How to find the x coordinates of the centre of mass?

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Homework Help Overview

The discussion revolves around finding the x-coordinate of the center of mass for a sphere with two smaller spherical cavities removed. The original sphere has mass M and radius R, while the cavities have a radius of R/4, positioned at R/4 and 3R/4 from the center of the original sphere.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the mass and volume of the original sphere and the removed cavities, questioning how to calculate the mass of the smaller spheres based on their dimensions.
  • Some participants express confusion regarding the application of the center of mass equation and the implications of the problem being framed in terms of spheres versus circles.
  • There is a discussion about the area and mass relationships, with attempts to clarify how the mass of the smaller cavities relates to the original sphere's mass.
  • Questions arise about the correct interpretation of the problem statement and whether a diagram affects the understanding of the three-dimensional nature of the objects involved.

Discussion Status

The discussion is ongoing, with participants actively questioning assumptions and clarifying concepts related to mass and volume. Some guidance has been offered regarding the relationship between the areas of the spheres and their respective masses, but no consensus has been reached on the correct approach to solving the problem.

Contextual Notes

There is uncertainty regarding the problem's framing, as participants note a potential typo in referring to spheres when a circle is depicted in the accompanying diagram. This has led to discussions about the implications of this discrepancy on the calculations and methods used.

Monkey D. Luffy
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Homework Statement


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.

Homework Equations


none, but the answer given is -1/62 x R

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]).
 
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Monkey D. Luffy said:

Homework Statement


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.

Homework Equations


none, but the answer given is -1/62 x R

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]).

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

Is it maybe M/4 (Pi) R^2? I just multiplied the mass by the area of the small cavity
 
Monkey D. Luffy said:
Is it maybe M/4 (Pi) R^2? I just multiplied the mass by the area of the small cavity
It is wrong. What is the area of the small circle of radius R/4?
 
ehild said:
It is wrong. What is the area of the small circle of radius R/4?
A=(Pi)r^2

So I guess it would be...

A= (Pi) (R^2)/16 ?
 
Monkey D. Luffy said:
A=(Pi)r^2

So I guess it would be...

A= (Pi) (R^2)/16 ?
Yes. The masses are related as the areas. So what is the mass of a small circle in terms of M?
 
ehild said:
Yes. The masses are related as the areas. So what is the mass of a small circle in terms of M?

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

But I am 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)
 
Monkey D. Luffy said:
I'm kind of confused as to what it would be in relation to one another...

But I am 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)[/QUOTE
Mass times area is not mass. The masses are related as the areas.
Monkey D. Luffy said:
Or since Radius is divided by 16 the mass can be divided by 16 as well...
The radius is divided by 4, not by 16, but the mass is divided by 16 as you wrote.
Monkey D. Luffy said:
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)
The ratio of the areas is 1:16, so the mass of the small circle is ?
 
  • #10

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
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. :frown:
 
  • #12
SteamKing said:
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. :frown:
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 let's 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
Monkey D. Luffy said:
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 let's 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...)
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
SteamKing said:
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.
Question 27
 

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  • #15
SteamKing said:
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.
I have added the diagram with the OP. Will i have to change my approach now?
 
  • #16
Monkey D. Luffy said:
I have added the diagram with the OP. Will i have to change my approach now?
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
SteamKing said:
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?

Well there is density (but we do not know it...) I don't think there are any others (i am solely going off of the formula sheet given)
 
  • #18
Monkey D. Luffy said:
Well there is density (but we do not know it...) I don't think there are any others (i am solely going off of the formula sheet given)
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
SteamKing said:
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.
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
Monkey D. Luffy said:
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?

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
SteamKing said:
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.
So set M/V = M'/V' and solve for M'...

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

That works?
 
  • #22
Monkey D. Luffy said:
So set M/V = M'/V' and solve for M'...

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

That works?
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
SteamKing said:
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.
Oh okay i think I understand now! thank you so much you guys really were a huge help
 

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