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Rotation & Energy problem

  1. Nov 3, 2013 #1

    vac

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    1. The problem statement, all variables and given/known data
    Torque = 3
    Time = 3.12 s
    Length of the each rod = 1 m so the radius = 0.500 m
    mass of each rod is = 0.500 kg
    [tex]M_1 = 4 kg[/tex]
    [tex]M_2 = 2 kg[/tex]
    [tex]M_3 = 4 kg[/tex]
    [tex]M_4 = 2 kg[/tex]
    http://imageshack.us/a/img27/5475/qu51.jpg [Broken]
    2. Relevant equations
    What is the energy of the object (please refer to the picture above) that has a mass fixed to each of its four corners? If it rotates for 3.12 s? Torque and time is given.
    3. The attempt at a solution
    [tex]K_E = \frac{1}{2} I \omega^2[/tex]
    [tex]I = \sum mr^2[/tex]
    [tex]I = (4+2+4+2+0.5+0.5)(0.5)^2[/tex]
    [tex]I = 3.25[/tex]
    [tex]\omega = \alpha t[/tex]
    Now we have "I" and Torque we can calculate alpha.
    [tex]Torque = I \alpha[/tex]
    [tex]3 = 2.25 \alpha[/tex]
    [tex]\frac{3}{3.25}= \alpha[/tex]
    so
    [tex]\omega = \alpha t[/tex]
    [tex]\omega = \frac{3}{3.25} 3.12 s = 2.88[/tex]
    substitute back in
    [tex]K_E = \frac{1}{2} I \omega^2[/tex]
    [tex]K_E = \frac{1}{2} 3.25 (2.88)^2[/tex]
    [tex]K_E = 13.5 J[/tex]
    Can you please tell me what is wrong with my answer and if there is an easier way to solve such problems.
    Thanks.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 3, 2013 #2

    haruspex

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    Pls explain how you get that. What is the value of Ʃm?
     
  4. Nov 3, 2013 #3

    vac

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    summation of masses times radius square: are (m1+m2+m3+m4+ mass of two rods )(r^2) = (4kg + 2kg + 4kg + 2kg + 0.5kg + 0.5 kg)(0.5 m)^2 = 3.25
     
  5. Nov 3, 2013 #4

    vac

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    Ok, do you see how I did that?
     
  6. Nov 3, 2013 #5

    haruspex

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    What about the horizontal rods?
     
  7. Nov 4, 2013 #6

    vac

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    Thank you for asking this question, it is the main thing that drove me to make this post.
    I think it should be (1/12) times mass times length of both rods ... (1/12)ML.
    But how about the vertical rods?
     
  8. Nov 4, 2013 #7

    vac

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    The moment of inertia for a rod is [itex]\frac{1}{12} ML^2[/itex]
     
  9. Nov 4, 2013 #8

    haruspex

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    Yes. You already included the vertical rods correctly. The 1/12 formula is for a rod rotating about its centre. Every part of each vertical rod is distance 0.5m from the axis, so 0.5m is right for those.
     
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