# Locating center of mass for hollowed out sphere

• astenroo
In summary: just move the fulcrum until it is past one of the masses. as you do so the distance associated with that mass will become smaller and smaller till it becomes negative. the other distance is still positiveand of course one of the masses is negative so the equation still works just fine.
astenroo

## Homework Statement

A uniform sphere of radius R has a spherical hole of radius r removed. The center of the hole is at a distance d from the center of the original sphere. Locate the CM relative to the center of the sphere.

## The Attempt at a Solution

Ok, so, the problem should be considered as two spheres, one with a negative mass (the hole). Due to symmetry the CM should be moved from the center of the original (x,y,z:0,0,0) sphere along one of the axes "backwards" (if hole is on positive x-axis, then CM moves from origin along the x-axis towards negative). My problem is, I'm not sure how to express r in terms of R (d is an irritant here ;)), and where to begin.

think in terms of 2 masses

what are the masses?

granpa said:
think in terms of 2 masses

what are the masses?

The masses are $$\rho$$V,
so for small sphere m=$$\rho$$*4/3$$\pi$$*r³
Mass for hollowed out sphere is $$\rho$$*4/3$$\pi$$*R³ - $$\rho$$*4/3$$\pi$$*r³

then from the equation d$$_2$$=M$$_{s}$$d$$_1$$ / M$$_{h}$$ where s=small and h=hollowed. But here I still have to contend with d1...

granpa said:
it looks to me like your distances will be -x and x+d

But I'm not sure

uhm...now I'm lost...

you have d1/d2=M2/M1

x = position of new center of mass

d1= x minus position of M1
d2= position of M2 minus x
d = position of M2 minus position of M1
d = position of M2

position of M1 = 0
position of M2 > 0
x < 0

it looks to me like your distances will be x and d-x (where x is negative)

But I'm not sure

astenroo said:
uhm...now I'm lost...

yeah. I got ahead of myself and got confused.

granpa said:
yeah. I got ahead of myself and got confused.

Heh, ok. So, your M1 would be the larger sphere and your M2 would be the "hole"? Just to make sure I'm following :)

yeah

I didnt know the formula but i took what you had and did i simple
thought experiment with a rectangle instead of a shere and created a hole
that eliminated one whole half of the rectangle.

astenroo said:

## The Attempt at a Solution

Ok, so, the problem should be considered as two spheres, one with a negative mass (the hole).
So why haven't you done this?

Backing up a bit, you did not cite the relevant equations. So what is the equation for the center of mass of two objects, each of which has some known mass and some known position? There is nothing per se in this particular equation that restricts objects to have positive mass. (Well almost nothing. The equation results in a division by zero if the total mass is zero.)

My problem is, I'm not sure how to express r in terms of R (d is an irritant here ;)), and where to begin.
Why do you feel you need to do so? (Hint: You don't.)

You are making this problem much harder than it is.

D H said:
So why haven't you done this?

Backing up a bit, you did not cite the relevant equations. So what is the equation for the center of mass of two objects, each of which has some known mass and some known position? There is nothing per se in this particular equation that restricts objects to have positive mass. (Well almost nothing. The equation results in a division by zero if the total mass is zero.)

Why do you feel you need to do so? (Hint: You don't.)

You are making this problem much harder than it is.

hi! Sorry for the late reply here.

Ok so,

equation for solving centre of mass is:

x(cm) = (m1x1 + m2x2)/(m1+m2). From earlier m2 is considered as negative mass, and the expressions of the masses have been posted earlier.

m1=rho*4/3*pi*R³

m2=rho*4/3pi*r³

x1=0, and x2=d. x(cm) then becomes: x(cm) = (-r^3*d)/(R^3-r^3). This expression could perhaps be further simplified, but I am ashamed to admit, that I do not know how...

looks like i made it more complicated than it really is. I do that sometimes.
Sometimes to solve a problem you just have to sleep on it and get a fresh perspective in the morning.

d1M1=d2M2

works fine. its just that one of the distances is negative

think of a fulcrum under the center of mass and then move the fulcrum until it is past one of the masses. as you do so the distance associated with that mass will become smaller and smaller till it becomes negative. the other distance is still positive

and of course one of the masses is negative so the equation still works just fine.

A quick thought experiment with rectangles shows that it works fine.

Last edited:
the first mass is the mass of the original sphere and the second mass is the negative mass of the hole.

the mass of the sphere minus the hole is irrelevant as far as I can see.

granpa said:
the first mass is the mass of the original sphere and the second mass is the negative mass of the hole.

the mass of the sphere minus the hole is irrelevant as far as I can see.

Ok, but I think the solution I posted recently is accurate. Now I just need to simplify the expression (-r^3*d)/(R^3 - r^3). It can be done, but I do not know how.

thank you for the help BTW :)

the equation you gave doesn't solve for the distance. it solves for the coordinate of the center of mass given the coordinates of the 2 masses

which is the coordinate for the center of mass

substituting
d1 = x0 - x1
and
d2 = x1 - x0
into
d1M1=d2M2

so the best way to reduce your equation is not to use it in the first place.

Last edited:
the whole question of how to reduce an equation with a sum in the denominator is a fascinating one. I would be very curious to know of any general equation for it.

astenroo said:
Ok, but I think the solution I posted recently is accurate. Now I just need to simplify the expression (-r^3*d)/(R^3 - r^3).

Your solution is exactly right, and it really isn't overly complex. You could multiply it by 1, with 1 written in the form

$$1=\frac{1/r^3}{1/r^3}$$

## 1. How do you locate the center of mass for a hollowed out sphere?

The center of mass for a hollowed out sphere can be found by using the formula: xcm = (x1m1 + x2m2 + ... + xnmn) / (m1 + m2 + ... + mn), where xcm is the center of mass, x1, x2, ... , xn are the positions of individual masses, and m1, m2, ... , mn are the corresponding masses.

## 2. What is the difference between center of mass and center of gravity?

The center of mass is the point at which the entire mass of an object is considered to be concentrated. It takes into account the mass and position of each individual component of the object. On the other hand, the center of gravity is the point at which the entire weight of an object is considered to be concentrated. It takes into account the weight and position of each individual component of the object, as well as the gravitational force acting on the object. In most cases, the two are located at the same point, but in some cases, such as when dealing with objects near a celestial body, they may differ slightly.

## 3. Can the center of mass be located outside of the object?

Yes, the center of mass can be located outside of the object. This typically occurs when the object has an irregular shape or when there are varying densities throughout the object. In these cases, the center of mass may be located at a point where there is no actual mass present.

## 4. How does the shape of a hollowed out sphere affect the location of its center of mass?

The shape of a hollowed out sphere can affect the location of its center of mass. For a uniform sphere, the center of mass will be located at its geometric center. However, for a hollowed out sphere with varying thicknesses or densities, the center of mass will likely be shifted from the geometric center towards the denser or thicker regions of the sphere.

## 5. Why is it important to locate the center of mass for a hollowed out sphere?

Locating the center of mass for a hollowed out sphere is important because it allows us to understand the behavior of the object when it is subjected to external forces. It also helps in determining the stability and balance of the object, and in predicting its motion and rotation.

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