# COM of a half circle

1. Jun 3, 2016

### Physgeek64

1. The problem statement, all variables and given/known data

Find the centre of mass of a semi-circle

2. Relevant equations

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

3. 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 :)

2. Jun 3, 2016

### BvU

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.

3. Jun 3, 2016

### Physgeek64

Hi- I did draw it out, but unfortunately cannot scan it in. The best way I can describe what Im 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....

4. Jun 3, 2016

### BvU

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

But: at first you mentioned vertical rectangles.
And you had already defined $\theta$; why change it ?

Last edited: Jun 3, 2016
5. Jun 3, 2016

### Staff: Mentor

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

6. Jun 3, 2016

### Staff: Mentor

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.

7. Jun 5, 2016

### Physgeek64

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

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8. Jun 5, 2016

### Staff: Mentor

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

9. Jun 5, 2016

### Physgeek64

Oh okay so it should be $dy= Rcos \theta d \theta$ in place of $R d \theta$ ? :)

10. Jun 5, 2016

### Physgeek64

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

11. Jun 5, 2016

### Staff: Mentor

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

12. Jun 5, 2016

### Physgeek64

Yes- thank you