Lesson 3 · 30 min
Systems of Particles
A rigid body is a system of particles held together by internal forces. Before we treat bodies, we add up the angular momenta of many particles and find two facts that carry all the way to 3D: internal forces never change the total, and the total splits neatly into the motion of the center of mass plus the motion about it.
Learning objectives
- Show that internal forces drop out, so \(\sum\Mvec_O = \dot{\Hvec}_O\) holds for a system about a fixed point.
- Use the moment equation about the center of mass, \(\sum\Mvec_G = \dot{\Hvec}_G\), even when \(G\) accelerates.
- Split the angular momentum as \(\Hvec_O = \rvec_G \times m\vvec_G + \Hvec_G\).
- Apply conservation of angular momentum to a system whose parts move relative to each other.
Internal forces cancel
For a system of \(n\) particles, the angular momentum about a fixed point \(O\) is the sum \(\Hvec_O = \sum \rvec_i \times m_i\vvec_i\). Each particle obeys Lesson 2's moment equation, with both external and internal forces:
\[ \rvec_i \times \Fvec_i + \rvec_i \times \sum_{j} \mathbf{f}_{ij} = \dot{\Hvec}_{O,i} \]Internal forces come in pairs: \(\mathbf{f}_{ij} = -\mathbf{f}_{ji}\), acting along the line joining the two particles (Newton's third law). Their two moments about any point are equal and opposite, so every pair cancels when we sum over the particles.
Moment equation for a system (\(O\) fixed)
\[ \sum\Mvec_O = \dot{\Hvec}_O, \qquad \Hvec_O = \sum_i \rvec_i \times m_i\vvec_i \]Only external forces contribute to \(\sum\Mvec_O\).
Angular momentum about the center of mass
Write each position as \(\rvec_i = \rvec_G + \boldsymbol{\rho}_i\), where \(\boldsymbol{\rho}_i\) is measured from the center of mass \(G\), so that \(\sum m_i\boldsymbol{\rho}_i = \mathbf{0}\). Substituting and using that sum makes the cross terms vanish:
Splitting the angular momentum
\[ \Hvec_O = \rvec_G \times m\vvec_G + \Hvec_G, \qquad \Hvec_G = \sum_i \boldsymbol{\rho}_i \times m_i\dot{\boldsymbol{\rho}}_i \] \[ \sum\Mvec_G = \dot{\Hvec}_G \quad \text{(valid even when } G \text{ accelerates)} \]The first term is the angular momentum of the whole mass concentrated at \(G\) ("orbital"); \(\Hvec_G\) is the angular momentum of the motion relative to \(G\) ("spin").
Why the moment equation also works about \(G\)
Differentiate the split: \(\dot{\Hvec}_O = \vvec_G \times m\vvec_G + \rvec_G \times m\mathbf{a}_G + \dot{\Hvec}_G = \rvec_G \times \sum\Fvec + \dot{\Hvec}_G\). Also \(\sum\Mvec_O = \sum(\rvec_G + \boldsymbol{\rho}_i) \times \Fvec_i = \rvec_G \times \sum\Fvec + \sum\Mvec_G\). Setting \(\sum\Mvec_O = \dot{\Hvec}_O\) leaves \(\sum\Mvec_G = \dot{\Hvec}_G\).
For an arbitrary moving point \(P\) the moment equation picks up an extra term, \(\sum\Mvec_P = \dot{\Hvec}_P + \vvec_P \times m\vvec_G\), where \(\Hvec_P\) is measured with velocities relative to \(P\). Use a fixed point or \(G\) and the extra term never appears.
Example 3.1 — Checking the split
Particle \(A\) (\(1\ \text{kg}\)) is at \((0,\ 0.5)\ \text{m}\) moving at \(\vvec_A = 2\,\ihat\ \text{m/s}\); particle \(B\) (\(3\ \text{kg}\)) is at \((1,\ 0)\ \text{m}\) moving at \(\vvec_B = -1\,\jhat\ \text{m/s}\). Find \(H_O\) directly, then check it with \(\rvec_G \times m\vvec_G + H_G\).
Show solution
Directly, with \(H = m(x v_y - y v_x)\): \(H_O = 1(0 - 0.5 \cdot 2) + 3(1 \cdot (-1) - 0) = -1 - 3 = -4\ \text{kg·m}^2/\text{s}\).
Center of mass. \(\rvec_G = \tfrac14[(0, 0.5) + 3(1, 0)] = (0.75,\ 0.125)\ \text{m}\), \(\vvec_G = \tfrac14[(2, 0) + 3(0, -1)] = (0.5,\ -0.75)\ \text{m/s}\).
\[ \rvec_G \times m\vvec_G = 4\left[0.75(-0.75) - 0.125(0.5)\right] = -2.5\ \text{kg·m}^2/\text{s} \]About \(G\). Relative positions and velocities: \(\boldsymbol\rho_A = (-0.75,\ 0.375)\), \(\dot{\boldsymbol\rho}_A = (1.5,\ 0.75)\); \(\boldsymbol\rho_B = (0.25,\ -0.125)\), \(\dot{\boldsymbol\rho}_B = (-0.5,\ -0.25)\).
\[ H_G = 1\left[(-0.75)(0.75) - 0.375(1.5)\right] + 3\left[0.25(-0.25) - (-0.125)(-0.5)\right] = -1.125 - 0.375 = -1.5 \]\(-2.5 + (-1.5) = -4\ \text{kg·m}^2/\text{s}\), as found directly.
Conservation for a system
If the external forces have no moment about \(O\) (or about \(G\)), the system's angular momentum about that point is constant, however the particles move relative to each other. Internal forces can redistribute angular momentum between the parts, but cannot create or destroy it.
Example 3.2 — Skaters on a rope
Two skaters, \(60\ \text{kg}\) and \(90\ \text{kg}\), hold the ends of a light \(6\ \text{m}\) rope and spin about their center of mass at \(0.5\ \text{rad/s}\) on frictionless ice. They pull themselves along the rope until they are \(2\ \text{m}\) apart. Find their new angular velocity and the work they do.
Show solution
The ice exerts no horizontal force, so \(H_G\) about the vertical axis is conserved. \(G\) divides the rope in the ratio \(90 : 60\), so the skaters are \(3.6\) and \(2.4\ \text{m}\) from \(G\):
\[ H_G = \left[60(3.6)^2 + 90(2.4)^2\right]\omega_1 = 1296(0.5) = 648\ \text{kg·m}^2/\text{s} \]At \(2\ \text{m}\) apart they are \(1.2\) and \(0.8\ \text{m}\) from \(G\): \(60(1.2)^2 + 90(0.8)^2 = 144\ \text{kg·m}^2\).
\[ \omega_2 = \frac{648}{144} = 4.50\ \text{rad/s}, \qquad T_1 = \tfrac12(1296)(0.5)^2 = 162\ \text{J}, \quad T_2 = \tfrac12(144)(4.5)^2 = 1458\ \text{J} \]The skaters' arms do \(1296\ \text{J}\) of work through the rope. The rope tension is an internal force of the two-skater system: it changes the kinetic energy, but not \(H_G\).
Check your understanding
Key takeaways
- Internal forces cancel: \(\sum\Mvec_O = \dot{\Hvec}_O\) uses external forces only.
- \(\Hvec_O = \rvec_G \times m\vvec_G + \Hvec_G\): orbital part plus spin part.
- \(\sum\Mvec_G = \dot{\Hvec}_G\) holds about the center of mass, even when it accelerates; take moments about a fixed point or about \(G\).
- With no external moment, \(H\) is conserved while internal forces redistribute it and change the kinetic energy.
- Next: Lesson 4 turns the sum \(\Hvec_G\) into \(I_G\omega\) for a rigid body.