「斑馬知識+挑戰」Proof of moment of inertia

Right, the answer can be easily found using Parallel Axis Theorem.

So,
I = (2/5) M R + M R = (7/5) M R

(note that I use the variable R here instead of r in the question to avoid confusion of the integrating variable r below)

But if we really want to calculate the moment of inertia by integration, here's the steps:

Consider the centre of the sphere is at the origin and the rotation axis is x=R, y=0

Considering a volume element in spherical coordinate
dV = r sinφ dr dφ dθ ,

then the infinitesimal mass is
dm = M / (4πR/3) dV = [3M/(4πR)] r sinφ dr dφ dθ

The perpendicular distance of the infinitesimal mass to the rotation axis is
d = √[(R- x) + y]

But as x = r sinφ cosθ and y = r sinφ sin θ
d = √[(R- r sinφ cosθ) + (sinφ sinθ)] = √(R - 2 R r sinφ cosθ + sinφ )

Therefore the moment of inertia is

I = ∫ (√(R - 2 R r sinφ cosθ + sinφ) dm

= ∫∫∫ (R - 2 R r sinφ cosθ + sinφ) [3M/(4πR)] r sinφ dr dφ dθ (integral for r:0->R, φ:0->π, θ:0->2π)

= [3M/(4πR)]∫∫∫(R r sinφ - 2 R r sinφ cosθ + r4 sinφ) dr dφ dθ

∫∫∫R r sinφ dr dφ dθ
= R [r / 3] [- cosφ][θ]
= 4πR5 / 3

∫∫∫- 2 R r sinφ cosθ dr dφ dθ
= 0 because the symmetry of cosθterm running from 0 to 2π

∫∫∫r4 sinφ dr dφ dθ
= [r5 /5 ] [cosφ/3 - cosφ] [θ]
= (8 / 15) πR5

So, the moment of inertia
= [3M/(4πR)] [ 4πR5 / 3 + (8 / 15) πR5]
= (7/5) M R


note:

∫ sinφ dφ = -∫ sinφ d cosφ = ∫ (cosφ - 1) d cosφ = cosφ/3 - cosφ

note: The approach is the same as my answer to another similar question.
http://hk.knowledge.yahoo.com/question/question?qid=7007102402012


2008-05-13 07:56:52 補充：
Correction, it seems that the square and cube in my calculation are gone. here is the correction

I = (2/5) M R² + M R² = (7/5) M R²

2008-05-13 07:58:02 補充：
dV = r² sinφ dr dφ dθ ,

dm = M / (4πR³/3) dV = [3M/(4πR³)] r² sinφ dr dφ dθ

d = √[(R- x)² + y²]

d = √[(R- r sinφ cosθ)² + (sinφ sinθ)²] = √(R² - 2 R r sinφ cosθ + sin²φ )

2008-05-13 07:59:12 補充：
I = ∫ (√(R² - 2 R r sinφ cosθ + sin²φ)² dm

= ∫∫∫ (R² - 2 R r sinφ cosθ + sin²φ) [3M/(4πR³)] r² sinφ dr dφ dθ (integral for r:0->R, φ:0->π, θ:0->2π)

= [3M/(4πR³)]∫∫∫(R² r² sinφ - 2 R r³ sin²φ cosθ + r4 sin³φ) dr dφ dθ

2008-05-13 07:59:20 補充：
∫∫∫R² r² sinφ dr dφ dθ
= R² [r³ / 3] [- cosφ][θ]
= 4πR5 / 3

∫∫∫- 2 R r³ sin²φ cosθ dr dφ dθ
= 0 because the symmetry of cosθterm running from 0 to 2π

∫∫∫r4 sin³φ dr dφ dθ
= [r5 /5 ] [cos³φ/3 - cosφ] [θ]
= (8 / 15) πR5

So, the moment of inertia
= [3M/(4πR³)] [ 4πR5 / 3 + (8 / 15) πR5]
= (7/5) M R²

2008-05-13 07:59:30 補充：
note:

∫ sin³φ dφ = -∫ sin²φ d cosφ = ∫ (cos²φ - 1) d cosφ = cos³φ/3 - cosφ

我自己認為唔洗再integration喇, 因為可以用Parallel Axis Theorem.....

你而家axis through centre個Moment of inertia係 (2/5)(Mr^2)

咁你而家要將條AXIS推出去, 唔係就係:

= (8 / 15) πR5

So, the moment of inertia
= [3M/(4πR)] [ 4πR5 / 3 + (8 / 15) πR5]
= (7/5) M R
∫ sinφ dφ = -∫ sinφ d cosφ = ∫ (cosφ - 1) d cosφ = cosφ/3 - cosφ
note: The approach is the same as my answer to another similar question.
http://hk.knowledge.yahoo.com/question/question?qid=7007102402012
Correction, it seems that the square and cube in my calculation are gone. here is the correction
I = (2/5) M R² + M R² = (7/5) M R²
dV = r² sinφ dr dφ dθ ,
dm = M / (4πR³/3) dV = [3M/(4πR³)] r² sinφ dr dφ dθ
d = √[(R- x)² + y²]
d = √[(R- r sinφ cosθ)² + (sinφ sinθ)²] = √(R² - 2 R r sinφ cosθ + sin²φ )
I = ∫ (√(R² - 2 R r sinφ cosθ + sin²φ)² dm
= ∫∫∫ (R² - 2 R r sinφ cosθ + sin²φ) [3M/(4πR³)] r² sinφ dr dφ dθ (integral for r:0->R, φ:0->π, θ:0->2π)
= [3M/(4πR³)]∫∫∫(R² r² sinφ - 2 R r³ sin²φ cosθ + r4 sin³φ) dr dφ dθ
∫∫∫R² r² sinφ dr dφ dθ
= R² [r³ / 3] [- cosφ][θ]
= 4πR5 / 3
∫∫∫- 2 R r³ sin²φ cosθ dr dφ dθ
= 0 because the symmetry of cosθterm running from 0 to 2π
∫∫∫r4 sin³φ dr dφ dθ
= [r5 /5 ] [cos³φ/3 - cosφ] [θ]
= (8 / 15) πR5
So, the moment of inertia
= [3M/(4πR³)] [ 4πR5 / 3 + (8 / 15) πR5]
= (7/5) M R²
∫ sin³φ dφ = -∫ sin²φ d cosφ = ∫ (cos²φ - 1) d cosφ = cos³φ/3 - cosφ

我自己認為唔洗再integration喇, 因為可以用Parallel Axis Theorem.....

你而家axis through centre個Moment of inertia係 (2/5)(Mr^2)

咁你而家要將條AXIS推出去, 唔係就係:

I = (2/5)(Mr^2) + Mr^2
= (7/5)(Mr^2)

I don't know