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 ) =[/math]?
- [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} \Righarrow \vec{r}_{ij} \times \vec{F}_{ij}=0[/math]
Moment of Inertia
Forest_UCM_MnAM#Angular_Momentum