### Derivation of moment of inertia of an uniform solid sphere

An uniform solid sphere has a radius R and mass M. calculate its moment of inertia about any axis through its centre.

**Note:** If you are lost at any point, please visit the beginner’s lesson or comment below.

First, we set up the problem.

- Slice up the solid sphere into infinitesimally thin solid cylinders
- Sum from the left to the right

Recall the moment of inertia for a solid cylinder:

$$I = \frac{1}{2} M R^{2}$$

Hence, for this problem,

$$dI = \frac{1}{2} r^{2} \: dm$$

Now, we have to find dm,

$$dm = \rho \: dV$$

Finding dV,

$$dV = \pi r^{2} \: dx$$

Substitute dV into dm,

$$dm = \rho \pi r^{2} \: dx$$

Substitute dm into dI,

$$dI = \frac{1}{2} \rho \pi r^{4} \: dx$$

Now, we have to force x into the equation. Notice that x, r and R makes a triangle above. Hence, using Pythagoras’ theorem,

$$r^{2} = R^{2} – x^{2}$$

Substituting,

$$dI = \frac{1}{2} \rho \pi (R^{2} – x^{2})^{2} \: dx$$

Hence,

$$I = \frac{1}{2} \rho \pi \int\limits_{-R}^{R} (R^{2} – x^{2})^{2} \: dx$$

After expanding out and integrating, you’ll get

$$I = \frac{1}{2} \rho \pi \frac{16}{15} R^{5}$$

Now, we have to find what is the density of the sphere:

$$\rho = \frac{M}{V}$$

$$\rho = \frac{M}{\frac{4}{3} \pi R^{3}}$$

Substituting, we will have:

$$I = \frac{2}{5} M R^{2}$$

And, we’re done!

**Derivation Of Moment Of Inertia Of Common Shapes:**

- Uniform Rigid Rod (“Beginners’ Lesson”)
- Hollow/solid Cylinder
- Thin Spherical Shell

Kajal - scribbled

Find moment of inertia of a solid sphere about it’s diameter.momemt of inertia of a sphere about a tangent is 7mr^2/5

Ahmed Mutawea scribbled

Correct

Devi sreee scribbled

This derivation is easy but I want detail derivation

Sean Arriate scribbled

What happened to pi at the end?

ak scribbled

See this.very simple proof

https://www.youtube.com/watch?v=_hD1GpbsMtY

Ted scribbled

Hello!

It seems to me a simpler derivation is given as follows:

Definition of I = ∑_i〖m_i r^2 〗 = ∫_body〖r^2 dm〗

For a body of uniform composition, dm = ρdV, where ρ is the density and dV is the change in volume.

For a sphere, dV = 4π/3 r^2 dr

Substitution gives: I= ∫body〖r^2 (4πρ/3 r^2 dr)〗

So, I= 4πρ/3 ∫(-R)^R r^4 dr=4πρ/3 ( r^5/5](-R)^R )= 4πρ/3 ((2R^5)/5) )

The mass of the object is given by M = 4πρ/3 R^3

So I = 2/5 MR^2

(I apologize that the formatting of the integrals isn’t coming out correctly.)

HARSH KUMAR scribbled

Why don’t we take dx as Rdθ like we did in case of a hollow sphere?

Nikhil scribbled

U can also do by that method

gizaw scribbled

why we not begin from I=MR**2 just like inertia of solid cylinder

sugi-bejo scribbled

thank you so much…

B scribbled

I think I’m starting to make sense of this, and this is how I’m thinking of it: I think the “normal” formula for moment of inertia, \(I=\int_V r^2dm\) assumes \(r\) is measured from the axis of rotation. The formula from Chandrasekhar’s book assumes the moments are measured as the mass elements rotate relative to the origin, instead of the axis of rotation.

The book is “ellipsoidal figures of equillibrium” by Chandrasekhar, on page 16 [I have a PDF if you need].

mystique scribbled

oh yes I do. I know it’s been more than a year but I really want the pdf if you could send it to my email which is agarwalrishita2000@gmail.com. Thanks!

b scribbled

I’m reading a text by Chandrasekhar, and he has this formula for the moment of inertia: $$I=\int_V\rho(x)|x|^2dx=\frac{3}{5}Mk^2$$, and $k$ is called the radius of gyration (radius of a sphere with the equivalent moment of inertia).

This seems to suggest that for a sphere of radius $R$, we should have $$I=\frac{3}{5}MR^2.$$ Also, when I compute $$\int_V\rho(x)|x|^2dx$$ using spherical coordinates (integrand $r^4sin(\phi)$, where $\phi$ is measured from the north pole), I’m also getting the same thing as Chandrasekhar. Yet all the web resources I’m finding give $\frac{2}{5}MR^2$… What’s going on here?

Mini Physics scribbled

Could you give me the title of the text and the relevant page number? I suspect that it is due to the density having a dependence on x. For the web resources, the density is treated as uniform throughout the sphere. And, radius of gyration of a uniform sphere is not equal to the radius of the sphere, it’s $\sqrt{\frac{3}{5}} R$.

I’ve did the integration for the equation that you provided with the assumption that density is constant.

Note:I’ve used $\theta$ as the angle measured from the north pole. (Old habits)$$\begin{aligned} I &= \int_{V} \rho |x|^{2} dx \\ &= \int \rho (r sin \, \theta)^{2} r^{2} sin \, \theta \, dr d \theta d \phi \\ &= \rho \frac{2}{5} \pi R^{5} \int\limits_{0}^{\pi} sin^{3} \, \theta \, d\theta \\ &= \frac{2}{5} MR^{2} \end{aligned}$$

Disclaimer: I did not calculate the radius of gyration of the sphere. I pulled it from some other web resource.

Yuri psm scribbled

In yourcalculation, you probably used the distance to the origin (center of the sphere), instead of the distance to the axis of rotation.

Raushan kumar scribbled

Hindi me jabab chahiye