Pendulum in polar coordinate system problem

[mkerikss]

Homework Statement

A pendulum consists of a particle of the mass m and a thread of the length l (we don't consider the threads mass). The acceleration caused by gravity is g. Solve the particles displacement and the force caused by the tension in the thread T in a polar coordinate system. The pendulums oscillations are of small amplitude. What is the period of the particle?

Homework Equations

NII: $$\sum$$F=ma
sin$$\theta$$=$$\theta$$
T=2$$\pi$$$$\sqrt{}$$(l/g)

The Attempt at a Solution

First of all I'd like to introduce myself. My name is Mikael, and I have recently started studying at Helsinki University of Technology. One thing that causes me a fair bit of difficulties is that Swedish is my native language, and all the lessons are in Finnish. I read the most advanced physics and maths courses the University offer, and it's been quite tough. Now here's my attempt at a solution:

I started by drawing a picture with all the forces and the components where needed, in this case I chose to keep T and divided mg into mgsin$$\theta$$ in the direction of x and mgcos$$\theta$$ in the direction of y. I applied NII so that

-mgsin$$\theta$$=ma_x.
m disappears and I wrote a as seen below, which leads to

d²x/dt²=-gsin$$\theta$$.
After integrating twice I had

x³/6=-1/2*gsin$$\theta$$t², but since

sin$$\theta$$=$$\theta$$

x=$$\sqrt[]{}3$$-3g$$\theta$$t²

At this point, I had a look at the tips and tricks one of our teachers gave us, and it looked nothing like this. He had used the vectors r and e$$\varphi$$, and lots of dots above them, that I don't even know what they mean. I started to think that maybe the polar coordinate system means I must write the answers with these vectors, but I have no idea how to do that, because the tips made no sense to me at all. Therefore, what I first of all want to know, is wath is the correct way to give the answer. Can I do as I have done, or is it completely wrong. If I'm wrong I appreciate all help because I'm not even sure what a polar coordinate system is. So some help to get me started would be great in that case. Thank you for your time!

 

[tiny-tim]

Hi mkerikss!

(have a theta: θ and a phi: φ and a pi: π and a square-root: √ and a sigma: ∑ )

… Solve the particles displacement and the force caused by the tension in the thread T in a polar coordinate system.
…
I started by drawing a picture with all the forces and the components where needed, in this case I chose to keep T and divided mg into mgsin$$\theta$$ in the direction of x and mgcos$$\theta$$ in the direction of y.
…
At this point, I had a look at the tips and tricks one of our teachers gave us, and it looked nothing like this. He had used the vectors r and e$$\varphi$$, and lots of dots above them, that I don't even know what they mean.

(dots are the same as dashes … they mean d/dt )

The question asks you to use a polar coordinate system …

so position is (r,θ), and velocity is measured along the local directions e_r and e_θ …

e_r is the unit vector along the r direction, and e_θ is the unit vector perpendicular to e_r, in the direction of increasing θ.

Also, (e_r)' = θ'e_θ, and (e_θ)' = -θ'e_r

Then, for example, r = re_r, and so r' = rθ'e_θ, and r'' = … ?

 

[willem2]

Polar coordinates:

http://en.wikipedia.org/wiki/Polar_coordinate_system"

instead of x and y you use r: distance from origin and $$\phi$$ angle from a fixed direction.

you use this with problems that deal with rotation. problems will often be easier when converted to polar coordinates

In the case of a pendulum, r always the length of the pendulum, so you only have to deal with a differential equation for phi. phi =0 corresponds to straight down.
you have to split the force of gravity on the particle in a component in the direction of the origin (the r direction) and a component in the direction tangential to the arc that the particle makes. Only this component works in the phi direction. you get

$$F_{tangential} = m r \frac {d^2\phi} {dt^2}$$

instead of F=ma

 

[mkerikss]

Thanks for all the help, I'll give it another try. At least now I understand what it's all about