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Dynamics - pendulum hanging on a cord, determine T in cord

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A pendulum of mass m and length L is released
    from rest at an angle Ф to the vertical. There is no
    friction and g = 9.81 m/s2.

    Determine the normal force in the cord when Ф =
    0°.

    http://img87.imageshack.us/my.php?image=pend.jpg

    Note that phi is giving in degrees.

    2. The attempt at a solution
    Decomposing in the radial direction, and taking T as the tension in the cord:
    [tex]\sum F_{r} = m*a_{r} = - T + mgcos\Phi = m(-L\omega[/tex]²) (1)

    Decomposing in transverse direction:
    [tex]\sum F_{\theta} = m*a_{\theta} = mgsin\Phi = m(L\alpha)[/tex]

    [tex]\alpha = d\omega/d\Phi * d\Phi/dt = d\omega/d\Phi * \omega [/tex]

    So:
    [tex]mgsin\Phi = m(L(d\omega/d\Phi)\omega)[/tex]

    [tex]sin\Phi d\Phi = (L/g) \omega d\omega[/tex]

    When the angle is phi, the angular velocity is 0. When the angle is 0, the angular velocity is omega.
    [tex](L/g) \int^{\omega}_{0}\omega d\omega = \int^{0}_{\Phi} sin\Phi d\Phi[/tex]

    [tex]\omega^{2} = -2(g/L)(1 - cos \Phi)[/tex]

    So I end up with a negative term for my angular velocity square. So I can't calculate my angular velocity. Even if I use this omega square and stick it into (1), the resulting equation for T is wrong.

    This was easily solvable with the principle of work and energy, however this method specifically is required.

    Can someone give me a headsup of where I've gone wrong? Thanks!
     
    Last edited: May 12, 2009
  2. jcsd
  3. May 12, 2009 #2
    I think the problem is a simple sign one.

    Your equation for Sum of F_Theta has mg*sinPHI = mL*ALPHA. But if you're measuring PHI from the downward-vertical, this equation is false. It should read:

    -mg*sinPHI = mL*ALPHA.

    The reason? Well, for PHI > 0, the pendulum is to the right of the lower diagonal, sinPHI is positive, but the pendulum bob will have negative acceleration.

    This resolves the problem and will yield the same answer as the energy method for OMEGA^2: 2m/L (1 - cos phi).

    Yeah, just a sign thing.
     
  4. May 12, 2009 #3
    edit: Oops ignore previous question, I understand it now. Thanks a lot for your help!
     
    Last edited: May 12, 2009
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