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Total moment of inertia of a two particle system

  1. Oct 24, 2009 #1
    1. Find the moment of inertia I_x of particle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).

    Particle a is located 3r from the y axis, and particle b is located r away.




    2. I = mr^2
    I = SUMM(m_i*r_i^2)




    3. I_a = 9mr^2
    I_b = mr^2

    Total I = 10mr^2

    No idea what I am doing wrong
     
  2. jcsd
  3. Oct 24, 2009 #2

    Doc Al

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    Give the complete coordinates and the mass of each particle.
     
  4. Oct 24, 2009 #3
    Particle a:

    distance from y axis = 3r
    distance from x axis = r

    Particle b:

    distance from y axis = r
    distance from x axis = -4r

    It doesn't say anything explicitly about mass, and so I think we assume each particle has mass m.
     
  5. Oct 24, 2009 #4

    Doc Al

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    Looks like you've found I_y. What about I_x?
     
  6. Oct 24, 2009 #5
    I_x would be m(r)^2 + m(-4r)^2 = mr^2 + 16mr^2 = 17mr^2.

    Am I simply adding that to I_y to get I_total? It asks for total I with respect to the y axis, though, why would I need I_x?
     
  7. Oct 24, 2009 #6
    Shoot, just realized I posted to wrong question. Here's what I am trying to answer:

    Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
     
  8. Oct 24, 2009 #7

    Doc Al

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    Assuming the diagram matches your description, your answer seems correct to me. (Post the diagram, if you can.)
     
  9. Oct 24, 2009 #8
  10. Oct 24, 2009 #9

    Doc Al

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    Looks good to me. I say your answer is correct.
     
  11. Oct 24, 2009 #10
    Ok, I requested the correct answer: 11mr^2

    WTF?!
     
  12. Oct 24, 2009 #11

    Doc Al

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    Sounds bogus to me. You might want to mention this to your instructor.
     
  13. Oct 24, 2009 #12
    Ok, now:

    Using the formula for kinetic energy of a moving particle K=\frac{_1}{^2}mv^2, find the kinetic energy K_a of particle a and the kinetic energy K_b of particle b.
    Express your answers in terms of m, omega, and r separated by a comma.

    I know I need to calculate linear speed, how do I do that? I know it's v=omega*r but I don't see how to put this together.....
     
  14. Oct 24, 2009 #13

    Doc Al

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    What's the relationship between tangential speed and angular speed for something going in a circle?
     
  15. Oct 24, 2009 #14
    Just edited before you replied: v=omega*r

    But what are omega and r here?!
     
  16. Oct 24, 2009 #15
    Particle A:

    Ka = 0.5m(3*omega*r)^2

    Kb = 0.5m(omega*r)^2

    Apparently Kb is wrong but I did it the same way as Ka.............
     
  17. Oct 24, 2009 #16
    Ok I got Kb = m(omega*r)^2

    What is total kinteic eneregy?
     
  18. Oct 25, 2009 #17

    Doc Al

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    These both look OK to me.

    How did you get that? (Are you posting this problem exactly as given, word for word?)

    Just add up the kinetic energy of each mass.
     
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