Angular Momentum in Classical Mechanics

Subjects: Classical Mechanics
Links: Linear Momentum in Classical Mechanics, Newton's Laws, Centre of Mass

We can consider the centripetal acceleration of a particle moving in a curved path. Let $v (t)$ be the velocity of a partivle moving on a curved path as a function of time. We get that $$\mathbf a = \frac{d\mathbf v(t)}{dt} = \frac{dv}{dt} \mathbf u_t + \frac{v^2}{r}\mathbf u_n,$$where $r$ is the radius of curvature, $u_{t}$ is the unit tangent vector, and $u_{n}$ is the principal normal vector. From this result, we can define the terms$$\mathbf a_t := \dot v \mathbf u_t \quad \text{and}\quad \mathbf a_c := \frac{v^2}{r}\mathbf u_n,$$and are called the tangential acceleration and normal/radial acceleration, respectively. In the special case where the particle is moving in circular motion, the radial acceleration is called centripetal acceleration. This is derived from the Frenet–Serret formulas.

By Newton's second law, the cause of acceleration is a net force acting on the object. The force, usually referred to as a centripetal force is $$\mathbf F_c = m \mathbf a_c = m \frac{v^2}{r} \mathbf u_n$$
We can consider the idea of angular velocity $ω$ which is related to the tangential velocity by the formula $v = ω r$ , so that $F_{c} = m r ω^{2}$ .

Single Particle

The angular momentum $ℓ$ of a single particle is defined as the vector

ℓ = r \times p

Here $r$ is the position vector relative to an origin $O$ , and its momentum $p$ . Since $r$ depends on $O$ , so does $ℓ$ . The angular momentum $ℓ$ depends on the choice of origin, and we should, strictly speaking, refer to $ℓ$ as the angular momentum relative to $O$ . We can get the rate of change as

\dot{ℓ} = \frac{d}{d t} r \times p = (\dot{r} \times p) + (r \times \dot{p})

since $p = m \dot{r}$ , then the first part cancels out by properties of the cross product. Then we can replace $\dot{p}$ by the net force $F$ on the particle, and we get

\dot{ℓ} = r \times F = Γ

Here $Γ$ denoted the net torque about $O$ on the particle defined as $r \times F$ , other popular symbols for torque are $τ$ and $N$

\dot{ℓ} = Γ

is the rotational equivalent to Newton's second law

In many one-particle we can choose the origin $O$ so that the net torque $Γ$ is zero, meaning constant angular momentum. In the case of a single planet orbiting the sun. The only force on the planet is gravitational $G m M / r^{2}$ of the sun, and this means that the force is parallel to $r$ , meaning $r \times F = 0$ . Thus if we choose our origin at the sun, the planet's angular momentum about $O$ is constant. Because $r \times p$ is constant, then $r$ and $p$ must remain in a fixed plane; meaning, the planet's orbit is confined to a single plane containing the sun, and the problem is reduced to two dimensions.

Kepler's Second Law: As each planet moves around the sun, a line drawn from the planet to the sun sweeps out equal areas in equal times.

If we consider the area function $A$ , this is equivalent to saying that $\dot{A}$ is constant. To prove this, we need to consider that for small times $d t$ , we can approximate the area as

d A = \frac{1}{2} ∥ r \times v ∥ d t

Then we can convert $v = p / m$ getting that

\frac{d A}{d t} = \frac{1}{2 m} ∥ r \times p ∥ = \frac{1}{2 m} ∥ ℓ ∥

since $ℓ$ is constant by our choice of coordinates, then $\dot{A}$ is constant.

Multiple Particles

A system of $N$ particles, $α = 1, \dots, N$ each with its angular momentum $ℓ_{α} = r_{α} \times p_{α}$ , with all $r_{α}$ measured from the same origin $O$ . We define the total angular momentum $L$ as

L = \sum_{α = 1}^{N} ℓ_{α} = \sum_{α = 1}^{N} r_{α} \times p_{α}

differentiating with respect to $t$ we get that

\dot{L} = \sum_{α} {\dot{ℓ}}_{α} = \sum_{α} r_{α} \times F_{α}

Then we can expand the forces to, as we did for the linear momentum, getting that

\dot{L} = \sum_{α} \sum_{β \neq α} r_{α} \times F_{α β} + \sum_{α} r_{α} \times F_{α}^{ext}

doing some algebra we get the first part is

\sum_{α} \sum_{β \neq α} r_{α} \times F_{α β} = \sum_{α} \sum_{β > α} (r_{α} - r_{β}) \times F_{α β}

Since $r_{α} - r_{β}$ is the vector from $r_{β}$ to $r_{α}$ , and the force $F_{α β}$ is in this direction, since the forces are central, then the we get that $(r_{α} - r_{β}) \times F_{α β} = 0$ . The remaining sum single sum is just the net external torque, and we conclude that $$
\mathbf {\dot L} = \Gamma^\text

**Principle of Conservation of Angular:** If the net external force of an $N$-particle system is zero, the system's total angular momentum $\bf L = \sum r_\alpha \times p_\alpha$ is constant. The validity of this principle depends on our two assumptions that all internal forces $\bf F_{\alpha \beta}$ are central and satisfy the third law. If a particle of mass $m$ is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where someone is holding it. Initially the particle is moving in a circle of radius $r_0$ with angular velocity $\omega_0$, but now they pull the string down through the hole until a length of $r$ remains between the hole and the particle. Since there's no torque applied to the system, then we know that angular momentum is conserved. Now, we know that $\ell = mr_0^2\omega_0$ is constant, then $\omega = (r_0/r)^2\omega_0$. ## Angular Momentum about CM The conservation of angular momentum and the more general result $\mathbf{\dot L} =\Gamma^\text{ext}$ were derived on the assumption that all quantities were measured in inertial frame, so that Newton's Second Law could be invoked. This required that both $\bf L$ and $\Gamma^\text{ext}$ be measured about an origin $O$ fixed in some inertial frame. The weird thing, is that we can get the same results also hold if $\bf L$ and $\Gamma^\text{ext}$ are measured about the centre of mass - even if $\text{CM}$ is being accelerated and so is not fixed in an inertial frame. That is

\frac{d}{dt} \mathbf L (\text{about CM}) = \Gamma^\text{ext} (\text{about CM})