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Conical pendulum circular motion question

  1. Aug 5, 2012 #1
    1. The problem statement, all variables and given/known data
    By resolving forces horizontally and vertically and using newton's second Law, find an expression for the angle swung out of a Chair-o-plane ride.
    Im just not really sure how to resolve the forces vertically and horizontally.


    2. Relevant equations

    So far I know that forces acting on the chair are its own weight, mg and the tension in the chair, T.
    Also that
    T cos ѳ = mg
    a= rω^2 ( ω being angular velocity).
    Force causing the motion = F = mrω^2
    = ma << newtons second law
    and horizontally T sin ѳ = mrω^2
     
  2. jcsd
  3. Aug 5, 2012 #2
    that's ok.does not that give you
    tanα=r(ω2)/g
     
  4. Aug 5, 2012 #3
    Yeah, is that vertical or horizontal.?
     
  5. Aug 5, 2012 #4
    with vertical.
     
  6. Aug 5, 2012 #5
    So is that expression for the angle swung?
     
  7. Aug 5, 2012 #6
    yes, but if the situation is much different from what you telling ,there can be changes.
     
  8. Aug 5, 2012 #7
    Cool thanks. :)
    That's the situation, the full questions are in a chair-o-plane 1. will a child swing out at a greater angle than a much heavier adult? 2. Will people on the inside swing out at the same angle as those on the outside? 3. Will empty chairs be a problem? and 4. What will happen as the speed increases. So by resolving forces horizontally and vertically to find the expression for the angle swung out we need to answer those four question. :)
     
  9. Aug 5, 2012 #8
    you can predict it from the formula just written which does not depend on mass.
     
  10. Aug 5, 2012 #9

    Simon Bridge

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    I think it is easy to get mixed up if you don't keep track of what you've done and what you know. It can help to draw the pictures of what is happening - you want one overhead and another from the side.

    You have already noticed: [itex]T\sin(\theta)[/itex] is the horizontal unbalanced (tension) force ... which must be the centripetal force. [itex]T\cos(\theta)[/itex] is the balanced vertical component ... so you have two equations with two unknowns. Number them.

    1. [itex]T\cos(\theta)=mg[/itex]
    2. [itex]T\sin(\theta)=mr\omega^2[/itex]

    The trouble is that you have to deal with trig functions right?
    But it solves itself - you can do it blindly: solve for T in your first equation ([itex]T=mg/ \cos(\theta)[/itex], then substitute into the second one ... realize that sin/cos=tan. OR just divide the two equations directly - the T cancels out.

    Now you just need the inverse tangent.

    Keep these things straight and the rest will follow [note: I took too long and you had other answers :) ]
     
  11. Aug 5, 2012 #10
    Thanks so much this has been the biggest help!
    Also how could I show that a heavier person will swing out at a greater angle than a small child if the equation cuts out mass. Does that mean that the weight wont affect the angle swung.?
     
  12. Aug 5, 2012 #11

    Simon Bridge

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    That's what it means - when the equation for angle does not include some factor - like temperature or color or (ahem) weight, then we say that the angle does not depend on that factor.
     
  13. Aug 5, 2012 #12
    Ok thank you so much, I think I can finish it now. :)
     
  14. Aug 5, 2012 #13

    Simon Bridge

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    However - the size of the chair's occupant will affect that r in your relation: it gets smaller.
    Can you see how?
     
  15. Aug 5, 2012 #14
    Umm not really, doesn't r= l sin theta. So how does the weight of the occupant affect r.? :)
     
  16. Aug 5, 2012 #15

    Simon Bridge

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    Didn't say "weight of the occupant", I said "size" ... there's a difference ;)

    Forinstance: on a regular swing that just goes back and forth, the swinging is faster for an adult than it is for a child on the same swing. It's also faster for the child if she stands up.

    I don't know if you are expected to take this effect into account ... but it should be worth bonus marks if not.
     
    Last edited: Aug 5, 2012
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