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Mechanics Pendulum

  1. Aug 22, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    http://img198.imageshack.us/img198/2900/mechanicspendulum1.jpg [Broken]

    2. Relevant equations

    [tex]w=\frac{d\theta}{dt}[/tex] (1)

    [tex]v=rw[/tex] (2)

    [tex]F=ma[/tex] (3)

    [tex]a=\frac{v^2}{r}[/tex] (4)


    3. The attempt at a solution

    I'm guessing that the horizontal component is using formula (4) where [tex]r=l.sin(\alpha)[/tex] and v is found from formula (2) such that [tex]v=l.sin(\alpha).w[/tex]

    Thus, [tex]a=\frac{(l.sin(\alpha).w)^2}{l.sin(\alpha)}=l.sin(\alpha).w^2[/tex]

    However, for the vertical component, I'm unsure how to begin. Oh and I'm not certain if I'm solving the horizontal correctly either, so don't hesitate to scold my mistakes :smile:
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 22, 2009 #2

    zcd

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    It's not really a pendulum problem as the mass doesn't oscillate through an energy valley. Consider an angle θ such that it lies on the circle around which the mass revolves. It can be said that [tex]\omega=\frac{d\theta}{dt}[/tex].

    The net force [tex]F_{net,y}[/tex] on the mass must be zero, and a free body diagram of the mass will note that [tex]F_{net,y}=0=T_{y}-mg[/tex].

    For the mass to revolve in a circle, there must be a centripetal force [tex]T_{x}=m\frac{v^{2}}{r}=m\omega^{2}r[/tex]; in this case, [tex]r=l\sin(\alpha)[/tex] and x component of tension can be rewritten as [tex]T_{x}=\omega^{2}lm\sin(\alpha)[/tex].

    Noting that [tex]\vec{T}=T_{x}\vec{i}+T_{y}\vec{j}[/tex], [tex]T_{x}=T\sin(\alpha)[/tex] and [tex]T_{y}=T\cos({\alpha})[/tex]. This gives [tex]T_{x}=T_{y}\tan(\alpha)[/tex].

    After some substitution, [tex]mg\tan(\alpha)=\omega^{2}lm\sin(\alpha)[/tex] and [tex]\omega^{2}=\frac{g}{l\cos(\alpha)}[/tex]
     
    Last edited: Aug 22, 2009
  4. Aug 22, 2009 #3

    Mentallic

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    Thanks, the Tx makes a lot of sense now!
    But I'm unfamiliar with the notations you've used here and so I couldn't follow it from there-on:

     
  5. Aug 23, 2009 #4

    zcd

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    I just separated it into components to form a right triangle. From the right triangle, you can see how each component is related to the other component and the tension force vector itself.
     
  6. Aug 23, 2009 #5

    Mentallic

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    Oh ok I see. While I still can't figure out what that vector notation is meant to represent (somehow, a right triangle), I can see how you got [tex]T_x=T_y.tan\alpha[/tex]

    Ok but now, what did you substitute and into which equations?
     
  7. Aug 23, 2009 #6

    ideasrule

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    There's this:

    And this:

     
  8. Aug 23, 2009 #7

    Mentallic

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    Aha now it all makes sense! So for most questions like these, to resolve the horizontal and vertical components, I should find the horizontal in terms of [tex]m,\omega,l,\alpha[/tex] and then the vertical in terms of the horizontal tension force.
     
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