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Homework Help: Confusing pendulum question.

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A mass at the end of a light rigid rod of length (r) is dropped from a position vertically above the pivot point. What is the tension in the rod when the mass reaches the lowest point of its swing?

    2. Relevant equations
    PEi +KEi = PEf +KEf and maybe also Acen = V^2/r, maybe fnet=ma

    3. The attempt at a solution

    Well I managed to solve a similar problem for angular velocity it came out to (2g/r)^1/2 but that was half the height. I know the tension isn't simply mg. I am trying to figure out if I need to start the equation with mg2r = (1/2)MV^2 + V^2/r. Honestly I am stuck, please help.

    (Update: for this problem I think W = (4g/r)^1/2 and V = (4gr)^1/2.)
    (Update #2: I said that at the bottom of the swing T= (V^2/r)+mg => 4g +mg, is this correct?)
    (Update #3: Sorry to waist any ones time. Fnet = ma (a in this case is centripetal) so T = (MV^2/r) + mg = 5mg! I believe this is correct.)
    Last edited: Nov 7, 2012
  2. jcsd
  3. Nov 7, 2012 #2
    I have another question which I am stumped on. If say the pendulum starts half-way then:


    Okay so assuming I did the algebra correctly. Can't I now say:

    If this is the case then how is this equivalent to w=(2g/r)^1/2 because if I start off by using a bridge equation and saying mgr=(1/2)m(w^2*r^2) then:

    (Has my algebra gone wrong somewhere or is w=((2gr)^1/2)/r = w=(2g/r)^1/2? Can I just cancel that r even though it is under the square root bracket? Or maybe I can't say that (2gr)^1/2=wr?)
  4. Nov 8, 2012 #3


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    5mg is correct for the first problem.
    Indeed it is. (√a)/b = (√a)/(√b2) = √(a/b2).
    But you didn't really need to get into sq roots for this problem.
    ΔE = mgh = mv2/2
    centripetal accn = mv2/r = mgh*2/r
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