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Rotation-Energy Question

  1. Feb 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A tall,cylindrical chimney falls over when its base is ruptured.Treat the chimney as a thin rod of lenght ##55.0m##.At the instant its makes an angle of ##35.0^°## with the vertical as it falls,what are the (a) ##a_r## of the top ? (b) ##a_t## of the top ? (Hint:Use energy considerations,not a torque )
    2. Relevant equations
    ##mgh=\frac 1 2Iw^2##
    3. The attempt at a solution
    ##L=55m##
    For tangential veloctiy I found ##\frac 1 2Mg(L-Lsin(35))=\frac 1 2(\frac 1 3ML^2)(\frac {(V_{tan})^2} {L^2})##

    from that I get ##V_{tan}=26.27 \frac m s##

    I stucked here... I thought ##\frac {V_{tan}} {rt}=∝##

    ##t## is the time that end point comes that angle.
    but it didnt worked out.

    ##a_r=\frac {(v_r)^2} {r}## I dont know how to find radial velocity...

    Thanks
     
  2. jcsd
  3. Feb 12, 2017 #2

    TSny

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    Check the trigonometry that you used to get the left side. (The 35o angle is measuresd from the vertical.)
    I don't understand your equation here. There is no need to worry about time in this problem.

    If a particle moves in a circle at constant speed, does it have any radial acceleration? Does it have any radial velocity? Thus, does your equation ##a_r=\frac {(v_r)^2} {r}## hold?
     
  4. Feb 12, 2017 #3
    sorry it will be cos35.
    its ##w-w_0=∝t##
    I think velocity is a vector and as the rod moves , the velocity vector will change,thus, there will be some acceleration
     
  5. Feb 12, 2017 #4

    TSny

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    Yes.
    Yes. If you review the basic formula for radial (or centripetal) acceleration of a particle moving in a circle, you will see that it does not involve radial velocity, it involves tangential velocity.
     
  6. Feb 12, 2017 #5
    so its ##a_r=\frac {(v_t)^2} {r}## ??
     
  7. Feb 12, 2017 #6

    TSny

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    Yes, except be careful with signs (i.e., directions).
     
  8. Feb 12, 2017 #7
    Ok I found ##a_r## correctly,what about ##a_t## ?
     
  9. Feb 12, 2017 #8

    TSny

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    Despite the wording of the problem, I think using torque is the way to go here.
     
  10. Feb 12, 2017 #9
    yep I found thanks again
     
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