# Calculating the moment of inertia of a uniform sphere [closed]

Currently trying to calculate the moment of inertia of a uniform sphere, radius R, I know the answer is $$\frac{2}{5}MR^2$$ but I keep getting $$\frac{1}{5}MR^2$$

Setup:

Assume mass per unit volume $$\rho$$

Centering the sphere on the $$z$$ axis I am first calculating the moment of inertia for one side then doubling the answer exploting the symmetry.

Splitting the hemisphere into circular discs of width $$dx$$:

$$dm = \rho\pi(R^2-x^2)dx$$

So wouldn't the moment of inertia be:

$$2\rho\pi\int_0^R(x^2(R^2-x^2))dx$$ Which comes out to $$\frac{1}{5}MR^2$$

any help appreciated.

## closed as off-topic by AccidentalFourierTransform, ACuriousMind♦May 20 at 16:27

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• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind May 20 at 16:27
• Well the answers showed a fatal flaw in my idea of calculating the moment of inertia so I can argue that this would be helpful to other students too. – PolynomialC May 20 at 17:30

I don't understand how are you splitting the hemisphere into circular discs. your $$dm$$ must have same distance from the axis to zeroth order everywhere inside it. Disc does not. If you want to split your sphere into discs, you need to compute the moment of inertia of a disc first ($$I=mr^2/2$$), and then you can compute: $$dI_z=dm(R^2-z^2)/2=\rho \pi (R^2-z^2)^2dz/2$$ leading to: $$I_z=\int_{-R}^{R}dm(R^2-z^2)/2=\int_{-R}^{R}\rho \pi (R^2-z^2)^2dr/2=\frac{8}{15}\rho\pi R^5=\frac{2}{5}M R^2$$ This assumes the knowledge of moment of inertia of a disc though.

I would split sphere into cylinder shells centered on z-axis of width $$dx$$, then $$dm=\rho\pi h ((x+dx)^2-x^2)=2\rho\pi h x dx$$ where $$h(x)$$ is the height of the cylinder: $$h=2\sqrt{R^2-x^2}$$. The cylinder shell has same distance from the axis everywhere, so i can continue to write: $$I_z=\int_0^R x^2dm=4\rho\pi\int_0^R \sqrt{R^2-x^2} x^3 dx=\frac{8}{15}\rho\pi R^5=\frac{2}{5}M R^2$$

Each slice (disk) is offset by $$x$$ from the origin and has radius $$r = \sqrt{R^2-x^2}$$. The contribution to MMOI of each disk is $${\rm d} I = \frac{\rho}{2} \pi r^4 {\rm d}x$$

Why? Either do another integral on a disk, or use $${\rm d} I = \tfrac{1}{2} r^2 {\rm d}m$$, with $${\rm d} m = \rho \left( \pi r^2 {\rm d}x \right)$$.

You also have $$m = \rho \left( \tfrac{4}{3} \pi R^3 \right)$$

Now do the integral,

$$I = 2 \int \limits_0^R \frac{\rho}{2} \pi (\sqrt{R^2-x^2})^4 {\rm d}x = \pi \rho \int \limits_0^R (R^2-x^2)^2 {\rm d}x$$

$$I = \pi \left( \frac{m}{ \tfrac{4}{3} \pi R^3 } \right) \left( \tfrac{8}{15} R^5 \right) = \tfrac{2}{5} m R^2$$

You mistaking in calculating moment of inertia of thin circular disc by multiplying 'dm' by x^2, instead of you have to calculate it through parallel axis theorem it means you know moment of inertia of disc which is 1/2 MR^2 but it's from centre of disc not from our reference point is that centre of uniform solid sphere hence for finding that use parallel axis theorem by this we get moment of inertia as 1/2 M(R^2-X^2) +MX^2.