# Elliptical Motion

1. Sep 15, 2007

### danago

Given the following displacement vector (and thus, also, acceleration vector):

$$\begin{array}{l} \overrightarrow r (t) = \left( {\begin{array}{*{20}c} {1 + 3\cos (\frac{{\pi t}}{2})} \\ {2 + 4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) \\ \overrightarrow a (t) = \left( {\begin{array}{*{20}c} {\frac{{ - 3\pi ^2 }}{4}\cos (\frac{{\pi t}}{2})} \\ { - \pi ^2 \sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) \\ \end{array}$$

I need to show that the body is undergoing eliptical motion. This is how i proceeded:
$$\begin{array}{l} \overrightarrow r (t) = \left( {\begin{array}{*{20}c} {1 + 3\cos (\frac{{\pi t}}{2})} \\ {2 + 4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {3\cos (\frac{{\pi t}}{2})} \\ {4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right) \\ \therefore\left( {\begin{array}{*{20}c} {3\cos (\frac{{\pi t}}{2})} \\ {4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \overrightarrow r (t) - \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right) \\ \\ \overrightarrow a (t) = \left( {\begin{array}{*{20}c} {\frac{{ - 3\pi ^2 }}{4}\cos (\frac{{\pi t}}{2})} \\ { - \pi ^2 \sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \frac{{ - \pi ^2 }}{4}\left( {\begin{array}{*{20}c} {3\cos (\frac{{\pi t}}{2})} \\ {4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \frac{{ - \pi ^2 }}{4}\left[ {\overrightarrow r (t) - \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right)} \right] \\ \end{array}$$

Is that how i should do it? By showing that the acceleration is proportional to the displacement vector?

Dan.

2. Sep 15, 2007

### danago

I just thought about using the cartesian equation of motion.

$$\begin{array}{l} x = 1 + 3\cos (\frac{{\pi t}}{2}) \\ \therefore\left( {\frac{{x - 1}}{3}} \right)^2 = \cos ^2 (\frac{{\pi t}}{2}) \\ \\ y = 2 + 4\sin (\frac{{\pi t}}{2}) \\ \therefore\left( {\frac{{y - 2}}{4}} \right)^2 = \sin ^2 (\frac{{\pi t}}{2}) \\ \\ \therefore\left( {\frac{{x - 1}}{3}} \right)^2 + \left( {\frac{{y - 2}}{4}} \right)^2 = 1 \\ \end{array}$$

Which is the cartesian equation of an ellipse. Is this method also mathematically valid?

3. Sep 15, 2007

### HallsofIvy

Staff Emeritus
Yes, that's the obvious way to go!

In fact, saying that "the acceleration is proporional to the displacement vector" would be saying that the acceleration vector points toward the center. That's true for circles but NOT for ellipses.

The planets move in ellipses and their acceleration is, of course, toward the sun- which is at a focus, not the center.

Last edited: Sep 15, 2007
4. Sep 16, 2007

### danago

In regards to my first post, how does it show that the acceleration is to the center? Wouldnt the $$- \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right)$$ part mean that the acceleration isnt towards the centre? Or were you referring to the comment i made about it being proportional to the displacement vector?

5. Sep 16, 2007

### HallsofIvy

Staff Emeritus
You are right. that $$- \left( {\begin{array}{*{20}c} 1 \\ 2 \\\end{array}} \right)$$ means it is not pointing toward the center.

You said "Is that how i should do it? By showing that the acceleration is proportional to the displacement vector?

For a vector to be proportional to another vector, i.e. a constant times the vector, they would have to be parallel or anti-parallel. I thought you were saying that you had shown that the acceleration vector was anti-parallel to the displacement vector- pointing toward the center.