What's the difference between rotational Energy/angular momentum and linear kinetic energy/linear momentum?

Question

Consider the point mass $M$ being hold by a massless string at distance $r$ and rotate at angular frequence $w$.

Energy:

To orter to obtain the energy of the point mass, we calculate as a rotational motion: $E_{rotational}=\frac{1}{2}Iw^2=\frac{1}{2}mr^2w^2=\frac{1}{2}mv^2=E_{kinetic}$.

Thus the rotational motion and kenatic energy is exactly equivelent(by dm).

Momentum:

$|L|=|r\times P|=|rmv|=|rP|$ so there was a difference between angular momentum and linear momentum, where from Mike Dunlavey's answer in (How can linear and angular momentum be different? ) and wiki (https://en.wikipedia.org/wiki/Angular_momentum#Discussion), I kind of get the picture that linear momentum is the dipict of conservition of the body's motion in general, where the angular momentum is the body's motion "at distance".

So when I look back at the energy part, it's saying that linear momentum with respect to the "center of the motion", where angular momentum was with respect to the mass at "length" or say "rotational arms".

However, I still have some questions:

1. I want to make sure that if there are some mistakes in the above statement.

2. Could you give me some idea about your understanding of the difference between angular momentum and linear momentum?

3. Is there any way to write linear momentum and angualr momentum by the same equation?(Unify them with a single integration/differential equaion)

Answer 1

Angular momentum is a different name from momentum (aka linear momentum). That's because angular momentum is a different quantity from momentum. The quantities are related by$$\vec{L}=\vec{r}\times \vec{p}.$$ In other words, $\vec{L}$ is the cross product of the linear momentum vector with the displacement, $\vec{r}$, of the particle from some reference point (O, say).

Note that this is exactly analogous to the way we define torque (or moment), $\vec{G}$, using $$\vec{G}=\vec{r}\times \vec{F}.$$in which the force, $\vec{F}$ acts at a point displaced from O by $\vec{r}$.

But when you calculate the kinetic energy of a particle rotating about a point, the name of what you're calculating is the same as the name of what you might calculate for a particle moving in a straight line. That's because the quantity you are calculating is the same, though for the rotating particle you might express it in terms of different variables – as you did in your question.

Note that $\vec{L}$ and $\vec{G}$ are related by $$\vec{G}=\frac{d\vec{L}}{dt}$$This equation is analogous to Newton's second law $$\vec{F}=\frac{d\vec{p}}{dt}$$A nice example of using $\vec{G}=\frac{d\vec{L}}{dt}$ is for planets orbiting the Sun. To a very good approximation each planet is acted upon by a force directed towards the centre of the Sun. So $\vec{G}=\vec{r}\times \vec{F}=0$, that is the planet has no torque on it about the centre of the Sun. Therefore $\vec{L}$ is constant. It follows quite easily from this that the planet's orbit is confined to one plane and that a line running between the planet and the centre of the Sun sweeps out equal areas in equal times (Kepler's Second Law).

Answer 2

They are different.

$\vec{p}=m\vec{v}\ \ $ and $\ \ \vec{L}=\vec{r}\times\vec{p}$

They are obviously different.

And their meaning is also different. It might help you to think about systems of particles. Although you can define angular momenta for single particles, angular momentum is especially useful for systems of particles that can rotate; especially rigid bodies.

Check that a spinning ball can have no net $\vec{p}$ because its center of mass stays the same. However, it does has a very clear angular momentum, as some of their particles have northwards velocity, others southwards, others eastwards, andso on. All of them with opposite radii, so they "cancel out" the translational effect, but the object is rotating.

So the fact that there are analogies between translation $\lbrace v, F, p, ...\rbrace$ and rotation $\lbrace \omega, \tau, L, ...\rbrace$ does not mean they are the same.

Finally, as for your last question, we must say that, however, the is one way to put them together. Lagangian/Hamiltonian formalism uses "generalized coordinates and momenta". That means you can use whatever you want as coordinates: $x$, $y$, $x+y$, $r\omega$... And hence their associated momenta (their time derivatives) can be linear or angular, depending on the coordinates.