Posted by joe on Monday, December 7, 2009 at 9:41am.
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.