# Moment of Intertia of Spherical Cap?

1. Apr 24, 2008

### CasualDays

[SOLVED] Moment of Intertia of Spherical Cap?

1. The problem statement, all variables and given/known data
I am having some serious trouble trying to setup this problem..

Find the rotational intertia about the z axis of a spherical cap cut from the sphere $$x^2$$+$$y^2$$+$$z^2$$=4 by the horizontal plane z=1.

2. Relevant equations
Suggestions for solving the problem:
$$I\check{z}$$= $$\int$$ $$\rho$$ $$r^2$$ dV
where $$\rho$$=$$1/r$$

3. The attempt at a solution
Can't post URL's till after 15 posts but it's there.

Here is the suggestion from this this thread:
$$I = \frac{1}{2}\rho\pi(\int_{z_0}^R R^4 dz -\int_{z_0}^R 2R^2z^2dz + \int_{z_0}^R z^4dz)$$

Is this the idea I should be following? Or is there a different method to calculating my problem?

It seems my book wants me to go with a triple integration, so here is how I have been trying to do it:

$$I = \frac{1}{r} \int_{z= 1}^h\int_{r= 0}^q\int_{\theta=0}^{2\pi}} r dr d\theta dz[/itex] where q=\sqrt{4-z^2} Last edited: Apr 24, 2008 2. Apr 24, 2008 ### jandl Take your spherical cap and cut it up into infinitesimal slices. Make your cuts perpendicular to the z axis. For any of the slices you should be able to find the moment of inertia based on the radius and density of the slice since it is simply a disk. Now to find the total moment of inertia of the cap, add up all of the slices. This leads to a single integral. However, the trick that is left to you is to determine how the radius of those slices depends on z since your integration will have to be with respect to z. So find r(z) and do an integral with dz. If you're having trouble finding r(z) from the equations alone, try drawing some pictures, they should be able to help a lot. 3. Apr 24, 2008 ### dynamicsolo jandl's suggestion is probably the most efficient approach. You would be correct in applying cylindrical coordinates in conducting an integration for moment of inertia, since it involves reference to an axial line. However, you really want to avoid having to work with spherical sections in cylindrical coordinates, if that is possible... 4. Apr 24, 2008 ### CasualDays Thinking through your method, I first tried converting [tex]x^2+y^2+z^2=4$$

to cylindrical and I got:

$$r(z)=\sqrt{4-z^2}$$

Then, I attempted a single integral with this equation:

$$I = \frac{1}{2}r^2\rho\int_{z_0}^r \sqrt{4-z^2}dz$$

where I assumed $$z_0=r-h$$ for h=1 and r=2

Which leads me to a solution, but it doesn't seem quite right. I ended up with an answer involving arctan. Are my limits of integration off?

5. Apr 24, 2008

### dynamicsolo

That looks OK...

Why is r^2 outside the integral? And what is inside it?

Each infinitesimal disk has a thickness dz and a radius r(z), so you would want to integrate disks that contribute

dI = (1/2) (dm) [r(z)^2] ,

where dm = (rho) (pi) [r(z)^2] dz , the mass of each infinitesimal disk.

What are the limits of integration for z then?

You will end up with something involving units of length^5 . If you are looking to end up with an expression for the moment of inertia of the form I = C x M(R^2) , you will also need to do an integration to find the mass of the spherical cap.

6. Apr 24, 2008

### CasualDays

So:

$$I_z = \frac{1}{2}\rho\pi\int_{z_0}^r (4-z^2)^2 dz$$

where $$z_0=1$$

7. Apr 24, 2008

### dynamicsolo

That looks good. And the mass would be

$$M = \rho\pi\int_{z_0}^r (4-z^2) dz$$ .

I guess this problem doesn't require you to find that out. But were you to work out expressions for both I_z and M , using the limits z_0 and R , you would have a way to check the validity of your result. For instance, setting z_0 = 0 , you would have the result for a uniformly dense hemisphere, which should have the same I_z as for a full sphere, I_z = (2/5)·M·(R^2) .