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# Moments of Inertia (1)

This note is less of a proof and more of a series of derivations. We will derive the following moments of inertia:

(1) Spherical, thin shell about any axis: $$\frac{2mr^2}{3}$$

(2) Solid sphere about any axis: $$\frac{2mr^2}{5}$$

(3) Cylindrical shell about z-axis: $$mr^2$$

(4) Solid disk about z-axis: $$\frac{mr^2}{2}$$

(5) Solid cylinder about z-axis: $$\frac{mr^2}{2}$$

In all cases, $$m$$ represents mass and $$r$$ represents radius. Also, all of the aforementioned rigid bodies are assumed to have completely homogenous (uniform) mass distribution (in other words, constant density throughout). Please note that this note will not cover all the details of the actual integration, as this would make the note excessively lengthy and most likely redundant to the reader.

Before we begin these derivations, we will establish two coordinate systems (in addition to the Cartesian coordinate system) and their respective Jacobians. This will help in finding our moments of inertia.

Cylindrical Coordinate System:

Let: $$x=rcos(\theta)$$

$$y=rsin(\theta)$$

$$z=z$$

We now take partial derivatives and form the Jacobian matrix:

$$\partial_{r}x = cos(\theta)$$ and $$\partial_{\theta} x = -rsin(\theta)$$

$$\partial_{r} y = sin(\theta)$$ and $$\partial_{\theta}y = rcos(\theta)$$

$$J = \left( \begin{array}{cc} cos(\theta) & -rsin(\theta) \\ sin(\theta) & rcos(\theta) \end{array} \right)$$

Taking the determinant:

$$det(J) = rcos^2(\theta) + rsin^2(\theta) = r = \mid r \mid$$

Our Jacobian determinant is clearly $$det(J) = r$$. We will include this in our integrals when we change coordinate systems.

(Note: For those that are unaware, the Jacobian determinant takes into account the change in volume due to a change of basis. For example, if the Jacobian determinant is 1, this means that no change in volume will occur with this transformation. If the Jacobian is $$r$$, then the volume will shift by a factor of $$r$$. If it is negative, this indicates a change in orientation in our new basis. Since we wish to find volume (most of the time), we will use the absolute value of the Jacobian, as orientation has no effect on volume.)

Spherical Coordinate System:

Let: $$x=\varrho sin(\phi) cos(\theta)$$

$$y= \varrho sin(\phi) sin(\theta)$$

$$z=\varrho cos(\phi)$$

Taking partial derivatives:

$$\partial_{\varrho} x = sin(\phi)cos(\theta)$$ $$\partial_{\theta} x=-\varrho sin(\phi) sin(\theta)$$ and $$\partial_{\phi}x = \varrho cos(\phi) cos(\theta)$$

$$\partial_{\varrho} y = sin(\phi) sin(\theta)$$ $$\partial_{\theta} y= \varrho sin(\phi)cos(\theta)$$ and $$\partial_{\phi}y = \varrho cos(\phi)sin(\theta)$$

$$\partial_{\varrho} z = cos(\phi)$$ $$\partial_{\theta} z= 0$$ and $$\partial_{\phi} z = -\varrho sin(\phi)$$

Organizing our matrix:

$$J= \left ( \begin{array}{ccc} cos(\phi) & 0 & - \varrho sin(\phi) \\ sin(\phi) sin(\theta) & \varrho sin(\phi) cos(\theta) & \varrho cos(\phi) sin(\theta) \\ sin(\phi) cos(\theta) & - \varrho sin(\phi) sin(\theta) & \varrho cos(\phi)cos(\theta) \end{array} \right)$$

And, omitting some algebraic details, the determinant can be found by cofactor expansion to be:

$$det(J) = \varrho^2 sin(\phi)$$

We will include this in our integrals as well.

(1) Spherical, thin shell:

The general formula for moment of inertia is: $$I = \displaystyle \int_{V} d(\vec{x}) r^2 \,dV$$

Where $$d(\vec{x})$$ is the density distribution within the body. Since we are dealing with completely homogenous bodies, we may write:

$$I = \frac{m \displaystyle \int_{V} r^2 \,dV}{V}$$

This will be our base formula for all of these. For a hollow sphere, we will change to spherical coordinates with the following bounds:

$$0 \leq \theta \leq 2\pi$$

$$0 \leq \phi \leq \pi$$

$$r_1 \leq \varrho \leq r_2$$

Now, before we integrate, we need to take note of the fact that the $$r^2$$ term in the formula represents the distance between any massive particle in our structure and the axis of rotation. In this case, this is:

$$r^2 = (\varrho sin(\phi))^2$$

Then, the integral with the included Jacobian is:

$$I= \displaystyle \lim_{r_1 \to r_2} \frac{m \int_{0}^{2\pi} \int_{0}^{\pi} \int_{r_1}^{r_2} \varrho^4 sin^3 (\phi) d \varrho d \phi d \theta}{\int_{0}^{2\pi} \int_{0}^{\pi} \int_{r_1}^{r_2} \varrho^2 sin (\phi) d\varrho d\phi d\theta}$$

I will leave the details of the actual integration to the reader (these integrals are generally very easy to evaluate). We arrive at:

$$I = \displaystyle \lim_{r_1 \to r_2} \frac{4m (r_2^5 - r_1^5)}{10(r_2^3 -r_1^3)} = \frac{0}{0}$$

This is an indeterminate form conducive to L'Hopital's Rule. Differentiating the numerator and denominator with respect to $$r_1$$, we arrive at:

$$I = \displaystyle \lim_{r_1 \to r_2} \frac{10m r_1^4}{15r_1^2} = \frac{2mr^2}{3}$$

Which is the intended result. Note that if we had a spherical pseudo-shell, that is, the shell was not infinitely thin, we would simply omit the limit to obtain our result.

(2) Solid Sphere:

We use the same formula, in of course spherical coordinates, with bounds:

$$0 \leq \varrho \leq r$$

$$0 \leq \theta \leq 2\pi$$

$$0 \leq \phi \leq \pi$$

Our integral is (with the same substitution for $$r^2$$ as before):

$$I= \frac{m \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{r} \varrho^4 sin^3 (\phi) d \varrho d \phi d \theta}{\int_{0}^{2\pi} \int_{0}^{\pi} \int_{r_1}^{r_2} \varrho^2 sin (\phi) d\varrho d\phi d\theta}$$

Simplifying, we obtain:

$$I = \frac{3mr^2}{10} \displaystyle \int_{0}^{\pi} sin^3(\phi) \,d\phi$$

After evaluating completely, we arrive at:

$$I = \frac{2mr^2}{5}$$

The remaining derivations will appear in "Moments of Inertia (2)"

Note by Ethan Robinett
3 years, 5 months ago

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We when prepare for IIT-JEE also do this series of derivation. But your method of deriving it is very cool.

- 3 years, 5 months ago

Thanks how would you go about deriving it?

- 3 years, 5 months ago

We would do the same thing ultimately but we do it in an informal way rather than using complete triple integrals. We would often use the results derived earlier to simplify the integrals.

- 3 years, 5 months ago

Oh I see, yeah deriving them this way is a lot of work, I didn't show all of it because the note would've been really long.

- 3 years, 5 months ago

@Michael Mendrin Sir, please could you help me with a doubt I have? I failed to understand the following result:: $$I = \frac{m \displaystyle \int_{V} r^2 \,dV}{V}$$



Sir, I don't see how this result was arrived at. If we take the density (mass per unit volume, which is uniform throughout the body) as $$\rho$$, then $$dm=\rho dV$$ where $$V$$ denotes volume.

Thus the moment of Inertia gets converted into  $I=\large{\dfrac{\rho \int _V r^2 dV}{V}}$ I fail to understand why instead it is $I = \frac{m \displaystyle \int_{V} r^2 \,dV}{V}$ Please could you explain why it is $$m$$ instead of $$\rho$$ in the numerator?

- 2 years, 3 months ago