7.2 KiB
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.