## Center of mass of half square without a half circle

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.

[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)*$\int\2\,x\,R-2\,x\,{\left( {R}^{2}-{x}^{2}\right) }^{0.5}$/extract_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
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 The CM of rectangle R×2R=CM half circle + CM shaded area.
 You should use the guyndanh formula for sheet

## Center of mass of half square without a half circle

 Quote by takudo_1912 You should use the guyndanh formula for sheet
I wonder what that is....

 Quote by azizlwl The CM of rectangle R×2R=CM half circle + CM shaded area.
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

 Quote by takudo_1912 You should use the guyndanh formula for sheet
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.

 Quote by SergioQE 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
That formula is the distance from the base(diameter) of the semi circle... And the semicircle is 'inverted', here
 you should also consider the mass as proportional to area and not just subtract the x's.

 Quote by SergioQE 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
That from top.
Require reference to x-axis
 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)
 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.

 Quote by andrien 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.
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.
 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

 Quote by Yukoel 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
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 :
$Xcm=[\frac{{R}^{3}-\frac{\pi\,{R}^{2}\,\left( R-\frac{4\,R}{3\,\pi}\right) }{2}}{2\,{R}^{2}-\frac{\pi\,{R}^{2}}{2}}]$

 Quote by SergioQE 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 : $Xcm=[\frac{{R}^{3}-\frac{\pi\,{R}^{2}\,\left( R-\frac{4\,R}{3\,\pi}\right) }{2}}{2\,{R}^{2}-\frac{\pi\,{R}^{2}}{2}}]$
Hello SergioQE
Your expression sounds correct to me and so do your calculations.Your answer seems correct too.
regards
Yukoel

 Quote by SergioQE 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.
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.

 Quote by andrien 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.
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
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