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    as title says, this chapter concerns about the stability of a mechanical system.
    Ex5A Q3.
    A uniform heavy rod AB, of mass m and length 4a, can turn in a vertical plane about one end A which is fixed. To the other end B is attached a light elastic string of natural length 3a and modulus 1/2 mg. The other end of the string is attached to a light ring which can slide on a smooth horizontal bar which is fixed at a height of 8a above A and in the vertical plane through AB. Find the equilibrium position of the rod and determine their nature.

    I tried to find the extension of the string, so that i can find the E.P.E. and hence the P.E. in that string. Afterwards, i can add the P.E. of the rod so that it forms a function V = f(theta), where V = P.E.. However, I can't seem to do the extension bit. Can someone help me?

    For some further note, this chapter's main focus is:
    In a system, possible position of equilibrium occur when the P.E. of the system has stationary values, which is d P.E. / d theta = 0.
    Minimum values of the P.E. correspond to stable equilibrium positions.
    Maximum values of the P.E. correspond to unstable equilibrium positions.
    (d^2 P.E. / d theta^2)

    Please, any help is appreciated. Thanks in advance.
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    the string must be vertical, right?
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    Doesn't have to, because it is linked to the smooth ring that is movable along the horizontal 8a above A rite?
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    if it's not vertical. how can it be equalibrium since no frictional force?
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    And I don't think there is a way to solve the problem without working this out first.
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    The answer given in the book says,
    if theta = the angle between upward vertical and the rod, then
    theta = 0 (unstable), pi/3 (stable), pi (unstable)
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    (Original post by kira18)
    Doesn't have to, because it is linked to the smooth ring that is movable along the horizontal 8a above A rite?
    Exactly. It's linked to a smooth ring mounted horizontally. So there can't be any horizontal force on the ring, because if there was, it would slide.
 
 
 

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