Coordinates of centre of mass of lamina

AI Thread Summary
The discussion focuses on calculating the mass and center of mass of an industrial tool made from a laminar material with variable density. The vertices of the region are identified as (1,2), (2,1), and (4,2), and the mass is derived from double integrals, leading to a result of M=5σ after corrections to initial calculations. Participants express confusion about finding the coordinates of the center of mass, emphasizing the need to sum both X and Y double integrals. Several users request clarification on integration steps and encourage sharing detailed calculations to identify errors. The conversation highlights the importance of accuracy in mathematical integration and the collaborative effort to resolve misunderstandings.
diorific
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An industrial tool is made from alaminar material occupyingthe region between thestraight lines y =2, y=1 2x and y =3−x. The density of thematerial varies and is given by thefunction σ(1 + x), where x is the horizontal distancefrom the y-axis and σ is aconstant. The mass of the tool andthe position of its centre of mass are sought.

(a) Sketch the region, and find the coordinates of the vertices.

(b) Calculate the mass of the tool in terms of σ.

(c)Find the coordinates of the centre of mass of the tool.a)

graph_zpsce5f92de.jpg


The vertices are the points (1,2), (2,1) and (4,2)

b)
The mass M is the sum of these double integrals

Integrlas_zps60d1efb1.jpg


When you do all the calculations M=4δ

c)
Now I'm confused on how I find the coordinates of the centre of mass.
Since the coordinates are
coordinates_zps6e7cafbd.jpg


But I have two double integrals, then do I need to sum both of the X double integrals to find the X-coordinate?
 
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Yes, and there will be two integrals involved in finding the y-coordinate of the C.O.M.as well.
 
diorific said:
An industrial tool is made from alaminar material occupyingthe region between thestraight lines y =2, y=1 2x and y =3−x. The density of thematerial varies and is given by thefunction σ(1 + x), where x is the horizontal distancefrom the y-axis and σ is aconstant. The mass of the tool andthe position of its centre of mass are sought.

(a) Sketch the region, and find the coordinates of the vertices.

(b) Calculate the mass of the tool in terms of σ.

(c)Find the coordinates of the centre of mass of the tool.


a)

graph_zpsce5f92de.jpg


The vertices are the points (1,2), (2,1) and (4,2)

b)
The mass M is the sum of these double integrals

Integrlas_zps60d1efb1.jpg


When you do all the calculations M=4δ

c)
Now I'm confused on how I find the coordinates of the centre of mass.
Since the coordinates are
coordinates_zps6e7cafbd.jpg


But I have two double integrals, then do I need to sum both of the X double integrals to find the X-coordinate?

ok, so this is what I did

coordinate_zpsf96e2734.jpg


But the result is 303/16≈18.94 and this clearly is not the X-coordinate of the centre of mass.

Can anyone help??
 
You are not showing the details of your calculations. If you post these, you might get some suggestions.
 
Ok, I think I did a mistake when I did the calculations. I've done it again and the X-coordinate is 89/16≈5.56

xcoordinate_zps06f5c67f.jpg
 
C.O.M. coordinates generally lie within the boundaries of the body. Your calculation shows the C.O.M well outside.
 
You don't show your working for M. I get 5, not 4.
In the calculation starting at (2), the coefficient of x3 is wrong in the second line.
 
Yes, you are right.
I've done the calculations again and the M=5
Also I've done the calculations again for the coordinates and everything looks correct since the coordinates lie within the shape.

Thank you for your help.
 
hi
i also stuck on this problem can you show me how do you get 5 sigma coz i try it many time but i got 4
thanks
 
  • #10
fatima123 said:
hi
i also stuck on this problem can you show me how do you get 5 sigma coz i try it many time but i got 4
thanks

Pls post your working
 
  • #11
haruspex said:
Pls post your working

Hi I'm stuck on the integral if sigma(1+x) it self as I every time I got different value can you please show me how to integrate this
Thank you
 
  • #12
fatima123 said:
Hi I'm stuck on the integral if sigma(1+x) it self as I every time I got different value can you please show me how to integrate this
Thank you

That is why you should post your working, so that we can point out to you where your mistake is. We are not mind readers.
 
  • #13
arildno said:
That is why you should post your working, so that we can point out to you where your mistake is. We are not mind readers.
integrate from 1 to 2 and from y= 3-x to y=2 for sigma (1+x) = sigma (1+x) y
Which gives (x+x^2/2)*y and tgen substitute the values of y
Then integrate again and substitute value of 1 and 2 this gives me 4 sigma
 
  • #14
fatima123 said:
integrate from 1 to 2 and from y= 3-x to y=2 for sigma (1+x) = sigma (1+x) y
Which gives (x+x^2/2)*y and tgen substitute the values of y
Then integrate again and substitute value of 1 and 2 this gives me 4 sigma
The range y= 3-x to y=2 gives ∫(1+x) (2 - (3-x))dx. If you don't understand POST ALL YOUR WORKING!
 

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