Center of Mass

• Dec 4th 2006, 02:02 PM
cyberdx16
Center of Mass
I have two problems here that i cant figure out

EDIT find the center of mass given that the mass is constant

1.)the "triangular" region in the first quadrant between the circle x^2+y^2=9 and the lines x=3 and y =3 (Hint: use geomentry to find the area).

2.) Find the center of mass of a thin plate covering the region between the x-axis and the curve y= 2/x^2. 1<=x<=2 if the plates density ate the point (x,y)
is delta(x)=x^2
• Dec 4th 2006, 04:45 PM
ThePerfectHacker
(1)You can view the the region as subtracting the rectangle from the quater-circle.

Now the centroid of the rectangle is trivial (1.5,1.5) und the area is 9.

The centroid of a quater-circle is located on the line $y=x$ (that is symettry) and is $\frac{4r}{3\pi}$ (a known formula). In this case $r=3$. Thus the centroid is $\frac{4(3)}{3 \pi}=\frac{4}{\pi}=1.27$. Thus, $(x,y)=(1.27,1.27)$. And has area $\frac{1}{4}\pi (3)^2=7.06$.

Thus, by the composite centroid formula we have,
$\bar x= \frac{(1.5)(9)-(1.27)(7.06)}{9+7.06}$

$\bar y=\frac{(1.5)(9)-(1.27)(7.06)}{9+7.06}$
• Dec 4th 2006, 06:02 PM
cyberdx16
hmmm.... my book say's i should get 2/(4-pie).... i think ur #### gives me the c of m of the semicircle... i want to find the c of m of the triangular shape made by the semicircle and x=3 and y=3