Spherical cap moment of inertia

André Verhecken

Hi, I'm trying to find the formula for the moment of inertia of a spherical cap relative to the axis perpendicular to its flat area. I know it should be done by (triple) integration, but it's a very long time since I've done that kind of math ! So, if someone out there knows the formula ....

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Andrew Mason

Homework Helper
André Verhecken said:
Hi, I'm trying to find the formula for the moment of inertia of a spherical cap relative to the axis perpendicular to its flat area. I know it should be done by (triple) integration, but it's a very long time since I've done that kind of math ! So, if someone out there knows the formula ....
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?

AM

André Verhecken

That's easy: half of that, of course. But further on ? Do you suggest that the Moment Of Inertia of the cap is then to be calculated from the half sphere volume and the relation between cap volume and half sphere volume ?
AV

Andrew Mason

Homework Helper
André Verhecken said:
That's easy: half of that, of course. But further on ? Do you suggest that the Moment Of Inertia of the cap is then to be calculated from the half sphere volume and the relation between cap volume and half sphere volume ?
AV
From your initial statement of the problem, I assumed that 'cap' referred to a portion of the solid sphere.

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

$$I = \frac{1}{2}\rho\pi \int_{0}^R (R^2 - z^2)^2 dz$$

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

AM

André Verhecken

As a retired chemist, it’s a very long time (40 years) that I haven’t done any integration calculus, so I forgot quite a lot of it. But I’ll try:

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

Andrew Mason

Homework Helper
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.
$$I = \frac{1}{2}\rho\pi \int_{z_0}^R (R^2 - z^2)^2 dz$$

$$z_0 = R-h$$

Expand:

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

Separate the integrals:

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

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

AM

André Verhecken

I did use exactly the same formulas as you gave now, but I found where I went wrong the other day: I calculated the separated integrals for R, and then presumed that the result for z° would we similar in form. So for this second calculation I forgot that R is not a variable.
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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