Spherical pendulum, linear approximation? HELP HELP HELP!

1. May 5, 2004

clumsy9irl

Hello there. I'm currently dead beat on this problem, maybe because I'm not sure I quite understand what it's asking (I'm taking my upper level mechanics course in germany, and I don't have any books, and it's the second week, and I'm up at 4am with 2 problem sets due tomorrow, each half done. ahhh!)

Anyway, here's what I interpret:

A pendulum of length L and mass M is in a Gravitationalfield, where it is displaced by a small angle, theta and is lightly nudged. The displacement r is small in comparison to the length, L (r << L). Here, let the motion be treated in the horizontal plane.

a) What are the equations of motion in cartesian coordinates? (hinte: write the gravitational forces on the mass in spherical coordinates, then use the approximation r <<L)

These equations of motion are equivalent to which already known problems?

b) Show that the mass traces out an elliptical pth. Solve here the equations of motion.

I'm lost. I've been working on these sets all day (and since Tuesday, when I had another one due), and I'm just.. my brain is fried.

Any help would be appreciated!!

2. May 6, 2004

Sitewinder

Hi,

at first, you need a drawing showing the relevant forces:

Due to the displacement of the pendulum, the gravitational force causes a resulting force (according to the drawing):

$$F_r=-F_G \sin(\theta)=Mg\sin(s/L)$$

where s is the displacement on the circular arc (because of the definition of radian).

If the angle theta is small, and that is the case, because L >> r, then the displacement on the cirular arc s is approximately equal to the horizontal deplacement, called r (the circle's bending nearly disappears for small distances). Then, you can also use the approach for small angles, that sin(theta) is approximately equal to theta itself.

The formula changes to:

$$F_r=-Mg\frac{r}{L}$$

Newton gave us the coherence F = ma (a = F/m):

$$a=-g\frac{r}{L}$$

Further, we now the following coherence:

$$a=\frac{dv}{dt}=\frac{d^2 r}{dt^2}$$

So, we have the differential equation:

$$\frac{d^2 r}{dt^2}=-\frac{g}{L}r$$

That reminds us of the harmonical oscillator and we solve the equation as follows:

$$r(t)=r_0\cos(\omega t)$$

$$v(t)=-\omega r_0\sin(\omega t)$$

$$a(t)=-\omega^2 r_0\cos(\omega t)$$

We need the previous formula to get omega:

$$\frac{d^2 r}{dt^2}=-\frac{g}{L}r$$

$$a=-\frac{g}{L}r$$

$$-\omega^2 r_0\cos(\omega t)=-\frac{g}{L}r_0\cos(\omega t)$$

$$\Rightarrow\omega=\sqrt{\frac{g}{L}}$$

As $$\omega=2\pi f$$: $$T=\frac{1}{f}=2\pi\sqrt{\frac{g}{L}}$$

Okay, I hope that was enough to make sure that you can make the rest by yourself.

Bye
Site

3. May 8, 2004

Theelectricchild

Site winder you help people out so much. Thanks.