# Finding center of mass of solid

• Draconifors
All the ##m(z)## formula does is find the area of the disc in terms of ##z##.In summary, the problem involves finding the center of mass of a solid B, which occupies the region of space above z=0 and between two spheres. The density of B is equal to the distance from its base, which is z=0, and the mass of the solid is given. The center of mass in x and y is at coordinate 0 due to the symmetry of the domain, so only the z-value needs to be calculated. This can be done by finding the mass of thin circular discs at various z-values and using the formula for center of mass.
Draconifors

## Homework Statement

A solid B occupies the region of space above ##z=0## and between the spheres ##x^2 + y^2 + z^2 = 16## and ##x^2+y^2+(z-1^2) = 1##. The density of B is equal to the distance from its base, which is ##z = 0##. The mass of the solid B is ##\frac{188\pi}{3}##. Find the coordinates of the center of mass.

## Homework Equations

Center of mass in z = ##\frac{1}{m} \iiint \limits_E z *\rho(x,y,z) dV##

## The Attempt at a Solution

The center of mass in x and y is at coordinate 0 because of the symmetry of the domain, so there's only the z-value to calculate.

So my integral is ## \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{2\cos\phi}^{4} \rho \cos\phi * \rho \cos\phi * \rho^2\sin\phi d\rho d\phi d\theta##, then divided by the mass, but the answer sheet (which is a student's graded paper) has ##\rho\cos\phi * \rho^2\sin\phi## as the term to integrate, and therein lies my question: why are they only integrating z once? Don't we have z (the density) * z (the term already present in the center of mass equation) as the term to integrate?

Last edited:
Draconifors said:

## Homework Statement

A solid B occupies the region of space above ##z=0## and between the spheres ##x^2 + y^2 + z^2 = 16## and ##x^2+y^2+(z-1^2) = 1##. The density of B is equal to the distance from its base, which is ##z = 0##. The mass of the solid B is ##\frac{188\pi}{3}##. Find the coordinates of the center of mass.

## Homework Equations

Center of mass in z = ##\frac{1}{m} \iiint \limits_E z *\ rho(x,y,z) dV##

## The Attempt at a Solution

The center of mass in x and y is at coordinate 0 because of the symmetry of the domain, so there's only the z-value to calculate.

So my integral is ## \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{2\cos\phi}^{4} \rho \cos\phi * \rho \cos\phi * \rho^2\sin\phi d\rho d\phi d\theta##, then divided by the mass, but the answer sheet (which is a student's graded paper) has ##\rho\cos\phi * \rho^2\sin\phi## as the term to integrate, and therein lies my question: why are they only integrating z once? Don't we have z (the density) * z (the term already present in the center of mass equation) as the term to integrate?

You are doing it the hard way, given the symmetry you already noticed. Between ##z## and ##z + \Delta z## you have a thin circular disk centered at ##(0,0,z)## whose mass you can get because you can figure out the radius of the disc. (There will be two separate formulas for the radius, depending on whether the disc extends out to the lower surface or the upper surface). If ##m(z) \, dz## is the mass of the thin circular disc of thickness ##dz## at ##z##, the CM is just ##\frac{1}{M} \int_0^4 z\, m(z) \, dz,## where ##M## is the mass of the whole solid.

## 1. What is the center of mass of a solid?

The center of mass of a solid is the point at which the mass of the object is evenly distributed in all directions. It is the average location of the mass of the object.

## 2. Why is it important to find the center of mass of a solid?

Finding the center of mass of a solid is important because it helps determine the stability and balance of the object. It also allows for accurate predictions of the object's motion and behavior.

## 3. How do you find the center of mass of a solid?

The center of mass of a solid can be found by dividing the object into smaller, simpler shapes and using the formula for center of mass for each shape. The center of mass of the entire object is then calculated by taking the weighted average of the individual center of mass values.

## 4. Can the center of mass of a solid be located outside of the object?

Yes, the center of mass of a solid can be located outside of the object if the object has an irregular shape or if there are holes or voids present in the object.

## 5. How does the distribution of mass affect the location of the center of mass?

The distribution of mass greatly affects the location of the center of mass. If the mass is evenly distributed, the center of mass will be located in the geometric center of the object. However, if there is more mass on one side of the object, the center of mass will be shifted towards that side.

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