Finding Mass and Center of Mass in a Solid Hemisphere

In summary, the problem involves finding the mass and center of mass of a solid hemisphere with radius a, where the density is proportional to the distance from the center of the base. Using spherical coordinates and the appropriate equations, the mass is found to be proportional to a^4 and the center of mass is located at (0,0,2a/5).
  • #1
squeeky
9
0

Homework Statement


Use Spherical Coordinates.
Let H be a solid hemisphere of radius a whose density at any point is proportional to its distance from the center of the base.
a) Find the mass of H.
b) Find the center of mass of H.


Homework Equations


[tex]M=\int\int_D\int\delta dV[/tex]
[tex]M_{yz}\int\int_D\int x \delta dV;M_{xz}\int\int_D\int y \delta dV;M_{xy}\int\int_D\int z \delta dV[/tex]
[tex]C.O.M.=(\bar{x},\bar{y},\bar{z})[/tex]
[tex]\bar{x}=\frac{M_{yz}}{M};\bar{y}\frac{M_{xz}}{M};\bar{z}\frac{M_{xy}}{M}[/tex]


The Attempt at a Solution


I think that if we place the hemisphere's center at (0,0,0), then the limit of theta is from 0 to 2pi, phi is from 0 to pi/2, and rho is from 0 to a, while the density is equal to rho. This gives me the equation:
[tex]M=\int^{2\pi}_0\int^{\pi/2}_0\int^a_0 \delta \rho^2 sin \phi d \rho d \phi d \theta[/tex]
Solving this, I get a mass of [tex]\frac{a^4 \pi}{2}[/tex], [tex]M_{xy}=\frac{a^5\pi}{5}[/tex], Mxz=Myz=0. Then the center of mass is (0,0,2a/5).
Is this right?
 
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  • #2
You are told only that the density is proportional to the distance from the center. How can you possibly get a specific number as the mass? What happened to the "proportionality"? Other than that, I think you are correct.
 

What are spherical coordinates and how are they used in mass calculations?

Spherical coordinates are a system of representing points in three-dimensional space using three coordinates: radius (r), inclination (θ), and azimuth (φ). They are commonly used in mass calculations because they allow for the efficient calculation of mass for spherical objects, such as planets or stars.

How do you convert between cartesian and spherical coordinates for mass calculations?

To convert between cartesian and spherical coordinates, you can use the following equations:

x = r * sin(θ) * cos(φ)

y = r * sin(θ) * sin(φ)

z = r * cos(θ)

where r is the distance from the origin, θ is the inclination angle, and φ is the azimuth angle. These equations can be used to convert between the two coordinate systems for mass calculations.

How do you calculate the mass of a spherical object using spherical coordinates?

The formula for calculating the mass of a spherical object using spherical coordinates is:

M = ρ * ∫∫∫ r² * sin(θ) dr dθ dφ

where ρ is the density of the object and the integral is taken over the volume of the object. This formula takes into account the variation of density with distance from the center of the object.

What are the advantages of using spherical coordinates for mass calculations?

One advantage of using spherical coordinates for mass calculations is that they are well-suited for spherical objects, which are common in many scientific fields. They also simplify the calculation process by reducing the number of variables needed. Additionally, spherical coordinates can provide a more intuitive understanding of the distribution of mass within a spherical object.

Are there any limitations or drawbacks to using spherical coordinates for mass calculations?

One limitation of using spherical coordinates for mass calculations is that they are not well-suited for objects with irregular shapes. In these cases, other coordinate systems, such as cylindrical or cartesian coordinates, may be more appropriate. Additionally, spherical coordinates can become more complex when dealing with non-uniform or non-spherical distributions of mass.

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