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Speed of a pendulum at certain angles

  1. Oct 1, 2006 #1
    A simple pendulum consists of a ball suspended by a string from the ceiling. (Treat the ball as a point particle.) The string, with its top end fixed, has negligible mass, and is non-stretchy. In the absence of air resistance, the system swings back and forth in a vertical plane. If the string is 2.2 m long, and is released from an initial angle of 30.8° with the vertical, calculate the speed of the particle when
    a)the ball is at the lowest part of its trajectory
    I used equation Kf + Uf = Ki + Ui
    1/2mvf^2 + mgyf = 1/2mvi^2 + mgyi
    1/2mvf^2 - m(9.8)(2.2) = 0 - m(9.8)(2.2cos30.8)
    this gave me 1/2mvf^2 = 21.56- 18.52
    Vf^2= 6.08
    Vf= 2.46m/s is this right?
    b)the string makes an angle of 15.4° with the vertical
    1/2mvf^2 + mgyf = 1/2mvi^2 + mgyi
    1/2mvf^2 - m(9.8)(2.2cos15.4)= 0 - m(9.8)(2.2cos30.8)
    Vf= 2.13m/s
    I thought this made sense since at this point the velocity is still increasing as it approaches it's lowest trajectory but it isn't right. I figured it's how I'm doing the displacement so can anybody point how where I went wrong?
     
  2. jcsd
  3. Oct 2, 2006 #2

    andrevdh

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    Homework Helper

    I think your answers are both correct.
     
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