Course:PHYS350/Energy and Angular Momentum

Let's start with some basic definitions. First m is the inertial mass of a particle -- we will consider only particles with constant masses (although the physics of rockets provides an important counter-example). A particle for us is something that can be described completely by its position $\overrightarrow{r}$ and mass (it doesn't have any important internal degrees of freedom).

Momentum and Force

We can define the velocity and momentum of the particle,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$ and $\overrightarrow{p}=m\overrightarrow{v}$.

Because the acceleration is the rate of change of velocity we have

$\overrightarrow{F}=m\overrightarrow{a}=m\frac{d\overrightarrow{v}}{dt}=\frac{d\overrightarrow{p}}{dt},$

so if the force vanishes, then the momentum is constant in time or conserved.

Angular Momentum and Torque

Let's also define the angular momentum and the torque,

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$ and $\overrightarrow{N}=\overrightarrow{r}\times \overrightarrow{F}=\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}.$

Let's calculate the rate of change of the angular momentum

$\frac{d}{dt}\overrightarrow{L}=\frac{d\overrightarrow{r}}{dt}\times \overrightarrow{p}+\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}=\overrightarrow{N}$

so if the torque vanishes, the angular momentum is conserved. Specifically if the force is always directed along the position vector, $\overrightarrow{r}$, the torque vanishes and the angular momentum is conserved. This is called a central force.

Work and Energy

We can define the work done on a particle while moving it from position #1 to position #2 to be

$W_{12}=\underset{1}{\overset{2}{\int }}\overrightarrow{F}\cdot d\overrightarrow{s}.$

Again let's assume that the mass of the particle is constant so

$W_{12}=\underset{1}{\overset{2}{\int }}\overrightarrow{F}\cdot d\overrightarrow{s}=m\underset{1}{\overset{2}{\int }}\frac{d\overrightarrow{v}}{dt}\cdot \overrightarrow{v}dt=\frac{m}{2}\underset{1}{\overset{2}{\int }}\frac{d}{dt}\left(v^2\right)dt=\frac{m}{2}(v_2^2-v_1^2)=T_2-T_1$

where we have defined the kinetic energy,

$T=\frac{1}{2}m{v}^2$.

If the integral $W_{12}$ depends only on the locations of the points 1 and 2 and not the path, then the force or system is conservative. In particular,

$\oint \overrightarrow{F}\cdot d\overrightarrow{s}=0$ for such a system.

This means that the force is the gradient of some other function

$\overrightarrow{F}=\overrightarrow{\mathrm{\nabla }}V(r)$

call the potential energy $V$. In this case we find that

$W_{12}=V_1-V_2=T_2-T_1$

so

$T_1+V_1=T_2+V_2$