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

Click For Summary
SUMMARY

The discussion focuses on calculating the center of mass (COM) of a non-uniform metal rod with varying densities. Participants emphasize that the COM shifts towards the denser section of the rod, and calculations should consider the mass distribution rather than equal lengths. The correct approach involves treating the rod as a two-body system, where the individual masses are calculated at their respective centers of mass. The final equation must account for the leverage of the two masses to determine the accurate position of the COM.

PREREQUISITES
  • Understanding of center of mass concepts
  • Familiarity with two-body systems in physics
  • Knowledge of mass distribution and density calculations
  • Basic algebra for solving equations involving mass and volume
NEXT STEPS
  • Learn how to calculate the center of mass for composite objects
  • Study the principles of mass-weighted averages in physics
  • Explore density variations in materials and their impact on COM
  • Practice solving two-body system problems with varying densities
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and center of mass calculations, as well as educators looking for practical examples of mass distribution in non-uniform objects.

VitaminK
Messages
46
Reaction score
4
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
 
Physics news on Phys.org
calculate the COM for each side then treat it as a 2-body system.
 
  • Like
Likes   Reactions: FactChecker
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)
 
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?
 
300lb/20 = 15 ?
 
VitaminK said:
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!
 
Last edited:
PeroK said:
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
 
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.
 
PeroK said:
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.
 
  • Like
Likes   Reactions: PeroK
  • #10
60DE137B-DA3F-4188-898D-D9305E1A2AA9.jpeg

Is this correct?
 
  • #11
VitaminK said:
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.
 
  • #12
VitaminK said:
View attachment 260260
Is this correct?
That doesn't look right.

Can you break down your calculation so we can see where you've gone wrong?
 
Last edited:
  • #13
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.
 
  • Like
Likes   Reactions: jbriggs444 and PeroK
  • #14
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.
 

Similar threads

Replies
10
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
16
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
2K