Forest UCM Ch3 AngMom

Definition of Angular Momentum

The angular momentum of a single particle is defined as

[math]\vec \ell = \vec r \times \vec p[/math]

An coordinate must be defined in order to express the vectors for the particles position and momentum. The resulting angular momentum is defined with respect to the origin (rotation point) of the particle.

Torque

If I take the derivative of angular momentum with respect to time I get

[math]\vec{\dot \ell} = \frac{d}{dt} \left ( \vec r \times \vec p \right )[/math]

[math]= \left ( \vec \dot r \times \vec p \right ) + \left ( \vec r \times \vec \dot p \right )[/math]

[math] \left ( \vec \dot r \times \vec p \right )= \left ( \frac{1}{m} \vec p \times \vec p \right ) =0 [/math] cross product of parallel vectors

[math] \left ( \vec r \times \vec \dot p \right )=\left ( \vec r \times \vec F \right )= \vec \mathcal T[/math] Definition of Torque

[math]\vec \mathcal T =\vec{\dot \ell} [/math] Newton's second law for angular motion

Kepler's second Law

Kepler's second Law

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

The above law is the result of the conservation of angular momentum when two bodies are attracted to eachother by a central force.

Consider a planet orbiting the Sun such that the sun if Fixed at the origin.

The line segment that joins the planet and the Sun is just the radius vector [math](\vec r )[/math] from the origin of a coordinate system where the sun is located to the position of the planet orbiting the sun.

The displacment vector for the planet is given as

[math]d \vec r = v dt[/math]

The differential area of the triangle representing a differential element of the orbital motion is given as

[math]dA = \frac{1}{2} \left | \vec r \times (\vec v dt ) \right | [/math]

as seen at Forest_UCM_NLM#Vector_.28_Cross_.29_product the area of a parallelogram is given by the cross product and the above triangle makes up half of the parallelogram

using the definitionof momentum

[math]dA = \frac{1}{2m} \left | \vec r \times (\vec p dt ) \right | [/math]

[math]\frac{dA}{dt} = \frac{1}{2m} \left | \vec r \times \vec p \right | [/math]

[math] = \frac{1}{2m} \left | \ell \right | [/math]

Since central forces are conserved forces, the angular momentum is conserved.

Additionally, there is no net Torque for this system.

As a result

[math]\frac{dA}{dt} =[/math] constant

Thus

Equal areas are swept out in equal times

Angular momentum for a system of particles

Definition

The total angular momentum for a system of particles is defined as

[math]\vec L = \sum_i^N \vec \ell_i= \sum_i^N \vec{r}_i \times \vec{p}_i[/math]

Conservation of Angular Momentum

Taking the derivative of the above definition

[math]\vec \dot L = \sum_i^N \left ( \vec{\dot r}_i \times \vec{p}_i \right ) + \sum_i^N \left ( \vec{r}_i \times \vec{\dot p}_i \right ) [/math]

[math]\left ( \vec{\dot r}_i \times \vec{p}_i \right ) = 0[/math] : parallel vectors

[math]\vec \dot L = \sum_i^N \left ( \vec{r}_i \times \vec{\dot p}_i \right ) [/math]

[math] = \sum_i^N \left ( \vec{r}_i \times \vec{F}_i \right ) [/math]

For a system of many particles we can write the Total force on one of the particles in terms of a sum of the Internal forces from all the other particles [math] (\vec{F}_{ij})[/math] and any external force [math]( \vec{F}_i)[/math] .

[math]\vec{F}_{i} = \sum_{i \ne j}^N \vec{F}_{ij} + \vec{F}_i^{\mbox{ext}}[/math]

[math]\vec \dot L = \sum_i^N \left ( \vec{r}_i \times \vec{F}_i \right ) [/math]

[math] = \sum_i^N \left ( \vec{r}_i \times \left [\sum_{i \ne j}^N \vec{F}_{ij} + \vec{F}_i^{\mbox{ext}} \right ] \right ) [/math]

[math] =\left ( \sum_i^N \vec{r}_i \times \sum_{i \ne j}^N \vec{F}_{ij} \right ) + \left ( \sum_i^N \vec{r}_i \times \vec{F}_i^{\mbox{ext}} \right ) [/math]

[math] = \sum_i^N \sum_{i \ne j}^N \left ( \vec{r}_i \times \vec{F}_{ij} \right ) + \sum_i^N \left ( \vec{r}_i \times \vec{F}_i^{\mbox{ext}} \right ) [/math]

The second term represents the external torques exerted on the system

[math] \sum_i^N \left ( \vec{r}_i \times \vec{F}_i^{\mbox{ext}} \right ) = \vec {\mathcal T}_{\mbox{ext}}[/math]

The first term represents the forces exerted on each particle by every other particle (internal forces)

[math] \sum_i^N \sum_{i \ne j}^N \left ( \vec{r}_i \times \vec{F}_{ij} \right ) =[/math]?

The terms in the above double sum can be regrouped

[math] \sum_i^N \sum_{i \ne j}^N \left ( \vec{r}_i \times \vec{F}_{ij} \right ) = \left ( \vec{r}_1 \times \vec{F}_{12} +\vec{r}_1 \times \vec{F}_{13} + \cdots \vec{r}_1 \times \vec{F}_{1N} \right ) + \left ( \vec{r}_2 \times \vec{F}_{21} +\vec{r}_2 \times \vec{F}_{23} + \cdots \vec{r}_2 \times \vec{F}_{2N} \right ) + [/math]

[math]\cdots \left ( \vec{r}_N \times \vec{F}_{N1} +\vec{r}_N \times \vec{F}_{N3} + \cdots \vec{r}_N \times \vec{F}_{N(N-1)} \right ) [/math]

[math] = \left [ \left ( \vec{r}_1 \times \vec{F}_{12} + \vec{r}_2 \times \vec{F}_{21} \right ) + \left ( \vec{r}_1 \times \vec{F}_{13} + \vec{r}_3 \times \vec{F}_{31} \right ) + \cdots + \left ( \vec{r}_1 \times \vec{F}_{1N} + \vec{r}_N \times \vec{F}_{N1} \right ) \right ] [/math]

[math] + \left [ \left ( \vec{r}_2 \times \vec{F}_{23} + \vec{r}_3 \times \vec{F}_{32} \right ) + \left ( \vec{r}_2 \times \vec{F}_{24} + \vec{r}_4 \times \vec{F}_{42} \right ) + \cdots + \left ( \vec{r}_2 \times \vec{F}_{2N} + \vec{r}_N \times \vec{F}_{N2} \right ) \right ] [/math]

[math]+ \cdots + \left ( \vec{r}_N \times \vec{F}_{N(N-1)} +\vec{r}_(N-1) \times \vec{F}_{N(N-1)} \right ) [/math]

[math] = \sum_{j\gt 1}^N \left ( \vec{r}_1 \times \vec{F}_{1j} + \vec{r}_j \times \vec{F}_{j1} \right ) [/math]

[math] + \sum_{j\gt 2}^N \left ( \vec{r}_2 \times \vec{F}_{2j} + \vec{r}_j \times \vec{F}_{j2} \right ) [/math]

[math]+ \cdots + \left ( \vec{r}_N \times \vec{F}_{N(N-1)} +\vec{r}_(N-1) \times \vec{F}_{N(N-1)} \right ) [/math]

[math] = \sum_{i=1}^N \sum_{j\gt i}^N \left ( \vec{r}_i \times \vec{F}_{ij} + \vec{r}_j \times \vec{F}_{ji} \right ) [/math]

Newton's Third law

[math]\Rightarrow \vec{F}_{ij} = - \vec{F}_{ji}[/math]

[math] \sum_i^N \sum_{i \ne j}^N \left ( \vec{r}_i \times \vec{F}_{ij} \right ) = \sum_{i=1}^N \sum_{j\gt i}^N \left ( \vec{r}_i \times \vec{F}_{ij} - \vec{r}_j \times \vec{F}_{ji} \right ) [/math]

[math] = \sum_{i=1}^N \sum_{j\gt i}^N \left ( \vec{r}_i - \vec{r}_j\right ) \times \vec{F}_{ij} [/math]

The vector difference is the vector pointing from particle [math]i[/math] to particle [math]j[/math] anti-parallel to the central Force.

[math] \vec{r}_i - \vec{r}_j = \vec{r}_{ij} [/math] = vector pointing from particle [math]i[/math] to particle [math]j[/math]

[math] \vec{r}_{ij} \parallel \vec{F}_{ij} \Rightarrow \vec{r}_{ij} \times \vec{F}_{ij}=0[/math]

[math] \sum_i^N \sum_{i \ne j}^N \left ( \vec{r}_i \times \vec{F}_{ij} \right ) =0 [/math]

[math]\vec \dot L = \sum_i^N \left ( \vec{r}_i \times \vec{F}_i^{\mbox{ext}} \right ) = \mathcal T_{\mbox{ext}} [/math]

If

the NET external torque on a system of particles is zero

Then

the angular momentum is constant

Moment of Inertia

Definiitons

for a system of particles

[math]I = \sum_i^N m_i r_i^2[/math]

for a Rigid Bodies

[math]I = \int \rho r^2 dV[/math]

Moment of Inertia of a Uniform disk

Find the center of mass for a disk of radius [math]R[/math] and area mass density [math]\sigma=m/A[/math]

[math]I = \int \sigma r^2 dA = \int \sigma r^2 r dr d\theta[/math]

[math]= \sigma \int_0^R r^3 dr \int_0^{2\pi} d\theta [/math]

[math]= \sigma \int_0^R r^3 dr (2 \pi)[/math]

[math]= 2\pi \sigma \frac{R^4}{4} [/math]

[math]=\frac{1}{2} \frac{M}{\pi R^2} R^4[/math]

[math]=\frac{1}{2} M R^2[/math]

Forest_UCM_MnAM#Angular_Momentum