Finding the center of mass in a system of pipes

In summary, the system S is comprised of three identical metal pipes, A, B, and C, with outer and inner radii of 0.51m and 0.50m, respectively. The pipes are placed so that A is on top of B and C, with B lying along the y-axis and all pipes oriented parallel to the z-axis. The pipes are made of iron with a density of 7870kgm^-3. For question 1, the X and Y coordinates of the center of mass for the system can be found using the equations X = ΣXiMi/ΣMi and Y = ΣYiMi/ΣMi, which gives the coordinates (1.02
  • #1
Freshtone
8
0

Homework Statement


System S is static and is comprised of 3 identical metal pipes A, B and C which are placed so that A lays ontop of B and C. B lays into the y axis. The length of the pipes are orientated parallel with the z axis. The outer and inner radius of each pipe is (Outer : 0.51m) (inner : 0.50m) They have a length of 2m. The pipes are made of iron with a density of 37870kgm^-3

1) Find the X and Y coordinates to the System S.

2) what is the momment of inertia about the Z-axis

Homework Equations


no relevant equations

The Attempt at a Solution


[/B]
As a mechanical engineering student i have learned the mechanical way of doing this which is similar but i get caught up on how to find the x or y position of the top pipe.

Any help would be fantastic
 
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  • #2
Hello Freshtone, :welcome:

Can you explain what you mean with
Freshtone said:
B lays into the y axis. The length of the pipes are orientated parallel with the z axis
and perhaps post a drawing of the arrangement ?
 
  • #3
Freshtone said:

Homework Statement


System S is static and is comprised of 3 identical metal pipes A, B and C which are placed so that A lays ontop of B and C. B lays into the y axis. The length of the pipes are orientated parallel with the z axis. The outer and inner radius of each pipe is (Outer : 0.51m) (inner : 0.50m) They have a length of 2m. The pipes are made of iron with a density of 37870kgm^-3

Iron has a density of about 7850 kg / m3.

http://www.engineeringtoolbox.com/metal-alloys-densities-d_50.html
 
  • #4
Ah yes sorry i miss-copied the iron density..it was actually 7870kgm^-3

Sorry about that!...ill copy a picture of the setup so you can see ! =) thank you for the replies already !
 
  • #5
Screen_Shot_2016_03_13_at_11_37_51.png


By saying B lays into the Y axis you can see here it is easier to see instead hehe.

By the z axis i think they mean that the pipes have a length of 2 meters going in the z axis which obviously is directly into the picture ...=)
 
  • #6
Good. So we have the problem statement amost complete. What I'm still missing is: what is meant with
Freshtone said:
the X and Y coordinates to the System S
but I'm prepared to guess that they mean "the X and Y coordinates of the center of mass of the System S". Approved ?
So for 1. we need some math, nothing more. Except perhaps the relevant equation for the center of mass. See below.
But for 2 we really need something to help us find the moment of inertia, so we do need about two relevant equations. Any idea ?

In the mean time, working on 1: The center of mass (let's do the axis parallel to the z-axis on which the c.o.m. lies) of B is a giveaway.
And for C it's also a giveaway. A is only half a giveaway: the x coordinate is pretty clear. For the y-coordinate you need the fact that all three diameters are the same. And there are various possible approaches: the distance to the z-axis for A and C could well be the same, or you see an equilateral triangle when you think of the three pipe axes.
Top it off with the c.o.m. definition, which look up.
 
  • #7
Hey thank you!

I have now "i think" worked out the first question at least...i used some trig to get the height on y-axis for A.

this was my work so far ... Xo=(7870*0.51+0.255*7870+0.765*7870) / 3(7870) and got Xo = 0.51m

my Yo was similar and i got the height from x up to A as 0.825m. so Yo = 0.445.

my x,y axis cords become (0.51 , 0.445) <-- this would be my answer to 1)

Can those sound right ? ...now I am working on finding second question and by so I am using Isz and working the area for the pipe so i can figure out the rest...Not sure i explainned that correct but maybe you see my point ?
 
  • #8
Before i write down the answer, i have to sorry about my english. For question a, i will use X = ΣXiMi/ΣMi and Y = ΣYiMi/ΣMi to calculate. The answer is X = 1,01 and Y ≈ 0,8. Actually, if three pipes are homogenous, you won't have to calculate its mass. But if not then you will have to calculate its mass. Base on your problem statement, you did not mention about it so i assume that three pipes are homogenous.
 
  • #9
i also think that for the second question we would use the equations as follows ...

Sx =Σ(A*y) = Yo*A
but in the z axis ?
 
  • #10
The Vinh said:
Before i write down the answer, i have to sorry about my english. For question a, i will use X = ΣXiMi/ΣMi and Y = ΣYiMi/ΣMi to calculate. The answer is X = 1,01 and Y ≈ 0,8. Actually, if three pipes are homogenous, you won't have to calculate its mass. But if not then you will have to calculate its mass. Base on your problem statement, you did not mention about it so i assume that three pipes are homogenous.

Im not sure how that can be correct as my whole systems x length is 1.02m ..if the mass in the x-axis is so far as you say then the system would very unbalanced as all pipes are the same mass , size and length !
 
  • #11
But the radius is 1.01 m
 
  • #12
The Vinh said:
But the radius is 1.01 m
oh wait i see i have treated the radius as the diameter ! you are right then ! let me just try to work it out myself quick =) so sorry about that ! too much maths has me not thinking straight !
 
  • #13
I have looked up google for inner and outer radius and the radius must be equal inner radius + outer radius, right ?
 
  • #14
right after adjusting all the values i have got the same values for x and Y cords ...(1.02, 0.89)

On to the second question and I am having a little trouble understanding what i need to find ..as the question states it wants me to find the inertia about the z axis ? any pointers to help ?
 
  • #15
If the three pipes are identical, then the center of mass is the geometrical center. And that would be the centroid of the equilateral triangle having the three centers of the pipes as vertices.
 
  • #16
Freshtone said:
right after adjusting all the values i have got the same values for x and Y cords ...(1.02, 0.89)

On to the second question and I am having a little trouble understanding what i need to find ..as the question states it wants me to find the inertia about the z axis ? any pointers to help ?
I think that the value of y = 0,79
 
  • #17
Having a little confuse, the second question, I have to calculate the center of mass or the moment of inertia ? ( your topic mentions about the center of mass)
 
  • #18
Freshtone said:
right after adjusting all the values i have got the same values for x and Y cords ...(1.02, 0.89)

On to the second question and I am having a little trouble understanding what i need to find ..as the question states it wants me to find the inertia about the z axis ? any pointers to help ?
Well, what do you understand about calculating the moment of inertia for a cylindrical pipe?
 
  • #19
SteamKing said:
Well, what do you understand about calculating the moment of inertia for a cylindrical pipe?
We will use Iz = mr^2/2
Ix = Iy = m(3r^2 + h^2)/12
Right ?
 
  • #21

1. What is the center of mass in a system of pipes?

The center of mass in a system of pipes is the point at which the entire mass of the system can be considered to be concentrated. It is the average position of all the individual masses in the system.

2. Why is finding the center of mass important in a system of pipes?

Finding the center of mass in a system of pipes is important because it helps determine the stability and balance of the system. It also allows engineers to calculate the forces acting on the system and make predictions about its behavior.

3. How is the center of mass calculated in a system of pipes?

The center of mass in a system of pipes can be calculated using the following formula:
xcm = (m1x1 + m2x2 + m3x3 + ... + mnxn) / (m1 + m2 + m3 + ... + mn)
where xcm is the x coordinate of the center of mass, mi is the mass of each pipe, and xi is the x coordinate of each pipe.

4. Can the center of mass be outside of the physical space of the pipes?

Yes, the center of mass can be outside of the physical space of the pipes. This can happen if the pipes are not evenly distributed or if there is a significant difference in the masses of the pipes.

5. How does the shape of the pipes affect the position of the center of mass?

The shape of the pipes can affect the position of the center of mass. Pipes with larger diameters or thicker walls will have more mass and therefore have a greater influence on the position of the center of mass. The orientation of the pipes can also affect the position of the center of mass.

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