Calculating Centers of Mass for Geometric Shapes

[Wigglyben]

I'm just trying to review for an exam and the only thing I don't really get is how to calculate centers of mass. I'll just use two problems I had in my homework that I didn't know how to do. I feel like these two problems should be really easy but I have idea how to do them.

Homework Statement

1. Find the center of mass of a uniform wire that subtends an arc θ if the radius of the arc is a. My terrible MSPaint picture to give you an idea:

2. Find the center of mass of this isoceles triangle:

Homework Equations

1/M [Integral] rdm

The Attempt at a Solution

In the first one I tried to use polar coordinates and substitute dm for density*dθ. Density would be M/θ. But when you get to the integral it's 1/θ so you would get the natural log of zero or a negative number depending on how you want to do it.

For the second one, I have no idea where to begin even. Would you have to do a double integral and find the equation of the line that forms the triangle or is there an easier way?

 

[HallsofIvy]

I'm just trying to review for an exam and the only thing I don't really get is how to calculate centers of mass. I'll just use two problems I had in my homework that I didn't know how to do. I feel like these two problems should be really easy but I have idea how to do them.

Homework Statement

1. Find the center of mass of a uniform wire that subtends an arc θ if the radius of the arc is a. My terrible MSPaint picture to give you an idea:

2. Find the center of mass of this isoceles triangle:

Homework Equations

1/M [Integral] rdm

The Attempt at a Solution

In the first one I tried to use polar coordinates and substitute dm for density*dθ. Density would be M/θ. But when you get to the integral it's 1/θ so you would get the natural log of zero or a negative number depending on how you want to do it.

No, the density is NOT M/\theta. i suspect you are using \theta to mean two different things! If you are using \theta to mean the central angle of the sector, then you cannot use it to mean the angle in polar coordinates. The area of a sector with central angle \theta and radius a is \theta a^2/2. If the mass is M, then the density is 2M/(\theta a^2), a constant.

By symmetry, the y-coordinate of the center of mass is 0. Since in polar coordinates x= r cos(\phi) (I am using \phi because \theta is the central angle of the sector) M\overline{x} is
\frac{2M}{\theta a^2}\int_{r= 0}^a\int_{\phi= -\theta/2}^{\theta/2} (r cos(\phi))r dr d\phi

For the second one, I have no idea where to begin even. Would you have to do a double integral and find the equation of the line that forms the triangle or is there an easier way?

Apparently each of the slant sides of the triangle has length a but what is the length of the base? By symmetry the x coordinate of the center of mass is 0 but the y coordinate will depend upon both the height and the base of the triangle. Given that the slant sides is a, one of those determines the other- but you have to know one or the other.

By the way, when the density is a constant, it will cancel when you divide the [math]\int x dm[/math] integral by the mass so you can always take the density to be 1. In that case the "center of mass" depends only on the geometric shape and is called the "centroid".

 

[Mark44]

To calculate the center of mass of something you need to find the two moments: moment about the x-axis and moment about the y-axis. Due to the symmetry in both your figures, you won't have to calculate one of these moments for each figure.

In the problem of the wire, the c.m. will be at (x_CM, 0) and for the triangle, the c.m. will be at (0, y_CM).

For the wire, polar coordinates might not be the way to go. Can you come up with the integral for the moment about the y-axis? For the triangle, what do you get for you integral for the moment about the x-axis?

 

[Wigglyben]

Well I didn't have any center of mass problems on my exam thankfully, it was actually pretty easy.

Thanks for the help, I knew I must have been doing something stupid. And I wrote the second problem wrong, it was a right isosceles triangle so the bottom vertices are both 45 degrees.