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Simple center of mass question - Thanks

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Hey all. Below is the question and picture related to it.

    "Three uniform metre sticks each of mass M are attached together as follows: Stick 1 lies along the y axis from y = 0 to y = 1.0m. Stick 2 lies along the x axis from x = 0 to x = 1.0m. Stick 3 makes an angle of 60 degrees with the y axis. Find the x and Y coordinates of the centre of mass of the system of three metre sticks, and marks its approximate location on the diagram of the three sticks."



    ----Sorry, its kinda upside down :)
    j1251k.jpg

    2. Relevant equations


    3. The attempt at a solution

    I am really confused with this one.

    Maybe lets start with the X

    The formula I know is: Xcm = m1x1 + m2x2 + m3x3 / m1+m2+m3

    But how do I input into this formula... ??
     
  2. jcsd
  3. Nov 14, 2011 #2

    ehild

    User Avatar
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    Replace every metre stick with a point mass of M at its own centre of mass.

    ehild
     
  4. Nov 14, 2011 #3
    How do I find each metre sticks own center of mass??
     
  5. Nov 14, 2011 #4
    assuming each stick is of uniform density, it will be in the physical center of the stick.

    [itex] x_{c} = \frac{\sum_{i = 1}^{N} m_{i} x_{i}}{\sum_{i = 1}^{N} m_{i}} [/itex]

    [itex] y_{c} = \frac{\sum_{i = 1}^{N} m_{i} y_{i}}{\sum_{i = 1}^{N} m_{i}} [/itex]

    these describe the center of mass, but assuming it is uniform, the masses are taken out and you are left with just the physical center:

    [itex]x_{c} = \frac{\sum_{i=1}^{N} x_{i}}{N}[/itex]

    [itex]y_{c} = \frac{\sum_{i=1}^{N} y_{i}}{N}[/itex]

    now plug that into the same equation for all three sticks:

    [itex]x_{cm} = \frac{x_{cm1} + x_{cm2} + x_{sm3}}{3}[/itex]

    [itex]y_{cm} = \frac{y_{cm1} + y_{cm2} + y_{sm3}}{3}[/itex]
     
    Last edited: Nov 14, 2011
  6. Nov 14, 2011 #5

    What about the stick at the 60 degree angle?

    Im actually having trouble with plugging the right info into the formula, lets say for X - what would lets say "Xcm1" be?
     
  7. Nov 14, 2011 #6
    You can resolve the center of mass of the angled stick into its components using trig ratios.

    [itex] x = \frac{1}{2} \cos (\frac{\pi}{6})[/itex]

    [itex] y = \frac{1}{2} \sin (\frac{\pi}{6})[/itex]

    and then proceed from there.

    by [itex]x_{cm1}[/itex] i mean the x component of the center of mass for stick one.
     
  8. Nov 14, 2011 #7
    is Xcm1 = .5 m?
     
  9. Nov 14, 2011 #8
    well that depends on what you choose to be your first stick. if you're talking about the bottom one then you'd be correct
     
  10. Nov 14, 2011 #9
    Ahhg, im not getting this. :(

    Thanks for trying tho
     
  11. Nov 15, 2011 #10
    keep at it, just try to understand what the formulas are saying.

    keep trying, but you can check it against this:

    [itex]x_{cm} = \frac{0 + \frac{1}{2}\cos(\frac{\pi}{6}) + 0.5}{3}[/itex]

    [itex]y_{cm} = \frac{0.5 + \frac{1}{2}\sin(\frac{\pi}{6}) + 0}{3}[/itex]
     
  12. Nov 15, 2011 #11
    Gordon or anyone, here is what I got. Can you tell me if the answer I got to this question is correct, including image (posted below)

    (.333 i + .168 j) m

    Image of where centre of mass roughly would be?

    55s2v7.jpg
     
    Last edited: Nov 15, 2011
  13. Nov 15, 2011 #12
    Can anyone confirm this for me please ? :)
     
  14. Nov 15, 2011 #13
    Two words my friend, radian mode.
     
  15. Nov 15, 2011 #14
    dammit :)

    How about now:

    0.31i, 0.25j

    Is that correct?
     
  16. Nov 15, 2011 #15
    yup, now i'd suggest going over the problem again or making up your own problem simmilar to this one to practice.
     
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