Moment of inertia of hemispherical shell

[mishima]

Homework Statement

The distribution of mass on the hemispherical shell:

z=(R²-x²-y²)^1/2

is given by

σ(x,y,z)=(σ₀/R²)(x²+y²)

where σ₀ is a constant. Find the moment of inertia about the z-axis of the hemispherical shell.

Homework Equations

I=∫r²dm

The Attempt at a Solution

r² is just R². ∫dm in spherical coordinates is:

σ₀R²∫∫sin³θ dθ d(phi)

with boundaries 0 to pi/2 for θ, and 0 to 2pi for phi. The completed definite integral representing the total mass is then:

(4pi/3) σ₀R².

I feel pretty confident about that, since it is given in the back of the book as the first step in the problem. What I don't understand is why the moment of inertia is anything more than just this value times another R² to give:

(4pi/3) σ₀R⁴.

The answer is supposed to be:

(16pi/15) σ₀R⁴.

So, I'm guessing there's something more to the calculation of moment of inertia. I've been looking through the different derivations of moments but can't see how to get the right factor. I could use a hint.

 

[mishima]

Oh, I understand why I can't do it that way. r in the moment of inertia formula is measured from the center of mass and is not equal to R in this case.

 

[LawrenceC]

What are the units of sigma0?

 

[mishima]

None given, but the sigma function is a mass per area so I guess it represents an average mass density, or "baseline" mass density that then varies according to the more general function.

 

[mishima]

Yay, got it.

∫∫r²σdA

r in the formula for moment of inertia can be put into spherical coordinates and integrated over the whole hemisphere (also in spherical).

σ₀R⁴∫∫sin⁵dθ d(phi)

The first integral makes the 8/15 factor needed and the second of course is 2 pi. I guess this was a fairly simple problem after all.

 

[sriracha]

Oh, I understand why I can't do it that way. r in the moment of inertia formula is measured from the center of mass and is not equal to R in this case.

I follow this, but do you mean to say rotational axis, not center of mass? The center of mass would be a point. If r is distance from the rotational axis then we can let r=p*sin[itex]\phi[/itex], or in this case r=R*sin[itex]\phi[/itex].

Yay, got it.

∫∫r²σdA

r in the formula for moment of inertia can be put into spherical coordinates and integrated over the whole hemisphere (also in spherical).

σ₀R⁴∫∫sin⁵dθ d(phi)

The first integral makes the 8/15 factor needed and the second of course is 2 pi. I guess this was a fairly simple problem after all.

I follow this and it makes sense, but when I integrated sin^5[itex]\phi[/itex], I got

-5 Cos[x] 5 Cos[3 x] Cos[5 x]
--------- + ----------- - ----------
8....48....80

which, when integrated from [o,∏/2], equals zero.