MadSci Network: Physics |
The centre of mass of a system of particles m1,m2,m3 is defined to be at Xcm = m1x1+m2x2+m3x3/m1+m2+m3 Ycm = m1y1+m2y2+m3y3/m1+m2+m3 where the coordinates xj, yj are measured from some fixed reference. Let us take the origin of a rectangular coordinate system to be centred at the O atom, the x axis to be horizontal and positive to the right and the y axis to be vertical and positive upwards. H H _ \ / | \ 105¤ / | \-------/ | D = .96A \ / | \ / | \ / | O-----------------------------------------> x axis - then Xo=0 Xh1= Dcot(37.5)=1.3032D=1.2511A Xh2= -1.2511A Yo=0 Yh1=D=.96A Yh2=D=.96A I call h1 to be the h atom on the right and h2 to be the one on the left. Xcm = Xo*Mo + Xh1*Mh1+ Xh2*Mh2 / Mo + Mh1 +Mh2 = 0 Ycm =Yo*Mo + Yh1*Mh1+ Yh2*Mh2 / Mo + Mh1 +Mh2 = 0 + D + D / 16 + 1 +1 = D/9 = .1067A So, the centre of mass is located .1067 A exactly above the O atom.
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