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5 Angular Momentum

5.1 Angular Momentum for a Point Particle​

Define the angular momentum L⃗S\vec{L}_S about the point SS as the cross product of the vector from SS to the point particle, r⃗S\vec{r}_S, and the momentum of the particle, p⃗\vec{p},

L⃗S=r⃗S×p⃗.\begin{align*} \vec{L}_S = \vec{r}_S \times \vec{p}. \end{align*}

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 r⃗\vec{r} and v⃗\vec{v} for the polar coordinate system,

L⃗=r⃗×p⃗=mr⃗×v⃗=mr2θ˙r^×θ^=mr2θ˙k^=mr2ω⃗,\begin{align*} \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{align*}

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

v⃗=v⃗r+v⃗θ=r˙r^+rθ˙θ^.\begin{align*} \vec{v} &= \vec{v}_r + \vec{v}_\theta \\ &= \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}. \end{align*}

5.2 Conservation of Angular Momentum​

Torque​

dL⃗Sdt=ddt(r⃗S×p⃗)=(dr⃗Sdt×p⃗)+(r⃗S×dp⃗dt).\begin{align*} \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{align*} dr⃗Sdt×p⃗=v⃗×mv⃗=0.\begin{align*} \frac{\mathrm{d}\vec{r}_S}{\mathrm{d}t} \times \vec{p} = \vec{v} \times m\vec{v} = 0. \end{align*}

The rate of angular momentum change about the point SS is then

dL⃗Sdt=r⃗S×dp⃗dt=r⃗S×F⃗=M⃗S.\begin{align*} \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{align*}

Theorem of Angular Momentum for a Particle​

The change in angular momentum is caused by the torque,

J⃗S=∫t1t2M⃗Sdt=L⃗S,t2−L⃗S,t1,\begin{align*} \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{align*}

where J⃗S\vec{J}_S is defined as the angular impulse about a point SS.

Theorem (Conservation of Angular Momentum about a Point). If the torque about point SS is zero,

M⃗S=dL⃗Sdt=0,\begin{align*} \vec{M}_S = \frac{\mathrm{d}\vec{L}_S}{\mathrm{d}t} = 0, \end{align*}

then L⃗S=const\vec{L}_S = \text{const}.

5.3 System of Particles​

Theorem of Angular Momentum​

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

L⃗S,j=r⃗S,j×p⃗j.\begin{align*} \vec{L}_{S,j} = \vec{r}_{S,j} \times \vec{p}_j. \end{align*} L⃗Ssys=∑j=1NL⃗S,j=∑j=1Nr⃗S,j×p⃗j.\begin{align*} \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{align*} dL⃗Ssysdt=ddt∑j=1NL⃗S,j=∑j=1N(dr⃗S,jdt×p⃗j+r⃗S,j×dp⃗jdt).\begin{align*} \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{align*}

we have

dL⃗Ssysdt=∑j=1N(r⃗S,j×dp⃗jdt)=∑j=1N(r⃗S,j×F⃗j)=∑j=1N(r⃗S,j×(F⃗jext+F⃗jint))=M⃗Sext+M⃗Sint=M⃗Sext.\begin{align*} \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{align*}

Theorem. Angular Momentum of a System of Particles about Different Points A and B satisfies

L⃗A=L⃗B+r⃗A,B×p⃗sys.\begin{align*} \vec{L}_A = \vec{L}_B + \vec{r}_{A,B} \times \vec{p}_{sys}. \end{align*}

Theorem (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,

M⃗Sext=0⇒L⃗Ssys=const.\begin{align*} \vec{M}_S^{ext} = 0 \Rightarrow \vec{L}_S^{sys} = \text{const}. \end{align*}

Center of Mass Reference Frame​

M⃗′=dL⃗′dt=∑ir⃗i′×f⃗i\begin{align*} \vec{M}' = \frac{\mathrm{d}\vec{L}'}{\mathrm{d}t} = \sum_{i} \vec{r}'_i \times \vec{f}_i \end{align*} L⃗′=∑imir⃗i′×v⃗i′.\begin{align*} \vec{L}' = \sum_{i} m_i \vec{r}'_i \times \vec{v}'_i. \end{align*} M⃗inertial′=∑ir⃗i′×(−mia⃗c)=∑i−mir⃗i′×a⃗c=0×a⃗c=0.\begin{align*} \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{align*}

Here (−mia⃗c)(-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:

L⃗=∑imir⃗i×v⃗i=∑imi(r⃗i′+r⃗c)×(v⃗i′+v⃗c)=∑imir⃗i′×v⃗i′+∑imir⃗c×v⃗c=L⃗′+r⃗c×p⃗c,\begin{align*} \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{align*}

since ∑imir⃗i′=0\sum_i m_i \vec{r}'_i = 0 and ∑imiv⃗i′=0\sum_i m_i \vec{v}'_i = 0.