# Find Centre of Mass of Semi-Circle: Solving Integrals

• Physgeek64
In summary, the conversation discusses the process of finding the center of mass of a semi-circle using the equation ##y_{cm}=\frac{1}{M} \int y dm ##. There are multiple methods proposed, including changing coordinates to polar coordinates and using rectangular strips. Ultimately, the correct method involves using rectangular strips with a length of 2‧R‧cosθ and a width of Rcosθdθ, which results in the correct answer of ##\frac {4}{3 \pi}##.
Physgeek64

## Homework Statement

Find the centre of mass of a semi-circle

## Homework Equations

##y_{cm}=\frac{1}{M} \int y dm ##

## The Attempt at a Solution

So ## y= R cos \theta ## where theta is measured from the vertical, and the base of the semi-circle is along the horizontal

Now apparently from here you can change coordinated to polar coordinates and replace ##dm## with ## \rho r dr d \theta## to obtain the correct answer of ##\frac {4}{3 \pi}## But I'm confused as to why you can't also replace ##dm## with ##2 \rho R sin \theta R d \theta## to split to 'dm' segment into rows perpendicular to the horizontal base?

Many thanks :)

With ##R\cos\theta## the height of such a slice, what would you have for the width ? I would guess ##\ R\cos\theta\ d\theta\ ## And the center of mass of such a slice would be at ##{R\over 2} \cos\theta##. It might work. Make a drawing to check what you are doing.

BvU said:
With ##R\cos\theta## the height of such a slice, what would you have for the width ? I would guess ##\ R\cos\theta\ d\theta\ ## And the center of mass of such a slice would be at ##{R\over 2} \cos\theta##. It might work. Make a drawing to check what you are doing.

Hi- I did draw it out, but unfortunately cannot scan it in. The best way I can describe what I am doing is by saying that I'm calculating it in the same way I would the moment of inertia by dividing the semicircle into rectangles of length ## 2 R sin \theta## (or ## sin \theta## depending on how you define your angles) and width ##Rd \theta## It does not appear to give the correct answer...

Actuallly it does give the correct answer. Could you show at least your calculation ?

But: at first you mentioned vertical rectangles.

Last edited:
R‧dΘ is the arc length, so it's a sloping distance perpendicular to neither axis.

I think it would be better to set this up in cartesian coordinates, and then to convert to polar coordinates in doing the trig substitution as part of the integration. If y is the distance above the base, the area of the differential "rectangle" between y and y + dy is ##2\sqrt{R^2-(R-y)^2}dy##. The moment of the area about the base is ##2y\sqrt{R^2-(R-y)^2}dy##. This can be integrated from y = 0 to y = R to get the total moment of the semi-circle.

BvU said:
Actuallly it does give the correct answer. Could you show at least your calculation ?

View attachment 101612

But: at first you mentioned vertical rectangles.

Hi- sorry if I was not being very clear. This is what I had meant :) but I can't see my mistake

#### Attachments

• image.jpeg
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The rectangular strip has its long side of length 2‧R‧cosθ
but its dimension perpendicular to this is ...

NascentOxygen said:
The rectangular strip has its long side of length 2‧R‧cosθ
but its dimension perpendicular to this is ...
Oh okay so it should be ##dy= Rcos \theta d \theta ## in place of ##R d \theta## ? :)

NascentOxygen said:
The rectangular strip has its long side of length 2‧R‧cosθ
but its dimension perpendicular to this is ...

And it comes out with that correction! Thank you so much :)

So you get the answer you quote in post #1 ?

NascentOxygen said:
So you get the answer you quote in post #1 ?

Yes- thank you

## What is the definition of the centre of mass of a semi-circle?

The centre of mass of a semi-circle is the point where the entire mass of the semi-circle can be considered to be concentrated. It is the average position of all the individual points within the semi-circle, taking into account their mass and distance from a reference point.

## How do you determine the centre of mass of a semi-circle?

To determine the centre of mass of a semi-circle, you need to solve an integral using the formula for calculating the centre of mass of a continuous distribution of mass. This involves integrating the distance of each infinitesimal mass element from the reference point, divided by the total mass of the semi-circle.

## What is the difference between the centre of mass and the centroid of a semi-circle?

The centre of mass and the centroid are two different terms used to describe the same point in a semi-circle. The centre of mass refers to the average position of all the individual masses within the semi-circle, while the centroid refers to the geometric centre or the point of balance of the semi-circle.

## Why is it important to find the centre of mass of a semi-circle?

The centre of mass of a semi-circle is important because it is the point at which the semi-circle will be in perfect balance when subjected to external forces. This is useful in understanding the stability and motion of objects with a semi-circular shape, such as wheels or curved beams.

## What are some real-world applications of finding the centre of mass of a semi-circle?

The centre of mass of a semi-circle has many real-world applications, such as in the design of vehicles with curved components, calculating the stability of objects with a semi-circular base, and determining the trajectory of projectiles with a semi-circular path. It is also used in fields such as architecture, engineering, and physics.

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