Understanding Angular Momentum and Equal Areas in Equal Times

  • Thread starter Benny
  • Start date
  • Tags
    Areas
In summary, the conversation is discussing the relationship between the displacement vector r, velocity vector v, and force vector F of a moving particle. The conservation of angular momentum is used to show that the rate at which the area is swept out by r is equal to half the norm of r cross v. This leads to the deduction that when a mass is subject to a central force directed towards the origin, the vector displacement r moves in a plane and sweeps out equal areas in equal times. The conversation ends with a clarification of the meaning of p, which stands for momentum.
  • #1
Benny
584
0
Hi, can someone help me do the following question? (I've cut out some details, leaving the results which might be of help)

Let the vector r represent the displacement from the origin to a moving particle of mass m which is subjected to a force F.

Results which I've been able to arrive at:

[tex]
\mathop H\limits^ \to = \mathop r\limits^ \to \times \mathop p\limits^ \to \Rightarrow \frac{{d\mathop H\limits^ \to }}{{dt}} = m\frac{d}{{dt}}\left( {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right) = \mathop r\limits^ \to \times \mathop F\limits^ \to = \mathop M\limits^ \to
[/tex]

v is velocity, F is force, p = mv is moment.

If F acts towards the origin then [itex]\frac{d}{{dt}}\left( {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right) = \mathop 0\limits^ \to [/itex].

Now here is what I am having trouble with.

Show that [itex]\left\| {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right\|[/itex] is twice the rate at which the area A is swept out by the vector r, that is,

[tex]
\left\| {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right\| = 2\frac{{dA}}{{dt}} = r^2 \frac{{d\theta }}{{dt}}
[/tex]

and hence deduce that, when a mass is subject to a central force directed towards the origin, the vector displacement r moves in a plane and sweeps out equal areas in equal times.

I don't know where to start. I can't visualise what is going on in this problem. Norm of r cross v is the area of a parallelogram...ok...but that doesn't tell me much. As for r, it is arbitrary so I can't try to make a picture. I just don't know what to do here. Can someone help me?
 
Physics news on Phys.org
  • #2
The equality between the first and the 3-rd part is obvious, however, it would help if you knew the exact expression for A...

Daniel.
 
  • #3
Thanks for your input but like I said, the main thing which is preventing me from being able to make any significant progress in this question is that I simply cannot picture what is going on.

Going back to basics I would consider an xyz coordinate system and a curve traced out by a position vector r, tail at the origin and head moving along a curve. I think that the area being referred to in the question is the area 'bounded' below by the position vector and above by the curve. Something like that. But I really can't think of a way to approach this problem.

Oh and p = mv is supposed to be momentum. I typed it as moment.
 
  • #4
Don't you have the answer right there?

As you said: [itex]\frac{1}{2}||\vec r \times \vec v ||=\frac{d A}{dt}[/itex] is the rate at which the area is swept out.

You also know that [itex]\frac{d}{dt}(\vec r \times \vec v)[/itex] is constant (conservation of angular momentum). So the rate at which the area is swept out is constant. It's staring you in the face.
 

What is the concept of "equal areas in equal times"?

The concept of "equal areas in equal times" is a fundamental principle in physics and astronomy which states that an object in motion will sweep out equal areas in equal amounts of time. This means that the speed of an object is not constant, but instead varies based on its distance from the center of motion.

How does "equal areas in equal times" relate to Kepler's Second Law?

Kepler's Second Law, also known as the Law of Equal Areas, is based on the principle of "equal areas in equal times". This law states that the line connecting a planet to the sun will sweep out equal areas in equal times, meaning that the planet's speed varies as it moves along its elliptical orbit.

Why is "equal areas in equal times" important in understanding planetary motion?

Understanding "equal areas in equal times" is crucial in understanding planetary motion because it allows us to accurately predict the speed and position of a planet at any given time. It also helps explain why planets move at different speeds at different points in their orbits, and why their orbits are not perfect circles.

How does the distance from the center of motion affect the speed of an object according to "equal areas in equal times"?

According to "equal areas in equal times", the closer an object is to the center of motion, the faster it will move. This is because as the object moves closer to the center, the distance it needs to cover to sweep out equal areas decreases, thus increasing its speed. The opposite is also true, as the object moves further away from the center, its speed decreases.

Can "equal areas in equal times" be applied to objects other than planets?

Yes, "equal areas in equal times" can be applied to any object in motion, as long as it follows a curved path. This principle is used in various fields such as physics, astronomy, and engineering to understand the motion of satellites, comets, and other celestial bodies. It can also be applied to objects on Earth, such as projectiles and pendulums.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
5K
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
1
Views
1K
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
902
  • Calculus and Beyond Homework Help
Replies
3
Views
488
  • Calculus and Beyond Homework Help
Replies
4
Views
97
  • Calculus and Beyond Homework Help
Replies
1
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top