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Centripetal things

  1. Nov 9, 2005 #1
    How do you find the place when the ball is about to "swing" out?

    A ball is attached to a vein, which has max tension of 300N; and Hercules is swinging the vein above his head. The length from vein to ball is .8meters. The ball has mass of 1.2kg..

    How do you find the velocity of the ball just before it swings out?


    Please advise... Thanks you!
     
  2. jcsd
  3. Nov 9, 2005 #2

    Doc Al

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    What's a vein? Do you mean chain?

    In any case, here's a hint: The max tension is also the max centripetal force that can be supplied.
     
  4. Nov 9, 2005 #3

    BobG

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    The ball will escape when it's 'centripetal force' is greater than the max tension of the vein.

    Hopefully, you know that Force = Mass * Acceleration.

    If you know your max 'centripetal force', then you can calculate your max centripetal acceleration.

    Knowing your max centripetal acceleration and your radius, you can calculate your max velocity.
     
  5. Nov 9, 2005 #4
    Centrepidal force is = mv^2/r,

    but i dunno my velocity... What is another way to find Centrepedal force?

    The only force in the x direction is the Tension force in the x component...

    But i dunno that.
     
  6. Nov 9, 2005 #5
    Is this right? :


    Fc = 300
    mv^2/r = 300
    v = sqrt (300R/m)
     
  7. Nov 9, 2005 #6

    Doc Al

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    You got it.
     
  8. Nov 9, 2005 #7
    aha! Thanks!!!! That was sooo much simpler than I thought...

    But... how can we do this based on FreeBody Diagram?

    There's 2 forces: Tension and Gravity right?

    And Tension has 2 components: x and y.

    ...

    Any advise on how I should draw the free body diagram?
     
  9. Nov 9, 2005 #8
    and at what position is the ball when it is about to fall out?

    Do i use V^2 = V0^2 + 2ax? What is its final speed? 0?
     
  10. Nov 9, 2005 #9

    Doc Al

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    I've been assuming that the ball travels in a horizontal circle. To give you a better answer, I'd need to see what this apparatus looks like. Do you have a picture?
     
  11. Nov 11, 2005 #10
    your assumption is right. it is just a plain circle.
     
  12. Nov 11, 2005 #11

    Doc Al

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    If the ball is travelling in a horizontal circle, and the tension in the "vein" (what's a vein?) is also horizontal, then the problem ignores gravity. You've already solved the problem.
     
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