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Pendulum Momentum Question

  1. Mar 18, 2008 #1
    [SOLVED] Pendulum Momentum Question

    1. The problem statement, all variables and given/known data

    Figure 8-34 shows a pendulum of length L = 1.25 m. Its bob has speed v0 when the cord makes an angle theta0 = 40 degrees with the vertical.

    What is the speed of the bob when it is in its lowest position if v0 = 8 m/s?

    The speed is 8.35 m/s.

    What is the least value that v0 can have if the pendulum is to swing down and then up to a horizontal position?

    2. Relevant equations

    W + KE1 + PE1 = KE2 + PE2

    3. The attempt at a solution

    W + KE1 + PE1 = KE2 + PE2:

    0 + (1/2)m(8.57 m/s)^2 + 0 = (1/2)m v^2 + m(9.8 m/s^2)1.25m

    v= 6 or 7 something. The answer in the book is 4.33 m/s
    Last edited: Mar 18, 2008
  2. jcsd
  3. Mar 20, 2008 #2
    Use 8.35 m/s.
  4. Mar 20, 2008 #3
    Sorry. I forgot to put the 8.35 in there. But, it still doesn't give me the correct answer. I get 6.74 m/s.
  5. Mar 20, 2008 #4


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    Homework Helper

    The final kinetic energy will be zero when the pendulum reaches the horizontal when considering the minimum initial velocity.

    The 8.35m/s does not apply here, since that value is based on the first part of the question and is irrelevant to this part. You are trying to find the velocity of the pendulum needed at 40 degrees to the vertical that will just get you to the horizontal position on the other side.
  6. Mar 20, 2008 #5
    Oh, hello! Man. I know I can work it now. I'll do it after I eat. lol.
  7. Mar 22, 2008 #6
    Got it. Thanks. Sometimes, I just read over the problem too quickly and assume I read it correctly.
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