Coordinates of centre of mass of lamina

In summary: 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.
  • #1
diorific
19
0
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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  • #2
Yes, and there will be two integrals involved in finding the y-coordinate of the C.O.M.as well.
 
  • #3
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??
 
  • #4
You are not showing the details of your calculations. If you post these, you might get some suggestions.
 
  • #5
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
 
  • #6
C.O.M. coordinates generally lie within the boundaries of the body. Your calculation shows the C.O.M well outside.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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!
 

Related to Coordinates of centre of mass of lamina

1. What is the definition of "centre of mass" in relation to a lamina?

The centre of mass of a lamina is the point where the entire mass of the lamina can be considered to be concentrated. It is the point at which the lamina would balance if it were placed on a sharp point.

2. How is the centre of mass of a lamina calculated?

The centre of mass of a lamina can be calculated by finding the average of the x-coordinates and y-coordinates of all the points that make up the lamina. This can be done using the formula x̅ = ∫x dm / m and y̅ = ∫y dm / m, where x and y are the coordinates of each point, dm is the mass element, and m is the total mass of the lamina.

3. Can the centre of mass of a lamina be located outside of the lamina itself?

Yes, it is possible for the centre of mass of a lamina to be located outside of the lamina. This occurs when the lamina is not evenly distributed and has a non-uniform density. In this case, the centre of mass may be located at a point where there is no actual mass present.

4. How does the shape and distribution of mass affect the location of the centre of mass of a lamina?

The shape and distribution of mass have a significant impact on the location of the centre of mass of a lamina. A symmetrical and uniform distribution of mass will result in the centre of mass being located at the geometric centre of the lamina. However, irregular shapes and uneven distributions of mass will cause the centre of mass to shift towards the heavier or denser regions of the lamina.

5. Are there any real-world applications of finding the centre of mass of a lamina?

Yes, there are many real-world applications of finding the centre of mass of a lamina. For example, in engineering, the centre of mass is an important factor in designing balanced and stable structures. In physics, the centre of mass is used to calculate the motion of objects in various systems. It is also used in sports, such as in gymnastics and diving, to determine the optimal position for performing stunts and dives.

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