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Pendulum angle problem

  1. Mar 28, 2007 #1
    1. The problem statement, all variables and given/known data
    1) A pendulum consists of a 2 kg stone swinging on a 4 m long string
    of negligible mass. The stone has a speed of 8 m/s when it passes its
    lowest point. a) What is the speed of the rock when it is at an angle of
    60 degrees to the vertical? b) What is the greatest angle with the vertical
    that the string will make during the stone's motion?


    2. Relevant equations
    Not really sure Just trying to do it with energy conservation..


    3. The attempt at a solution

    Well I figured out that at 60 degree's the height of the stone from the ground is 2m

    Thne I figured I should use E = k+mgy ...
    and came up with Einitial = (1/2)(2kg)(8m/s)^2 + (2kg)(-9.8m/s^2)(4m)

    Efinal = (1/2)(2kg)(v)^2 + (2kg)(-9.8m/s^2)(2m)


    Where to go Next... no clue
     
  2. jcsd
  3. Mar 28, 2007 #2

    Dick

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    "energy conservation". Set Einitial=Efinal and solve for v.
     
  4. Mar 28, 2007 #3

    Dick

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    Be careful about the sign and stuff on your gravitational terms though. It's mgh, where h is the height above the ground and there is no minus sign on the g.
     
  5. Mar 28, 2007 #4
    Really... can you tell me what that is derived from like ive seen deltaE = E final - E initial..
     
  6. Mar 28, 2007 #5

    Dick

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    There's no energy going into or out of the system. So deltaE=0. Right?
     
  7. Mar 28, 2007 #6
    Im getting v @ 60 degrees = sqrt(55.2) You?
     
  8. Mar 28, 2007 #7
    Okay for part B im going to assume that Einitial will remain the same... But Efinal will be = 0 + (2kg)(9.8m/s^2)(4m(1-cos(theta))
     
  9. Mar 28, 2007 #8
    Then again Im assuming I can set them equal to each other.. cause no energy is gained or lost. so That gives me 94.4 = (2kg)(9.8m/s^2)(4m(1-cos(theta))

    94.4 = 19.6(4m(1-cos(theta))

    94.4 = 78.4(1-cos(theta))

    .204081633 = -cos(theta)

    -.204081633 = cos(theta)

    cos^(-1)(-.2040816330) = theta
     
  10. Mar 28, 2007 #9
    im getting (theta) = 101.775744 degrees

    101.8 Degrees
     
  11. Mar 28, 2007 #10

    Dick

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    Sorry. No, I don't get that. Can you post details? And you didn't post part B either. So I don't know what you are working on.
     
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