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Centripetal Acceleration question

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a conical pendulum that consists of a bob
    on one end of a string of negligible mass with the other end of the
    string attached to a point on the ceiling, as shown. Given the proper
    push, this pendulum can swing in a circle at a given angle, maintaining
    the same distance from the ceiling throughout its swing. If the mass of
    the bob is , the length of the string is d, and the angle at which it
    swings is θ, what is the speed (v) of the mass as it swings? [Hint: Find
    the vertical and inward radial components of the string’s tension.]

    2. Relevant equations

    FR = (mv2)/r

    3. The attempt at a solution

    For [tex]\Sigma[/tex]Fy I got FT = mg/cos θ

    I said r = dsinθ

    And for [tex]\Sigma[/tex]FR I got FTdsin2 θ = v2

    Plug in mg/cos θ for FT and sqrt(gd tanθ sin θ)= v

    Is that right?
     

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  3. Nov 1, 2009 #2

    Delphi51

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    Homework Helper

    I don't follow your equation for the radial forces. You have the Ft*sinθ component equaling the centripetal force, don't you? Surely there has to be an m in it. Why is the sin squared?
     
  4. Nov 2, 2009 #3
    At the start I have
    FTsin θ = mv2/r

    Put in dsin θ for r and move to the other side

    FTsin2 θ = mv2

    Put in mg/cos θ from the y forces for FT

    mgdsin2 θ/cos θ = mv2

    The masses cancel and 1 of the sin2/cos becomes tangent

    gd tan θ sin θ = v2

    sqrt(gdtan θ sin θ) = v

    Did I do something wrong? r is equal to dsin θ right?
     
  5. Nov 2, 2009 #4
    :( from your image, I find that [tex]sin \Theta = \frac{d}{R}[/tex]
     
  6. Nov 2, 2009 #5
    sin is opp/hyp right? r would be opp and d would be hyp. sin θ = r/d So dsin θ = r
     
  7. Nov 2, 2009 #6

    Doc Al

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    Staff: Mentor

    No, your solution is perfectly fine. And yes, r = d sinθ. (In your first post you did leave off an m, as Delphi51 points out.)
     
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