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Pendulum in polar coordinate system problem

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    NII: [tex]\sum[/tex]F=ma
    sin[tex]\theta[/tex]=[tex]\theta[/tex]
    T=2[tex]\pi[/tex][tex]\sqrt{}[/tex](l/g)

    3. 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[tex]\theta[/tex] in the direction of x and mgcos[tex]\theta[/tex] in the direction of y. I applied NII so that

    -mgsin[tex]\theta[/tex]=max.
    m disappears and I wrote a as seen below, which leads to

    d2x/dt2=-gsin[tex]\theta[/tex].
    After integrating twice I had

    x3/6=-1/2*gsin[tex]\theta[/tex]t2, but since

    sin[tex]\theta[/tex]=[tex]\theta[/tex]

    x=[tex]\sqrt[]{}3[/tex]-3g[tex]\theta[/tex]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[tex]\varphi[/tex], 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!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi mkerikss! :smile:

    (have a theta: θ and a phi: φ and a pi: π and a square-root: √ and a sigma: ∑ :wink:)
    (dots are the same as dashes … they mean d/dt :wink:)

    The question asks you to use a polar coordinate system …

    so position is (r,θ), and velocity is measured along the local directions er and eθ

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

    Also, (er)' = θ'eθ, and (eθ)' = -θ'er

    Then, for example, r = rer, and so r' = rθ'eθ, and r'' = … ? :smile:
     
  4. Sep 30, 2009 #3
    Polar coordinates:

    http://en.wikipedia.org/wiki/Polar_coordinate_system" [Broken]

    instead of x and y you use r: distance from origin and [tex]\phi[/tex] 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

    [tex] F_{tangential} = m r \frac {d^2\phi} {dt^2} [/tex]

    instead of F=ma
     
    Last edited by a moderator: May 4, 2017
  5. Sep 30, 2009 #4
    Thanks for all the help, I'll give it another try. At least now I understand what it's all about :smile:
     
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