Moment of Non-Uniform Sphere

A solid sphere has a radius of 1 1 unit. The sphere parametrization is:

x = r c o s θ s i n ϕ y = r s i n θ s i n ϕ z = r c o s ϕ 0 r 1 0 θ 2 π 0 ϕ π x = r \, cos \theta \, sin \phi \\ y = r \, sin \theta \, sin \phi \\ z = r \, cos \phi \\ 0 \leq r \leq 1 \\ 0 \leq \theta \leq 2 \pi \\ 0 \leq \phi \leq \pi

The volume-mass-density is as follows:

ρ = r θ \rho = r \, \theta

The moment of inertia about the z z -axis can be expressed as:

I = a b π 2 I = \frac{a}{b} \, \pi^2

If a a and b b are coprime positive integers, determine a + b a + b


The answer is 13.

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2 solutions

Steven Chase
Mar 6, 2018

x = r c o s θ s i n ϕ y = r s i n θ s i n ϕ z = r c o s ϕ 0 r 1 0 θ 2 π 0 ϕ π x = r \, cos \theta \, sin \phi \\ y = r \, sin \theta \, sin \phi \\ z = r \, cos \phi \\ 0 \leq r \leq 1 \\ 0 \leq \theta \leq 2 \pi \\ 0 \leq \phi \leq \pi

Volume element:

d V = r 2 s i n ϕ d r d θ d ϕ dV = r^2 \, sin \phi \, dr \, d \theta \, d \phi

Differential mass:

d m = ρ d V = r 3 θ s i n ϕ d r d θ d ϕ dm = \rho \, dV = r^3 \, \theta \, sin \phi \, dr \, d \theta \, d \phi

Distance from z-axis:

R 2 = x 2 + y 2 = r 2 s i n 2 ϕ R^2 = x^2 + y^2 = r^2 \, sin^2 \phi

Contribution to moment:

d I = d m R 2 = r 5 θ s i n 3 ϕ d r d θ d ϕ dI = dm \, R^2 = r^5 \, \theta \, sin^3 \phi \, dr \, d \theta \, d \phi

Total moment:

I = 0 1 r 5 d r 0 2 π θ d θ 0 π s i n 3 ϕ d ϕ = 1 6 2 π 2 4 3 = 4 9 π 2 I = \int_0^1 r^5 \, dr \, \int_0^{2 \pi} \theta \, d \theta \, \int_0^{\pi} sin^3 \phi \, d \phi \\ = \frac{1}{6} \, 2 \pi^2 \, \frac{4}{3} \\ = \frac{4}{9} \pi^2

matrix = FullSimplify [ Block [ { x , y , z } , x = r cos ( θ ) sin ( ϕ ) ; y = r sin ( θ ) sin ( ϕ ) ; z = r cos ( ϕ ) ; ( y 2 + z 2 x y x z x ( y ) x 2 + z 2 y z x ( z ) y ( z ) x 2 + y 2 ) ] ] ( r 2 ( cos 2 ( ϕ ) + sin 2 ( θ ) sin 2 ( ϕ ) ) r 2 cos ( θ ) sin ( θ ) sin 2 ( ϕ ) r 2 cos ( θ ) cos ( ϕ ) sin ( ϕ ) r 2 cos ( θ ) sin ( θ ) sin 2 ( ϕ ) r 2 ( cos 2 ( ϕ ) + cos 2 ( θ ) sin 2 ( ϕ ) ) r 2 cos ( ϕ ) sin ( θ ) sin ( ϕ ) r 2 cos ( θ ) cos ( ϕ ) sin ( ϕ ) r 2 cos ( ϕ ) sin ( θ ) sin ( ϕ ) r 2 sin 2 ( ϕ ) ) \text{matrix}=\text{FullSimplify}\left[\text{Block}\left[\{x,y,z\},x=r \cos (\theta ) \sin (\phi );y=r \sin (\theta ) \sin (\phi );z=r \cos (\phi );\left( \begin{array}{ccc} y^2+z^2 & -x y & -x z \\ x (-y) & x^2+z^2 & -y z \\ x (-z) & y (-z) & x^2+y^2 \\ \end{array} \right)\right]\right] \Rightarrow \\ \left( \begin{array}{ccc} r^2 \left(\cos ^2(\phi )+\sin ^2(\theta ) \sin ^2(\phi )\right) & -r^2 \cos (\theta ) \sin (\theta ) \sin ^2(\phi ) & -r^2 \cos (\theta ) \cos (\phi ) \sin (\phi ) \\ -r^2 \cos (\theta ) \sin (\theta ) \sin ^2(\phi ) & r^2 \left(\cos ^2(\phi )+\cos ^2(\theta ) \sin ^2(\phi )\right) & -r^2 \cos (\phi ) \sin (\theta ) \sin (\phi ) \\ -r^2 \cos (\theta ) \cos (\phi ) \sin (\phi ) & -r^2 \cos (\phi ) \sin (\theta ) \sin (\phi ) & r^2 \sin ^2(\phi ) \\ \end{array} \right)

jacobian = CoordinateTransformData [ Spherical Cartesian , MappingJacobianDeterminant ] [ { r , ϕ , θ } ] \text{jacobian}=\text{CoordinateTransformData}[\text{Spherical}\to \text{Cartesian},\text{MappingJacobianDeterminant}][\{r,\phi ,\theta \}]

inertiaMatrix = 0 π ( 0 2 π ( 0 1 jacobian matrix ( θ r ) d r ) d θ ) d ϕ ( 4 π 2 9 π 9 0 π 9 4 π 2 9 0 0 0 4 π 2 9 ) \text{inertiaMatrix}=\int_0^{\pi } \left(\int_0^{2 \pi } \left(\int_0^1 \text{jacobian}\ \text{matrix}\ (\theta\ r) \, dr\right) \, d\theta \right) \, d\phi \Rightarrow \\ \left( \begin{array}{ccc} \frac{4 \pi ^2}{9} & \frac{\pi }{9} & 0 \\ \frac{\pi }{9} & \frac{4 \pi ^2}{9} & 0 \\ 0 & 0 & \frac{4 \pi ^2}{9} \\ \end{array} \right)

Eigensystem [ inertiaMatrix ] ( 1 9 π ( 1 + 4 π ) 4 π 2 9 1 9 π ( 1 + 4 π ) { 1 , 1 , 0 } { 0 , 0 , 1 } { 1 , 1 , 0 } ) \text{Eigensystem}[\text{inertiaMatrix}] \Rightarrow \\ \left( \begin{array}{ccc} \frac{1}{9} \pi (1+4 \pi ) & \frac{4 \pi ^2}{9} & \frac{1}{9} \pi (-1+4 \pi ) \\ \{1,1,0\} & \{0,0,1\} & \{-1,1,0\} \\ \end{array} \right)

Since the z z axis is one of the principal axes of the object, it was easy to read out the moment of inertia.

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