# Homework Help: 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

21. Oct 9, 2011

### vela

Staff Emeritus
I have to admit, I'm somewhat at a loss to figure out where you're running into difficulty. You seem to have everything to set up the integral(s) and the crank it out.

There's really no need to use subtraction, but it works out either way. Can you show the details of you calculations so I can see what you're doing?

22. Oct 9, 2011

### coolingwater

I ran into problems trying to find the mass, i was under the impression that the only way to get the mass was to subtract the values, but i didn't know what limits to use haha, and thank you very much vela! I now too realize there is no use for subtraction.

I'll just write out a long equation for the mass in the overlapping circle:

m = ∫∫ (Kr) r dr dθ where the limits are 0 ≤ r ≤ 2 cos θ, -π/3 ≤ θ ≤ π/3
= K/3 ∫∫ r3 dr dθ
= K/3 ∫ (8 cos3 θ) dθ
= 16K/3 ∫ (cos3 θ) dθ for 0 ≤ θ ≤ π/3
= 16K/3 ∫(cos2 θ)(cos θ) dθ
= 16k/6 ∫ (cos θ + cos2 3θ) dθ
= 16K/6 ∫ (cos θ + 1/2 + 1/2 cos 6θ) dθ
= 16K/6 (sin θ + 1/2θ + 1/12 sin 6θ) for 0 ≤ θ ≤ π/3
= 16K/6 (√3/2 + π/6)
= 4√3K/3 + 4Kπ/9

Yeap, i realize im doing something wrong somewhere, but that's roughly what i have at the moment, not quite complete yet.

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

### vela

Staff Emeritus
Oh, I see what you did. Either you calculate the entire mass of circle 2 and then subtract out the overlap, or you calculate the mass of just the part of circle 2 that doesn't overlap circle 1. You're kind of doing a mixture of the two, which is why things aren't working out.

To get the total mass of circle 2, you need to integrate letting the angle range from -pi/2 to pi/2 and the radius range from 0 to 2 cos θ. Those ranges cover the entire circle. Then you calculate the overlap as we discussed before, and subtract that result from the mass of the entire circle.

If you want to find the mass directly, you want the part of circle 2 that doesn't overlap circle 1. In this case, the angle will run from -pi/3 to pi/3, and r has to go from circle 1 to circle 2. Do you see what you change in your integral to achieve this?

24. Oct 9, 2011

### coolingwater

OHHHHH! Hot damn, no wonder the answer looks similar like yours but just doesn't quite add up lol, didn't realize i was doing a mixture of it. Now i see what i am doing wrong, been scratching my head day and night haha.

Thank you very much for your assistance vela! I can't thank you enough, very much appreciated!