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Tension in rotating rod.

  1. Apr 24, 2016 #1
    a thin uniform rod of length l and density d is rotating with angular velocityω about an axis passing through one of its ends and perpendicular to it.find the tension int the rod as a function of x.(x= dist from axis of rotation).area of cross section = a

    attempt at solution:

    consider a small of length dx at dist x from axis.
    upload_2016-4-24_21-45-30.png
    for equilibrium of rod T = T-dT + dmω^2 x
    dT = dm ω^2 x
    integrating
    T(x) =(d * a* ω^2 (x^2 - l^2))/2
    this is negative of answer.where am i wrong?
     
  2. jcsd
  3. Apr 24, 2016 #2
    I guess that [itex]a[/itex] and [itex]d[/itex] are the cross-sectional area and the density of the rod, respectively.

    Your second equation, the one for [itex]dT[/itex], seems ok to me. I do not understand how can you obtain your last result from it. What are the limits of your integral?
    Seems to me you're missing a sign in the integration, or, equivalently, you are swapping the integration limits.
     
  4. Apr 24, 2016 #3

    haruspex

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    Further to @FranzDiCoccio 's comments, since it is a thin rod I would interpret "density" as meaning mass per unit length. You do not need a variable for the cross sectional area.
     
  5. Apr 24, 2016 #4

    TSny

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    I think the problem is associated with the minus sign in writing T - dT for the tension at the outer end of dx. The change in the function T(x) as you go from x to x + dx should be written as dT not - dT. The change in the function will be a negative quantity. So, the value of T at x+dx is T + dT where dT is a negative quantity.

    By writing the tension at x + dx as T - dT, your symbol dT does not represent the change in the function T(x), it's the negative of the change in T(x).
     
    Last edited: Apr 24, 2016
  6. May 13, 2016 #5
    oh!
    thanks!
     
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