top does not slip; in the gyrocompass the ship is stationary, the deviation is small and time is sped up 600 times; the T-handle is two solid steel cylinders.

Did you know?

In 1985, during a repair expedition to the Salyut-7 station, cosmonaut Vladimir Dzhanibekov noticed that a wing nut that had been unscrewed flipped through 180° once every few seconds. The phenomenon bears his name, although it follows from Euler's equations for a rigid body.
The Earth is also a giant top: because of the torque exerted by the Sun and the Moon on the Earth's equatorial bulge, the Earth's axis precesses about once every 26 000 years. That is why the Pole Star will not remain the “pole” star forever.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Non-inertial frames of reference — Grade 10
Inclined plane and Atwood machine — Grade 10
Free fall and air — Grade 10
Jet propulsion and the rocket — Grade 10
Viscosity and flow — Grade 10
Lagrangian mechanics and chaos — Year 2
Hamiltonian mechanics — Year 2