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Pendulum problem, need to be checked

  1. Dec 4, 2012 #1
    A pendulum of length 0.6m swings so that at the bottom of its swing, it contacts a peg located 0.4m from the pivot point (0.2m from the CG). At what angle from vertical must the bob be released so that it will just make a revolution around the peg?

    F = Mg = M*V^2/r



    Mg (l-l cos x) = 1/2 MV^2
    g (l-l cos x) = 1/2 gr
    l (1- cos x) = 1/2 r
    1- cos x = r/2l
    1-r/2l = cos x
    x = cos^-1 (1- .2/(2*.6) = 33.6

    I tried an improvised impromptu experiment and wasn't getting anywhere near the angle 33.6 to work.
     
  2. jcsd
  3. Dec 4, 2012 #2

    mfb

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    Try to work backwards: Based on the condition "makes a revolution around the peg" (this is not equivalent to "has enough energy to reach that height"), find the required velocity at the lowest point first, and determine the angle afterwards.
     
  4. Dec 5, 2012 #3
    At the top of its motion the tension in the string is zero, but the centripetal accelerating force is its weight, so its speed there is not zero (otherwise it will just fall down and not complete the revolution.)
     
  5. Dec 5, 2012 #4
    The angle then, needs to be greater than 33.6?
     
  6. Dec 5, 2012 #5

    mfb

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    You will need more speed (and a larger angle), right.
     
  7. Dec 5, 2012 #6
    I am working on solving the problem with speed, but my examples from book have mass in the bob given for kinematics, still looking for energy examples. This type of problem kicks me good.
     
  8. Dec 5, 2012 #7

    mfb

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    You don't need the mass, it cancels in the equations anyway. If you like to work with energy, you can assume "m" as mass, or even any specific value you like (like 1kg).
     
  9. Dec 6, 2012 #8
    It can be done by just considering the energy. You just need to get the speed at the bottom. This requires you to analyze the circular motion of the bottom part of the string. Try and get its speed at the top of the revolution by assuming that the tension in the string is zero there, that is it isb just its weight that drives the circular motion.
     
  10. Dec 6, 2012 #9
    I think I figured it out. Ft at top = 0, gives speed of 1.4 m/s, using conservation of energy gets speed at bottom then use conservation of energy again to figure out height to drop bob from with 0 velocity. Then do some trig, yay! Anyway, I came up with an 80.5 degree angle. I can post equations and work tomorrow. Have to go to work then study for calc test tonight. Thanks.
     
  11. Dec 7, 2012 #10
    I also got the same angle.
     
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