The actual span of the base of the dome is 143 feet.
1) Use cylindrical coordinates to write the surface of the dome as a function of the
distance from the center of the base; that is find z = f (r) .
2) Use your function to find the height of the dome; that is what is f (0) ?
3) Find the volume of the inside of the dome.
4) Assume the density is one, that is
P(r,"theta",z)=1 :
a. Find the moment around the x-y plane.
b. Use 3) and a) above to find the center of mass for the dome.
5) The density of the sides decreases as the dome gets higher. Assume that the
density is P(r,"theta",z)=k∗(f(0)−z) where k is a constant.
a. Find the mass of the dome.
b. Find the moment around the x-y plane.
c. Find the center of mass of the dome.
The word dome suggests that it is hollow. However, question 4b suggests using the previously calculated volume to calculate centre of mass would imply that the "dome" refers to a solid, or rather a hemisphere. The remaining calculations will be based on a solid hemisphere.
Actual "span" = 143 ft.
Radius, R = 143/2=71.5 ft.
A general point on the surface of the dome in cylindrical coordinates would be P(r, θ, z), where z is the vertical axis, and the r-θ plane corresponds to the x-y plane.
1.
z = f(r) = √(R²-r²)
2. height of dome, H
H = f(0) = √(R²-0²)
=R
3. Volume
The volume can be found by integrating thin slices over the height of the dome. In order to do so, we must express the radius of the dome as a function of z, i.e.
from z=f(r)=√(R²-r²), we obtain
r=f(z)=√(R²-z²)
(Side note: the inverse of f(r) equals f(z) by the fact that the first quadrant of the dome is symmetrical with respect to the line y=x)
Volume
=∫πr²dz
=∫π(R²-z²)dz
=2πR³/3 [z from 0 to R]
4. Uniform density, ρ(r,θ,z)=1
a. Moment
If we subdivide the dome/hemisphere into thin horizontal slices, the moment about the x-y plane would be the mass of each slice multiplied by the distance from the plane.
Moment
=∫ ρzdV
=∫ ρz*πr²dz
=ρ ∫ z*π(R²-z²)dz
=ρπR⁴/4
4b. Centre of mass, ρ=1
Centre of mass, z̄
=moment/volume
=3R/8
5. &rho(r,θ,z)=k(R-z)
5a. mass
Mass
=∫ ρdV
=∫ k(R-z)*πr²dz
=5πkR⁴/12
5b. Moment
The moment can again be calculated by multiplying the elemental disk by the distance from the x-y plane:
moment = ∫ ρz dV
=7πkR5/60
5c. Centre of mass
z̄
= moment/mass
= 7R/25
Check:
7R/25 = 0.28R < 3R/8 = 0.375R
This checks since the density is higher near the bottom.
1) To write the surface of the dome as a function of the distance from the center of the base (r), we can use cylindrical coordinates. The equation of a dome can be represented as z = f(r).
Given that the actual span of the base of the dome is 143 feet, we can assume that the base of the dome lies on the x-y plane. We can express the equation of the dome using cylindrical coordinates as follows:
r^2 + z^2 = 143^2
Simplifying the equation, we get:
z = sqrt(143^2 - r^2)
Therefore, the surface of the dome can be written as z = f(r) = sqrt(143^2 - r^2).
2) To find the height of the dome, we need to evaluate f(0). Plugging in r = 0 into the equation, we get:
z = sqrt(143^2 - 0^2) = 143
Therefore, the height of the dome, or f(0), is 143 feet.
3) To find the volume of the inside of the dome, we need to integrate the surface area over the entire dome. We can use the equation z = f(r) = sqrt(143^2 - r^2) and integrate it over the appropriate range of r and z to find the volume. However, since the specific range of integration is not provided, it is not possible to provide an exact volume without the given boundaries.
4) a) To find the moment around the x-y plane, we need to integrate the density function P(r, "theta", z) = 1 multiplied by the squared distance from the x-y plane (z = 0) to each infinitesimally small element of volume within the dome.
The moment around the x-y plane can be calculated using the integral:
M_x = ∫∫ r^2 * P(r, "theta", z) * r dz dr d"theta"
b) To find the center of mass for the dome, we can use the moment calculated in part a) and the volume calculated in part 3) (assuming the density is uniform). The center of mass for the dome (x̄, ȳ, z̄) can be calculated using the equations:
x̄ = M_x / V
ȳ = M_y / V
z̄ = M_z / V
Where M_x is the moment around the x-axis, V is the volume of the dome, and M_y and M_z are the moments around the y-axis and z-axis, respectively.
5) a) To find the mass of the dome, we need to integrate the density function P(r, "theta", z) = k * (f(0) - z) over the entire volume of the dome. Assuming the density is uniform, we can calculate the mass using the equation:
M = ∫∫∫ P(r, "theta", z) * r dz dr d"theta"
b) Once we have found the mass, we can use the result from part a) along with the moment around the x-y plane calculated in part 4a) to find the moment around the x-y plane for the variable density case.
c) To find the center of mass of the dome with variable density, we can use the equations provided in part 4b) and calculate the center of mass using the moments and the total mass obtained in part 5a).