Lesson 7 · 35 min

Euler's Equations of Motion

\(\sum\Mvec = \dot{\Hvec}\) is easy to write and awkward to use: the inertia tensor is constant only in axes that turn with the body, but \(\dot{\Hvec}\) must be taken in fixed axes. Euler's trick is to differentiate in the rotating axes and add a correction for their rotation. The result explains gyroscopic moments, wobbling rotors and why a tossed book flips.

Learning objectives

Differentiate a vector expressed in rotating axes: \(\dot{\Hvec} = (\dot{\Hvec})_{xyz} + \Wvec \times \Hvec\).
Write and apply Euler's equations in principal body axes.
Use the modified form with \(\Wvec \ne \wvec\) for axisymmetric bodies, and compute gyroscopic moments.
Explain the stability of free rotation about the maximum and minimum principal axes, and the instability about the intermediate axis.

Rates of change in rotating axes

Let axes \(x, y, z\) rotate with angular velocity \(\Wvec\). A vector written in them, \(\mathbf{A} = A_x\ihat + A_y\jhat + A_z\khat\), changes for two reasons: its components change, and the unit vectors turn (\(\dot{\ihat} = \Wvec \times \ihat\), and so on). So

Derivative in a rotating frame

\[ \dot{\mathbf{A}} = (\dot{\mathbf{A}})_{xyz} + \Wvec \times \mathbf{A}, \qquad (\dot{\mathbf{A}})_{xyz} = \dot A_x\ihat + \dot A_y\jhat + \dot A_z\khat \]

Applied to angular momentum about \(G\) (or a fixed point \(O\)): \(\sum\Mvec = (\dot{\Hvec})_{xyz} + \Wvec \times \Hvec\).

The natural choice is axes fixed in the body, \(\Wvec = \wvec\): then the inertia tensor in those axes never changes. Choosing them along the principal axes also removes the products of inertia.

Euler's equations

With \(\Wvec = \wvec\) and principal body axes, \(\Hvec = (I_x\omega_x,\ I_y\omega_y,\ I_z\omega_z)\) and \((\dot{\Hvec})_{xyz} = (I_x\dot\omega_x,\ I_y\dot\omega_y,\ I_z\dot\omega_z)\). Expanding \(\wvec \times \Hvec\):

Euler's equations (principal axes fixed in the body)

\[ \begin{aligned} \sum M_x &= I_x\dot\omega_x - (I_y - I_z)\,\omega_y\omega_z \\ \sum M_y &= I_y\dot\omega_y - (I_z - I_x)\,\omega_z\omega_x \\ \sum M_z &= I_z\dot\omega_z - (I_x - I_y)\,\omega_x\omega_y \end{aligned} \]

Moments about \(G\), or about a fixed point \(O\) with principal moments about \(O\). The components are along the body axes at the instant considered.

For plane motion (\(\omega_x = \omega_y = 0\)) the third equation reduces to \(\sum M_z = I_z\dot\omega_z\): the familiar \(\sum M_G = I_G\alpha\). The extra product terms are what 3D adds.

Example 7.1 — The moments behind a tumbling motion

A body has principal moments \(I_x = 10\), \(I_y = 15\), \(I_z = 20\ \text{kg·m}^2\) about \(G\). At an instant \(\wvec = (1,\ 2,\ 3)\ \text{rad/s}\) and \(\dot{\wvec} = (0.5,\ 0,\ -0.2)\ \text{rad/s}^2\) in body axes. Find the moment about \(G\) that produces this motion.

Show solution

\[ \begin{aligned} M_x &= 10(0.5) - (15 - 20)(2)(3) = 5 + 30 = 35\ \text{N·m} \\ M_y &= 15(0) - (20 - 10)(3)(1) = -30\ \text{N·m} \\ M_z &= 20(-0.2) - (10 - 15)(1)(2) = -4 + 10 = 6\ \text{N·m} \end{aligned} \]

A moment of \(-30\ \text{N·m}\) about \(y\) is needed even though \(\dot\omega_y = 0\): keeping \(\omega_y\) steady while the body turns about \(x\) and \(z\) takes effort.

Axisymmetric bodies: letting the frame lag the body

For a wheel, rotor or gyroscope, any axes through the symmetry axis are principal, whether or not they spin with the body. It is then convenient to let the axes follow the axle but not the spin. The frame turns at \(\Wvec\), the body at \(\wvec = \Wvec + \omega_s\khat\), and

\[ \sum\Mvec = (\dot{\Hvec})_{xyz} + \Wvec \times \Hvec, \qquad \Hvec = I_x\omega_x\ihat + I_y\omega_y\jhat + I_z\omega_z\khat \]

For steady motion (every component constant in the rotating axes) the first term vanishes and \(\sum\Mvec = \Wvec \times \Hvec\).

Example 7.2 — The moment on a turning axle

The disk of Lesson 6 (Example 6.1) spins at \(30\ \text{rad/s}\) on its axle while the axle turns about the vertical at a steady \(2\ \text{rad/s}\). What moment about the pivot \(O\) is needed to keep this motion going?

Show solution

Use axes turning with the axle, \(\Wvec = 2\khat\), with \(x\) along the axle. From Example 6.1, \(\Hvec_O = 2.40\,\ihat + 2.08\,\khat\), constant in these axes.

\[ \sum\Mvec_O = \Wvec \times \Hvec_O = 2\khat \times (2.40\,\ihat + 2.08\,\khat) = 4.80\,\jhat\ \text{N·m} \]

The moment is horizontal and perpendicular to the axle. Only the spin part of \(\Hvec\) contributes: \(M = I_s\omega_s\Omega = 0.08(30)(2)\). This is a gyroscopic moment. Lesson 8 finds the precession rate at which gravity alone supplies it.

Free rotation and the intermediate axis

With no moments, Euler's equations still allow interesting motion. Spin a body about principal axis \(x\) at \(\omega_x \approx n\), with tiny wobbles \(\omega_y\), \(\omega_z\). To first order the second and third equations give

\[ \ddot\omega_y = \frac{(I_z - I_x)(I_x - I_y)}{I_y I_z}\,n^2\,\omega_y \]

If \(x\) is the axis of maximum or minimum moment, the factor is negative: \(\omega_y\) oscillates, and the wobble stays small. If \(x\) is the intermediate axis, the factor is positive: the wobble grows exponentially and the body flips. \(\Hvec\) is conserved throughout (no moments), so it is \(\wvec\) and the body that wander.

Figure 7.1 A book-shaped block (\(24 \times 16 \times 4\ \text{cm}\), \(1\ \text{kg}\)) tossed spinning at \(10\ \text{rad/s}\), simulated by integrating Euler's equations. Choose the spin axis. About the maximum axis (\(z\), normal to the cover) or the minimum axis (\(x\), the long one) the block spins steadily; about the intermediate axis (\(y\)) a 1% disturbance grows until the block flips, again and again. The red \(\colH{\Hvec}\) never moves; the green \(\colW{\wvec}\) does.

Example 7.3 — How fast does the book flip?

For the block of Figure 7.1, \(I_x = 2.267\), \(I_y = 4.933\), \(I_z = 6.933\) (all \(\times 10^{-3}\ \text{kg·m}^2\)). Spinning about \(y\) at \(n = 10\ \text{rad/s}\), how fast does a small wobble grow? And spinning about \(z\), at what rate does a wobble oscillate?

Show solution

About the intermediate axis \(y\) (swap the roles of \(x\) and \(y\) in the equation above):

\[ \lambda^2 = \frac{(I_z - I_y)(I_y - I_x)}{I_x I_z}\,n^2 = \frac{(2.000)(2.667)}{(2.267)(6.933)}(100) = 33.9\ \text{s}^{-2}, \qquad \lambda = 5.83\ \text{s}^{-1} \]

A wobble grows by a factor \(e\) every \(0.17\ \text{s}\): a 1% disturbance becomes the whole spin in under a second.

About the maximum axis \(z\): \(\dfrac{(I_x - I_z)(I_z - I_y)}{I_x I_y}n^2 = -83.5\ \text{s}^{-2}\), a negative factor, so the wobble oscillates at \(\sqrt{83.5} = 9.14\ \text{rad/s}\) and stays small.

Check your understanding

Key takeaways

\(\sum\Mvec = (\dot{\Hvec})_{xyz} + \Wvec \times \Hvec\) in axes rotating at \(\Wvec\).
Euler's equations: \(\sum M_x = I_x\dot\omega_x - (I_y - I_z)\omega_y\omega_z\), and cyclically, in principal body axes.
For axisymmetric bodies let the frame follow the axle; in steady motion \(\sum\Mvec = \Wvec \times \Hvec\), the gyroscopic moment \(I_s\omega_s\Omega\).
Torque-free spin is stable about the maximum and minimum principal axes and unstable about the intermediate one.
Next: Lesson 8 puts this to work on gyroscopes and tops.