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Period of swing of a rope with two fixed ends

  1. Dec 30, 2011 #1
    Friends:

    If you take a rope and fix the top end and allow the rope to swing, then it is just a pendulum and the period of swing is widely known (and on wikipedia) as approx. T = 2*pi*(L/g)^0.5 where L is the length of the pendulum and g is gravity.

    My question is: What if I take both ends of the rope and fix them (like tying them to an overhead horizontal metal bar), like a playground swing. How do you calculate the period of swing then? In this case the "length of the pendulum" would vary - it would go from 0 at the fixed ends to the maximum at the saggiest part of the rope.

    This problem seems really simple but I can't figure it out!!! Any help would be greatly appreciated. Thanks a lot!
     
  2. jcsd
  3. Dec 30, 2011 #2
    You need to be careful using the equation
    T = 2pi (l/g)^0.5
    if you have a rope (or a rod) on it's own then l is the distance to the centre of mass of the rope or rod (l/2 if uniform).
    In a simple pendulum it is assumed that the string has zero mass.
    Hope this helps with your rope fixed at both ends (l/4 ????)
     
  4. Dec 30, 2011 #3

    Doc Al

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    Staff: Mentor

    Careful. Whenever you have a distributed mass acting as a pendulum, you cannot use that simple formula. The period of such a 'physical pendulum' (as opposed to a 'simple pendulum') depends on the rotational inertia as well as the distance of the center of mass to the pivot. (It's not enough to just replace L by L/2.)
    Again, this must be treated as a physical pendulum. You'll need to know the rotational inertia and the location of the center of mass.

    See: Physical Pendulum
     
  5. Dec 30, 2011 #4
    Doc Al
    I have looked up your reference and it works out that the term inside the square root is 2L/3
    Not L/2 for a uniform rod.
    Thank you for picking that up.
     
  6. Dec 30, 2011 #5
    You'll need to find the moment of inertia about the axis joining the two ends of the rope, this means integrating the square of the distance of each part of the rope from this axis over the length of the rope (along the curve it makes which is a catenary). From this you solve
    [tex]\tau = -I\alpha[/tex]
    Where [itex]\tau[/itex] is the torque on the system, [itex]I[/itex] is the moment of inertia and [/itex]\alpha[/itex] is the second derivative of the angle made between the vector of the center of mass of the rope relative to a plane which lies along the line of the axis of the swinging rope and is perpendicular to the ground.
     
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