1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Foucault Pendulum

  1. May 8, 2012 #1
    This problem is about the Foucault pendulum. It is a mass, m, attached to a cable that does not restrict the motion. So it is sensitive to centrifugal and coriolis forces.

    The first part states that it has a potential ## V = \frac{m \omega_0^2}{2}(x^2 + y^2)## and to find the force associated with that. To me this is a hint that you are going to use the equation

    ## m \ddot{\vec{r}}= \vec{F} - 2 m \vec{\omega} \times (\vec{\omega} \times \vec{r})-m \vec{\omega}\times\vec{\dot{r}} ##

    The earth rotates at ## \omega## ccw. The pendulum is located at the North pole and the co-ordinate axes are such that x points 90 degrees longitude and y 180 degrees longitude. I don't really understand how this makes a difference.

    I decided that ## \vec{\omega}=\omega \vec{\hat{k}} ## which I am unsure about.

    So for the above equation, since ## \vec{F} = -\nabla V ## and after some computation/approximation, i.e. ## \dot{z} ## is small.

    ## m \ddot{\vec{r}}= \vec{F} - 2 m \vec{\omega} \times (\vec{\omega} \times \vec{r})-m \vec{\omega}\times\vec{\dot{r}} \Rightarrow##

    ## \ddot{x} = \omega_0 x + \omega^2 x + 2\omega \dot{y} ##

    ## \ddot{y} = \omega_0 y + \omega^2 y - 2 \omega \dot{x} ##

    Which seem OK for now.

    Then we're given the hint to use ## \zeta(t) = x(t) + i y(t) ## to simplify things and hopefully turn it into a SHO problem with damping. Multiply the second equation by ##i##

    ## \ddot{x} + i\ddot{y} = \omega_0(x+i y) +\omega^2(x+i y) + 2\omega(\dot{y}-i\dot{x}) ##

    ## \ddot{\zeta} =(\omega_0^2+\omega^2) \zeta - 2\omega i \dot{\zeta} ##

    Then plug in ## \zeta = \exp(\alpha t) ## get ##\alpha= \omega-\sqrt{\omega_0^2-\omega^2}## and from here it doesn't really get anywhere near the answer given.

    First I assume that ##\omega^2 ## is really small.

    ## \zeta(t) = c_1 \exp(i (\omega + \omega_0) t) + c_2 \exp(i \omega - \omega_0)t) ##

    But we are required to show that:

    ## \zeta(t) = e^{-i \omega t}\left[ \zeta(0) \left( \cos(\omega_0 t) + i\frac{\omega }{\omega_0}\sin(\omega_0 t)\right)+\frac{\dot{\zeta(0)}}{\omega} \sin (\omega_0 t) \right] ##

    But I have no idea how to do this. I think that possibly I have made a mistake with the orientation of my axes or the angular velocity components. Any help would be appreciated!
     
    Last edited: May 8, 2012
  2. jcsd
  3. May 8, 2012 #2
    I don't like bumping but this is the third problem I have posted with hundreds of views and no replies. Any ideas? Need more information?
     
  4. May 11, 2012 #3

    Cleonis

    User Avatar
    Gold Member

    I suggest you check out Joe Wolfe's derivation of the foucault pendulum precession

    Like in the hints you've been given Joe Wolfe uses complex number notation to express the two coupled differential equations as a single equation. (In fact, I suspect the hints you've been given were inspired by that very derivation.)
    By comparing you may well find where you go awry.

    Joe Wolfe's derivation is for the more general case of any latitude. For the polar pendulum case the equations are simpler.


    Some general remarks.
    A frictionless pendulum will have an equation of motion that is a function of position and acceleration, but no term with velocity.
    When friction is approximated as proportional to velocity then the equation of motion will include a term with velocity.

    The two coupled equations (of motion) of a foucault pendulum have likewise a term with velocity. (But in the foucault pendulum case the associated acceleration is at right angles to the velocity vector.)
     
    Last edited: May 11, 2012
  5. May 12, 2012 #4
    Our solutions are identical up until the point where

    ## \ddot{\zeta} + 2i \omega \dot{\zeta } +\omega^2 \zeta = 0 ##

    But I have no idea how to get to the final answer.
     
  6. May 12, 2012 #5

    Cleonis

    User Avatar
    Gold Member

    As it turns out that was not good advice at all.
    The last expressions on that page are messed up terribly.

    I will use the following notation: the uppercase Omega (Ω) for the angular velocity of the Earth, and the lowercase omega (ω) for the period of the pendulum.

    ## \ddot{\zeta} + 2i \Omega \dot{\zeta } +\omega^2 \zeta = 0 ##

    In the case where Ω=0 the equation reduces to the equation of motion for a pendulum with natural frequency ω.

    In the above equation of motion only the Coriolis term is a function of the Earth's angular velocity. The term with ω squared relates to the period of the pendulum.

    As to the final form you want/need to arrive at:
    Clearly that has been elaborated to explicitly include starting conditions.
    That expression becomes simpler for the case of releasing the pendulum parallel to the y-axis, and with zero initial velocity.

    I haven't checked whether there are mistakes in that expression. I guess you shouldn't rely on it. If there are mistakes in it you'll have to find the correct form.

    For comparison material I guess you need to dig around. Maybe with combinations of search terms such as 'foucault pendulum' 'derivation' 'complex', to try and find other derivations that use this method of casting the equation in complex number notation.
     
    Last edited: May 12, 2012
  7. May 18, 2012 #6
    I don't really understand how to put the initial conditions in and get that final answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Foucault Pendulum
  1. A Pendulum (Replies: 8)

  2. Quantum pendulum (Replies: 8)

Loading...