# Equal areas in equal times

1. Mar 8, 2006

### Benny

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:

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

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

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

Now here is what I am having trouble with.

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

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

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?

2. Mar 9, 2006

### dextercioby

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. Mar 10, 2006

### Benny

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. Mar 10, 2006

### Galileo

Don't you have the answer right there?

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

You also know that $\frac{d}{dt}(\vec r \times \vec v)$ 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.