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Stability physics problem

  1. Feb 25, 2007 #1
    Just come across a question and I'm at a point where i see no further.

    A uniform rod AB, of mass m and length 2a, is free to rotate in a vertical plane, about the end A. A light elastic string of modulus kmg and natural length a, has one and attached to B and the other end to a fixed point O which is vertically above A with OA = 2a. Show that when AB makes an angle [tex]\theta[/tex] with the downward vertical, the potential energy, V, of the system when the string is stretched is given by

    [tex]V = mga[(4k - 1)\cos\theta -4k\cos\frac{\theta}{2}] + constant[/tex]

    I drew the following diagram:-

    I equated the GPE of the uniform rod to be
    [tex]AB = -mg\cos\theta[/tex]

    And when it comes to calculating the energy of the elastic string i tried:

    [tex]OC = \frac{\lambda x^2}{2a}[/tex]

    Where x is the extension in the string, which i calculated to be:

    [tex]x = (2a\sin\theta - a)[/tex]

    [tex]OB = \frac{kmg(2a\sin\theta - a)^2}{2a}[/tex]

    [tex]OB = \frac{kmga(4\sin^2\theta -4\sin\theta + 1)}{2}[/tex]

    [tex]OB = 2kmgasin^2\theta -2kmg\sin\theta + \frac{kmg}{2}[/tex]

    But since:
    [tex]OB = \frac{kmg}{2}[/tex] is a constant, we can take it out of the equation.

    Using the identity: [tex]\sin^2\theta = 2\sin\theta\cos\theta[/tex]

    [tex]4kmga\sin\theta\cos\theta -2kmga\sin\theta[/tex]

    However i can't get the 2nd term in the above equation to equal [tex]4kmga\cos\frac{\theta}{2}[/tex]

    Any help would be really appreciated.

    Thank you.
    Last edited: Feb 25, 2007
  2. jcsd
  3. Feb 26, 2007 #2


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    Homework Helper

    Hey Bob, how can you be sure the angle made by AB and OB is 90 degrees?
  4. Feb 26, 2007 #3
    Hmm. Since the triangle OAB is isosceles, could i split it into 2 right angled triangles to get the length of the extension in terms of [tex]\cos\frac{\theta}{2}[/tex]?

    I'm not too clear why the sine of the angle theta can't just be used. Is there a fundamental flaw in my understanding of mathematics?

    Thank you.
  5. Feb 26, 2007 #4
    Just ran through it trying splitting the triangle and i got

    [tex]OB = kmg(4a\cos\theta -4a\cos\frac{\theta}{2})[/tex]

    As this is equal to the GPE of the rod.

    [tex]mga\cos\theta = kmg(4a\cos\theta -4a\cos\frac{\theta}{2})[/tex]

    [tex]V = mga[(4k - 1)\cos\theta - 4k\cos\frac{\theta}{2}][/tex]

    Which is the correct answer.
  6. Feb 26, 2007 #5


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    Homework Helper

    Yea, you are right. Good work!.
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