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Quick Pendulum Question

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A pendulum with a mass of 10kg and a length of 4.1 meters is pulled back 26 centimeters to the right from the vertical and released. As it is released, an additional amount of work equal to 5x10^-1 joules is done on the pendulum in the tangential direction toward the left. What is its period in seconds? Assuming a perfect pendulum (No more work is done on it by any force), what is the total energy of the pendulum at all times? What is the maximum distance that the pendulum can ever be from the vertical after it is released?

    Since there's a perfect pendulum now, does that mean I don't add that extra .5 joules to the total energy...or does that still apply? I'm pretty sure that I don't add it now but I'm not sure. Also, does that extra work affect the period? No right? Furthermore, in regards to the maximum distance, is that simply the distance where the angle of the pendulum doesn't exceed 90 degrees? Thank you very much for your help.
     
  2. jcsd
  3. Apr 28, 2013 #2

    Doc Al

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    The additional work adds to the total energy. The pendulum wasn't merely released, but was pushed.

    How would you find the period of a pendulum? Does it have anything to do with energy?

    No. How high can the pendulum swing?
     
  4. Apr 28, 2013 #3
    Got the first part, and period is 2piSQRT(L/g) so no it doesn't. Is how high the pendulum can swing simply the amplitude of the pendulum?
     
  5. Apr 28, 2013 #4

    Doc Al

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    Good.

    Yes. But note that the question asks for distance from the vertical, so you'll have to figure that out.
     
  6. Apr 28, 2013 #5
    can I simply do L(1-cos(θ)) ?
     
  7. Apr 28, 2013 #6

    Doc Al

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    That will give you the height above the lowest point as a function of angle. That might prove useful as a step towards the answer.

    Hint: Use the total energy to find the maximum height.
     
  8. Apr 28, 2013 #7
    Oh so I would use the formula for PE which is mgh, and since its at the top of its swing the PE is the TE so TE = mgh, then isolate for h, convert that to L(1-cos(θ)) and have the new equation L(1-cos(θ))= TE/mg, then further isolate the cos(θ) to cos(θ) = -1 + TE/MGL. then i get the angular amplitude, convert that to linear by doing Lsin(θ) ?
     
  9. Apr 28, 2013 #8

    Doc Al

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    Looks good to me.
     
  10. Apr 28, 2013 #9
    Thank you very much for your help.
     
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