Need to find center of mass of a rod

  • Thread starter Oomair
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  • #1
Oomair
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[SOLVED] Need to find center of mass of a rod

Homework Statement


Where is the center of mass of a uniform, L-shaped iron rod of sides X = 0.8 m and Y = 0.5 m, respectively? Take the corner to be at (x,y) = (0,0), with the X and Y sides along those axes, respectively. (Assume that the rod is so narrow that the dimensions of the outer bend are the same as those of the inner bend of the L.)



Homework Equations





The Attempt at a Solution



i don't know if this involves integration, but it doesn't look like it, i think that xcm = 1/length total * ( (.8)(0) +(.8)(.8), but it does not work out
 

Answers and Replies

  • #2
Doc Al
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Start by finding the center of mass of each side separately. Then use that to find the center of mass of the complete object.
 
  • #3
Oomair
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how do i know the center of mass if I am not given the mass?

i broke it into 2 parts

(ycm)(Mt) = (m1)(Y1) +(m2)(Y2) Y1 = 0 Y2 = .5, but the unknowns are mass

(xcm)(Mt) = (m1)(X1) + (m2)(X2) X1 = 0 X2 = .8
 
  • #4
Doc Al
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45,447
1,907
how do i know the center of mass if I am not given the mass?
The mass of each side is proportional to its length. That's all you need.

i broke it into 2 parts

(ycm)(Mt) = (m1)(Y1) +(m2)(Y2) Y1 = 0 Y2 = .5, but the unknowns are mass

(xcm)(Mt) = (m1)(X1) + (m2)(X2) X1 = 0 X2 = .8
Careful here: X1, Y1 should be the center of one side; X2, Y2, the center of the other.
 
  • #5
Oomair
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but should'nt X1 and Y1 be treated as the reference line?
 
  • #6
Doc Al
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but should'nt X1 and Y1 be treated as the reference line?
Not sure what you mean. X & Y are the coordinates of the centers of each side. (Try drawing yourself a diagram.)
 
Last edited:
  • #7
stingray78
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Break it and sum. Everything is linear when talking about center of mass
 
  • #8
stingray78
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Check this to understand the center of mass better:
http://www.tubepolis.com/play.php?q=center%20of%20mass&title=Center%2Bof%2BMass%2C%2BConservation%2Bof%2BLinear%2BMomentum%2BPart%2B1&engine=1&id=ROYYNJkhG4g&img=http%253A%252F%252Fi.ytimg.com%252Fvi%252FROYYNJkhG4g%252Fdefault.jpg [Broken]
 
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