Center of mass of a sphere with cavity removed

In summary: Another way to look at it is to take the object X as having a mass ##M_X## and a centre of mass, ##(x,0,0)## then add the smaller sphere, which has a known mass ##m## and centre of mass ##(R/2, 0, 0)## and the result is the large sphere, with a known mass ##M## and centre of mass.This gives an equation in terms of positive masses. If you move the terms for the small sphere to the other side of the equation, this is a negative term. You could interpret this as adding a negative mass, but that sounds a bit dramatic to me.
  • #1
1v1Dota2RightMeow
76
7

Homework Statement


A solid sphere of density ##ρ## and radius ##R## is centered at the origin. It has a spherical cavity in it that is of radius ##R/4## and which is centered at ##(R/2, 0, 0)##, i.e. a small sphere of material has been removed from the large sphere. What is the the center of mass ##R_{cm} = (x_{cm}, y_{cm}, z_{cm})## of the large sphere, including the cavity?

Homework Equations


##R=1/M \int \rho r dV##, where ##dV=dxdydz## and ##\rho = dm/dV## and ##M=## total mass

The Attempt at a Solution


##R=(1/M) \int r dm = (1/M) \int \rho r dV##

##M##= total mass = ##\rho V = \rho(V_{total} - V_{cavity})##

##R=(1/(\rho(V_{total} - V_{cavity})) \int \rho r dV##

Here, I see that the ##\rho##'s cancel. But now I'm stuck wondering what ##r## is.
 
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  • #2
Hi DRM,

Like before, ##\vec r## is the position of ##dV##. The ##\rho## do not cancel; they aren't even the same thing: the first one is the ##\rho## of the material only. The second one (within the integral) is zero where the cavity is and equal to the other one where the material is.

However, the integral you are left with in this approach is cumbersome, to say the least. Personally I'm not in favour of using tricks, but for this exercise some lateral thinking might save you a lot of work. Do you know how to calculate the center of mass of two solid bodies with mass ##m_1## and ##m_2## centered at ##\vec r_1## and ##\vec r_2 \ \ ## :smile: ?
 
  • #3
BvU said:
Hi DRM,

Like before, ##\vec r## is the position of ##dV##. The ##\rho## do not cancel; they aren't even the same thing: the first one is the ##\rho## of the material only. The second one (within the integral) is zero where the cavity is and equal to the other one where the material is.

However, the integral you are left with in this approach is cumbersome, to say the least. Personally I'm not in favour of using tricks, but for this exercise some lateral thinking might save you a lot of work. Do you know how to calculate the center of mass of two solid bodies with mass ##m_1## and ##m_2## centered at ##\vec r_1## and ##\vec r_2 \ \ ## :smile: ?

I do, and I've seen this trick done elsewhere. But in regards to it - why is this allowed, mathematically? Why is this equivalent to finding the CM via the integrals?

For reference: http://physics.stackexchange.com/qu...-in-finding-the-center-of-mass-of-this-sphere
 
  • #4
Hedging your bets with the competition, eh :smile: ?

Nothing wrong with that; never mind, all for the good cause. So:

Anything unclear about the answer by Prasad ? All I can do is rephrase, unless you give us a clue what's the step you aren't comfortable with ...

If you have to integrate the value 0 over the small sphere you can first integrate ##\rho## (and that's the same ##\rho## as in the solid part of the big sphere) and then correct by integrating ##-\rho## over the small sphere (the cavity).
 
  • #5
1v1Dota2RightMeow said:
I do, and I've seen this trick done elsewhere. But in regards to it - why is this allowed, mathematically? Why is this equivalent to finding the CM via the integrals?

Another way to look at it is to take the object X as having a mass ##M_X## and a centre of mass, ##(x,0,0)## then add the smaller sphere, which has a known mass ##m## and centre of mass ##(R/2, 0, 0)## and the result is the large sphere, with a known mass ##M## and centre of mass.

This gives an equation in terms of positive masses. If you move the terms for the small sphere to the other side of the equation, this is a negative term. You could interpret this as adding a negative mass, but that sounds a bit dramatic to me.
 

Related to Center of mass of a sphere with cavity removed

1. What is the center of mass of a sphere with a cavity removed?

The center of mass of a sphere with a cavity removed is the point at which the mass of the object is evenly distributed in all directions. It is the average position of all the particles that make up the sphere.

2. How is the center of mass of a sphere with a cavity removed calculated?

The center of mass of a sphere with a cavity removed can be calculated by finding the average position of all the particles that make up the object. This can be done by dividing the object into smaller, simpler shapes, finding the center of mass of each shape, and then using the weighted average method to find the overall center of mass.

3. Does the size of the cavity affect the center of mass of the sphere?

Yes, the size of the cavity does affect the center of mass of the sphere. The larger the cavity, the more it will shift the center of mass towards its location. This is because the mass of the cavity is considered in the calculation of the center of mass.

4. Can the center of mass of a sphere with a cavity removed be outside of the object?

No, the center of mass of a sphere with a cavity removed will always be located within the object. This is because the center of mass is determined by the distribution of mass within the object, and there is no mass outside of the object to affect its position.

5. How does the shape of the cavity affect the center of mass of the sphere?

The shape of the cavity can affect the center of mass of the sphere. If the cavity is irregularly shaped, it may shift the center of mass towards its location. However, if the cavity is symmetrically placed within the sphere, it may not have a significant effect on the center of mass. In general, the more asymmetrical the cavity, the more it will affect the center of mass of the sphere.

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