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X and y coordinates, integration, semicircular plate (masteringphysics)

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Use equations x[itex]_{cm}[/itex]=[itex]\frac{1}{M}[/itex][itex]\int x dm[/itex] and y[itex]_{cm}[/itex]=[itex]\frac{1}{M}[/itex][itex]\int y dm[/itex] to calculate the x- and y-coordinates of the center of mass of a semicircular metal plate with uniform density [itex]\rho[/itex] and thickness t. Let the radius of the plate be R. The mass of the plate is thus M=[itex]\frac{1}{2}[/itex][itex]\rho\pi[/itex]a[itex]^{2}t[/itex].

    Use the coordinate system indicated in the figure.

    YF-08-51.jpg

    1. Calculate the x-coordinate of the center of mass of a semicircular metal plate. Express your answer in terms of the variables a, ρ and t.

    2. Calculate the y-coordinate of the center of mass of a semicircular metal plate. Express your answer in terms of the variables a, ρ and t.

    2. Relevant equations

    I think these:

    [itex]\vec{r_{cm}}[/itex]=[itex]\frac{m_{1}\vec{r_{1}}+m_{2}\vec{r_{2}}+...}{m_{1}+m_{2}}[/itex]

    But instead of the sum I need to integrate, right?
    Does this equation work in 3D?

    3. The attempt at a solution

    I'm not sure how to use the equation and what information to use where.

    To find x-coordinate:

    x[itex]_{cm}[/itex]=[itex]\frac{Mx_{cm}}{M}[/itex]=x[itex]_{cm}[/itex] ??

    y[itex]_{cm}[/itex]=[itex]\frac{My_{cm}}{M}[/itex]
     
  2. jcsd
  3. Oct 12, 2011 #2
    Well I can tell you a couple of things. Because of symmetry, you don't need to use the z coordinate, you already know the z coordinate of centre of mass. I would also say the same thing for the x coordinate. So the only coordinate that you need to iron out is the y coordinate.

    EDIT: You will have to put dm in terms of something else I believe.
     
    Last edited: Oct 12, 2011
  4. Oct 12, 2011 #3
    But how do I find the x coordinate in terms of rho, a and t?
     
  5. Oct 12, 2011 #4
    Is x=0 by symmetry?
     
  6. Oct 12, 2011 #5
    yes sir, because you go from -R to R.

    The best day to find y is ysqrt(r^2-y^2) and do a substitution
     
  7. Oct 12, 2011 #6
    I got the answer [itex]\frac{4a}{3\pi}[/itex] from a friend but I want to know how to get there!

    How does that work with [itex]y=y\sqrt{r^{2}-y^{2}}[/itex]?

    I haven't seen that in my textbook.
     
  8. Oct 12, 2011 #7
    darn i forgot that textbook kind of sucks :(. maybe forget that method since you won't be able to reference it easily
     
  9. Oct 12, 2011 #8
    Thanks anyway... :) :/
     
  10. Oct 13, 2011 #9
    dm can be written in terms of rho dV. this rho will cancel out which gives you the clue that you're headed in the right direction. You just have to perform dV properly.
     
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