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Center of Mass Calculation: x-bar, y-bar

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to find the center of mass of the given figure attached. I am given the density (7850 kg/m^3) and a thickness of 0.3 cm (not sure where this goes)



    2. Relevant equations

    Just some things I need to verify:

    When finding the x-bar of the function y=x3 using the equation:

    [itex] \bar{x} [/itex]= [itex]\frac{∫ \tilde{x} dm}{∫dm}[/itex],

    Is my [itex] \bar{x} [/itex] going to be the x distance(which will just be "x") times x3dm?

    Also, will x3 be in the denominator just before "dm"?

    If I am also given a density ρ, will that stay in front of each integrand or in front of ([itex]\frac{∫ \tilde{x} dm}{∫dm}[/itex])? And would I calculate the thickness with the [itex]\bar{z}[/itex]?

    Is the same process is done for the [itex]\tilde{y}[/itex] or are there differences?

    I know this may be a lot but I'd like to make sure of the method.
    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2

    Simon Bridge

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    dm as the amount of mass in a volume element at point (x,y) with width dx, height dy, and thickness t ... that would be ##dm = \rho(x,y) dV = \rho tdxdy##
    ...that help clarify things?

    Of course you can simplify the calculation a lot by understanding it.
     
    Last edited: Nov 9, 2012
  4. Nov 10, 2012 #3

    rude man

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    I suggest paying very little attention to the (uniform) density and not too much time on the (uniform) thickness ...
     
  5. Nov 10, 2012 #4

    Simon Bridge

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    Well, me too ... though it comes out in the wash :) It's part of the "you can simplify the calculation a lot by understanding it" thing.

    It is also easier to do the integration against y instead of x and part of the integration can be done just by knowing the formula for the area of a triangle - but I don't know how much aaronfue understands vs memorizes equations. Do need to hear from him before we can continue methinks.
     
  6. Nov 11, 2012 #5

    rude man

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    Big 10-4, Simon. Not having my textbook at the ready right now, I would approach the problem as finding the axis of zero net torque under gravity and arbitrary orientation, in which case integration of a differential wedge dθ might make sense. As an EE I'll probably pass on doing the actual work, so talk's cheap! :biggrin:
     
  7. Nov 11, 2012 #6

    Simon Bridge

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    Who needs text books when you have google ;)
     
  8. Nov 11, 2012 #7

    rude man

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    True, true. So chalk it up to indolence on my part ... :blushing:
     
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