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Homework Help: Center of Mass question

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    A 2.80 kg particle has the xy coordinates (-2.00 m, 0.950 m), and a 3.30 kg particle has the xy coordinates (0.672 m, -0.0440 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 3.39 kg particle such that the center of mass of the three-particle system has the coordinates (-0.663 m, -0.527 m)?


    2. Relevant equations



    3. The attempt at a solution
    I don't even know what equation I have to use to find this, I looked in the book and I can't find anything about it. Can anyone help me with this?
     
  2. jcsd
  3. Oct 21, 2007 #2

    Doc Al

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    You need to understand the basic definition of center of mass: Center of Mass
     
  4. Oct 21, 2007 #3
    How do I find the coordinates of the center of mass if they already gave them to me though? I now know what equation to use but I don't know how to apply it to this problem.
     
  5. Oct 21, 2007 #4

    arildno

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    You must solve two equations for the coordinates of the final particle.
     
  6. Oct 21, 2007 #5
    Now I'm confused which two equations do I use?

    xcom=(m1x1+m2x2)/M and...?
     
  7. Oct 21, 2007 #6

    Doc Al

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    You'll need one equation for the x-component and another for the y-component.
     
  8. Oct 21, 2007 #7

    arildno

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    You have THREE particles here, not two!!

    YOu have, for the x's:
    [tex]x_{c.m}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}[/tex]

    Now, in this expression, what quantities are KNOWN, which quantity (or quantities) is unknown?
     
  9. Oct 21, 2007 #8
    what's unknown is the ycom and the xcom... so i put my information in that equation and solved it and got xcom= -0.593 and ycom=0.0767 and they are wrong. i have no idea what i'm doing wrong here.
     
  10. Oct 21, 2007 #9

    Doc Al

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    What's unknown are the coordinates of the third particle: x_3 and y_3. The coordinates of the center of mass are known.
     
  11. Oct 21, 2007 #10

    arildno

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    Remember that your exercise can be equivalently rephrased into:
    (First some blah&info about the two first particles)
    ..The common centre of mass of the three-particle system IS (-0.663,-0.527), and the mass of the third particle IS 3.39kg.
    Question:
    What is the position of the third particle?
     
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