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Trapeze performer

  1. Jul 31, 2014 #1
    1. The problem statement, all variables and given/known dataA trapeze performer swinging in a vertical circle can be modeled as a rotating bar pivoted at one end, as shown in figure 8.26. Let the mass of the bar be 70 kg with the center of mass located 1.2 m form the axis of rotation. If the angular velocity at the top of the swing is 3 rad/s, what is the value of w at the bottom of the swing? (hint use the principle of conservation of energy.)



    2. Relevant equations




    3. The attempt at a solution
    I am not really sure where to begin with this problem. I was thinking that I could use kinetic and potential energy to solve this problem and basically convert the potential energy at the top to kinetic energy to find the velocity but it doesn't say how high the bar is.
     
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  3. Jul 31, 2014 #2
  4. Jul 31, 2014 #3

    Nathanael

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    It tells you that "the center of mass located 1.2 m from the axis of rotation"

    So how much does the height (of the center of mass) change from the top of the swing to the bottom?
     
  5. Jul 31, 2014 #4
    2.2 meters
     
  6. Jul 31, 2014 #5

    Nathanael

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    Well no, it would be 2.4 meters, but I think that you just made a typo.

    So you can now use the (change in) gravitational potential energy (along with conservation of energy) to solve the problem.
     
  7. Jul 31, 2014 #6
    OK well I did that and I am getting very close to the right answer. So
    PE = mgy
    PE = 70(9.8)(2.4)
    PE = 1646.4

    1646.4 = 1/2mv^2
    v = 6.85857
    v = rw
    w = 5.72 rad/s
    The correct answer is supposedly 5.79 rad/s
     
  8. Jul 31, 2014 #7

    Nathanael

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    You used the equation for linear kinetic energy.

    Are you familiar with the equation for rotational kinetic energy?
     
  9. Jul 31, 2014 #8
    Yes but how do I find the moment of inertia of a bar?
     
  10. Jul 31, 2014 #9

    Nathanael

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    Two options, Calculus or Google :)

    The moment of inertia is [itex]\frac{1}{3}MR^2[/itex]

    EDIT: By "R" I meant "L" as in the length of the bar
     
    Last edited: Jul 31, 2014
  11. Jul 31, 2014 #10
    using that I got 8.4 which is still the wrong answer
    1646.4 = 1/3IW^2
    W = the sqaure root of 70.56
    W = 8.2
     
  12. Jul 31, 2014 #11

    Nathanael

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    If you want to do it by calculus, it would be like something like this:

    [itex]I=\frac{M}{L}(^{L}_{0}∫x^2.dx)=\frac{1}{3}ML^2[/itex]

    If you really really want, I can explain that equation, but I'm not great at explaining.
    (I've barely just learned calculus, so I'm not the best at explaining it :tongue:)
     
  13. Jul 31, 2014 #12

    Nathanael

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    No, it should be [itex]\frac{1}{2}I\omega ^2=\frac{1}{2}\frac{ML^2}{3}\omega ^2=\frac{1}{6}ML^2\omega ^2[/itex]


    But another thing, you forgot to include the rotational energy that it already had at the top of the swing (the problem said it was moving at 3 rad/s)
     
  14. Jul 31, 2014 #13
    OK now I got the total energy to be 1797.6
    so
    1797.6 = 1/6MR^2w^2
    1797.6 = 1/6(70)(1.2^2)w^2
    w = 10.344 which is not correct
     
  15. Jul 31, 2014 #14

    Nathanael

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    I think you're very close, you just made a mistake about the length of the object. The length of the object is not 1.2m

    The problem does not tell you the length of the object, but if you assume that the object is uniform (uniform in mass) then you can deduce that the length of the object is 2.4m (because the center of mass would be at the midpoint, which was 1.2m)

    So the energy is not 1796.6 (that would be the energy if the length was 1.2m)

    Use the correct length of the object and see if you get the correct answer
     
  16. Aug 1, 2014 #15
    OK thanks! I got it right!
     
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