Lesson 6 · 35 min
Angular Momentum in Three Dimensions
In the plane, angular momentum and angular velocity always point the same way, out of the page. In three dimensions they part company: a wheel spinning on an axle that is itself turning has an angular velocity in one direction and an angular momentum in quite another. The inertia tensor is what connects them.
Learning objectives
- Compute the angular momentum of a rigid body in 3D as \(\Hvec_G = \Imat_G\wvec\) or, about a fixed point, \(\Hvec_O = \Imat_O\wvec\).
- Use the principal-axis form \(\Hvec = I_x\omega_x\ihat + I_y\omega_y\jhat + I_z\omega_z\khat\).
- Add angular velocities (spin plus precession) and find the resulting \(\Hvec\).
- Compute the kinetic energy and apply the impulse–momentum principles in 3D.
From a sum to a matrix
For a rigid body turning at \(\wvec\) about its center of mass, each particle moves relative to \(G\) at \(\wvec \times \boldsymbol\rho_i\), so
\[ \Hvec_G = \sum \boldsymbol\rho_i \times m_i(\wvec \times \boldsymbol\rho_i) = \int \boldsymbol\rho \times (\wvec \times \boldsymbol\rho)\,dm \]Expanding the triple product component by component (as in Lesson 6 of the Mass Moments of Inertia module) gives a matrix times \(\wvec\):
Angular momentum of a rigid body
\[ \Hvec_G = \Imat_G\,\wvec, \qquad \begin{Bmatrix} H_x \\ H_y \\ H_z \end{Bmatrix} = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{yy} & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{zz} \end{bmatrix} \begin{Bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{Bmatrix} \]About a fixed point \(O\) of the body: \(\Hvec_O = \Imat_O\,\wvec\). In general: \(\Hvec_P = \rvec_{G/P} \times m\vvec_G + \Hvec_G\), exactly as in Lesson 3.
Principal axes
If \(x, y, z\) are principal axes of the body at the point, every product of inertia vanishes and
\[ \Hvec = I_x\omega_x\,\ihat + I_y\omega_y\,\jhat + I_z\omega_z\,\khat \]Each component of \(\Hvec\) is simply the matching component of \(\wvec\) times its principal moment. Unless the three moments happen to be equal, the unequal scaling tilts \(\Hvec\) away from \(\wvec\): they are parallel only when \(\wvec\) lies along a principal axis.
Example 6.1 — A disk on a turning axle
For the disk of Figure 6.1, \(\omega_s = 30\ \text{rad/s}\) and \(\omega_p = 2\ \text{rad/s}\). At the instant the axle lies along \(x\), find \(\Hvec_G\) and \(\Hvec_O\).
Show solution
\(\wvec = 30\,\ihat + 2\,\khat\ \text{rad/s}\). The disk's axis \(x\) and every diameter are principal axes at \(G\): \(I_x = \tfrac12 mr^2 = 0.08\), \(I_y = I_z = \tfrac14 mr^2 = 0.04\ \text{kg·m}^2\).
\[ \Hvec_G = 0.08(30)\,\ihat + 0.04(2)\,\khat = 2.40\,\ihat + 0.08\,\khat\ \text{kg·m}^2/\text{s} \]About the pivot, the transverse moments grow by \(ma^2 = 4(0.5)^2 = 1\ \text{kg·m}^2\) (parallel-axis theorem), and \(x, y, z\) stay principal at \(O\):
\[ \Hvec_O = 0.08(30)\,\ihat + 1.04(2)\,\khat = 2.40\,\ihat + 2.08\,\khat\ \text{kg·m}^2/\text{s} \]\(\wvec\) is \(3.8^\circ\) above the axle; \(\Hvec_G\) is only \(1.9^\circ\) above it, while \(\Hvec_O\) is \(40.9^\circ\) above it. Check \(\Hvec_O = \rvec_G \times m\vvec_G + \Hvec_G\): \(\vvec_G = 2\khat \times 0.5\ihat = 1\jhat\), and \(0.5\ihat \times 4(1\jhat) = 2\khat\). ✓
Kinetic energy
Kinetic energy of a rigid body
\[ T = \tfrac12 m v_G^2 + \tfrac12\wvec^\mathsf{T}\Imat_G\wvec \qquad \text{or, about a fixed point,} \qquad T = \tfrac12\wvec^\mathsf{T}\Imat_O\wvec = \tfrac12\wvec\cdot\Hvec_O \]In principal axes: \(\tfrac12\wvec^\mathsf{T}\Imat\wvec = \tfrac12(I_x\omega_x^2 + I_y\omega_y^2 + I_z\omega_z^2)\).
Example 6.2 — Energy of the disk
Find the kinetic energy of the disk of Example 6.1.
Show solution
About the fixed pivot: \(T = \tfrac12\left[0.08(30)^2 + 1.04(2)^2\right] = \tfrac12(72 + 4.16) = 38.08\ \text{J}\).
Or with \(G\): \(T = \tfrac12(4)(1)^2 + \tfrac12\left[0.08(30)^2 + 0.04(2)^2\right] = 2 + 36.08 = 38.08\ \text{J}\).
Impulse and momentum in 3D
Lesson 5's principles carry over with vectors: \(m\vvec_{G1} + \sum\int\Fvec\,dt = m\vvec_{G2}\) and \(\Hvec_{G1} + \sum\int\Mvec_G\,dt = \Hvec_{G2}\) (or about a fixed point \(O\)). The angular equation is a vector equation, three scalar equations in all.
Example 6.3 — Firing a spacecraft's thrusters
A spacecraft with principal moments \(I_x = 200\), \(I_y = 300\), \(I_z = 400\ \text{kg·m}^2\) about its center of mass is at rest. Its thrusters deliver an angular impulse of \(10\,\ihat + 20\,\khat\ \text{N·m·s}\) about \(G\). Find its angular velocity afterward.
Show solution
\(\Hvec_{G2} = \mathbf{0} + (10,\ 0,\ 20)\ \text{kg·m}^2/\text{s}\). In principal axes each component divides by its own moment:
\[ \wvec = \left(\frac{10}{200},\ 0,\ \frac{20}{400}\right) = (0.05,\ 0,\ 0.05)\ \text{rad/s} \]\(\Hvec\) and \(\wvec\) point in different directions (\(\Hvec\) at \(63.4^\circ\) from \(x\), \(\wvec\) at \(45^\circ\)), because \(I_x \ne I_z\).
Check your understanding
Key takeaways
- \(\Hvec_G = \Imat_G\wvec\); about a fixed point \(\Hvec_O = \Imat_O\wvec\); in general \(\Hvec_P = \rvec_{G/P} \times m\vvec_G + \Hvec_G\).
- In principal axes, \(\Hvec = (I_x\omega_x,\ I_y\omega_y,\ I_z\omega_z)\): parallel to \(\wvec\) only along a principal axis.
- Angular velocities add as vectors: spin plus precession.
- \(T = \tfrac12 m v_G^2 + \tfrac12\wvec^\mathsf{T}\Imat_G\wvec\), and \(\Hvec_{G1} + \sum\int\Mvec_G\,dt = \Hvec_{G2}\).
- Next: Lesson 7 differentiates \(\Hvec\) in a rotating frame to get Euler's equations.