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Homework Help: Modeling of pendulum with external horizontal force

  1. Sep 18, 2007 #1
    1. The problem statement, all variables and given/known data

    In general, I know that the dynamic equation of pendulum is

    theta'' + (k/m)*thata' + (g/l)*sin(theta) = 0

    , where k=friction co, m=mass, l=length of string, g=gravity.

    But if the pendulum is placed in a constant draft, the equation has to be changed.

    Assuming that the pendulum is suspended in a constant horizontal wind,

    imparting a constant force w on the bob, the equation is

    theta'' + (k/m)*thata' + (g/l)*sin(theta) - (w/m)*cos(theta)= 0

    I cannot understand the term of (w/m)*cos(theta). If I rewrite the eq,

    m*l*theta'' + (k*l*theta') + (m*g*sin(theta)) - (w*l*cos(theta)) = 0
    ( inertia )------(friction)--------( gravity )---------( what ?? )

    My question is why the length of string, ' l ' affects the external force term.

    In my oppinion, external force term is just ( w * cos(theta) ), but the answer is not :(.

    Please somebody help me ......
  2. jcsd
  3. Sep 18, 2007 #2


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    For a cursory look at the problem, if one has (g/l)*sin(theta), then the term (w/m)*cos(theta) should include length to get (w/ml)*cos(theta), so that both have a term of acceleration/length.
  4. Sep 18, 2007 #3


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    Using purely Newtonian mechanics, by what means do you get the equation of motion for the angle variable [itex] \theta [/itex] ...?
  5. Sep 18, 2007 #4

    D H

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    Any force on the pendulum bob can be split into radial and normal components. All the radial term will do is change the tension on the string. Assuming an ideal unbreakable and unstretchable string, the radial component doesn't do anything. The normal term is the sole contributor to the dynamics, and hence the [itex]\sin\theta[/itex] factor for the vertical gravitational force and the [itex]\cos\theta[/itex] factor for the horizontal wind force.

    Another way to look at this is as a rotational dynamics problem: work with torques rather than forces. The longitudinal components of forces don't contribute to the torque.

    BTW, Astronuc is right. The wind contribution is w/ml, not just w/m. Its a good idea to do a quick dimensional analysis to ensure that the equations of motion are dimensionally correct. Dimensional correctness doesn't mean the EOM are correct, but if the dimensions are wrong the EOM most certainly are wrong.
  6. Sep 18, 2007 #5

    D H

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    By working in polar coordinates. There is nothing in Newtonian mechanics that restricts their use to cartesian coordinates.
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