Find Centre of Mass of Semi-Circle: Solving Integrals

[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 :)

 

[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.

 

[Physgeek64]

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...

 

[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 ?

 

[NascentOxygen]

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

 

[Chestermiller]

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.

 

[Physgeek64]

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

View attachment 101612

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

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

 

[NascentOxygen]

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

 

[Physgeek64]

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 $R d \theta$ ? :)

 

[Physgeek64]

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 :)

 

[NascentOxygen]

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

 

[Physgeek64]

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

Yes- thank you