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Moment of inertia of a sphere question.

Hi, I keep getting stuck on this seemingly simple question and I cannot figure out why. Firstly, I understand that it's possible to calculate this by finding the moment of inertia of a disc and then integrating across the radius - however I'm unsure why my method fails. If somebody could point out the error I'd be extremely grateful.

So, "Show that the moment of inertia of a solid sphere of uniform density is I=25MR2 I = \frac{2}{5}MR^2 "

Taking the upward axis of the sphere as the z-axis we have the integral form of moment of intertia as: ∫0Mz2dm \int^M_0 z^2 dm

Let the radius of the sphere be R R .
Create infinitesimally thin discs of mass dm dm and width dz dz which are at a height z z .

dm=ρπr2dz dm = \rho \pi r^2dz

substituting this into the integral gives:
2ρπ∫0Rz2r2dz 2\rho\pi \int^R_0 z^2r^2dz The factor of 2 is because we're only integrating across one half of the sphere.
Using Pythagoras we can get r2=R2−z2 r^2 = R^2-z^2

Therefore the integral is 2ρπ∫0Rz2(R2−z2)dz 2\rho\pi \int^R_0 z^2(R^2-z^2)dz
Computing this gives 415ρπR5 \frac{4}{15} \rho \pi R^5
But ρ=3M4πR3 \rho=\frac{3M}{4\pi R^3}
Plugging this in gives
I=15MR2 I=\frac{1}{5}MR^2
Which obviously isn't the correct answer. If anybody knows what I'm missing I'd be very grateful!
Original post by Aiden223

I=15MR2 I=\frac{1}{5}MR^2
Which obviously isn't the correct answer. If anybody knows what I'm missing I'd be very grateful!


Your approach seems to be completely wrong, I'm afraid. When you compute ∫z2dm\int z^2 dm, you are adding up the contributions due to mass dmdm that is a distance zz from the axis of rotation.

With your method, the distance from the axis of the mass of the discs is not constant, as far as I can see.
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Aiden223
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Ooh, so you're not meant to use the distance to the centre of mass? But rather the distance to the centre of mass axis? Which in this case is the z axis?
Original post by Aiden223
Ooh, so you're not meant to use the distance to the centre of mass? But rather the distance to the centre of mass axis? Which in this case is the z axis?


There are two ways to compute MOI via integration:

a) You consider an infinitesimal mass dmdm all of which is the same distance rr from the axis of rotation - then you compute ∫r2dm\int r^2 dm.

In your method, your dmdm is spread out over flat discs, whose centres are on the axis of rotation. This means that some of your dmdm is closer to the axis than other bits of it. So you wont find the correct MOI as the condition above is not satisfied.

You can fix this by considering cylinder shells of mass dmdm and radius rr, whose centre line is the axis of rotation - then all of the mass is the same distance from the axis.

b) You can use an already known result for the MOI of a particular thin body, then integrate over some range that adds up the MOIs of copys of that body that equate to the object that you are considering e.g. you can use the result the MOI of a thin spherical shell radius rr, thickness drdr - this has infinitesimal MOI

dI=f(m,r)drdI = f(m,r) dr

where f(m,r)f(m,r) is a function that you can look up. You then compute

I=∫dI=∫r=0r=Rf(m,r)drI = \int dI = \int_{r=0}^{r=R} f(m,r) dr

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