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Center of Mass bounded by Equations

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

    xcm= [itex]\frac{1}{M}[/itex]∫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>[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
     
  2. jcsd
  3. Mar 8, 2012 #2

    gneill

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    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.
     
  4. Mar 8, 2012 #3
    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!
     
  5. Mar 8, 2012 #4

    gneill

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    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!
     
  6. Mar 8, 2012 #5
    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.
     
  7. Mar 8, 2012 #6

    gneill

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    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.
     
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