Center of Mass bounded by Equations

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Homework Statement



I have equations that are y1= 2sin([itex]\frac{3}{2}[/itex]x) and y2= [itex]\frac{1}{3}[/itex]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= [itex]\frac{1}{M}[/itex]∫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>[itex]\frac{1}{3}[/itex](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
 

Answers and Replies

  • #2
gneill
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2,886

Homework Statement



I have equations that are y1= 2sin([itex]\frac{3}{2}[/itex]x) and y2= [itex]\frac{1}{3}[/itex]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= [itex]\frac{1}{M}[/itex]∫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>[itex]\frac{1}{3}[/itex](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.
 
  • #3
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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
gneill
Mentor
20,945
2,886
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!
 
  • #5
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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 [itex]\frac{1}{area between the curves}[/itex]∫(y1+y2)/2dx is this right? assuming M and dm are the same as above.
 
  • #6
gneill
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20,945
2,886
(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.
 

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