Recent content by SergioQE

  1. S

    Center of mass of half square without a half circle

    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...
  2. S

    Center of mass of half square without a half circle

    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.
  3. S

    Center of mass of half square without a half circle

    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)
  4. S

    Center of mass of half square without a half circle

    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.
  5. S

    Center of mass of half square without a half circle

    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
  6. S

    Center of mass of half square without a half circle

    Homework Statement 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)]Homework Equations Rcm=(1/M)*∫rdm The Attempt at a Solution By symmetry we know Ycm=0. For de...
Back
Top