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Horizontal circular motion problem, what am doing wrong?

  1. Feb 26, 2016 #1
    1. The problem statement, all variables and given/known data

    The County Fair Swing carries the mass of riders and chairs in a circular path in a horizontal plane while suspended by cables or chains. Let's assume that:

    Each chair with riders is supported by a single cable

    The tension in the cable equals 2 x the total weight riders and chair

    The speed of the center of mass at the end of the cable is 10 m/s.

    Determine the radius of the circular path in meters.


    Selected Answer:
    50

    2. Relevant equations

    T=mv^2/r

    3. The attempt at a solution

    T=mv^2/r
    so r=10^2/2
    100/2=50
    50=r
     
  2. jcsd
  3. Feb 26, 2016 #2
    Hi Sherlock:

    I am not sure I understand the problem statement correctly, but it seems to me you are assuming that the chains are horizontal. I visualize a different picture with the chains at an angle. If that is correct, then the tension has a horizontal and a vertical component.

    Hope this helps.

    Regards,
    Buzz
     
  4. Feb 26, 2016 #3

    jbriggs444

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    If the cables are purely horizontal, what force supports the chairs and riders against gravity?

    [i.e. -- what Buzz said]
     
  5. Feb 26, 2016 #4
    this is all the information I was given, I imagine the angle too, but I have no value for theta. is that something I can solve for with the information given?
     
  6. Feb 26, 2016 #5
    Hi Sherlock:

    The vertical component of the tension balances what force? Visualize a right triangle with the chain as the hypotenuse.
    Regards,
    Buzz
     
  7. Feb 26, 2016 #6

    BvU

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    What do you do with that information ?
    Do I read T/m = 2 here ? I always thought weight = mg so I am missing a factor g -- not to mention the dimension of g !

    Oh, and -- in your relevant equation -- what do you mean with ##\vec T## ?
     
  8. Feb 26, 2016 #7
    Sherlock,

    Have you drawn a free body diagram for the combination of rider and chair, or do you feel like you have advanced beyond the point where you need to use free body diagrams?
     
  9. Feb 26, 2016 #8

    BvU

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    Well now, all this response and not even one :welcome: -- let me make up for that: :welcome: ! :smile:
     
  10. Feb 26, 2016 #9
    t=tension. and I have drawn diagrams, and i just, tried v^2/2g, and it was also incorrect, am i using the right Equations?
     
  11. Feb 26, 2016 #10
    thanks!
     
  12. Feb 26, 2016 #11

    Nidum

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    It is always much easier to sort these types of problem out if there is something to look at .
    pf01-jpg.95812.jpg

    Mark in the forces which you think are acting and any other information you consider relevant .
     
  13. Feb 26, 2016 #12
    Let's see what your free body diagram looks like
     
  14. Feb 26, 2016 #13
    does this help
     

    Attached Files:

  15. Feb 26, 2016 #14
    Hi Sherlock:

    Your diagram is a good start, but you need to add the horizontal and vertical components of the tension.

    Regards,
    Buzz
     
  16. Feb 26, 2016 #15
    Yes. OK If T is the tension, what are its components in the x and y directions, in terms of the angle the rope makes with the horizontal?
     
  17. Feb 26, 2016 #16
    I want to say
    (T Cos(theta)i, T sin(theta)j) where the y component would be T Sin(theta)=mg..?
    so t/mg= 2, so sine (0) =0, so it is in the horizontal plane, the alternative T Cos(0)= T*1= T.
     
  18. Feb 26, 2016 #17
    so all the velocity would be in the horizontal plane?
     
  19. Feb 26, 2016 #18
    Hi Sherlock:

    If you visualize the right triangle, you are given the facts regarding the vertical component, and the tension is along the hypotenuse. You don't need the angle to calculate the horizontal component. You can use the Pythagorean theorem.

    Regards,
    Buzz
     
  20. Feb 26, 2016 #19
    If T = 2mg, then ##2mg \sin\theta=mg##, then what is ##\sin \theta## and what is ##\theta##?
     
  21. Feb 26, 2016 #20

    BvU

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    Pity T is missing in the side view in post #13...
     
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