Chapter 3 - Momentum and Angular Momentum

Section 3.1 - Conservation of Momentum

Consider a system of \(N\) particles, indexed with \(\alpha \in {1, 2, \ldots, N}\). We found that as long as Newton's third law applies, \(\mathbf{P} = \mathbf{p}_1 + \ldots + \mathbf{p}_N = \sum \mathbf{p}_\alpha\) is determined entirely by external forces. That is, \(\dot{\mathbf{P}} = F^{ext}\).

Section 3.2 - Rockets

Consider a rocket with mass \(m\) traveling in the \(\hat{\mathbf{x}}\) direction and exhausting fuel at a rate of \(dm\) at speed \(v_{ex}\). At time \(t\), the rocket's momentum is \(P(t) = mv\). Then, \(P(t + dt) = (m + dm)(v + dv)\), where the fuel ejected in time \(dt\) has mass \(-dm\) and velocity \(v - v_{ex}\) relative to the ground. Thus, the total momentum becomes \(P(t + dt) = (m + dm)(v + dv) - dm(v - v_{ex}) = mv + m dv + dm v_{ex}\), as \(dm dv\) becomes infinitesimally small.

Then, \(dP = P(t + dt) - P(t) = m dv + dm v_{ex}\). If there is a net external force, this would equal \(F^{ext}dt\). In this case, we assume \(F^{ext} = 0\). Then, \(m dv = -dm v_ex\), or \(m \dot{v} = -\dot{m} v_ex\). This is Newton's second law, where \(F = m \dot{v} = -\dot{m} v_ex\).

Definition. Conventionally, we say thrust is \(-\dot{m} v_{ex}\).

We can solve this via separation of variables. If \(v_{ex}\) is constant, we see \(v - v_0 = v_{ex} \ln\frac{m_0}{m}\).

Section 3.3 - The Center of Mass

Definition. Given a system of \(\alpha \in {1, \ldots, N}\) particles, we define the center of mass of the system with respect to the origin O is

\[\mathbf{R} = \frac{1}{M} \sum_{\alpha = 1}^{N} m_\alpha \mathbf{r}_\alpha\]

This is effectively a weighted average of the positions with masses as weights. Using this definition, we can rewrite the total momentum as

\[\mathbf{P} = \sum_\alpha \mathbf{p}_\alpha = \sum_\alpha m_\alpha \dot{\mathbf{r}}_\alpha = M \dot{\mathbf{R}}\]

Notably, this says that the total momentum of a mass of particles can be replaced by one particle with the same momentum of the center of mass.

We know that \(\dot{\mathbf{P}} = \mathbf{F}^{ext}\), so we can then say that \(\mathbf{F}^{ext} = M \ddot{\mathbf{R}}\). That is, the center of mass moves as if it was a single particle with mass \(M\) subject to the net external force. This is why we can approximate bodies as single point masses.

Note that when mass is distributed evenly, we see that if

\[\mathbf{R} = \frac{1}{M} \int \mathbf{r} dm = \frac{1}{m} \int \rho \mathbf{r} dV\]

Section 3.4 - Angular Momentum for a Single Particle

Definition. Angular momentum for a single particle is defined as \(\mathbf{\ell} = \mathbf{r} \times \mathbf{p}\). Note that as \(\mathbf{r}\) is origin-dependent, so is \(\mathbf{\ell}\). With a vector identity, we see that

\[\dot{\mathbf{\ell}} = \frac{d}{dt}(\mathbf{r} \times \mathbf{p}) = (\dot{\mathbf{r}} \times \mathbf{p}) + (\mathbf{r} \times \dot{\mathbf{p}})\]

Notably, we can replace \(\mathbf{p} = m \dot{\mathbf{r}}\), so we say that \(\dot{\mathbf{r}} \times \mathbf{p} = \dot{\mathbf{r}} \times m \dot{\mathbf{r}} = 0\), so

\[\dot{\mathbf{\ell}} = \mathbf{r} \times \dot{\mathbf{p}} = \mathbf{r} \times m \ddot{\mathbf{r}} = \mathbf{r} \times \mathbf{F} = \mathbf{\Gamma}\]

Here, \(\mathbf{\Gamma}\) represents the net torque about the origin \(O\) on the particle, defined as \(\mathbf{r} \times \mathbf{F}\). This is often described as the rotational form of Newton's second law.

Note that in many one-particle systems, the origin \(O\) is deliberately chosen such that \(\mathbf{\Gamma} = 0\), so that the angular momentum is constant.

Definition. For a central force between objects, \(\mathbf{F}\) is parallel to the vector between the two objects.

We will now attempt to prove Kepler's Second Law. Consider a planet at position \(\mathbf{r}\) orbiting a star at the origin \(O\). Then, we define \(d\mathbf{r} = \mathbf{v} dt\) as the planet moves from position \(P\) to position \(Q\) in its orbit. The area swept out by the planet when orbiting thus becomes \(dA = \frac{1}{2} |\mathbf{r} \times d\mathbf{r} = \frac{1}{2} |\mathbf{r} \times \mathbf{v} dt|\). As \(\mathbf{p} = m\mathbf{v}\), we can say that \(\mathbf{v} = \frac{\mathbf{p}}{v}\), so that \(dA = \frac{1}{2m} |\mathbf{r} \times \mathbf{p}| = \frac{\ell dt}{2m}\), and we can say that \(\frac{dA}{dt} = \frac{\ell}{2m}\). This is constant.

An alternative proof shows that \(\ell = mr^2 \omega\) where \(\omega = \dot{\phi}\), so that \(\frac{dA}{dt} = \frac{1}{2} r^2 \omega\).

Section 3.5 - Angular Momentum for Several Particles

Next, define a system with \(\alpha\) particles, where each particle has angular momentum \(\mathbf{\ell}_\alpha = \mathbf{r}_\alpha \times \mathbf{p}_\alpha\), with common origin \(O\). We can then define total angular momentum as

\[\mathbf{L} = \sum_{\alpha=1}^N \mathbf{\ell}_\alpha = \sum_{\alpha = 1}^N \mathbf{r}_\alpha \times \mathbf{p}_\alpha\]

We can then differentiate to see that

\[\dot{\mathbf{L}} = \sum_{\alpha=1}^N \dot{\mathbf{\ell}}_\alpha = \sum_{\alpha = 1}^N \mathbf{r}_\alpha \times \mathbf{F}_\alpha\]

Then, we see the rate of change of angular momentum is simply net torque on the system.

We can then state that \(\mathbf{F}_alpha = \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} + \mathbf{F}_\alpha^{ext}\). Then,

\[\dot{\mathbf{L}} = \sum_\alpha \sum_{\beta \neq \alpha} \mathbf{r}_\alpha \times \mathbf{F}_{\alpha \beta} + \sum_\alpha \mathbf{r}_\alpha \times \mathbf{F}_\alpha^{ext}\]

Now, we cam convert the force between particles in the double sum by pairing forces.

\[\sum_\alpha \sum_{\beta \neq \alpha} \mathbf{r}_\alpha \times \mathbf{F}_{\alpha \beta} = \sum_\alpha \sum_{\beta > \alpha} (\mathbf{r}_\alpha \times \mathbf{F}_{\alpha \beta} + \mathbf{r}_\beta \times \mathbf{F}_{\beta \alpha})\]

Then, if we assume all internal forces obey the third law such that \(\mathbf{F}_{\alpha \beta} = -\mathbf{F}_{\beta\alpha}\), we can write that

\[\sum_\alpha \sum_{\beta > \alpha} (\mathbf{r}_\alpha \times \mathbf{F}_{\alpha \beta} + \mathbf{r}_\beta \times \mathbf{F}_{\beta \alpha}) = \sum_\alpha \sum_{\beta > \alpha} (\mathbf{r}_\alpha \times \mathbf{F}_{\alpha \beta} - \mathbf{r}_\beta \times \mathbf{F}_{\alpha\beta}) = \sum_\alpha \sum_{\beta > \alpha} (\mathbf{r}_\alpha - \mathbf{r}_\beta) \times \mathbf{F}_{\alpha\beta}\]

We can denote \(\mathbf{r}_{\alpha}{\beta} = \mathbf{r}_\alpha - \mathbf{r}_\beta\) as the vector pointing from \(\beta\) to \(\alpha\). Thus,

\[\sum_\alpha \sum_{\beta \neq \alpha} \mathbf{r}_\alpha \times \mathbf{F}_{\alpha \beta} = \sum_\alpha \sum_{\beta > \alpha} (\mathbf{r}_\alpha - \mathbf{r}_\beta) \times \mathbf{F}_{\alpha\beta}\]

If the internal forces are all central, the cross product becomes zero. Then, the rate of change of angular momentum becomes

\[\dot{\mathbf{L}} = \sum_\alpha \mathbf{r}_\alpha \times \mathbf{F}_\alpha^{ext} = \mathbf{\Gamma}^{ext}\]

where \(\mathbf{\Gamma}^{ext}\) is the net external torque. Notably, if this value is \(0\), the total angular momentum is a constant.

Definition. About any axis of rotation \(\hat{\mathbf{k}}\), if an object is rotating at angular velocity \(\omega\), then \(L_k\), the angular momentum in the \(\hat{\mathbf{k}}\)-direction, can be written as \(L_k = O \omega\), where \(I\) is the moment of inertia of the object. For any multi-particle system, \(I = \sum m_\alpha \rho_\alpha^2\), where \(\rho_\alpha\) is the distance from particle \(\alpha\) to the axis of rotation.