Centre of mass

1. Oct 3, 2011

coolingwater

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

Find the centre of mass of the 2-dimensional plate which occupies the region inside the circle x^2+y^2=2x, but outside x^2+y^2=1, and for which the density is proportional to its distance from the origin.

2. Relevant equations

Centre of mass for x coordinate: 1/m ∫∫ x ρ(x,y) DA

3. The attempt at a solution

I have already calculated the mass which turned out to be 8/3 by integrating and subtracting the two circle to obtain the area where they intersects. I am now trying to find the centre of mass, the y-coord is 0 as it is symmetrical about the x-axis.

To calculate the centre of mass i am using this domain in polar coord:

1 ≤ r ≤ 2 cos θ and π/6 ≤ θ ≤ 5π/6

With the domain above my equation is:

3/8 ∫∫ x(Kr) r dr dθ
= 3/8 ∫∫ (r cos θ)(Kr) r dr dθ

Is this correct? I was doing the working, i couldn't get any answer because it didn't make much sense to me, and i was sure i am doing something wrong somewhere. Do you guys mind advising me on where i have gone wrong? Thank you very much!

Last edited: Oct 3, 2011
2. Oct 3, 2011

vela

Staff Emeritus
I don't get this for the mass. For one thing, it has to have the constant of proportionality K in it since the mass depends on the density.
The limits for θ are wrong, but your set-up is otherwise fine.

3. Oct 3, 2011

coolingwater

Oh let me clarify for the mass, I used polar coordinates: ∫∫ (Kr) r dr dθ and the domains are, for circle 1: 0 ≤ r ≤ 2 cos θ and -π/2 ≤ θ ≤ π/2 and for circle 2: 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π, I integrated them and subtracted them. I'm kinda lost as to which θ I should to find the centre of mass, any tips for thos restless soul? Thank you very much!

4. Oct 3, 2011

vela

Staff Emeritus
That won't work. You should only subtract out the part where the two circles overlap from the total mass of circle 1. Also because the mass has to depend on K, if K doesn't appear in your answer, it can't possibly be correct.

It would be the most straightforward to calculate the mass using
$$m = \iint \rho\, dA = \iint \rho(r,\theta) r\,dr\,d\theta$$with the same limits you use to calculate the first moment. (The only difference when you calculate the first moment is you throw in a factor of x or y into the integrand.)
The equation for circle 2 is r=1, so the two circles intersect when r=1 and r=2 cos θ are satisfied simultaneously.

5. Oct 3, 2011

coolingwater

Thank you very much for your help vela, much appreciated. So if i got this correctly, my mass should be:

m =∬ρ(r,θ)rdrdθ, where the domain is 1 ≤ r ≤ 2 cos θ and π/6 ≤ θ ≤ 5π/6 and p(r,θ) = Kr?

And when i get the mass, i follow the steps below to get the centre of mass:

1/m ∫∫ x(Kr) r dr dθ
= 1/m ∫∫ (r cos θ)(Kr) r dr dθ

the domain for r is similar to the one above, but the θ is different....right? This is severely confusing, sorry for troubling you with my questions..

6. Oct 3, 2011

vela

Staff Emeritus
The integrals are exactly the same except for the factor of x in the second one. In both cases, you want to integrate over the entire shape, so you want to use the same limits.

7. Oct 3, 2011

coolingwater

Ahha! Its much clearer to me now, thank you very much vela. However I'm still wondering which I couldn't integrate and subtract the two circles to get the mass...

8. Oct 4, 2011

coolingwater

So i finally calculated the values and the x-coord is 15/2π + 15/4

is this correct?

9. Oct 4, 2011

vela

Staff Emeritus
Hmm, I got a different answer. What limits did you use for θ?

10. Oct 4, 2011

coolingwater

When calculating the moment i used -π/2 ≤ θ ≤ π/2

Gah, just when i thought i had it all figures out...bummer

11. Oct 4, 2011

coolingwater

for the mass i i took the total mass of circle 1 (ie: 0 ≤ r ≤ 2 cos θ, -π/2 ≤ θ ≤ π/2) minus the part where the two circles overlap (ie: 1 ≤ r ≤ 2 cos θ, -π/2 ≤ θ ≤ π/2) and i got

mass= Kπ/3

than i used these limits 1 ≤ r ≤ 2 cos θ, -π/2 ≤ θ ≤ π/2 to calculate the moment and i got

moment= 5K/2 + 5Kπ/4

12. Oct 4, 2011

vela

Staff Emeritus
You're not getting the limits for r correct for the overlap. Try sketching the overlap area and drawing lines for fixed values of θ.

13. Oct 4, 2011

coolingwater

I just sketched the overlapping area, and is the values of θ between π/6 ≤ θ ≤ 5π/6?

Last edited: Oct 4, 2011
14. Oct 4, 2011

vela

Staff Emeritus
No. I've attached a plot of r=1 and r=2 cos θ for 0 ≤ θ ≤ pi/2.

Note that when θ is less than a certain value, r is bounded by 0 and circle 2 in the overlap region. When θ is greater than that value, r is bounded by 0 and circle 1.

Attached Files:

• plot.png
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15. Oct 4, 2011

coolingwater

Thank you very much vela, okay so i used the coordinates at the intersection points to try and get the angle, and based on what you said above when the angle is between 0 and 60 degrees, r is bounded by 0 and circle 2.

So does that mean if the angle is more than 60 degree, r will be bounded by 0 and circle 1?

16. Oct 4, 2011

vela

Staff Emeritus
Reread what I said in my previous post.

17. Oct 5, 2011

coolingwater

Haha, i will retry it again and get back to you! Thank you very much!

18. Oct 9, 2011

coolingwater

Okay so i am confused as to which limits of θ i should be using, it will either be π/3 or π/2, but based on the diagram above i should be using -π/3 and π/3 because that angle is bounded where r=0 and r=1.

After subtracting the masses the final mass is 4Kπ/9 + 8K/3

Is this correct? Because that's the only way i can see it working out...

Alternatively is used -π/2 and π/2, i will be getting Kπ/3 + 8K/3

Im also wondering if f my final mass should have the term 4Kπ/9, it feels like my answer is wrong =x

Last edited: Oct 9, 2011
19. Oct 9, 2011

vela

Staff Emeritus
I find the mass to be
$$M = K\left(2\sqrt{3}-\frac{2\pi}{9}\right)$$

20. Oct 9, 2011

coolingwater

Okay, my answer is wayy different than yours, i can see a square root of 3 and from that i know i must should be using r = 2 cos θ and θ = π/3 for my calculation