Where is the mass of the joined squares?

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The discussion focuses on calculating the center of mass (COM) for a system of joined squares. Participants clarify the use of coordinates and the importance of identifying the COM for individual plates, emphasizing that the COM for uniform squares is at their geometric center. The conversation highlights the need to treat each plate as a particle located at its COM to simplify calculations. A formula for the overall COM is suggested, which involves summing the position vectors of the individual masses. The final consensus is to apply the definition of the center of mass using the weighted position vectors of the three parts.
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I did Xcm=Lx/(xyz) Ycm=Ly/(xyz) Zcm=Lz/(xyz)

I modeled it from the equation CM=(mx+...)/M
 

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Sneakatone said:
I did Xcm=Lx/(xyz) Ycm=Ly/(xyz) Zcm=Lz/(xyz)

Is this your answer?
 
yes that is my answer
 
Sneakatone said:
yes that is my answer

What are x, y and z in these expressions?
 
I believe they are axis coordinates, but cas they also be mass?
 
Your answer shouldn't involve these variables.
Go back a bit - can you find the centre of masses of the individual plates?
 
L*L if your talking about one of the faces.
 
Sneakatone said:
L*L if your talking about one of the faces.
That's not the COM of the face.

Have a look at the bottom face - since it is a uniform square, then the COM must be at its geometric centre. The same for the other 2 faces.
Then replace each of the faces by a single particle at their COM - you can then combine these to get the overall COM
 
so would the faces turn to xy,xz,yz ?
 
  • #10
Sneakatone said:
so would the faces turn to xy,xz,yz ?
not sure what you mean?

Hint : The COM of the bottom piece is the point ( L/2, L/2, 0 )
 
  • #11
so the xz piece would be (L/2 0 L/2)
and the xy piece would be (0 L/2 L/2)
 
  • #12
Sneakatone said:
so the xz piece would be (L/2 0 L/2)
and the xy piece would be (0 L/2 L/2)
The second one here is the yz piece.

So, now just regard the mass of each piece as being concentrated at its COM and you end up with a system of 3 particles.
Then use the definition of the COM to find the overall COM of these 3 particles.
 
  • #13
will the combined mass be just L.
or will each axis (x y z) be the mass?
 
  • #14
Sneakatone said:
will the combined mass be just L.
or will each axis (x y z) be the mass?

?

Hint : in the xy plane, you can replace the piece of mass m with a particle of mass m at the point (L/2, L/2, 0).
Do the same for the other 2 pieces.
 
  • #15
xy=xL/2+yL/2+x0
xz=xL/2+y0+zL/2
yz=x0+yL/2+zL/2
 
  • #16
Sneakatone said:
xy=xL/2+yL/2+x0
xz=xL/2+y0+zL/2
yz=x0+yL/2+zL/2

I'm afraid this is not right :(
Why are you using x, y, and z?

What is the definition of centre of mass?
 
  • #17
center of mass is where the weighted position vectors to point sum is zero
 
  • #18
So, the CM is given by

\mathbf r_{CM} =\frac{1}{M} \sum m_i \mathbf r_i

You have the position vectors ri for each of the 3 parts. Call the mass of each piece m and then apply the above equation.
 

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