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Center of mass of half square without a half circle

  1. Jun 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the position of the center of mass for a thin sheet and homogeneous, with sides R and 2R ,from which has been subtracted a half circle of radius R.
    35mkv42.png

    [Xcm=(2/3)*R*(4-pi)]


    2. Relevant equations

    Rcm=(1/M)*∫rdm

    3. The attempt at a solution

    By simmetry we know Ycm=0.
    For de calculus of Xcm we rewrite the equation: Xcm=(1/Area)*∫xdA

    dA=dx*y; \[y=2\,R-2\,{\left( {R}^{2}-{x}^{2}\right) }^{0.5}\]
    For determinate the dA we take a
    Xcm=(1/A)*[itex]\int\2\,x\,R-2\,x\,{\left( {R}^{2}-{x}^{2}\right) }^{0.5}\[/itex]
    -R≤x≤R

    \[Xcm=\frac{{R}^{3}}{3\,A}\]

    \[A=\frac{4\,{R}^{2}-\pi\,{R}^{2}}{2}\]

    \[Xcm=\frac{6\,R}{4-\pi}\]

    I´ve been attempting in a lot of ways to solve the problem but i didn't get the result. This is an exercise from my exams in Physics I of 1º grade chemical engineering .


    PS: I don't know how to make the formulas visible in LaTex
     
    Last edited: Jun 23, 2012
  2. jcsd
  3. Jun 23, 2012 #2
    The CM of rectangle R×2R=CM half circle + CM shaded area.
     
  4. Jun 23, 2012 #3
    You should use the guyndanh formula for sheet
     
  5. Jun 23, 2012 #4
    I wonder what that is....:uhh:
     
  6. Jun 23, 2012 #5
    Thank you, but when I try doesn't match with the correct answer.

    Xcmrectangle= R/2
    Xcmcircle= (4/(3*pi))*R

    Xcm=(R/2)-(4/(3*pi))*R=0.0755*R

    Correct answer: Xcm=(2/3)*R*(4-pi)=0.055*R
     
  7. Jun 23, 2012 #6
    I don't know what is that, but i´m pretty sure that the problem can be solved just with the center of mass definition.
     
  8. Jun 23, 2012 #7
    That formula is the distance from the base(diameter) of the semi circle... And the semicircle is 'inverted', here :wink:
     
  9. Jun 23, 2012 #8
    you should also consider the mass as proportional to area and not just subtract the x's.
     
  10. Jun 23, 2012 #9
    That from top.
    Require reference to x-axis
     
  11. Jun 23, 2012 #10
    Okey, so the center of mass for the semicircle should be R-(4/(3*pi))*R?

    Asr=Area of semirectangle, Asc=Area of semicircle.

    Then: Xcm=[Asr*(R/2)-Asc*( R-(4/(3*pi))*R)]/(Asr-Asc);

    Xcm=0.223*R;(I've reviewed my calculations)
     
  12. Jun 27, 2012 #11
    Xcm=∫(R/2+√(R^2-X^2)/2)(R-√R^2-X^2)dX/∫(R-√R^2-X^2)dX
    limits are from 0 to R OR from -R to R that does not matter because of symmetry.It gives that.verify this yourself.
     
  13. Jun 27, 2012 #12
    Okay so the dA= (R-√R^2-X^2)dx, I understand that. But why you take x=(R/2+√(R^2-X^2)/2)?
    And I tried to solve your equation with Maxima but i didn't get the result.
     
  14. Jun 28, 2012 #13
    Hello SergioQE,
    You can handle your question without appealing to calculus, provided you know :
    (1) Center of mass position of semicircle (removed part) and center mass of full rectangle.
    (2) The basic expression for center of mass.
    Now the semicircle and the truncated mass (shown figure) form such a system whose center of mass lies on the position of full figure.(use a suitable coordinate system).Please note that they have different masses so you need to be careful.As a matter of fact any calculus derivation of your problem employs the same method.
    Does this help?

    regards
    Yukoel
     
  15. Jul 3, 2012 #14
    So why this expression is wrong?
    "Asr=Area of semirectangle, Asc=Area of semicircle.

    Then: Xcm=[Asr*(R/2)-Asc*( R-(4/(3*pi))*R)]/(Asr-Asc);"

    I've considered the different masses, taking lambda= Mass(sr)/Area(sr), and the same for the sc.

    And the result still wrong. I've used the coordinate system shown in the figure.

    Thanks for answer.

    PS: I add the final expression in LaTex for comfort :
    [itex]Xcm=[\frac{{R}^{3}-\frac{\pi\,{R}^{2}\,\left( R-\frac{4\,R}{3\,\pi}\right) }{2}}{2\,{R}^{2}-\frac{\pi\,{R}^{2}}{2}}][/itex]
     
    Last edited: Jul 3, 2012
  16. Jul 4, 2012 #15
    Hello SergioQE
    Your expression sounds correct to me and so do your calculations.Your answer seems correct too.
    regards
    Yukoel
     
  17. Jul 4, 2012 #16
    what do you mean by maxima,just evaluate the integral.in the upper integral,after opening the brackets you will have x^2/2 left whose integral is R^3/6(for 0 to R).the bottom integral is R^2-(∏R^2/4).so center of mass =(R/6)/(1-(∏/4))=4R/6*(4-∏)or 2R/3(4-∏) WHICH IS THE ANSWER YOU GOT THAT.it is now up to you to verify it.
    edit:4-∏ is in denominator.
     
    Last edited: Jul 4, 2012
  18. Jul 4, 2012 #17
    Hello andrien ,
    You have used the inverted coordinate system I think(x=0 at the center of the semicircle).Your answer is correct with respect to that.SergioQE's value of x(cm)
    and your value add up to R if my calculations are correct .This shows that you have chosen the center as reference for x=0.Correct me if I am wrong.

    regards
    Yukoel
     
  19. Jul 5, 2012 #18
    of course.
     
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