Angular Momentum

5.1 Angular Momentum for a Point Particle

Define the angular momentum \(\vec{L}_S\) about the point \(S\) as the cross product of the vector from \(S\) to the point particle, \(\vec{r}_S\), and the momentum of the particle, \(\vec{p}\), \[ \begin{aligned} \vec{L}_S = \vec{r}_S \times \vec{p}. \end{aligned} \]

Now, let us check out the relation between angular momentum and angular velocity. For simplicity, choose the fixed point to be the origin and the plane spanned by \(\vec{r}\) and \(\vec{v}\) for the polar coordinate system,

\[ \begin{aligned} \vec{L} &= \vec{r} \times \vec{p} \\ &= m \vec{r} \times \vec{v} \\ &= m r^2 \dot{\theta} \hat{r} \times \hat{\theta} \\ &= m r^2 \dot{\theta} \hat{k} \\ &= m r^2 \vec{\omega}, \end{aligned} \]

where we use the fact that in the polar coordinate system,

\[ \begin{aligned} \vec{v} &= \vec{v}_r + \vec{v}_\theta \\ &= \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}. \end{aligned} \]

5.2 Conservation of Angular Momentum

\[ \begin{aligned} \frac{\mathrm{d}\vec{L}_S}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(\vec{r}_S \times \vec{p}) = \left( \frac{\mathrm{d}\vec{r}_S}{\mathrm{d}t} \times \vec{p} \right) + \left( \vec{r}_S \times \frac{\mathrm{d}\vec{p}}{\mathrm{d}t} \right). \end{aligned} \]

\[ \begin{aligned} \frac{\mathrm{d}\vec{r}_S}{\mathrm{d}t} \times \vec{p} = \vec{v} \times m\vec{v} = 0. \end{aligned} \]

The rate of angular momentum change about the point \(S\) is then

\[ \begin{aligned} \frac{\mathrm{d}\vec{L}_S}{\mathrm{d}t} = \vec{r}_S \times \frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = \vec{r}_S \times \vec{F} = \vec{M}_S. \end{aligned} \]

5.2.1 Theorem of Angular Momentum for a Particle

The change in angular momentum is caused by the torque,

\[ \begin{aligned} \vec{J}_S = \int_{t_1}^{t_2} \vec{M}_S \mathrm{d}t = \vec{L}_{S,t_2} - \vec{L}_{S,t_1}, \end{aligned} \]

where \(\vec{J}_S\) is defined as the angular impulse about a point \(S\).

Theorem 5.1 (Conservation of Angular Momentum about a Point) If the torque about point \(S\) is zero, \[ \begin{aligned} \vec{M}_S = \frac{\mathrm{d}\vec{L}_S}{\mathrm{d}t} = 0, \end{aligned} \]

then \(\vec{L}_S = \text{const}\).

5.3 System of Particles

5.3.1 Theorem of Angular Momentum

The angular momentum \(\vec{L}_{S,j}\) of the \(j^{th}\) particle is

\[ \begin{aligned} \vec{L}_{S,j} = \vec{r}_{S,j} \times \vec{p}_j. \end{aligned} \]

\[ \begin{aligned} \vec{L}_S^{sys} = \sum_{j=1}^{N} \vec{L}_{S,j} = \sum_{j=1}^{N} \vec{r}_{S,j} \times \vec{p}_j. \end{aligned} \]

\[ \begin{aligned} \frac{\mathrm{d}\vec{L}_S^{sys}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \sum_{j=1}^{N} \vec{L}_{S,j} = \sum_{j=1}^{N} \left( \frac{\mathrm{d}\vec{r}_{S,j}}{\mathrm{d}t} \times \vec{p}_j + \vec{r}_{S,j} \times \frac{\mathrm{d}\vec{p}_j}{\mathrm{d}t} \right). \end{aligned} \]

we have

\[ \begin{aligned} \frac{\mathrm{d}\vec{L}_S^{sys}}{\mathrm{d}t} &= \sum_{j=1}^{N} \left( \vec{r}_{S,j} \times \frac{\mathrm{d}\vec{p}_j}{\mathrm{d}t} \right) \\ &= \sum_{j=1}^{N} (\vec{r}_{S,j} \times \vec{F}_j) \\ &= \sum_{j=1}^{N} (\vec{r}_{S,j} \times (\vec{F}_j^{ext} + \vec{F}_j^{int})) \\ &= \vec{M}_S^{ext} + \vec{M}_S^{int} = \vec{M}_S^{ext}. \end{aligned} \]

Theorem 5.2 Angular Momentum of a System of Particles about Different Points A and B satisfies \[ \begin{aligned} \vec{L}_A = \vec{L}_B + \vec{r}_{A,B} \times \vec{p}_{sys}. \end{aligned} \]

Theorem 5.3 (Conservation Law of Angular Momentum) For a system of particles, when the net external torque equals zero, the angular momentum of the system becomes constant, \[ \begin{aligned} \vec{M}_S^{ext} = 0 \Rightarrow \vec{L}_S^{sys} = \text{const}. \end{aligned} \]

\[ \begin{aligned} \vec{M}' = \frac{\mathrm{d}\vec{L}'}{\mathrm{d}t} = \sum_{i} \vec{r}'_i \times \vec{f}_i \end{aligned} \]

\[ \begin{aligned} \vec{L}' = \sum_{i} m_i \vec{r}'_i \times \vec{v}'_i. \end{aligned} \]

\[ \begin{aligned} \vec{M}'_{inertial} = \sum_{i} \vec{r}'_i \times (-m_i\vec{a}_c) = \sum_{i} -m_i\vec{r}'_i \times \vec{a}_c = 0 \times \vec{a}_c = 0. \end{aligned} \]

Here \((-m_i\vec{a}_c)\) is the inertial force. The inertial torque equals 0 since the overall effect of inertial force acts on the center of mass.

The angular momentum of the system in an inertial reference frame is: \[ \begin{aligned} \vec{L} = \sum_{i} m_i \vec{r}_i \times \vec{v}_i = \sum_{i} m_i (\vec{r}'_i + \vec{r}_c) \times (\vec{v}'_i + \vec{v}_c) = \sum_{i} m_i \vec{r}'_i \times \vec{v}'_i + \sum_{i} m_i \vec{r}_c \times \vec{v}_c = \vec{L}' + \vec{r}_c \times \vec{p}_c, \end{aligned} \]

since \(\sum_i m_i \vec{r}'_i = 0\) and \(\sum_i m_i \vec{v}'_i = 0\).