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Tension in vertical circular motion

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data
    If you are spinning an object of mass 3.25 kg on a 0.8 m long chain at 20 rpm (vertical cirular motion), a) what is the tension at the top b) 43° from the top and c) at the bottom?


    2. Relevant equations
    String tension: T = Fc - mg cosθ
    Tension at top: T = (mv^2/r) - mg
    Tension at bottom: T = mv^2/r + mg
    Average velocity: v = r × RPM × 0.10472

    3. The attempt at a solution
    I've found the average velocity given the rpm, which is 1.58 m/s. I have all of the tension formulas that I need. The only problem is that I don't know how to find velocity for a given point on the circular path.
    Or...should I just assume that the object is being spun at a constant velocity?
     
  2. jcsd
  3. Dec 4, 2012 #2

    ehild

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    The rpm is given as a constant. The radius is constant, so is the speed.


    ehild
     
  4. Dec 4, 2012 #3

    Simon Bridge

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    It does not have to be a constant velocity does it?
    I think that has to go by context - it could well be that you are supposed to do all this with the constant average speed.

    [ehild will probably tell me off for this ...]
    ... but don't you have conservation of energy to give you some limits on what v can be at the top - given the rpm? i.e. there is a minimum speed it has to be going to stay on a circular path ... but if it goes too fast at the top (slowest part of the trajectory?) then it will complete a circle in less than the 3 seconds needed for 20rpm.

    (A constant rpm may imply a constant angular velocity or it could imply that you have 20 cycles in each minute - i.e. it is an average.
    I think this is a judgement call.)

    aside:
    "##v = r \times RPM \times 0.10472##" is better written as :
    ##v=r\omega_{[rpm]}\pi/30## ... says you know what the numbers mean.
    (It sounds pedantic but this sort of thing is worth marks.)
     
  5. Dec 4, 2012 #4
    Yeah, I guess I'll just assume that velocity is constant.
    Now for part b) T = Fc - (3.25 kg)(9.8m/s/s) cos(43)
    How do I find centripetal force in this situation?
     
  6. Dec 4, 2012 #5

    ehild

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    How do you get the centripetal force in general?

    ehild
     
  7. Dec 4, 2012 #6

    ehild

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    It would be a nice problem to find the speed along the circle assuming conservation of energy, so the object makes a certain revolutions per minute.
    The problem does not say that the object spins by itself. There can be some external torque to make it spin with uniform angular velocity.

    ehild
     
    Last edited: Dec 5, 2012
  8. Dec 5, 2012 #7
    Exactly
     
  9. Dec 5, 2012 #8
    Yeah, it was a stupid question: Fc = (mv^2)/r
     
  10. Dec 5, 2012 #9

    ehild

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    Exactly :smile:

    ehild
     
  11. Dec 5, 2012 #10
    I got tension at top to be -20.384 N, at 43° to be -11.83 N, and at bottom to be 43.32 N. Does this look good?
     
  12. Dec 5, 2012 #11
    You don't really need the -signs.
    As for the answers, generally speaking, tension is always greater at the bottom of a rotation (I think) due to gravitational potential energy being converted into kinetic energy - judging from this, I would say your answers are right.
     
  13. Dec 5, 2012 #12

    ehild

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    But is it possible to spin that object with a chain at so low rpm? Will it not fall on your head?

    ehild
     
  14. Dec 5, 2012 #13

    Simon Bridge

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    It does turn it into a "how do you handle the un-spoken assumptions?" problem, yes.
     
  15. Dec 5, 2012 #14

    ehild

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    Well, that means push instead of pull except the bottom point. Can a chain push? Or will that object fall on your head?
    But if the object is fixed to a rod, it looks correct.

    ehild
     
  16. Dec 5, 2012 #15

    PhanthomJay

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    Pretty tough to spin an object at constant speed in a vertical circle using a chain. I mean this is not a ferris wheel where a constant speed can be maintained with an external torque. I believe with a chain you have to move your hand in some sort of elliptical path to maintain constant speed throughout. Poorly worded question if you ask me.
     
  17. Dec 5, 2012 #16

    ehild

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    I agree.

    ehild
     
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