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anonymous

  • one year ago

I need help with this circular motion problem

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  1. anonymous
    • one year ago
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  2. BAdhi
    • one year ago
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    so what have you tried so far?

  3. anonymous
    • one year ago
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    idk, I've tried so many different methods that i'm just lost and I have no idea what i'm doing anymore.

  4. anonymous
    • one year ago
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    I know that Ac is toward the center

  5. anonymous
    • one year ago
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    |dw:1433905966296:dw|

  6. Michele_Laino
    • one year ago
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    here are the external forces acting on each seat: |dw:1433906060256:dw|

  7. anonymous
    • one year ago
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    why is centripital force outward and not toward the center?

  8. Michele_Laino
    • one year ago
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    the magnitude of the centrifugal force is: \[m\frac{{{v^2}}}{R}\] where R is the radius of the rotating platform

  9. Michele_Laino
    • one year ago
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    since I measure those forces with a reference system located outside of the rotating platform

  10. anonymous
    • one year ago
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    ah ok, got it

  11. anonymous
    • one year ago
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    oh ok, so you need to find Fc, centripetal force

  12. anonymous
    • one year ago
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    Then you can find your V

  13. Michele_Laino
    • one year ago
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    |dw:1433906399481:dw|

  14. Michele_Laino
    • one year ago
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    it is simple, since, using trigonometry, we ca write: \[\Large m\frac{{{v^2}}}{R} = mg\tan \theta \]

  15. Michele_Laino
    • one year ago
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    after a simplification, we get: \[\Large v = \sqrt {gR\tan \theta } \]

  16. anonymous
    • one year ago
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    ah ok i see

  17. anonymous
    • one year ago
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    |dw:1433906745625:dw|

  18. anonymous
    • one year ago
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    so It would look like this then?

  19. Michele_Laino
    • one year ago
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    no, since the direction of the centrifugal force, not centripetal, is horizontal |dw:1433906906053:dw|

  20. anonymous
    • one year ago
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    Oh, ok I see

  21. Michele_Laino
    • one year ago
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    |dw:1433906970477:dw|

  22. Michele_Laino
    • one year ago
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    |dw:1433907024481:dw|

  23. Michele_Laino
    • one year ago
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    please, more precisely R is given by the subsequent equation: \[\Large R = \frac{d}{2} + l\sin \theta \]

  24. Michele_Laino
    • one year ago
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    Since we have no motion along the direction of a single chain, then we can apply the first principle of Newton, and we can write: \[\Large \tau = \sqrt {{{\left( {m\frac{{{v^2}}}{R}} \right)}^2} + {{\left( {mg\tan \theta } \right)}^2}} \] where \tau is the requested tension

  25. BAdhi
    • one year ago
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    @Michele_Laino. You are solving this problem with reference to a person who is on the the seat itself. Not on an inertial frame of reference. Am i right? Otherwise youll have to use centripetal force instead of centrifugal force

  26. Michele_Laino
    • one year ago
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    I have applied the theorem of Pitagora to this triangle: |dw:1433907439233:dw|

  27. anonymous
    • one year ago
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    yeah this is a centripetal force problem

  28. Michele_Laino
    • one year ago
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    no, if I solve that problem using the seat as reference frame, then as you well said my new reference system is not inertial, so, I can consider it again inertial provided that I introduce a fictious force , well that force is the centrifugal force of the rotating platform, at the position of the seat

  29. Michele_Laino
    • one year ago
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    the centripetal force is the force applied by the chain on the seat

  30. anonymous
    • one year ago
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    I got V = 3.6156 m/s

  31. Michele_Laino
    • one year ago
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    for example, if I cut the chain, then the seat will go outside the rotating platform, along a radial direction

  32. anonymous
    • one year ago
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    oh ok

  33. Michele_Laino
    • one year ago
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    the requested value of V, is: \[\Large \begin{gathered} v = \sqrt {g\left( {\frac{d}{2} + l\sin \theta } \right)\tan \theta } = \hfill \\ \hfill \\ = \sqrt {9.81\left( {4.13 + 3.9\sin \left( {17.9} \right)} \right)\tan \left( {17.9} \right)} \hfill \\ \end{gathered} \]

  34. BAdhi
    • one year ago
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    From what i understand what we do is making the frame of the guy seated on that seat an inertial frame of reference by introducing a centrifugal force. and thats why i said you are looking at the problem wrt to the seated person. am I wrong? Also only the person seated will see himself moving along the radial axis when the chain is broken while an observer outside will see him moving tangently.. please correct me if im wrong..

  35. Michele_Laino
    • one year ago
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    you are right! @BAdhi

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