Center of Mass bounded by Equations

Homework Statement

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.

Homework Equations

xcm= $\frac{1}{M}$∫x dM

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

gneill
Mentor

Homework Statement

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.

Homework Equations

xcm= $\frac{1}{M}$∫x dM

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

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.

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!

gneill
Mentor
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!

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!

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!

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.

gneill
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.