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Kinetic energy around an axis - two methods?

  1. Nov 20, 2007 #1
    1. The problem statement, all variables and given/known data
    I have to find the kinetic energy of the total system; b has mass 2m and a has mass m. They rotate around the y-axis and and through the origin (0,0).

    There's two ways of finding the kinetic energy of the system around the y-axis.

    1) I find moment of inertia for each particle around the y-axis (and through the origin) and add - so it's 2*m*r^2+9*m*r^2 = 11*m*r^2. Then I use this in K = ½*I*w^2.

    2) I find the center of mass: r_cm = (2*m*r+m*3*r)/(3m) = (5/3)r. I use this in I = m*r^2 and then use K = ½*I*w^2.

    #2 doesn't work - why?
     

    Attached Files:

  2. jcsd
  3. Nov 20, 2007 #2
    Wait a minute.. when I use #2, I find the moment of inertia I_cm, not I_y, right?
     
  4. Nov 21, 2007 #3
    I have another question.

    If I want to find the kinetic energy, I can do it like this:

    K = ½*I_total*w^2, where w^2 = v^2/r^2 - what radius are we talking about here? Is it the distance from center of mass to the point on the axis, on which they are turning?

    Or can I only use K = ½*I_single*v^2/r^2 when looking at the particles individually, and then add the two energies? Here r is the perpendicular distance to the axis.
     
    Last edited: Nov 21, 2007
  5. Nov 21, 2007 #4

    Doc Al

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    Staff: Mentor

    When analyzing rotation of an extended body, you can't just replace the body by a point mass at its center of mass. (Imagine a body rotating about an axis through its center of mass. It has some rotational inertia. But replace it by a point mass at that point and it has none.)

    r is always the perpendicular distance to the axis. In the first case, for some reason you aren't given w but have the speed of some point, so you use that point's distance from the axis to calculate w.
     
  6. Nov 21, 2007 #5
    Ok, so I can't find the kinetic energy of the total system by using

    K = ½ * I_{total} * v^2/r^2? If I use this equation, I have to find it for each particle, and then add it?

    My problem here is: In "K = ½ * I_{total} * v^2/r^2", what is r for the total system?
     
  7. Nov 21, 2007 #6

    Doc Al

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    Staff: Mentor

    What you want to use is:
    [tex] K = 1/2 I_{total} \omega^2[/tex]

    How you find [itex]\omega[/itex] depends on what you're given.

    If you happen to have the speed of one particular point of the rotating system, you can use it to find omega. r would not be for "the total system", but just for that particular point.
     
  8. Nov 21, 2007 #7
    Ok, I still don't believe my question has been answered.

    If I want to calculate the kinetic energy of the total system, I do like this:

    K_total = ½*I_a*w_a + ½*I_b*w_b

    The total moment of inertia is 11*m*r^2, and we call this I_total. The total kinetic energy, K_total, can also be found as

    K_total = ½*I_total*w_system.

    In w_system, what is the distance r? That is my question.
     
  9. Nov 21, 2007 #8
    Ok, when reading your replies, my question has been answered.

    Thank you very much!
     
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