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Y coordinate of the system's center of mass?

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    In Fig. 9-39, three uniform thin rods, each of length L = 28 cm, form an inverted U. The vertical rods each have a mass of 12 g; the horizontal rod has a mass of 31 g. What are (a) the x coordinate and (b) the y coordinate of the system's center of mass? (Give your answer in cm)

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c09/fig09_37.gif


    2. Relevant equations
    xcom = (m1x1 + m2x2)/M


    3. The attempt at a solution
    I got the x coordinate alright at 14cm but I can't find the y coordinate. I took the center of mass between the two vertical rods with y1 = 14cm and its m1 = 24g as total of the vertical rods. I then compared that center of mass to the center of mass of the horizontal rod with y2 = 28cm and its m2 = 31g. I thus found y = 21.89, but this is apparently wrong, could somebody please tell me what it is that I'm doing wrong?
     
  2. jcsd
  3. Feb 16, 2009 #2

    LowlyPion

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    Homework Helper

    Apply the same formula in the y direction.

    2 vertical rods with com = L/2 and mass m = 2*(m*L/2) = mL

    The horizontal rod at the top is m*L from the bottom, so the sum is 2m*L and with M = 3*m then

    Com-y = 2*m*L/3*m = 2/3 L
     
  4. Feb 16, 2009 #3
    What you said doesn't completely agree with the information, because the mass of the horizontal rod is different from the mass of the vertical rod. Also, where are you getting the mass of the rod with m = 2*(m*L/2)? Your explanation confuses me. I'm thinking that the mass is relative to the two masses so the m1 = 2(12g) and the y1 = L/2 which is 14cm. Is this wrong?
     
  5. Feb 16, 2009 #4

    LowlyPion

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    Homework Helper

    Sorry. Yes the horizontal rod has a different mass. I missed that. But the treatment is the same.
    The vertical rods - 2 of them - each have the same mass and each have a Com at L/2 ... hence 2*(m(L/2))

    So putting numbers to it then

    m*L = .012*.28

    And the horizontal rod is contributing a moment of .031*.28

    Total then is (.012 + .031)*.28 /M = .033(.28)/.055 = 3/5(.28)
     
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