Help Me Solve My Diagram Problem: Calculating Center of Mass

In summary, the conversation discusses the calculation of the center of mass for a system of two identical triangles. The conversation includes equations and formulas for calculating the center of mass, as well as a discussion on using the symmetry of the problem to simplify the calculation. The final answer is determined to be 2.5, as it is the midpoint between the two triangles due to their identical nature.
  • #1
jisbon
476
30
Homework Statement
Two identical uniform triangular metal played held together by light rods. Caluclate the x coordinate of centre of mass of the two mass object. Give that mass per unit area of plate is 1.4g/cm square and total mass = 25.2g
Relevant Equations
-
Not sure what I went wrong here, anyone can help me out on this? Thanks.
EDIT: Reformatted my request.
Diagram:
1567423208963.png

So as far as I know to calculate the center of mass for x, I have to use the following equation:
COM(x):
##\frac{1}{M}\int x dm##
And I also figured that to find center of mass, I will have to sum the mass of the 2 plates by 'cutting' them into stripes, giving me the following formula:
##dm = \mu * dx * y## where ##\mu## is the mass per unit area.
So subbing in the above equation into the first, I get:
##\frac{1}{M}\int x (\mu * dx *y) ##
##\frac{\mu}{M}\int xy dx##
Since the 2 triangles are identical, I can assume triangle on the left has equation ##y = 1/4x +4##
This is the part where I'm not sure. Do I calculate each of the triangle's center of moment, sum them and divide by 2? Or am I suppose to use another method?
Regardless of what, supposed I am correct:

COM for right triangle:
##\frac{\mu}{M}\int_{4}^{16}x(\frac{1}{4}x+4) dx## = 8 (expected)

COM for left triangle:
##\frac{\mu}{M}\int_{-11}^{1}x(-\frac{1}{4}x+4) dx## = 5.63...

Total COM = ##8+5.63/2## which is wrong :(

Thanks
 
Last edited:
Physics news on Phys.org
  • #2
Your equations are hard to read (writing them here would help) and you don't explain what you do either.
If the x-axis is horizontally why don't you just use the symmetry of the problem?
 
  • #3
If the mass density of the two triangles is uniform, then there is a pure geometric method to find the center of mass, that depends only on the shape of the triangles. But I am not sure whether you are allowed to use this method. Anyway for uniform mass density the center of mass is the so called centroid or barycenter of the triangle, which is the point that the three medians of the triangle intersect.
 
  • #4
mfb said:
Your equations are hard to read (writing them here would help) and you don't explain what you do either.
If the x-axis is horizontally why don't you just use the symmetry of the problem?
Delta2 said:
If the mass density of the two triangles is uniform, then there is a pure geometric method to find the center of mass, that depends only on the shape of the triangles. But I am not sure whether you are allowed to use this method. Anyway for uniform mass density the center of mass is the so called centroid or barycenter of the triangle, which is the point that the three medians of the triangle intersect.
Updated my question :)
 
  • #5
Certainly there is something wrong with that 5.63... Do the integral again.
and I believe the integral must be $$\int_{-11}^{1}x(x/4+\frac{11}{4})dx$$
 
  • #6
Delta2 said:
Certainly there is something wrong with that 5.63... Do the integral again.
and I believe the integral must be $$\int_{-11}^{1}x(x/4+\frac{11}{4})dx$$
Realized I used the wrong equation, answer turns out to be -7.16666...
It's still wrong though :/
 
  • #7
Use the integral I posted at post #5 the correct y(x) for the left triangle is ##y=\frac{x+11}{4}##. You ll find the x-coordinate for the center of mass of the left triangle to be -3.
 
  • Like
Likes jisbon
  • #8
Delta2 said:
Use the integral I posted at post #5 the correct y(x) for the left triangle is ##y=\frac{x+11}{4}##. You ll find the x-coordinate for the center of mass of the left triangle to be -3.
Oh yes the y intercept :/ I 'think' too much about identical, even the intercepts :/ Thanks tho :) Answer is 2.5 :)
 
  • Like
Likes Delta2
  • #9
jisbon said:
Answer is 2.5
Right.
Note that 2.5 is exactly the middle between the two triangles. The setup is symmetric, it can't be anywhere else. Finding symmetries like this can often save lengthy calculations.
 

FAQ: Help Me Solve My Diagram Problem: Calculating Center of Mass

What is the center of mass?

The center of mass is the point at which the mass of a body is considered to be concentrated and the body can be balanced or rotated.

Why is calculating center of mass important?

Calculating center of mass is important because it helps determine the overall motion and stability of an object. It is also used in various fields such as physics, engineering, and biomechanics to analyze and design systems.

How do I calculate the center of mass of a diagram?

To calculate the center of mass, you need to first determine the mass and position of each individual component or object in the diagram. Then, use the equation:
Center of mass = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn)
Where m is the mass and x is the position of each component.

Can the center of mass be outside of the object?

Yes, the center of mass can be outside of the object. This is possible if the object is irregularly shaped or has a non-uniform distribution of mass.

What are some real-life applications of center of mass calculation?

Center of mass calculation is used in various real-life situations, such as designing stable structures like buildings and bridges, analyzing the flight of airplanes, determining the balance of vehicles, and understanding the body movements of athletes and dancers.

Similar threads

Back
Top