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Conservation of energy of a pendulum

  1. Oct 28, 2008 #1
    An 80cm long pendulum with a .60kg bob is released from rest at an initial angle of Theta with the vertical. At the bottom of the swing, the speed of the bob is 2.8m/s.

    a)what is Theta?
    b)What angle does the pendulum makes with the vertical when the speed of the bob is 1.4m/s?


    Revelant equation:
    Ui + Ki = Uf + Kf

    I converted 80cm to .8m.
    mgh + 0 = 0 + 1/2 m v^2
    .60kg * 9.81 m/s * h = 1/2 .60kg * (2.8m/s)^2
    I need to solve for h which is .3995922528 approx .4m

    Length is .8m
    h = L - Lcos Theta
    .4m = .8m - Lcos Theta
    .5 = Cos Theta
    Cos^-1 = 60


    Is this correct?
    and Part b is the same procedure correct?
     
  2. jcsd
  3. Oct 28, 2008 #2
    Looks OK so far.
    And you can solve part b in a similar way.
     
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