König's theorem for angular momentum

Like kinetic energy, the angular momentum of a system of several bodies also decomposes into a “CM” contribution and an “internal” one.

Law — König's theorem for angular momentum

$\ev{\vec{L}_{\text{tot}} = \underbrace{\vec{r}_{\text{CM}}\times\vec{p}_{\text{tot}}}_{\vec{L}_{\text{CM}}} + \underbrace{\sum_i \vec{r}_i'\times m_i\vec{v}_i'}_{\vec{L}_{\text{int}}}}$ where $\vec{r}_i' = \vec{r}_i - \vec{r}_\text{CM}$ and $\vec{v}_i' = \vec{v}_i - \vec{v}_\text{CM}$ are the position and velocity of each body relative to the CM.

The two parts have distinct interpretations. $\vec{L}_\text{CM}$ (the orbital angular momentum): geometrically it is the angular momentum of a fictitious point mass placed at $\vec{r}_\text{CM}$ carrying the whole mass $M_\text{tot}$ and the whole momentum $\vec{p}_\text{tot}$; practically, if the system is not rotating about its own CM, the whole of $\vec{L}_\text{tot}$ reduces to $\vec{L}_\text{CM}$.

$\vec{L}_\text{int}$ (the spin angular momentum): geometrically it is “how the system’s internal parts rotate” about the CM — think of a yo-yo in flight, whose $\vec{L}_\text{int}$ is the yo-yo’s spin. For a rigid body rotating with angular velocity $\vec{\omega}$ about its own CM, $\vec{L}_\text{int} = I_\text{CM}\,\vec{\omega}$, where $I_\text{CM}$ is the moment of inertia about the CM.

If the total moment of the external forces is zero, the total angular momentum is conserved:

$\sum\vec{M}_{\text{ext}} = \vec{0} \;\Rightarrow\; \vec{L}_{\text{tot}} = \text{constant}$

Context — Earth's dual angular momentum

The Earth possesses both an orbital angular momentum ($\vec{L}_\text{CM}$: motion around the Sun) and a spin angular momentum ($\vec{L}_\text{int}$: rotation about its own axis). Both are conserved, which explains why the Earth keeps rotating: no significant external force applies a braking moment.

Example — Two balls in motion

$m_1 = 4\;\text{kg}$ at $\vec{r}_1 = (-1, 1)$ with $\vec{v}_1 = (0, 3)\;\text{m/s}$; $m_2 = 5\;\text{kg}$ at $\vec{r}_2 = (2, -2)$ with $\vec{v}_2 = (6\cos 45^\circ, -6\sin 45^\circ) = (4{,}24;\; -4{,}24)$.

Centre of mass: $\vec{r}_\text{CM} = \frac{4(-1,1) + 5(2,-2)}{9} = \frac{(6,\, -6)}{9} = (0{,}67;\; -0{,}67)$ $\vec{v}_\text{CM} = \frac{4(0,3) + 5(4{,}24;\,-4{,}24)}{9} = \frac{(21{,}2;\; -9{,}2)}{9} = (2{,}35;\; -1{,}02)$

$\vec{L}_\text{CM}$ about the origin ($\vec{p}_\text{tot} = M_\text{tot}\vec{v}_\text{CM} = (21{,}2;\, -9{,}2)$): $L_\text{CM} = x_\text{CM}\,p_y - y_\text{CM}\,p_x = 0{,}67\cdot(-9{,}2) - (-0{,}67)\cdot 21{,}2 = -6{,}16 + 14{,}2 = 8{,}04$

$\vec{L}_\text{int}$: from the relative velocities $\vec{v}_1' = (0,3) - (2{,}35;\,-1{,}02) = (-2{,}35;\, 4{,}02)$ and $\vec{v}_2' = (4{,}24;\,-4{,}24) - (2{,}35;\,-1{,}02) = (1{,}89;\, -3{,}22)$, and from the relative positions, you compute the cross products $\vec{r}_i'\times m_i\vec{v}_i'$ and sum them.

The centre of mass and angular momentum are the tools that let us pass from point-mass mechanics to the mechanics of rigid bodies and rotating systems, the subject of the next chapter.

Links

Topics: Centre of mass Concepts: Angular momentum · Conservation of angular momentum · Moment of inertia Skills: Choice of pole

Related exercises: The rotating diver · Problem — Angular momentum of the Earth’s own rotation · Problem — Ranking ω for equal angular momentum