The center of mass of a semicircular arc of non-negligible width

In summary: The differential area would be dA=-R2^2-R1^2.In summary, the goal of this attempt was to find the center of mass of a system of parts by approximating the center of mass of each sub-arc with a titled rectangle and solving for the radius of the centroid. However, the method proved to be incorrect when the parts had a thickness of zero.
  • #1
Hamza M khan
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Homework Statement
The goal: finding the center of mas a semicircular wire/disk of on non negligible width, with the inner radius being R1 and out radius being R2.

My attempt:

1) Im gonna start this with a goal of setting up a reimann sum. First I divide the "arc"(?) of angle pi into n sub-arcs of equal angle Δθ

2) The total center of mass can be found if centers of mass of parts of the system are known. In each circular arc interval, I choose a height, Hi, approximating the height of the center os mass of each sub arc, hoping that the error goes to 0 in the limit as n goes to infinity, and multiply this by the mass of the sub arc. Pushing this through the limiting process, I set up the integral of H w.r.t m

3) finding Hi . Now, as Δθ goes to 0, the sector-difference region formed by each sub-arc should get closer and closer to a tilted rectangle. Assuming that to be true, the center of mass of each sub-arc( being approximated by a titled rectangle) would be a distance Hi=(R1+R2)sin(θ)/2 above the origin

4) lastly, since the shape has a constant mass per unit area, the differential mass and total mass can be replaced by differential area and total area. Using the sector area formula for each subinterval, the differntial area, dA, should be equal to 0.5dθ (R2^2-R1^2)

solving this gives me ycom=(R1+R2)/pi which upon looking up is clearly wrong. It is interesting thought that it gives the correct result when R1=R2 ( 0 thickness). What is the error in my reasoning?
Relevant Equations
Ycom=m1y1+m2y2+....miyi
My attempt:

1) I am going to start this with a goal of setting up a reimann sum. First I divide the "arc"(?) of angle pi into n sub-arcs of equal angle Δθ

2) The total center of mass can be found if centers of mass of parts of the system are known. In each circular arc interval, I choose a height, Hi, approximating the height of the center os mass of each sub arc, hoping that the error goes to 0 in the limit as n goes to infinity, and multiply this by the mass of the sub arc. Pushing this through the limiting process, I set up the integral of H w.r.t m

3) finding Hi . Now, as Δθ goes to 0, the sector-difference region formed by each sub-arc should get closer and closer to a tilted rectangle. Assuming that to be true, the center of mass of each sub-arc( being approximated by a titled rectangle) would be a distance Hi=(R1+R2)sin(θ)/2 above the origin

4) lastly, since the shape has a constant mass per unit area, the differential mass and total mass can be replaced by differential area and total area. Using the sector area formula for each subinterval, the differntial area, dA, should be equal to 0.5dθ (R2^2-R1^2)

solving this gives me ycom=(R1+R2)/pi which upon looking up is clearly wrong. It is interesting thought that it gives the correct result when R1=R2 ( 0 thickness). What is the error in my reasoning?
 

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  • #2
Hamza M khan said:
3) finding Hi . Now, as Δθ goes to 0, the sector-difference region formed by each sub-arc should get closer and closer to a tilted rectangle. Assuming that to be true, the center of mass of each sub-arc( being approximated by a titled rectangle) would be a distance Hi=(R1+R2)sin(θ)/2 above the origin
No matter how small you make ##\Delta \theta##, the center of mass of the sub-arc will not be at a radial distance of ##(R_1+R_2)/2##. The width of the sub-arc is greater at ##R_2## than at ##R_1##.

You could find the correct location of the CM of a sub-arc and then proceed as you did. However, it might be easier not to bother with the sub-arcs.
 
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  • #3
Since the object has a constant thickness and a symmetrical shape, using a polar coordinate system would simplify calculations very much.

I agree with @TSny, your section has more area or mass above a line of arithmetic mean.
Basically, to exactly calculate the radius of the centroid for any infinitesimal section, you have two circular sectors to consider, the smaller one to be voided.

Since that section is symmetrical, the centroid will be located over the symmetry line of that section.
Please, see:
https://en.m.wikipedia.org/wiki/List_of_centroids

The location of the centroid of any shape of that kind must be dependent on the angle of the arc (which seems to be π in your schematic).
 
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  • #4
A different approach would be to find the center of mass of a semicircular slab. That's straightforward to do even in cartesian coordinates. Then consider the arc as a slab of radius R2 minus the slab of radius R1.
 
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What is the center of mass of a semicircular arc of non-negligible width?

The center of mass of a semicircular arc of non-negligible width is the point at which the entire mass of the arc can be considered to be concentrated. It is the average position of all the individual masses that make up the arc.

How is the center of mass of a semicircular arc of non-negligible width calculated?

The center of mass of a semicircular arc of non-negligible width can be calculated by finding the average position of all the individual masses that make up the arc. This can be done using the formula: x̄ = ∑(m_ix_i)/M, where x̄ is the center of mass, m_i is the mass of each individual component, x_i is the position of each individual component, and M is the total mass of the arc.

Why is the center of mass of a semicircular arc of non-negligible width important?

The center of mass of a semicircular arc of non-negligible width is important because it helps us understand the overall motion and stability of the arc. It is also used in many engineering and physics applications, such as calculating the torque and rotational motion of the arc.

How does the width of the semicircular arc affect its center of mass?

The width of the semicircular arc does not significantly affect its center of mass. As long as the width is not negligible, the center of mass will still be located at the same point. However, if the width becomes very large, the center of mass may shift slightly towards the wider end.

Can the center of mass of a semicircular arc of non-negligible width be outside of the arc?

No, the center of mass of a semicircular arc of non-negligible width will always be located within the boundaries of the arc. This is because the center of mass is calculated based on the positions of the individual masses that make up the arc, and these masses are all located within the arc.

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