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Circular motion & tension

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A rock is whirled at the end of a rope in a vertical circle

    Find a general expression fo the Tension

    What is the magnitude of the total acceleration

    2. Relevant equations



    3. The attempt at a solution

    [tex]\sum[/tex]Fr = ma = T + mgcos[tex]\theta[/tex]

    T = ma - mgcos[tex]\theta[/tex] not sure

    a = (T + mgcos[tex]\theta[/tex])/m not sure
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2009 #2

    rock.freak667

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    At an angle θ to the vertical, what is the component of the weight along the line of tension?
     
  4. Oct 13, 2009 #3
    not sure what u mean
     
  5. Oct 13, 2009 #4

    rock.freak667

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    Draw the mass at angle θ to the vertical.


    Can you split the weight mg into two components?
     
  6. Oct 14, 2009 #5
    into two components, would that be the tangental compenent in the direction of the velocity and the second pointing towards teh center
     
  7. Oct 14, 2009 #6

    rock.freak667

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    actually it points opposite to the velocity and away from the center. What are these two components in terms of the angle?
     
  8. Oct 14, 2009 #7
    Component along line of tension ma = mgcosQ

    how do i do tat for the velocity, do i integrate that
     
  9. Oct 14, 2009 #8

    rock.freak667

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    the component is mgcosθ.

    So the T points towards the center and the mgcosθ points away from the center. What is the resultant force equal to ?
     
  10. Oct 15, 2009 #9
    Fr = T - mgcosQ = 0
     
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