Moment(s) of Inertia

Earlier in mechanics, we saw that mass is a mathematical and physical quantity that represents the amount of "inertia" an object has; i.e. resistance to a change in motion. There is a similar concept in rotational dynamics. The math is almost the same but the concepts vary. Of course, in rotational dynamics, mass doesn't cease to exist, but we have this additional property. Guess what it's called?

Of course, moment of inertia! First, let us compare the definitions of mass and moment of inertia.

Mass is a property of a physical body which determines the body's resistance to being accelerated by a force and the strength of its mutual gravitational attraction with other bodies.

vs.

Moment of inertia is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation.

As a mathematical analogy, we can say mass is to force as moment of inertia is to the rotational equivalent of force, namely torque. Before we write out some of the rigor, let us define some notation. As usual, let $m$ be mass. Let $a$ be acceleration and $\alpha$ be the angular acceleration. Let $F$ be force, and let the torque be $\tau$. Finally, let's name our new quantity. Take a guess!

Great, now that we have some notation, consider the following. In the linear case, we had $F=ma$. Now let us say we did a direct translate from the linear framework to the rotational framework. Imagine that there was a Google Translate that translates from "Linear" to "Rotational" $-$ now, what would the machine spit out if we typed in $F=ma$? Think about it and then click continue for a hint.

Hint. For example, the rotational equivalent for force is $\tau$. Click continue for the answer.

Since $F$ maps to $\tau$, $m$ maps to $I$, and $a$ maps to $\alpha$, we get $\boxed {\tau = I \alpha}$, and this happens to be correct!

Now, let's check that you're awake. Pop quiz question for you.

So mass was a seemingly simple subject. Furthermore, it seemed very intuitive. However, moment of inertia is something we are usually less familiar with. We usually don't go on a scale to check our moment of inertia $-$ it's usually mass. In fact, the moment of inertia for a human is quite complex. There is one key thing which defines the moment of inertia for any given object. Click Continue to find out what it is.

Geometry! Yes, really. We will now delve into computing moment of inertia for a given object. Let's take an easy case. Any ideas?

Let us consider, like a good easy-case engineer, the easiest case ever for an object $-$ the point particle.

Moment of inertia is not a property that is intrinsic of the object. We need to find it with respect to something. This something is not a particular point or another object. It's a line $-$ we call this line the axis. You can pick any axis but the moment of inertia will change depending on which axis to pick. After all, we have to rotate objects about specified axes. Now, I can pose a more well-defined question.

As it turns out, the correct answer was $mr^2$, and this will be the basis for all moment of inertia computations you will have to do in your life.

Now I ask $-$ what if I have many particles and I wanted the moment of inertia of the whole system with respect to the axis, given that particle $i$ has mass $m_i$ and is at a distance $r_i$ away from the axis of interest? Click away for the answer.

Turns out the answer is $\sum_{i} m_i r_i^2$; i.e. the sum of the individual moment of inertias. If you thought that was intuitive, you're right.

Before we continue to rigorize this notion, let's ask a question that you might have asked by now.

You'd think that moment of inertia is a scalar. It does not have a direction associated with it, so it's definitely not a vector. It also sums up just like scalars, and it exhibits the same mathematical properties as other physical quantities that are also scalars. However, moment of inertia is best represented by a quantity called a second rank tensor, which is basically a fancy name for a $3 \times 3$ vector of quantities. You can read more about tensors here.

Let's continue our exploration of computing moments of inertia.

Let us say we have a bunch of particles in a straight line and we want the moment of inertia about an axis. Well, we just did that. Simply add the individual moments of inertia. Now let's say we add a lot more particles. We can just keep adding the individual moments of inertia, and no matter how many we add, as long as the number is finite, we can keep adding the moments of inertia. However, what if we start building a continuous, rather than discrete, system? For example, let us say we build a rod of mass $M$ and radius $R$ and we want to find the moment of inertia of the rod about an axis perpendicular to the rod crossing one end of the rod. Of course, assume the rod has negligible thickness.

Now, how do we add up infinite elements in the rod?

Simple! We resort to some infinite summing; i.e. calculus. We can get a quite simple integration formula for the moment of inertia after a bit of thought. Think about it and click ahead if you think you want a hint.

Hint. Remember the discrete sum was of the form $\displaystyle\sum_i m_i r_i^2$. Make this continuous using an integral. Don't forget to consider infinitesimal elements of mass! Answer on the next click.

Answer. $$I = \displaystyle\int r^2 \, \mathrm{d}m$$ Your reaction might be of course! The idea is that you want to sum $mr^2$ for every tiny bit of mass, represented by $\mathrm{d}m$.

Let us apply this to a rod of total mass $M$ and length $L$.

First, let $\lambda$ be the mass density, per unit length. In other worse, if the length of the rod is $L$, we have $\lambda = \frac {M}{L}$. More generally, we have $\lambda = \frac {\mathrm{d}m}{\mathrm{d}r}$. Think about why this is, then move on once you have it!

Now, note that $$\lambda = \frac {\mathrm{d}m}{\mathrm{d}r} \implies \mathrm{d}m = \lambda \, \mathrm{d}r,$$ so we can plug this in to our integral: $$I = \displaystyle\int r^2 \cdot \left( \lambda \, \mathrm{d}r \right) = \lambda \cdot \displaystyle\int r^2 \, \mathrm{d}r.$$

Clearly, if we are integrating with respect to $r$, we need $r$ to range from $r=0$ and $r=L$. (Don't forget what axis we are considering!)

Now we can finish: $$\begin {align*} I &= \lambda \cdot \displaystyle\int_0^L r^2 \, \mathrm{d}r \\&= \lambda \cdot \left[ \frac {r^3}{3} \right]_{0}^{L} \\&= \lambda \cdot \frac {L^3}{3}. \end {align*}$$However, we can realize that $M = \lambda L$, so we can put this in a slightly nicer form: $$I = \boxed{\frac {ML^2}{3}},$$ and we have our answer. $\blacksquare$

And that's the idea behind computing continuous moments of inertia. We finish our lesson with the most important theorem to compute general moments of inertia. We will first state the theorem.

Suppose an object of mass $m$ is made to rotate about an axis $z$ passing through the body's center of mass. The object has a moment of inertia $I_{\text{cm}}$ with respect to the axis $z$. If the object is made to rotate about a new axis $z'$ which is parallel to $z$ and is of a distance $d$ away from $z$, then the moment of inertia $I$ with respect to the new axis is related to $I_{\text{cm}}$ by $$I = I_{\text{cm}} + md^2.$$

That was a pretty important theorem. Let's allot it a name. We actually don't have to $-$ it already has a common name.

Indeed, this is called the parallel axis theorem. We will let you absorb that and leave you with the idea that the integral definition of moment of inertia and the parallel axis theorem can support you through plenty geometries for which you might want the moment of inertia about particular axes. We'll leave you with two exercises with solutions. Once you are done, we highly recommend you have some fun with trying to verify the results on Wikipedia. Don't quite peek yet; finish the below exercises first!

Solution. We can use some integration again but there's an easier way. We already know the moment of inertia if the axis is at the end of the rod. Call this $I'$. Let $I$ be the moment of inertia if the rod crosses the center. Clearly, from the parallel axis theorem, we can set $d=\frac{L}{2}$ and we get $$I' = I + M \cdot \left( \frac {L}{2} \right)^2,$$ and we know that $I' = \frac {ML^2}{3}$, so $$I = \frac {ML^2}{3} - \frac {ML^2}{4} = \boxed {\frac {ML^2}{12}},$$ which is our answer. $\blacksquare$

Wasn't that relatively easy? Moment of inertia is a simple concept which can get really nice. Study more physics on Expii to unravel some more interesting concepts in mechanics.

This explanation will go through a proof of the Parallel Axis Theorem, likely the most important theorem used to calculate moments of inertia. The theorem states the following:

Let $I$ denote the moment of inertia of an object about an axis passing through point $P.$ Let $I_{CM}$ denote the moment of inertia of the same object about an axis passing through the center of mass of the object and parallel to the original axis. Then: $$I = I_{CM} + Md^2$$ where $M$ is the mass of the object and $d$ is the distance between $P$ and the center of mass of the object.

To prove this, divide the object into small mass elements $m_i$. For each such element, let $\vec{r_i}$ denote the position vector relative to the original axis, and let $\vec{u_i}$ denote the position vector relative to the axis through the center of mass. Let the scalars $r_i$ and $u_i$ denote the magnitudes of these vectors, respectively. Let $\vec{r_{CM}}$ denote the position vector of the center of mass relative to the original axis.

The following are easy to see: $$\left|\vec{r_{CM}}\right| = d$$ $$\vec{r_i} = \vec{r_{CM}} + \vec{u_i}$$

Computing the moment of inertia and doing some algebra, $$I = \sum m_i r^2_i$$ $$I = \sum m_i(\vec{r_i}\cdot\vec{r_i})$$ $$I = \sum m_i(\vec{r_{CM}}+\vec{u_i})\cdot(\vec{r_{CM}}+\vec{u_i})$$ $$I = \sum m_i(\vec{r_{CM}}\cdot\vec{r_{CM}}) + \sum 2m_i(\vec{r_{CM}}\cdot\vec{u_i}) + \sum m_i(\vec{u_i}\cdot\vec{u_i})$$ $$I = (\sum m_i)d^2 + 2\vec{r_{CM}}\cdot\left(\sum m_i\vec{u_i}\right) + \sum m_i u^2_i$$ By the definition of the center of mass of an object, $$\sum m_i\vec{u_i} = \vec{0}$$ Also obvious is the following: $$\sum m_i = M$$ Finally, by the definition of moment of inertia, $$\sum m_i u^2_i = I_{CM}$$ From the above equations, it follows that: $$I = (Md^2) + 0 + I_{CM}$$ $$I = I_{CM} + Md^2$$ This completes the proof. The interesting thing is that this proof can be imitated in proving many results that take a similar form; for instance, one can relate the kinetic energy of a system in the center-of-mass reference frame to the kinetic energy in any other inertial reference frame in a very similar manner.

Compute the moment of inertia of a thin disk of uniform density, with axis of rotation through the center, perpendicular to the plane that the disk lies in. Express your answer in terms of the mass, $M$, and the radius, $R$.

This will require a double integral. Let $\rho$ be the surface density. Then $\rho = \frac{m}{A}$, so $m = \rho A$. Therefore, $dm = \rho dA$. So now compute $$\int_{D}r^2dm = \int_{D} r^2\rho dA=\rho\int_Dr^2dA,$$where $D$ is the (2-dimensional) region of the disk. One way to do this is with polar coordinates.

The computation is easier with polar coordinates, though Cartesian ones will work, too. $$\rho\int_Dr^2dA = \rho\int_{\theta=0}^{2\pi}\int_{r=0}^{R}r^2 \left(\frac{1}{2}rdrd\theta\right)= \rho\left(\frac{R^4}{4}\right)(2\pi) = \frac{1}{2}(\rho\pi R^2)R^2 = \frac{1}{2}MR^2.$$

Correct! Using the Parallel Axis Theorem, we can see that $I=MR^2/2+M(R/2)^2=3MR^2/8$

If you don't know how to compute a double integral like this one, it's okay! The Calculus section on Expii can help you with that.

A uniform annulus with mass $M$ has an inner radius of $R_1$ and an outer radius of $R_2.$ Compute the moment of inertia of the annulus about its axis of symmetry.

When you get your answer, check the following two cases: $$R_1 = R_2 = R$$ $$R_1 = 0, R_2 = R$$ What does the annulus look like in these cases? Does your answer agree with what you already know about these special cases?

Hint: Define the density (in this case, mass per unit area) as follows: $$\sigma = \frac{M}{\pi\left(R^2_2 - R^2_1\right)}.$$

Answer: $I = \frac{1}{2}M\left(R^2_1 + R^2_2\right)$

Solution: Define the density (in this case, mass per unit area) as follows: $$\sigma = \frac{M}{\pi\left(R^2_2 - R^2_1\right)}.$$

Consider a thin ring of radius $r$ and thickness $dr.$ Its mass is $dm = \sigma 2\pi r dr.$ Integrating, we get: $$I = \int r^2 dm$$ $$I = \int_{R_1}^{R_2} \sigma 2\pi r^3 dr$$ $$I = 2\pi\sigma \left(\frac{1}{4}R^4_2 - \frac{1}{4}R^4_1\right).$$ Rewriting $\sigma$ in terms of our original quantities, $$I = 2\pi \frac{M}{\pi\left(R^2_2 - R^2_1\right)} \frac{1}{4} \left(R^4_2 - R^4_1\right).$$ Simplifying, we get out answer: $$I = \frac{1}{2} M \left(R^2_1 + R^2_2\right).$$

To check our answer, we can use two special cases:

If we set $R_1 = R_2 = R,$ then all of the mass is at the edge, so we simply have a simple ring of radius $R.$ We see that the moment of inertia is indeed $I = MR^2.$

If we set $R_1 = 0$ and $R_2 = R,$ then we just have a uniform disk of radius $R.$ We see that the moment of inertia is indeed $I = \frac{1}{2}MR^2.$

Compute the moment of inertia $I$ of a hollow uniform sphere of mass $M$ and radius $R$ about an axis through the center of the sphere.

Hint (Method 1): Divide the sphere into uniform rings (a uniform ring has a very simple moment of inertia) and integrate.

Hint (Method 2): Recall that a uniform ring of mass $M$ and radius $R$ has moment of inertia $MR^2,$ since every small mass element is a distance $R$ from the axis of rotation.

For the hollow sphere, the mass elements are not all equidistant from the axis of rotation, but they are still all equidistant from a fixed point (namely, the center). Can you find a clever way to exploit this?

Answer: $\frac{2}{3}MR^2$

Solution (Method 1): Center the sphere at the origin of a $3$-dimensional Cartesian coordinate space, and let the desired axis be the $x$-axis. Split the sphere into thin rings.

For a particular value of $x,$ the radius of the sphere is $r = \sqrt{R^2-x^2}.$

To find $dm$ for a ring, take the mass density $\sigma = \frac{M}{4\pi R^2}$ and multiply it by the surface area of the ring: $$dm = \sigma 2\pi r \sqrt{dx^2 + dr^2}.$$ Since $r = \sqrt{R^2-x^2},$ we have: $$dr = -\frac{x}{\sqrt{R^2-x^2}}.$$ It follows that: $$dm = \sigma 2\pi\sqrt{R^2-x^2} \frac{R}{\sqrt{R^2-x^2}} dx$$ $$dm = 2\pi\sigma R dx.$$ Finally, we integrate: $$I = \int r^2 dm$$ $$I = 2\pi\sigma R\int_{-R}^{R} \left(R^2 - x^2\right) dx$$ $$I = 2\pi \frac{M}{4\pi R^2} R \times \frac{4}{3}R^3$$ $$I = \frac{2}{3}MR^2.$$

Solution (Method 2): Center the sphere at the origin of a $3$-dimensional Cartesian coordinate space, and let the desired axis be the $z$-axis. Then, the moment of inertia is: $$I = \int r^2 dm = \int \left(x^2 + y^2\right) dm$$ By the symmetry of the sphere, $$\int x^2 dm = \int y^2 dm = \int z^2 dm.$$ Therefore, $$\int \left(x^2 + y^2\right) dm = \frac{2}{3} \int \left(x^2 + y^2 + z^2\right) dm.$$ Notice that $x^2 + y^2 + z^2 = R^2$ for all points on the hollow sphere. Thus, $$I = \frac{2}{3}\int R^2 dm$$ $$I = \frac{2}{3} MR^2.$$