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Homework Help: Where is the mass of the joined squares?

  1. Mar 29, 2013 #1
    I did Xcm=Lx/(xyz) Ycm=Ly/(xyz) Zcm=Lz/(xyz)

    I modeled it from the equation CM=(mx+.....)/M

    Attached Files:

  2. jcsd
  3. Mar 29, 2013 #2
    Is this your answer?
  4. Mar 29, 2013 #3
    yes that is my answer
  5. Mar 29, 2013 #4
    What are x, y and z in these expressions?
  6. Mar 29, 2013 #5
    I belive they are axis coordinates, but cas they also be mass?
  7. Mar 29, 2013 #6
    Your answer shouldn't involve these variables.
    Go back a bit - can you find the centre of masses of the individual plates?
  8. Mar 29, 2013 #7
    L*L if your talking about one of the faces.
  9. Mar 29, 2013 #8
    That's not the COM of the face.

    Have a look at the bottom face - since it is a uniform square, then the COM must be at its geometric centre. The same for the other 2 faces.
    Then replace each of the faces by a single particle at their COM - you can then combine these to get the overall COM
  10. Mar 29, 2013 #9
    so would the faces turn to xy,xz,yz ?
  11. Mar 29, 2013 #10
    not sure what you mean?

    Hint : The COM of the bottom piece is the point ( L/2, L/2, 0 )
  12. Mar 29, 2013 #11
    so the xz peice would be (L/2 0 L/2)
    and the xy peice would be (0 L/2 L/2)
  13. Mar 29, 2013 #12
    The second one here is the yz piece.

    So, now just regard the mass of each piece as being concentrated at its COM and you end up with a system of 3 particles.
    Then use the definition of the COM to find the overall COM of these 3 particles.
  14. Mar 29, 2013 #13
    will the combined mass be just L.
    or will each axis (x y z) be the mass?
  15. Mar 29, 2013 #14

    Hint : in the xy plane, you can replace the piece of mass m with a particle of mass m at the point (L/2, L/2, 0).
    Do the same for the other 2 pieces.
  16. Mar 29, 2013 #15
  17. Mar 29, 2013 #16
    I'm afraid this is not right :(
    Why are you using x, y, and z?

    What is the definition of centre of mass?
  18. Mar 29, 2013 #17
    center of mass is where the weighted position vectors to point sum is zero
  19. Mar 30, 2013 #18
    So, the CM is given by

    [tex]\mathbf r_{CM} =\frac{1}{M} \sum m_i \mathbf r_i[/tex]

    You have the position vectors ri for each of the 3 parts. Call the mass of each piece m and then apply the above equation.
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