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    A uniform rod AB, of length 2a and mass 2m, rests with one end A in contact with a smooth vertical wall so that the rod lies in a plane perpendicular to the wall. The rod is supported by a smooth fixed horizontal rail which is parallel to the wall and at a disctance c from it. A particle of mass m is attached to the end B of the rod. Show that when the rod makes an angle \theta with the upward vertical the potential energy of the system is

    mg(4acos\theta - 3ccot\theta) + constant
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    (Original post by Bengaltiger)
    A uniform rod AB, of length 2a and mass 2m, rests with one end A in contact with a smooth vertical wall so that the rod lies in a plane perpendicular to the wall. The rod is supported by a smooth fixed horizontal rail which is parallel to the wall and at a disctance c from it. A particle of mass m is attached to the end B of the rod. Show that when the rod makes an angle \theta with the upward vertical the potential energy of the system is

    mg(4acos\theta - 3ccot\theta) + constant
    See image attached.

    P.E of mass at B = mg(acos\theta-x)
    P.E of uniform rod = -2mgx
    (All potential energies relative to fixed point)

    Simple trig
    tan\theta = c/(x+y)
    But y = acos\theta

    => c/tan\theta = x+y
    => c.cot\theta = x+y
    => c.cot\theta = x + acos\theta
    => x = c.cot\theta - acos\theta

    Therefore total potential energy of system
    V = P.E of mass at B + P.E of uniform rod
    V = mg(acos\theta-x) - 2mgx + constant
    V = mg(acos\theta-(c.cot\theta - acos\theta)) - 2mg(c.cot\theta - acos\theta) + constant
    V = 2mgacos\theta - mgc.cot\theta - 2mgc.cot\theta + 2mgacos\theta + constant
    V = 4mgacos\theta - 3mgc.cot\theta + constant
    V = mg(4acos\theta - 3ccot\theta) + constant
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    Thanks!

    Definatly deserve rep.

    Just a question, if i were to take my potential energy from the bottom of the rod, ie, where it is in contact with the wall, would it still work? Or do you always have to take it relative to a fixed point?
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    (Original post by Bengaltiger)
    Thanks!

    Definatly deserve rep.

    Just a question, if i were to take my potential energy from the bottom of the rod, ie, where it is in contact with the wall, would it still work? Or do you always have to take it relative to a fixed point?
    Yeah always to a fixed point. Cos if not, the apparent fixed point would move and make all your calculations wrong.
 
 
 
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