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 ${\vec {r}}$ 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,

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

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

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

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,

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

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

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

so if the torque vanishes, the angular momentum is conserved. Specifically if the force is always directed along the position vector, ${\vec {r}}$, 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

$W_{12}=\int _{1}^{2}{\vec {F}}\cdot d{\vec {s}}.$

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

$W_{12}=\int _{1}^{2}{\vec {F}}\cdot d{\vec {s}}=m\int _{1}^{2}{\frac {d{\vec {v}}}{dt}}\cdot {\vec {v}}dt={\frac {m}{2}}\int _{1}^{2}{\frac {d}{dt}}\left(v^{2}\right)dt={\frac {m}{2}}\left(v_{2}^{2}-v_{1}^{2}\right)=T_{2}-T_{1}$

where we have defined the kinetic energy,

$T={\frac {1}{2}}mv^{2}$.

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,

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

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

${\vec {F}}={\vec {\nabla }}V(r)$

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

$W_{12}=V_{1}-V_{2}=T_{2}-T_{1}$

so

$T_{1}+V_{1}=T_{2}+V_{2}$