Centre of mass of of an isosceles Triangle

**gbenguse78** · January 22nd 2011, 04:48 PM

Hi
I am really stuck on this problem. It reads as such:
A uniform triangular lamina is isocceles and has the line y= 4 as its axis of symmetry. One of the vertices of the triangle is the point (2,1). Given that the x-co-ordinate of the centre of mass of the lamina is -3, find the co-ordinates of the other two vertices.
I am not sure where to begin. Any hints or guides please?

**dwsmith** · January 22nd 2011, 04:51 PM

Originally Posted by gbenguse78

Hi
I am really stuck on this problem. It reads as such:
A uniform triangular lamina is isocceles and has the line y= 4 as its axis of symmetry. One of the vertices of the triangle is the point (2,1). Given that the x-co-ordinate of the centre of mass of the lamina is -3, find the co-ordinates of the other two vertices.
I am not sure where to begin. Any hints or guides please?

Draw the picture, then represent it as a function. I tend to use absolute value since it is isosceles triangle. Let us know when you get that much down.

**dwsmith** · January 22nd 2011, 04:59 PM

Here is the integration.

$\displaystyle \bar{y}=\frac{\iint\rho(x,y)ydydx}{\iint\rho(x,y)d ydx}$

**skeeter** · January 22nd 2011, 05:13 PM

center of mass is located at the triangle's centroid.

**gbenguse78** · January 22nd 2011, 05:52 PM

Originally Posted by dwsmith

Draw the picture, then represent it as a function. I tend to use absolute value since it is isosceles triangle. Let us know when you get that much down.

I am not sure how to draw it online. How can I do that?

**gbenguse78** · January 22nd 2011, 05:55 PM

Originally Posted by dwsmith

Here is the integration.

$\displaystyle \bar{y}=\frac{\iint\rho(x,y)ydydx}{\iint\rho(x,y)d ydx}$

Can you kindly clarify your integral? Thanks

**gbenguse78** · January 22nd 2011, 06:03 PM

Originally Posted by skeeter

center of mass is located at the triangle's centroid.

So for one of the vertex,
-3=(2+2+x)/3; [Where x is the unknown x co-ordinate of the 3rd vertex]. So, the co-ordinates are (2,7), (-13,4). Is that correct? Please check it. One of the co-ordiante will lie on the axis of symmetry.

**dwsmith** · January 22nd 2011, 06:11 PM

Originally Posted by gbenguse78

I am not sure how to draw it online. How can I do that?

Ask Skeeter how to do that.

**dwsmith** · January 22nd 2011, 06:12 PM

Originally Posted by gbenguse78

Can you kindly clarify your integral? Thanks

What do you want clarified about the integral?

**gbenguse78** · January 22nd 2011, 06:25 PM

Originally Posted by dwsmith

What do you want clarified about the integral?

I will like to know what I am integrating. There are also seems to be a symbol for density. Please explain how that is relevant to this problem.

**dwsmith** · January 22nd 2011, 06:44 PM

Originally Posted by gbenguse78

I will like to know what I am integrating. There are also seems to be a symbol for density. Please explain how that is relevant to this problem.

I think you may only need a single integral.

$\displaystyle\bar{y}=\frac{\int_a^b y(f(y)-g(y))dy}{\int_a^b(f(y)-g(y))dy}$

You need to define the isosceles triangle by an equation.

$x = 2$

Represent the triangle by an absolute value function then subtract 2 from it and integrate from a to b.

I defined the functions in y due to the orientation.

**gbenguse78** · January 22nd 2011, 07:35 PM

Originally Posted by dwsmith

I think you may only need a single integral.

$\displaystyle\bar{y}=\frac{\int_a^b y(f(y)-g(y))dy}{\int_a^b(f(y)-g(y))dy}$

You need to define the isosceles triangle by an equation.

$x = 2$

Represent the triangle by an absolute value function then subtract 2 from it and integrate from a to b.

I defined the functions in y due to the orientation.

Okay. So, y= x-2. Will the co-ordinates define my limits?

**dwsmith** · January 22nd 2011, 07:55 PM

Originally Posted by gbenguse78

Okay. So, y= x-2. Will the co-ordinates define my limits?

x = 2 is g(y)

You need to model f(y).

**Krahl** · January 22nd 2011, 07:57 PM

There are two ways to go about this question gbenguse78.

dwsmith has given you the first method. skeeter has given you the second.

The question tells you that the triangle has uniform mass. So set the density function to unity in the integrals. In this case you could use skeeters method to get the answer quickly.

The two methods should complement each other.

Also, notice the "y" variable in the top integral? using that method will give you the y coordinate of the centre of mass. If you instead use the "x" variable, it will give you the x coordinate of the centre of mass.

You can set the y limits in your integral to (the equation of the bottom diagonal line in skeeters drawing) to (the eqn of the top diagonal line) in terms of x.
i.e. $y = \frac{1}{5}x + ...$ etc...

then set the x limits to -13 to 2 and integrate.

Also the denominator in dwsmiths equation is just the area of the triangle, so you don't even need to use integration for that when dealing with a triangle

**HallsofIvy** · January 23rd 2011, 04:33 AM

There is no itegration required here. Have you not noticed that skeeter gave the answer? Since the density is uniform, you are really looking for the "centroid". And the centroid of a triangle has coordinates that are the arithmetic average of coordinates of the vertices. Since the axis of symmetry is 4, one vertex is at (x, 4) for some x. The two other vertices are symmetrically placec around that line of symmetry so, since one is given as (2, 1), 3 units below the axis of symmetry, the other is at (2, 4+ 3)= (2, 7).

Of course, (4+ 1+ 7)/3= 4. All you need to do is solve (2+ 2+ x)/3= -3 for x to find the third point.

gbenguse78, the density "function" appears in the integrals for moment and area (the numerator and denominator). Since, in this problem, it is a constant, it can be taken out of each integral and will cancel. You can just ignore it.

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