# Centroid of a semicircular arc

• bigevil
In summary, the conversation was about finding the y-coordinate centroid of a semicircular arc using polar coordinates. The first method involved considering a circle of radius a centered at the origin and using integrals to find the y-centroid. The second method used a circle with equation r=2a cos theta and a right angle triangle inside it. The person made a mistake in their final integral but eventually solved the problem. The y-coordinate centroid should be the same for both methods.
bigevil

## Homework Statement

Here, I have two ways of finding the y-coordinate centroid of a semicircular arc using polar coordinates.

First one is considering a circle of radius a, centred at the origin. What I have done is $$\int ds = \int_0^{\pi} a d\theta = \pi a$$ and then $$\int y ds = \int r^2 sin \theta d\theta = a^2 \int_0^{\pi} sin \theta d\theta = 2a^2$$, then y-centroid is 2a/π. I'm quite sure this is the right answer.

Second is a more roundabout method. It uses a circle with equation $$r=2a cos \theta$$. A right angle triangle fits inside the circle and the angle between the side passing through the origin and the x-axis is theta. But I do not get a similar answer. Here my method is to use $$\int y ds = \int r sin\theta \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta = \int_0^{\frac{\pi}{2}} 2a cos \theta (2a) d\theta = 4a^2$$, where the limits of the integral run from 0 to half pi due to how the angle is set up. (From my working, the messy square roots part becomes just 2a.) The y-centroid for the semicircle is 4a/π, which is wrong. Can someone advise me where I have gone wrong?

The y-coordinate centroid should be the same for both as the difference between the two figures is that one is shifted a-units along the x-axis.

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Sorry, I've solved this... careless... final integral should have been 2a cos theta sin theta.

## 1. What is the definition of centroid of a semicircular arc?

The centroid of a semicircular arc is the point at which the semicircle would balance if it were cut out of a sheet of uniform density material.

## 2. How do you calculate the centroid of a semicircular arc?

To calculate the centroid of a semicircular arc, use the formula (4r)/(3π) where r is the radius of the semicircle.

## 3. What is the significance of the centroid of a semicircular arc?

The centroid of a semicircular arc is important in structural engineering, as it is used to determine the stability and balance of structures with semicircular components.

## 4. Can the centroid of a semicircular arc be outside the arc itself?

Yes, it is possible for the centroid of a semicircular arc to be outside of the arc itself. This typically occurs when the semicircle is imbalanced or when the arc is not a perfect semicircle.

## 5. Are there any real-world applications of the centroid of a semicircular arc?

Yes, the centroid of a semicircular arc has many real-world applications, such as in the design of bridges, arches, and other curved structures. It is also used in the calculation of moments of inertia and center of mass for semicircular objects.

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