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The total tension acting on a rotating rod

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the total tension acting on a rod rotating about its end with an angular velocity of w as a function of its length x(length)

    2. Relevant equations

    F = ma


    3. The attempt at a solution

    Let the function be T(x) where x is the length of the rod.
    Considering an interval between x and x + dx
    For an increment in dx, let the function change by dT(x)
    dT(x) = T(x+dx) - T(x)
    T(x+dx) is the tension acting on a rod of length x + dx
    T(x) is the tension acting on a rod of length x
    Therefore, dT(x) is the tension acting on the tiny element of length dx

    dT(x) = dm(w^2) x
    dm = ρdx

    ∫dT(x) = ρ(w^2)∫ xdx from T(x) = 0 at x = 0 to T(x) at x = x.
    This gives me tension as a function of the length of the rod

    Is what I've done correct & rigorous? Is my general approach to calculus alright or am I viewing things wrong?
     
  2. jcsd
  3. Feb 11, 2017 #2

    TSny

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    Since the right side is positive, this implies that dT is positive. Should the tension increase as x increases?

    Where should the tension be zero?
     
  4. Feb 11, 2017 #3
    I'm not trying to find tension as a function of x. I'm finding the total tension acting on a rod of length x
     
  5. Feb 11, 2017 #4

    TSny

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    OK. But the tension in the rod varies along the length of the rod. So, when you say you need the "total tension", what does that mean?

    (The notation is a little confusing. You use "x" as a variable in your analysis, but you also use "x" for the total length of the rod.) [EDIT: I understand now that you are considering rods of variable length.]
     
    Last edited: Feb 11, 2017
  6. Feb 11, 2017 #5
    By total tension I mean the sum of all tensions acting at every point along the rod. I'm finding T(x), and here by x i mean the length of that rod. I basically want this function to give me the tension acting on a rod of any length for eg. T(5) would be the total tension acting on a 5m rod.
     
  7. Feb 11, 2017 #6

    TSny

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    In your analysis, you didn't sum the tensions, T. Instead, you summed the infinitesimal changes in tension dT.

    If you take an arbitrary cross-section of the rod, there will be a certain value of tension, T, at that cross section. That means that the material to the left of the cross section is exerting a force T toward the left on the material to the right of the cross section, while the material to the right of the cross section is exerting the same amount of force T toward the right on the material to the left of the cross section. I'm not sure I understand what it means to sum all tensions at every point along the rod.
     
  8. Feb 11, 2017 #7
    dT(x) = T(x+dx) - T(x)
    T(x+dx) is the total tension acting on a rod of length x + dx
    T(x) is the total tension acting on a rod of length x
    Therefore, dT(x) is the tension acting on the tiny element of length dx
    I've simply summed these up from 0 to x
     
  9. Feb 11, 2017 #8

    TSny

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    At what cross section is the tension T(x+dx) acting?

    Could it be that the question is asking for the tension T in the rod at the left end of the rod as shown below?
    upload_2017-2-11_12-9-59.png
     
  10. Feb 11, 2017 #9

    TSny

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    If T(x) is the tension at the left end of the rod when the rod has a total length x, then I think your analysis in your first post is OK. So, when you switch from x to x + dx, you are actually talking about two different rods (as you clearly said):
    upload_2017-2-11_13-4-0.png


    The difference in T(x+dx) and T(x) must provide the centripetal force for the extra mass dm of the longer rod. Your work looks correct if this was your way of looking at it. So, T(x) does in fact increase as x increases. And T(x) = 0 when x = 0.

    Sorry for not seeing this sooner.

    You don't need to approach it this way. In fact, you don't need calculus at all if you know the principle that the center of mass of an object has an acceleration acm given by
    Fnet, external = Macm.
     
  11. Feb 12, 2017 #10
    Okay, thanks!
     
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