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Circular Motion Help Please

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A 40kg gymnast swings in a vertical circle on a bar. Her centre of mass is 1.20m from the bar. At the highest point her centre of mass is moving at 1.0ms^-1)
    ai) How fast is she moving when her centre of mass is level with the bar?
    aii) How much force does she have to exert on the bar in order to hang on at the lowest point in the swing?


    2. Relevant equations
    Answer to ai) is 5ms^-1 and aii) is 1.6kN


    3. The attempt at a solution
    I got no idea where to start. ><
     
  2. jcsd
  3. Mar 5, 2012 #2

    tiny-tim

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    Hi Kotune! :smile:

    (try using the X2 button just above the Reply box :wink:)

    Start with ai)

    Use conservation of energy …

    what do you get? :smile:
     
  4. Mar 5, 2012 #3
    Hello! :)
    I get ai) now ty! I got 4.8ms-1 :)
     
  5. Mar 5, 2012 #4
    For Problem (a) just start by conserving the changes in kinetic and potential energies... Since as the gymnast moves from the highest point to the level of the bar.... There is increase in potential equal to MGH.. Got it?
     
  6. Mar 5, 2012 #5

    tiny-tim

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    ok, now for aii) first find the speed at the bottom (same method), then find the centripetal acceleration, then find the "tension" …

    what do you get? :smile:
     
  7. Mar 5, 2012 #6
    And for (b) am afraid you neglected self weight and you only calculated the centripetal force required to hold the gymnast at that position..... But so far... You just have to calculate the velocity of the gymnast center of mass as procedure in (a) then calculate the required force by adding both her weight and the centripetal force...
     
  8. Mar 5, 2012 #7
    I'm confused to as how to find the speed at the bottom because isn't the speed constant?
     
  9. Mar 5, 2012 #8
    Yes speed should remain constant.
     
  10. Mar 5, 2012 #9

    tiny-tim

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    uhh? :redface: you've already found that the speed is increasing, in part ai) :confused:

    use conservation of energy again …

    then get some sleep! :zzz:​
     
  11. Mar 5, 2012 #10
    for the mgh. Mass is still 40kg, gravity is still 9.8, that means the height has to change? But it can't be 0 ><.
     
  12. Mar 5, 2012 #11

    tiny-tim

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    the height difference is the diameter :wink:
     
  13. Mar 5, 2012 #12
    Does that mean h is 1?
    .5mv2 = mgh
    .5(40)v2 = 40 x 9.8 x 1
    then the speed at the lowest point is 4.4?

    Edit: wait no since its diameter then the h is 2.4?
     
  14. Mar 5, 2012 #13

    tiny-tim

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    Yup! :biggrin:
     
  15. Mar 5, 2012 #14
    Sorry dumb question but why is h the diameter? xD

    .5(40)v2=40 x 9.8 x 2.4
    v = 6.86

    Is tension = centripetal force + weight?
    = then mv2/r + m x 9.8
    = 40 x 6.85 2 / 2.4 + (40 x 9.8)
    = 1176N which is wrong ><
     
  16. Mar 5, 2012 #15

    tiny-tim

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    (h is the difference between the top and bottom positions of the centre of mass)

    did you include the 1.0 m/s when you used conservation of energy?
     
  17. Mar 5, 2012 #16
    ? then is v = 7.85?
     
  18. Mar 5, 2012 #17

    tiny-tim

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    You can't just add the 1.0, you have to do all the squaring and the unsquaring.
     
  19. Mar 5, 2012 #18
    I don't get it. :frown:

    The equation would then be
    .5 x (40) x (v+1)2 = (40) x (9.8) x (2.4)?
     
  20. Mar 5, 2012 #19

    tiny-tim

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    1/2 m (v2 - 1.02) = mgh
     
  21. Mar 6, 2012 #20
    thanks I got it ^_^
     
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