# Homework Help: How to find the x coordinates of the centre of mass?

1. Dec 9, 2015

### Monkey D. Luffy

1. The problem statement, all variables and given/known data
A sphere of mass M and Radius R had two spheres of R/4 removed. the centres of cavities are R/4 and 3R/4 from the centre of the original sphere (at x=0). what is the x coordinate of the centre of mass of this object?
there is a drawing next to the question literally showing the cavities on the right side of the circle nothing else so i did not include it with the question.

2. Relevant equations
none, but the answer given is -1/62 x R

3. The attempt at a solution
I have absolutely no idea where to start with this problem. I used the Xcm equation but it didnt work and I have done a similar problem but that one had mass and radius given so now this confuses me. any help or procedure would be helpful (this is a practice final [final is in 2 days]).

2. Dec 9, 2015

### ehild

You are given the mass of the original sphere: it is M. The radii of the spheres removed are R/4. What is the mass of a sphere of radius R/4?

3. Dec 9, 2015

### Monkey D. Luffy

Is it maybe M/4 (Pi) R^2? I just multiplied the mass by the area of the small cavity

4. Dec 9, 2015

### Monkey D. Luffy

Is it maybe M/4 (Pi) R^2? I just multiplied the mass by the area of the small cavity

5. Dec 9, 2015

### ehild

It is wrong. What is the area of the small circle of radius R/4?

6. Dec 9, 2015

### Monkey D. Luffy

A=(Pi)r^2

So I guess it would be...

A= (Pi) (R^2)/16 ?

7. Dec 9, 2015

### ehild

Yes. The masses are related as the areas. So what is the mass of a small circle in terms of M?

8. Dec 9, 2015

### Monkey D. Luffy

I'm kind of confused as to what it would be in relation to one another....

But im thinking since for the original circle it was mass M and radius R...

It can be either the mass times the area (so M(Pi)(R^2)/16)

Or since Radius is divided by 16 the mass can be divided by 16 as well...

I'm not really sure if there's an equation to relate them but these are the 2 possibilities I see?

Unless if since Radius went from R to R^2/16
Mass went from M to M^2/16 as well (I don't think this one is right though)

9. Dec 9, 2015

### ehild

10. Dec 9, 2015

### Monkey D. Luffy

Well then the mass should be M/16 if they are related to the area...

How did you deduce that they are related though? Kind of lost there because I see how the radius is divided by 4 ( half way through the circle twice ) and then when squared is /16 but not how the mass is divided by 16 out of that... Sorry this unit is really hard for me :/ I also do not get where to go from there... I thought the mass times the area would give the mass over that area and that was my original plan

11. Dec 9, 2015

### SteamKing

Staff Emeritus
I'm confused. The original problem talked about a sphere of mass M and radius R with two smaller spherical cavities removed, each with radius R/4.

Now, we're talking about a circle of radius R with two smaller circles removed. Which is the correct problem?

I suspect that you can't use a circle in place of a sphere and get the answer for x-bar as disclosed in the OP.

12. Dec 9, 2015

### Monkey D. Luffy

Oh i see what you mean... the diagram next to the problem shows a 2-D Circle not a sphere yet the problem lists sphere...

In that case i guess lets assume it was a typo and the original was a circle. If the answer is incorrect i should be able to follow the same format for the solution albeit with different formulas for area (which would change M as well i guess...)

My prof has a habit of messing up his problem questions haha

13. Dec 9, 2015

### SteamKing

Staff Emeritus
I've never see this diagram. It's not attached to any of your posts.

The answer in the OP for the location of x-bar can be shown for spherical cavities within a larger sphere. I suspect the same answer cannot be shown when using circles in place of spheres.

14. Dec 9, 2015

### Monkey D. Luffy

Question 27

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15. Dec 9, 2015

### Monkey D. Luffy

I have added the diagram with the OP. Will i have to change my approach now?

16. Dec 9, 2015

### SteamKing

Staff Emeritus
Yes. The problem statement clearly says 'sphere' not 'circle'.

You're taking the diagram much too literally. If you want to show a sphere in a two-dimensional sketch, most people understand that a circle can be used in place of, IDK, a hologram.

Now to get back on track with solving this problem, in Post #2, ehild suggested how you might determine the amount of mass removed from each of the small cavities in the large sphere. You are given the radius of the large sphere as R and its mass before the cavities removed as M. Is there some physical property of 3-dimensional bodies which involves mass and volume that could be used to calculate the amount of mass removed from the cavities?

17. Dec 9, 2015

### Monkey D. Luffy

Well there is density (but we do not know it...) I dont think there are any others (i am solely going off of the formula sheet given)

18. Dec 9, 2015

### SteamKing

Staff Emeritus
Well, what do you need to know to calculate the density of an object? The density of an object does not always have to be furnished.

19. Dec 9, 2015

### Monkey D. Luffy

You need Mass and Volume

So use mass and volume for the original and sub those into a density formula... subtract the one for the small circles?

20. Dec 9, 2015

### SteamKing

Staff Emeritus
Well assuming the original sphere has a homogeneous density based on its mass and volume, you should be able to calculate the amount of mass removed from each cavity, using that density and the volume of the cavity. Mind you, this is mostly an exercise in algebra, since you know no numerical values for R or M.

21. Dec 9, 2015

### Monkey D. Luffy

So set M/V = M'/V' and solve for M'...

Take 0- Xcm of the two circles / M-M'

That works?

22. Dec 9, 2015

### SteamKing

Staff Emeritus
It's more like ρ = M / V = density of the sphere before the cavities are removed.

Then, after calculating V' = volume of each cavity, M' = ρ ⋅ V' = mass removed from each cavity.

Once you know M' for each cavity, you are given the centers of each, so you can calculate the location of Xcm knowing the net mass of the sphere with the two cavities removed and by calculating the moment of the net mass of the sphere about the y-axis.

23. Dec 9, 2015

### Monkey D. Luffy

Oh okay i think I understand now! thank you so much you guys really were a huge help