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Find the center of mass of a plate

  1. Mar 16, 2019 at 7:21 PM #1
    1. The problem statement, all variables and given/known data
    Find the center of mass of the next plate if:
    A) Is homogeneous
    B) Its density per unit mass is ##\sigma=Axy##, where ##A## is a constant.
    Capture.PNG

    2. Relevant equations
    ##X_{cm}=\frac{\int\sigma x dA}{\int\sigma dA}##
    ##Y_{cm}=\frac{\int\sigma y dA}{\int\sigma dA}##
    3. The attempt at a solution
    I did part A) as is shown in the picture below:
    Image.jpg
    Image (2).jpg
    Am I correct?
    My main problem is with part B), I tried to do it as I did part A), I found ##X_{cm} without any problems, like this:
    Image (2)).jpg
    But when I tried to find ##Y_{cm}## I got stucked with this:

    ##Y_{cm}=\frac{\int_{0}^7 \sigma ydA_1+ \int_{1}^7 \sigma ydA_2}{\int_{0}^7 \sigma dA_1 + \int_{1}^7 \sigma dA_2}=\frac{A\left[ \int_{0}^7 y^2\sqrt{y-1}dy + \int_{1}^7 y^2(y-1)dy\right]}{A\left[ \int_{0}^7 y\sqrt{y-1}dy + \int_{1}^7 y(y-1)dy\right]}##, but when I tried to integrate ##\int_{0}^7 y^2\sqrt{y-1}dy## and ##\int_{0}^7 y\sqrt{y-1}dy##, I ended up with imaginary numbers, what am I doing wrong? Hope you can help me.
     
  2. jcsd
  3. Mar 16, 2019 at 7:32 PM #2

    haruspex

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    Think more carefully about what your ∫(x2+1).dx and ∫(x2+1)x.dx are actually measuring.
     
  4. Mar 16, 2019 at 7:38 PM #3
    ##\int (x^2+1)dx## is measuring the total area of the plate, while ##\int (x^2+1)xdx## is measuring the area of a small piece of the total area.
     
  5. Mar 17, 2019 at 12:12 AM #4

    haruspex

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    No.
    Look at your diagram for the x>0 section. What is the length in the y direction of the strip width dx?
     
  6. Mar 17, 2019 at 12:19 AM #5
    I think it would be ##7-(x^2+1)##, right?
     
  7. Mar 17, 2019 at 12:20 AM #6

    haruspex

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    Right, so what should your integrals be?
     
  8. Mar 17, 2019 at 12:32 AM #7
    Oh, I see know, I was calculating the area underneath the curve, so I need to write ##7-x^2-1## instead to calculate the are between the line ##y=7## and the curve.
    I think the integrals would be like this, am I correct?
    Image (4).jpg

    How would it be with ##Y_{cM}##?
     
  9. Mar 17, 2019 at 12:53 AM #8

    haruspex

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    Yes, that answer looks better.
    Please post a revised attempt. As a check, should it be more or less than 3.5?
     
  10. Mar 17, 2019 at 1:01 AM #9
    I'm trying to solve it. The length in the x direction of the strip width dy would be ##x##, right? And ##x=\sqrt{y-1}##.
    I think that it should be less than 3.5, but not too much.
     
    Last edited: Mar 17, 2019 at 1:49 AM
  11. Mar 17, 2019 at 2:02 AM #10

    haruspex

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    Consider the two parts separately. How would the two y coordinates of the mass centres compare with 3.5 individually?
     
  12. Mar 17, 2019 at 2:24 AM #11
    Well, the coordinates of the mass center of the rectangle would be ##(2.5,3.5)## and for the parabola, I guess the would be less.
     
  13. Mar 17, 2019 at 3:06 AM #12
    @haruspex Am I correct with this?
     
  14. Mar 17, 2019 at 3:13 AM #13

    haruspex

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    Yes, but only for y in a certain range.
    The y coordinate would be less? Doesn't look that way to me.
     
  15. Mar 17, 2019 at 3:15 AM #14
    Yes, y would be between 1 and 7. So is my calculation of ##Y_{cm}## correct?
    Well, then I think I didn't understand the question.
     
  16. Mar 17, 2019 at 3:51 AM #15

    haruspex

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    Consider the complete rectangle, 2x7, with lower left corner at the origin. The y coordinate of its mass centre would be 3.5, yes? To get the actual shape you have to trim off the piece below the parabola. Wouid that raise or lower the mass centre?
     
  17. Mar 17, 2019 at 3:58 AM #16
    That would raise the center of mass, so the y component would be greater than 3.5.
     
  18. Mar 17, 2019 at 4:30 AM #17

    haruspex

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    Right.
     
  19. Mar 17, 2019 at 10:50 AM #18
    Ok, so am I correct with how I calculated ##Y_{cm}## initially?
    Also, how do I calculate the center of mass when the density isn't constant, i.e. part B)? I tried to solve it as I did with part A, but ##Y_{cm}## gives me imaginary numbers.
     
    Last edited: Mar 17, 2019 at 1:36 PM
  20. Mar 17, 2019 at 2:45 PM #19

    haruspex

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    Since you got a value <3.5, no. The method looks ok so there must be an arithmetical error.
    By the way, for both x and y coordinate of CoM, the denominator is the area of the plate, so you don’t need to calculate that twice.
     
    Last edited: Mar 17, 2019 at 2:55 PM
  21. Mar 17, 2019 at 2:55 PM #20
    I think I know the problem. Do you see that I got a different value in the total mass integral? The problem is in the parabola, since I got 35 for the rectangle for both ##X_{cm}## and ##Y_{cm}##. The problem is that the equatio of the parabola ##y=x^2+1## isn't complete with ##x\in[0,2]##. So when I integrate ##\int_{0}^2 6-x^2 dx## I'm not getting the whole area, while when I integrate ##\int_{1}^7 \sqrt{y-1}dy## I'm indeed getting the whole area. This can be seen here:
    bfcd8436-46ac-4eaf-9ed5-a862dab945b1.jpg
    2073807a-8b11-494b-be4e-e068a795857e.jpg
    How can I fix this? The only way I can think of would be to change the upper limit on ##X_cm## with 2.449 instead of 2.
     
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