Calculate the center of mass of a non-uniform metal rod

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  • #1
VitaminK
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Homework Statement:
A metal rod consists of two parts. one part is a 1.2m metal with the density of 5kg/m3. The other part is a 1,8m metal with the density of 7kg/m3. Calculate the center of mass for this rod.
Relevant Equations:
Density=m/v
45819C6E-7A28-4850-83BF-68DE04A45A58.jpeg

I know that if they had the same density they would have the center of mass at 1,5 m. But now that they don't the center of mass will be shifted towards the part of the rod with higher density. they will have their center of mass where they
have equal mass
p1*v=p2*v

now i don't know how to move forward
 

Answers and Replies

  • #2
phinds
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calculate the COM for each side then treat it as a 2-body system.
 
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  • #3
VitaminK
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Do you mean
1.2/2=0.6m
1.8/2= 0.9m
(0.6m+0.9m)

im not familiar with 2-body system (high school physics)
 
  • #4
phinds
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If you have a 100lb man standing at -10 on the x axix and a 200lb man standing at +10 on the x axis, where is the center of gravity of the two men considered as one system?
 
  • #5
VitaminK
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300lb/20 = 15 ?
 
  • #6
PeroK
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Do you mean
1.2/2=0.6m
1.8/2= 0.9m
(0.6m+0.9m)

im not familiar with 2-body system (high school physics)
Alternatively, try to find the point ##x## along the rod where there is equal mass on both sides.

PS Ignore this post!
 
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  • #7
VitaminK
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Alternatively, try to find the point ##x## along the rod where there is equal mass on both sides.


you mean p1*v=p2*v --> 5*pi*r^2*l=7*pi*r^2*l now I am confused about the length l
 
  • #8
FactChecker
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There are rod two parts, each with a uniform density, where the center of mass of each part is simple to determine. The original problem can be converted to a problem with the entire mass of the two parts located at their individual centers of mass. That problem is then simple to solve.
 
  • #9
jbriggs444
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Alternatively, try to find the point ##x## along the rod where there is equal mass on both sides.
That won't work. You are not after the point where the mass on both sides is the same. You need the leverage of the two masses to be the same.

Worded differently, you are not after the median [mass-weighted] position. You are after the mean [mass-weighted] position.
 
  • #10
VitaminK
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60DE137B-DA3F-4188-898D-D9305E1A2AA9.jpeg

Is this correct?
 
  • #11
PeroK
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you mean p1*v=p2*v --> 5*pi*r^2*l=7*pi*r^2*l now I am confused about the length l
Ignore what I said! I wasn't thinking.
 
  • #13
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That is not the answer I get. To make your calculations more clear, I think you should specifically calculate the two masses, ##M_1, M_2## at two locations, ##x_1,x_2##. Then put them into a final equation to find the center of mass of two masses.

EDIT: Sorry, I see that you can not calculate specific numbers for the masses. You must leave the density in. But I think you can make your calculations more clear. I still think that there is a mistake in your calculations.
 
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  • #14
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I see the problem. You need to be more careful about the volume. Assuming that the diameter of both parts is identical, the two parts do not have the same volume.
 

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