# What is angular momentum

1. Jul 23, 2014

### Greg Bernhardt

Definition/Summary

Angular momentum (or "moment of momentum") of a point particle is position "cross" momentum: $\mathbf{L}\ =\ \mathbf{r}\times m\mathbf{v}$

Angular momentum of a rigid body about its centre of mass or centre of rotation equals moment of inertia tensor "times" angular velocity: $\mathbf{L}\ =\ I\,\mathbf{\omega}$

About any other point, angular momentum = angular momentum about centre of mass plus angular momentum of centre of mass: $\mathbf{L}_P\ =\ I_{c.o.m.}\,\mathbf{\omega}\,+\, \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}$

Net torque on a rigid body about any point equals rate of change of angular momentum: $\mathbf{\tau}_P\ =\ d\mathbf{L}_P/dt$. In particular, when net torque is zero, angular momentum is constant.

Angular momentum is measured relative to a point, and is additive in the sense that the angular momentum of a body is the sum (or integral) of the angular momentum of its parts, measured relative to the same point: $\mathbf{L}\ =\ \int\int\int\mathbf{r}\times \rho\mathbf{v}\ dxdydz$.

Surprisingly, angular momentum of a rigid body is not generally parallel to (instantaneous) angular velocity: this is obvious from the formula $\mathbf{L}\ =\ \int\int\int\rho \mathbf{r}\times (\mathbf{\omega}\times\mathbf{r})\ dxdydz$. For this reason, angular velocity is not generally constant (there is precession), even when angular momentum is.

Angular momentum is a vector (strictly, a pseudovector), with dimensions of mass times distance squared per time ($ML^2/T$), and measured in units of kg m²/s (or N.m.s or J.s).

Equations

ANGULAR MOMENTUM (POINT MASS):

$$\mathbf{L}\ =\ \mathbf{r}\,\times\,m\mathbf{v}$$

ANGULAR MOMENTUM (RIGID BODY):

about centre of rotation or centre of mass:

$$\mathbf{L}_{c.o.r.}\ =\ \tilde{I}_{c.o.r.}\mathbf{\omega}\ \ \ \mathbf{L}_{c.o.m.}\ =\ \tilde{I}_{c.o.m.}\mathbf{\omega}$$

about point P, where $\mathbf{v}_P$ is the velocity of the part of the body at position P:

$$\mathbf{L}_{P}\ =\ \tilde{I}_{P}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_P$$

$$=\ \tilde{I}_{c.o.m.}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_{c.o.m.}$$

FUNDAMENTAL EQUATION OF MOTION:

$$\mathbf{\tau}_{net,P}\ =\ d\mathbf{L}_P/dt$$

$$\mathbf{\tau}_{net,c.o.r.}\ =\ d(\tilde{I}_{c.o.r.}\mathbf{\omega})/dt$$

$$\mathbf{\tau}_{net,c.o.m.}\ =\ d(\tilde{I}_{c.o.m.}\mathbf{\omega})/dt$$

ROTATIONAL AND TOTAL KINETIC ENERGY:

$$KE_{rot}\,=\,\frac{1}{2}\ \mathbf{\omega}\cdot\mathbf{L}_{c.o.m.}\,=\,\frac{1}{2}\ \mathbf{\omega}^T\mathbf{L}_{c.o.m.}\,=\,\frac{1}{2}\ \mathbf{\omega}^T\tilde{I}_{c.o.m.}\mathbf{\omega}$$

$$KE_{total}\,=\,KE_{rot}\,+\,\frac{1}{2}m\mathbf{v}_{c.o.m.}^2\,=\,\frac{1}{2}\ \mathbf{\omega}^T\tilde{I}_{c.o.r.}\mathbf{\omega}$$

Extended explanation

Angular velocity:

A rigid body has an instantaneous axis of rotation, and an instantaneous angular velocity $\mathbf{\omega}$ parallel to that axis.

The position of the axis depends on the (inertial) frame of reference, but $\mathbf{\omega}$ itself does not.

Where the axis passes through the body, all points on the axis are instantaneously stationary.

Each point $\mathbf{r}$ on the body has instantaneous velocity $\mathbf{v}\ =\ \mathbf{\omega}\times(\mathbf{r}-\mathbf{a})$, where $\mathbf{a}$ is any point on the axis.

Centre of mass:

The centre of mass of a rigid body with density $\rho$ and total mass $m\ =\ \int\int\int\rho\,dxdydz$ is the point $\mathbf{r}_{c.o.m.}$ (fixed in the body but not fixed in space) such that $\int\int\int\rho (\mathbf{r}-\mathbf{r}_{c.o.m.})\ dxdydz\ =\ 0$, ie $\int\int\int\rho\,\mathbf{r}\ dxdydz\ =\ m\,\mathbf{r}_{c.o.m.}$

Ordinary (linear) momentum and orbital angular momentum:

The ordinary (linear) momentum is $\mathbf{p}\ =\ \int\int\int\rho\,\mathbf{v}\ dxdydz$$\ =\ d/dt \int\int\int\rho\,\mathbf{r}\ dxdydz\ =\ m\,\mathbf{v}_{c.o.m.}$.

The orbital angular momentum is the angular momentum calculated as if the whole body was concentrated at its centre of mass: $\mathbf{L}_{orbit}\ =\ \mathbf{r}_{c.o.m.}\times \mathbf{p}\ =\ \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}$.

Spin:

The spin is the angular momentum measured relative to the centre of mass: $\mathbf{L}_{spin}\ =\ I_{c.o.m}\,\mathbf{\omega}$.

Spin is not additive, since the spins of the parts of a rigid body are measured relative to the different centres of mass of each part. For example, if a "dumbell" of two uniform spheres rigidly joined by a light rod rotates about an axis, each sphere has angular momentum parallel to the axis, as measured relative to its own centre of mass, but not generally as measured relative to the other sphere's centre of mass, and so the total spin is not generally parallel to the axis, even though the individual spins always are.

Therefore angular momentum equals spin plus orbital angular momentum:

$\mathbf{L}\ =\ \int\int\int\rho\,\mathbf{r}\times\mathbf{v}\ dxdydz$

$=\ \int\int\int\rho (\mathbf{r}\ -\ \mathbf{r}_{c.o.m.})\times\mathbf{v}\ dxdydz$$\ \ +\ \ \mathbf{r}_{c.o.m.}\times\int\int\int\rho\mathbf{v}\ dxdydz$

$=\ \mathbf{L}_{spin}\ +\ \mathbf{L}_{orbit}$

$=\ I_{c.o.m}\,\mathbf{\omega}\ +\ \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}$.

Fundamental equation of motion (rotational):

total torque = rate of change of angular momentum (about the same point):
τnet = dL/dt

This equation for a system of particles comes from the same equation for a point particle (which in turn comes from "crossing" Newton's second law with r):

τ = r x F = r x d(mv)/dt = dl/dt​

Since the interactions (internal forces) between individual particles are (by Newton's third law) in equal and opposite pairs, when we add the above equation for all particles, all the internal forces cancel, leaving:

τnet = ∑ r x Fexternal = ∑ dl/dt = dL/dt​

Calculating dL/dt:

dL/dt is easiest to calculate about either the centre of mass (C) or (in a "two-dimensional case" where rotation stays parallel to a principal axis of the body) the centre of rotation (R) … in those cases, it is simply the moment of inertia "times" the angular acceleration:
τnet,c.o.m. = dLc.o.m./dt = Ic.o.m.α
τnet, c.o.r. = dLc.o.r./dt = Ic.o.r.α

Sometimes a more general point P is needed, and then:
LP = Ic.o.m.ω + m PC x vc.o.m.

Where rotation stays parallel to a principal axis, so that L stays parallel to ω, then m RC x vc.o.m. = RC x (ω x RC) = m RC2ω, which, from the parallel axis theorem, is (Ic.o.r. - Ic.o.m.)ω, so Lc.o.r = Ic.o.r.ω

This applies, for example, to a sphere or a cylinder rolling over a step, but not to a cone rolling on a plane, or a wheel rolling on a curved rail.

If PC is constant:
τnet,P = dLP/dt = Ic.o.m.α + m PC x ac.o.m.

Precession:

When net torque is zero, a rigid body has constant angular momentum: $d\mathbf{L}/dt\ =\ 0$.

However, angular momentum is parallel to angular velocity only when it is parallel to a principal axis of the body.

In all other cases, the angular velocity vector ($\mathbf{\omega}$) will rotate about the (fixed) angular momentum vector ($\mathbf{L}$): this is precession. See moment of inertia.

Moment of inertia tensor:

A tensor converts one vector to a different vector. The two vectors are parallel only if they are eigenvectors of the tensor.

The moment of inertia tensor converts the angular velocity vector of a rigid body into the angular momentum vector: $\mathbf{L}\ =\ I\,\mathbf{\omega}$.

The moment of inertia tensor is a symmetric tensor. Its eigenvectors are the principal axes of the rigid body, and its eigenvalues are the principal moments of inertia, and so in a coordinate system aligned with the those axes, its matrix is diagonal:
$$\tilde{I}\ =\ \left(\begin{array}{ccc} I_{11} & 0 & 0\\ 0 & I_{22} & 0\\ 0 & 0 & I_{33} \end{array}\right)$$
Published lists of moments of inertia always include only moments of inertia about principal axes.
In any other coordinate system, its matrix is:
$$\tilde{I}\ =\ \left(\begin{array}{rrr} I_{xx} & -I_{xy} & -I_{zx}\\ -I_{xy} & I_{yy} & -I_{yz}\\ -I_{zx} & -I_{yz} & I_{xz} \end{array}\right)$$
where $I_{xx}\ =\ \int\int\int\rho\,(y^2+z^2)\,dxdydz\$, $\ I_{yz}\ =\ \int\int\int\rho\,yz\,dxdydz$, and so on.

The words "moment of inertia" usually refer to the diagonal elements of the tensor, the moments of inertia about particular axes.

The tensor Î is fixed in the body, but is not generally fixed in space, and so is not generally constant in the fundamental equation of motion $\mathbf{\tau}\ =\ d\mathbf{L}/dt\ =\ d(I\,\mathbf{\omega})/dt$.

However, Î is constant in Euler's equations ($\tau_1\ =\ I_{11}\,d\omega_1/dt\ +\ (I_{33}\ -\ I_{22})\omega_2\omega_3$ etc), which are the fundamental equations of motion relative to coordinates fixed in the body. See Euler's equations.

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