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The moment of a full sphere is 2MR^2/5. How would the moment of inertia of 1/2 a sphere be related to the moment of inertia of the full sphere?André Verhecken said:

AM

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AV

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From your initial statement of the problem, I assumed that 'cap' referred to a portion of the solid sphere.André Verhecken said:

AV

Look at the integral for calculating the moment of inertia for a half sphere:

[tex]I = \frac{1}{2}\rho\pi \int_{0}^R (R^2 - z^2)^2 dz[/tex]

To find the moment of a portion you just integrate from z_0 rather than 0.

AM

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Let’s consider a sphere (radius R, density ρ) from which a cap is cut with a height (= the “arrow”) h and a radius of the cut circle: r. Then I name H = R – h.

Integrating your formula for a half sphere for z = 0 to R gives the familiar I = m R2 / 5

Integration for z = 0 to H (which is the same as z = 0 to R – h) gives I = m (R – h)2 / 5

So that I (cap) = m (R2 – (R- h)2 ) / 5

= m h (2R – h) / 5

In my problem, R is unknown; only r and h are measured. But R is easily calculated as:

R = (r2 + h2) / 2 h

Substituting R in the formula for I(cap) then yields : I(cap) = m r2 / 5

This may be correct or it may be wrong (I cannot grasp it intuitively): the moment of inertia of a thin cap with cut circle radius r should be the same as that of a half sphere with radius r; and moreover, h has disappeared ! This would mean that any cap with radius of cut circle = r, cut from any sphere with radius > R, would obey this formula, and this seems impossible to me (although the formula is correct when r = R). I fear there is something fishy in my reasoning or my calculus.

I would appreciate if you would point out my error !

AV

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[tex]I = \frac{1}{2}\rho\pi \int_{z_0}^R (R^2 - z^2)^2 dz[/tex]André Verhecken said:This may be correct or it may be wrong (I cannot grasp it intuitively): the moment of inertia of a thin cap with cut circle radius r should be the same as that of a half sphere with radius r; and moreover, h has disappeared ! This would mean that any cap with radius of cut circle = r, cut from any sphere with radius > R, would obey this formula, and this seems impossible to me (although the formula is correct when r = R). I fear there is something fishy in my reasoning or my calculus.

[tex]z_0 = R-h[/tex]

Expand:

[tex]I = \frac{1}{2}\rho\pi(\int_{z_0}^R (R^4 -2R^2z^2 + z^4)dz)[/tex]

Separate the integrals:

[tex]I = \frac{1}{2}\rho\pi(\int_{z_0}^R R^4 dz -\int_{z_0}^R 2R^2z^2dz + \int_{z_0}^R z^4dz)[/tex]

Simplifying that is a bit of a chore but it should give you the result in terms of R and h ([itex]z_0 = (R-h)[/itex]). You then have to work out the volume of the cap and substitute for [itex]\rho = M/V[/itex]. It is a non-trivial exercise. This is the kind of thing that Maple was created for!

AM

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I did the calculation again, and avoided the simplifying by entering the somewhat complex formula as such in Excel : it works !

Thanks for your stimulating help !

AV

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