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What is the pendulum's length?

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that a simple pendulum consists of a 50.0 g bob at the end of a cord of negligible mass. The angle (theta) between the cord and the vertical is given by: (theta)=(0.01000rad)cos[(5rad/s)t ], where t is time. A. What is the pendulum's length? B. What is the pendulums maximum potential energy? C. What is the maximum tension in the cord?

    2. Relevant equations
    PE=mgh
    KE=1/2mv^2

    3. The attempt at a solution
    I got part a. Part b is throwin me off. I know how to find the maximum kinetic energy (KE=5rad/s x .392m x .1 rad) Im confused as to how to find potential, and even more lost on the tension
     
  2. jcsd
  3. Apr 27, 2008 #2
    I have feel like mgh should equal KE because KE is energy in motion and the max potential of a pendulum would mean that its not moving right? As its about to swing the other way?
     
  4. Apr 27, 2008 #3

    tiny-tim

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    Hi BoostAdiction! :smile:

    Yes, that 's right … or, to put it in more mathematical language, PE + KE = constant, so PE is maximum when KE is minimum … in this case, when KE = 0! :smile:

    Hint: tension = radial component of weight minus acceleration … and the acceleration is … ? :smile:
     
  5. Apr 27, 2008 #4
    Woot! So in this case...PE=KE and KE = 5rad/s x .392m x .1 rad which equals, .196 J meaning PE does as well..Hopefully. Im gonna assume the acceleration is.... 0.5 rad/s^2?
     
  6. Apr 27, 2008 #5

    tiny-tim

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    Can't work out how you got that … you should be using the formula for radial acceleration in circular motion … and it should depend on the angle. :smile:
     
  7. Apr 27, 2008 #6
    Just made a guess on it :D. I cant find the radial acceleration in circular motion...all i found was the derivative of omega/derivative of t
     
  8. Apr 28, 2008 #7

    tiny-tim

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    Have you been taught acceleration = v²/r = ω²r? :smile:
     
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