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Center of mass problem.

  1. Oct 22, 2005 #1
    A thin rectangular plate of uniform area density = 2.86 kg/m^2 has a length of 42.0 cm and a width of 23.0 cm. The lower left hand corner is located at the origin, (x,y)=(0,0) and the length is along the x-axis. There is a circular hole of radius 7.00 cm with center at (x,y)= (15.00, 9.50) cm in the plate. Calculate the mass of the plate.
    I know that density= mass/volume
    so i found the volume of the plate to be .0966 m^2.
    I multiplied 2.86 * .0966 to get .276 kg for the mass.
    This wasn't right. Do I need to use the information about the circle for this problem?
  2. jcsd
  3. Oct 22, 2005 #2


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    Your calculations are correct for the mass of the entire plate. If that was the wrong answer then I imagine they want the mass of what remains of the plate after the 7 cm radius circle is cut out of it.
    Last edited: Oct 23, 2005
  4. Oct 22, 2005 #3
    Ok I figured out the mass of the plate to be .232 kg.
    The second part says calculate the x-coordinate of the CM of the plate
    I used the formula
    x-cm= m1x1+m2x2/ m1 + m2

    xcm= (.276* .42) - (.044* .15) /.232
    I got .471 m which wasn't right. Can someone tell me what I'm doing wrong?
  5. Oct 22, 2005 #4
    x = m1x1+m2x2/ m1+m2 gives the center of mass of two masses located at x1 and x2. Your plate is composed of an infinite amount of masses, so you'll need to integrate over the length of the plate to find the center of mass.
  6. Oct 22, 2005 #5
    I'm confused about how I would intregrate the equation.
  7. Oct 22, 2005 #6
    In this case, your density is constant, so you can say that the center of mass is the expectation of x = <x> = integral of x*f(x) from a to b

    a, b are the endpoints of your metal slab.
    f(x) is the normalized function that describes how much mass is at every interval dx.

    The only tricky part of the problem is finding the f(x) for the region that has the circle cut out. To do this, express the circle in x,y coordinates. You should have two equations corresponding to the top half and bottom half of the circle. Integrate x * the function describing the bottom half of the circle and add that to the integral of x*(height of the metal slab - top half of the circle).
  8. Oct 22, 2005 #7
    Ok I know that the x,y coordinates of the circle are (15.00, 9.50) cm. I don't have any equations cooresponding to the circle though. The only equation that I can think of that I would be able to integrate is the area of the circle... but I don't think that's what I'm supposed to do.

    I'm sorry... this is really confusing me.
  9. Oct 22, 2005 #8
    Ok, here's a step by step walk-through. I didn't do the algebra but you should get the main idea.

    First we want a function that describes how much mass there is per small increment dx along your metal slab. Since your metal slab is of uniform density it is obvious that
    f(x) = 23 for 0<x<8 and 22<x<42

    This leaves the region of 8<x<22, the region that has that annoying circle in it.

    Note that the equation of a circle is (x-x0)^2+(y-y0)^2=r^2, where x0, y0 is your center and r is your radius

    So 49 = (x - 15)^2 + (y - 9.5)^2 is the equation of the circle. Doing some algebra, you get
    y = 0.5(19-2*Sqrt[(22-x)(x-8)]) as the equation for the bottom half of the circle and
    y = 0.5(19+2*Sqrt[(22-x)(x-8)]) as the equation for the top half of the circle

    So to find the "height" of the slab at each point x, you subtract the top half of the circle from the real height (23). Then add the bottom half of the circle. So between 8 and 22,
    f(x) = 23-0.5(19+2*Sqrt[(22-x)(x-8)])+0.5(19-2*Sqrt[(22-x)(x-8)])
    This simplifies to f(x) = 23 - 2*Sqrt[(22-x)(x-8)]

    Now you have a piece-wise definition of f from 0<x<42. You have to normalize it by dividing by the area of your metal slab so that the integral of p(x) = f(x)/areaslab from 0 to 42 = 1. The area of the slab is 812.062

    So now, you can finally find the center of mass. <x> = Integral of x*f(x)/areaslab from 0 to 42.
    Therefore 812.062 * <x> = Integral of x*f(x) dx from x=0 to x=42

    You can now easily find the integral of the piece-wise function that I defined for you. Solving for <x> you should find that the center of mass = 22.1374 cm from your defined x=0
  10. Oct 23, 2005 #9


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    mcah5, you don't need to do it like that. Punchline girl already had the correct formula with xcm = (m1x1 + m2x2)/(m1+m2) but had applied it wrongly.

    Think of the problem as being one of two masses and having to find their COM.
    Mass1 (m1): A circle, masss = 0.04403 kg, com = (15, 9.5) = (x1,y1)
    Mass2 (m2): A rectangle with a circle cut out of it, mass = 0.2322 kg, com = (?, ?) = (x2, y2)
    Mass3 (m3): the two masses, m1 and m2, put together to give a solid rectangle, mass = 0.27623 kg, com = (21, 11.5) = (xcm, ycm)
    Now plug those values into your original formula and solve for x2

    (x2, y2) is the com of the plate without the circle.
    (xcm, ycm) is the com of the plate with the circle.
    Last edited: Oct 23, 2005
  11. Oct 23, 2005 #10
    Thank you both for your help. I see what I was doing wrong now.
    The third part of the problem says calculate the distance of the plate's CM from the origin.
    I know the x coordinate is .221, and the y coordinate is .256. I tried to find the slope of the line between the two, so .256/.221 to get 1.15, but this wasn't right.
    Is that even how I would try to solve it?
  12. Oct 23, 2005 #11


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    Use pythagoras' theorem.

    x² + y² = r²
  13. Oct 23, 2005 #12
    (.221)^2 + (.256)^2 = r^2
    gave me .338 which wasn't right....
    when you say r do you mean the radius of the circle?
  14. Oct 23, 2005 #13


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    r is just a distance, the hypotenuse of the right angled triangle with x as one side and y as the other side.
    I haven't worked out the coords of COM. I'll go do that now.
    You don't have the correct answer for the distance, do you ?
  15. Oct 23, 2005 #14


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    Typo in an earlier post.
    (x3,y3) = (21,11.5) not (21,23)
  16. Oct 23, 2005 #15
    Ok that makes sense. I got it. Thank you so much!
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