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Need Help - Regarding the mixture of circular motion and hooke's law

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data

    A 1.01 kg mass is attached to a spring of force constant 9.5 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.485 m at a rate of 2.14 revolutions per second?

    What i was thinking was using the following equation:
    Fc= Kx - mg
    Which would turn into
    a=kx-g
    (4pi^2(R)(F)^2 +g)/K=x
     
  2. jcsd
  3. Oct 17, 2012 #2

    ehild

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    Is it a horizontal surface the mass moves on?

    ehild
     
  4. Oct 17, 2012 #3
    It does not state......
     
  5. Oct 17, 2012 #4

    ehild

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    I guess the surfaceis horizontal, the angle of inclination would be stated otherwise. So the only force in the plane of motion is the force of the spring. Gravity cancels with the normal force of the surface.

    ehild
     
  6. Oct 17, 2012 #5
    Would u like to show me that using the equation...
    Kx=Mg somewhere you also need to include the acceleration i guess
     
  7. Oct 17, 2012 #6

    ehild

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    Mg is a vertical force. The object moves horizontally, along a circle. What force is needed to make it move along that circle of radius 0.485 m with 2.14 revolutions per second?

    ehild
     
  8. Oct 17, 2012 #7
    m= 1.01 kg
    K= 9.5 N/cm = 950 N/m
    r= 0.485 m
    f= 2.14 Hz

    Kx= ma
    kx= m x 4pi^2 x r x f^2
    950 N/mx = 88.472 N
    x= 0.0931 m
     
  9. Oct 18, 2012 #8

    ehild

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    Your solution is correct.

    ehild
     
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