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Pendulum force experienced at the top of the string?

  1. Nov 4, 2004 #1
    In a demo, a pendulum was used. (See my pic.) It showed that the object experiences 3 g's at the bottom of the swing, according to F = (mv^2)/L. (where L, length of string = r) A peg is located a distance D directly below the point of attachment of the cord.

    + = path of circular motion
    O = object

    First off, for a pendulum that starts at rest at a 90 degree angle to the vertical, energy conservation requires the speed to be consistent, so (mv^2)/2 = mgL or v^2 = 2gL. Now the force experienced at the bottom of the string is F = mg + (mv^2)/L, and after plugging in v^2 = 2gL, you get 3 mg (3 g).

    Now my teacher gave us mg + T = (mv^2)/L. Is this for the force experienced at the top of the string? I don't understand how this is correct or what to do with it.

    We are supposed to derive the relationship between D and L, where D ends up as (3/5)L. I read somewhere that the max. acceleration an object can endure is a = 5g. How can this be proven and how can I tie in this fact so the g's cancel out and I am left with D = (3/5)L?

    Please help me. Thanks.
  2. jcsd
  3. Nov 4, 2004 #2
    Soaring Crane

    Physics is not my field, but I'll give this a shot.
    If you draw a right-angle triangle, using the change in height (D) as one side, and the length L where it is horizontal to D as the Hypotenuse, then you are looking for x, which is the other side. So, L^2= D^2 + x^2
    Now, you have 3g=v^2/2, so g=v^2 (3/2), so L= 3/2
    You have D= 3/5L, so D= 9/10
    L^2 = D^2 + x^2
    1.5^2= .9^2 + x^2
    2.25 - .81 = x^2
    x^2 = 1.44

    I hope this is true, and I hope it helps!

  4. Nov 4, 2004 #3
    Soaring Crane - my typo

    My typo - sorry.
    x is not .12
    x is 1.2

  5. Nov 4, 2004 #4
    Why do you solve for x?

    Why is 3g = (v^2)/2? Wouldn't g then = (v^2)/6?
    Last edited: Nov 4, 2004
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