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**Definition/Summary**Angular momentum (or "moment of momentum") of a point particle is position "cross" momentum: [itex]\mathbf{L}\ =\ \mathbf{r}\times m\mathbf{v}[/itex]

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

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

Net torque on a rigid body about any point equals rate of change of angular momentum: [itex]\mathbf{\tau}_P\ =\ d\mathbf{L}_P/dt[/itex]. 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: [itex]\mathbf{L}\ =\ \int\int\int\mathbf{r}\times \rho\mathbf{v}\ dxdydz[/itex].

Surprisingly, angular momentum of a rigid body is not generally parallel to (instantaneous) angular velocity: this is obvious from the formula [itex]\mathbf{L}\ =\ \int\int\int\rho \mathbf{r}\times (\mathbf{\omega}\times\mathbf{r})\ dxdydz[/itex]. 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 ([itex]ML^2/T[/itex]), and measured in units of kg m²/s (or N.m.s or J.s).

**Equations**ANGULAR MOMENTUM (POINT MASS):

[tex]\mathbf{L}\ =\ \mathbf{r}\,\times\,m\mathbf{v}[/tex]

ANGULAR MOMENTUM (RIGID BODY):

about centre of rotation or centre of mass:

[tex]\mathbf{L}_{c.o.r.}\ =\ \tilde{I}_{c.o.r.}\mathbf{\omega}\ \ \ \mathbf{L}_{c.o.m.}\ =\ \tilde{I}_{c.o.m.}\mathbf{\omega}[/tex]

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

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

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

FUNDAMENTAL EQUATION OF MOTION:

[tex]\mathbf{\tau}_{net,P}\ =\ d\mathbf{L}_P/dt[/tex]

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

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

ROTATIONAL AND TOTAL KINETIC ENERGY:

[tex]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}[/tex]

[tex]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}[/tex]

**Extended explanation****Angular velocity:**

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

The

*position*of the axis depends on the (inertial) frame of reference, but [itex]\mathbf{\omega}[/itex] itself does not.

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

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

**Centre of mass:**

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

**Ordinary (linear) momentum and orbital angular momentum:**

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

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

**Spin:**

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

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:

[itex]\mathbf{L}\ =\ \int\int\int\rho\,\mathbf{r}\times\mathbf{v}\ dxdydz[/itex]

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

[itex]=\ \mathbf{L}_{spin}\ +\ \mathbf{L}_{orbit}[/itex]

[itex]=\ I_{c.o.m}\,\mathbf{\omega}\ +\ \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}[/itex].

**Fundamental equation of motion (rotational):**

total torque = rate of change of angular momentum (about the same point):

**τ**_{net}= d**L**/dtThis 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(m

**v**)/dt = d

**l**/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:

**τ**

_{net}= ∑

**r**x

**F**

_{external}= ∑ d

**l**/dt = d

**L**/dt

**Calculating dL/dt:**

d

**L**/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:

**τ**

_{net,c.o.m.}= d

**L**

_{c.o.m.}/dt = I

_{c.o.m.}

**α**

**τ**

_{net, c.o.r.}= d

**L**

_{c.o.r.}/dt = I

_{c.o.r.}

**α**

Sometimes a more general point P is needed, and then:

**L**

_{P}= I

_{c.o.m.}

**ω**+ m

**PC**x

**v**

_{c.o.m.}

*Where rotation stays parallel to a principal axis*, so that

**L**stays parallel to

**ω**, then m

**RC**x

**v**

_{c.o.m.}=

**RC**x (

**ω**x

**RC**) = m

**RC**

^{2}

**ω**, which, from the parallel axis theorem, is (I

_{c.o.r.}- I

_{c.o.m.})

**ω**, so

**L**

_{c.o.r}= I

_{c.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:

**τ**

_{net,P}= d

**L**

_{P}/dt = I

_{c.o.m.}

**α**+ m

**PC**x

**a**

_{c.o.m.}

**Precession:**

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

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 ([itex]\mathbf{\omega}[/itex]) will rotate about the (fixed) angular momentum vector ([itex]\mathbf{L}[/itex]): 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: [itex]\mathbf{L}\ =\ I\,\mathbf{\omega}[/itex].

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:

[tex]\tilde{I}\ =\ \left(\begin{array}{ccc}

I_{11} & 0 & 0\\

0 & I_{22} & 0\\

0 & 0 & I_{33}

\end{array}\right)[/tex]

Published lists of moments of inertia

*always*include only moments of inertia about principal axes.

In any other coordinate system, its matrix is:

[tex]\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)[/tex]

where [itex]I_{xx}\ =\ \int\int\int\rho\,(y^2+z^2)\,dxdydz\ [/itex], [itex]\ I_{yz}\ =\ \int\int\int\rho\,yz\,dxdydz[/itex], 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 [itex]\mathbf{\tau}\ =\ d\mathbf{L}/dt\ =\ d(I\,\mathbf{\omega})/dt[/itex].

However,

*Î*

*is constant*in Euler's equations ([itex]\tau_1\ =\ I_{11}\,d\omega_1/dt\ +\ (I_{33}\ -\ I_{22})\omega_2\omega_3[/itex] etc), which are the fundamental equations of motion

*relative to coordinates fixed in the body*. See Euler's equations.

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!