# Center of Mass bounded by Equations

1. Mar 8, 2012

### nat1

1. The problem statement, all variables and given/known data

I have equations that are y1= 2sin($\frac{3}{2}$x) and y2= $\frac{1}{3}$x the point where they intersect is called "a" (about x≈1.88). Find the center of mass where M is the total mass of the object.

2. Relevant equations

xcm= $\frac{1}{M}$∫x dM

3. The attempt at a solution

I found the vertical center of mass by using the area between the curves and setting

dm= density*thickness*(y1 - y2)dx

and setting

M=density*thickness*area

area is between the curves. As im trying to find the CM along the horizontial of the object i cant figure out dm because when y>$\frac{1}{3}$(a) y1 has two values, how can i correct for this? my first thought is to rearrange the equations to find f(y)=x and subtract the x values but i still run into the same problem.

Nat

2. Mar 8, 2012

### Staff: Mentor

You should be able to find both the x and y centers of mass by integrating along the x-axis. Consider that your dm "strips" bounded by y2 and y1 in addition to having a position along the x-axis, will also each have a vertical center where the actual center of mass of that strip lies.

3. Mar 8, 2012

### nat1

im still not sure i understand %100:

So each dm will have a vertical center of (y1-y2)/2 and this is where the center of mass for each strip lies. so then i would jsut need to take the average of these over the interval [0, a]

Thanks!

4. Mar 8, 2012

### Staff: Mentor

Careful, the midpoint between two numbers y1 and y2 is (y1 + y2)/2. And you still need to do the weighted sum of these positions (the dm strips all have different masses).

Good luck!

5. Mar 8, 2012

### nat1

oops, my fault! i would do the integral of $\frac{1}{area between the curves}$∫(y1+y2)/2dx is this right? assuming M and dm are the same as above.

6. Mar 8, 2012

### Staff: Mentor

(y1 + y2)/2 gives you the y-position of the cm for the given strip. The strip's mass is still dm, for which you previously obtained a formula. The ycm of the object is still given by the weighted sum:

$y_{cm} = \frac{1}{M} \int y\;dm$

where you use your new y-position in place of y, and use your dm expression as before.