Rotational Inertia

Rotational Inertia Definition

We'll start by defining inertia.

We usually measure inertia with mass, which makes sense; you already have a conceptual understanding of inertia because you know that heavier things are harder to move. For example, a boulder shows more resistance to motion than a piece of paper does. But what happens if the object is not moving on a line but instead it is spinning? Then, we need to talk about rotational inertia.

Mass is how we "measure" inertia in a sense. But experience tells us that spinning on a chair can be easier or harder depending on how we position ourselves on the chair. Therefore, rotational inertia is related to the mass and where that mass distributes relatively to the axis of rotation.

Also, even though we referred to an object above, a better term is a rigid system.

For example, you could push a piece of jello, and it can all stay connected, but it may be bent out of place at some spots; this is not a rigid system. Whereas someone could push a makeshift 3rd-grade solar system model at a planet such as Jupiter, and all it would do is spin: its shape would remain unchanged, the planets would all still align around the sun, and it would have only spun a little bit.

Rotational Inertia Formulas

We express rotational inertia mathematically by taking into account the mass and how that mass distributes around the axis of rotation for a single particle:

$$I=mr^2$$

where \(I\) is the rotational inertia, \(m\) is the mass, and \(r\) is the distance away from the axis that the object is perpendicularly rotating to.

Fig. 2 - This image shows the top and vertical view of the parameters of the rotational inertia formula. Notice how \(r\) is the distance from the axis of rotation.

Rotational Inertia Summation

The total rotational inertia of a rigid system is found by adding up all of the individual rotational inertias of the particles forming the system; the mathematical expression

$$I_\text{tot} = \sum I_i = \sum m_i r_i ^2,$$

conveys this concept where \(I_\text{tot}\) is the total rotational inertia, \(I_i\) is each value for the rotational inertia of each object, and \(m_i\) and \(r_i\) are each value for the mass and the distance from the axis of rotation for each object.

Rotational Inertia of a Solid

By implementing integrals, we can calculate the rotational inertia of a solid composed of many different differential masses \(\mathrm{d}m\).

$$I=\int r^2 \mathrm{d}m$$

is the equation we can use, with \(\mathrm{d}m\) as each little bit of mass and \(r\) as the perpendicular distance from each \(\mathrm{d}m\) to the axis upon which the solid is rotating.

Rotational Inertia and Rigid Systems

As the mass gets closer to the axis of rotation, our radius \(r\) gets smaller, drastically decreasing the rotational inertia because \(r\) is squared in our formula. This means that a hoop with the same mass and size as a cylinder would have more rotational inertia because more of its mass is situated farther away from the axis of rotation or center of mass.

One of the key concepts that you need to learn about rotational inertia is that a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass. And if we know the moment of inertia with respect to the axis going through the center of mass, we can find the moment of inertia with respect to any other axis parallel to it by using the following result.

Let's see an example.

Rotational Inertia Examples

Okay, we've done a lot of talking and explaining but little application, and we know that you need a lot of application in physics. So, let's do some examples.

Example 1

First, we'll do an example using the formula

$$I=mr^2\mathrm{.}$$

How difficult would it be to rotate a \(5.00\,\mathrm{kg}\) tether ball that is attached by a \(0.50\,\mathrm{m}\) rope to a center pole? (Assume the rope is massless).

Find the rotational inertia of the tether ball to see how hard it would be to move.

Fig. 4 - We can find the rotational inertia of the ball at the end of a tether ball rope.

Recall our rotation inertia equation,

$$I=mr^2\mathrm{,}$$

and use it to plug in the values

$$m=5.00\,\mathrm{kg}$$

and

$$\begin{align*} r &= 0.50\,\mathrm{m}\mathrm{:} \\ I &= 5.00\,\mathrm{kg}(0.50\,\mathrm{m})^2 \\ \end{align*}$$

giving us an answer of

$$I=1.25\,\mathrm{kg\,m^2.}$$

Therefore, the ball would be \(1.25\,\mathrm{kg\,m^2}\) difficult to rotate. That may be weird for you to hear because we never talk about things being difficult to move with that kind of unit. But, in reality, that is how rotational inertia and mass work. They both give us a gauge of how much something resists motion. Therefore, it is not inaccurate to say that a boulder is \(500\,\mathrm{kg}\) difficult to move or that a tether ball is \(1.25\,\mathrm{kg\,m^2}\) difficult to rotate.

Example 2

Now, let's use our knowledge of rotational inertia and summations to solve the next problem.

Rotational Inertia of a Disk

We can calculate the rotational inertia of a disk by using our normal rotational inertia equation but with a \(\frac{1}{2}\\\) in front.

$$I_\text{disk}=\frac{1}{2}\\mr^2.$$

Applications of Rotational Inertia

How do all of our formulas tie together? How can we use our knowledge to actually prove something? The following deep dive has a derivation that will answer these questions. It is probably beyond the scope of your AP Physics C: Mechanics course.

One can derive the formula for the rotational inertia of a disk by implementing integrals. Recall the equation

$$I=\int r^2 \mathrm{d}m\mathrm{,}$$

which describes the rotational inertia of a solid composed of many different tiny elements of mass \(\mathrm{d}m\).

If we treat our disk as many different infinitely thin rings, we can add the rotational inertia of all those rings together to get the total rotational inertia for the disk. Recall that we can add infinitely small elements together using integrals.

Assuming that the mass is evenly distributed, we can find the surface density dividing the mass over the area \(\frac{M}{A}\). Each of our tiny rings would be composed of a length of \(2\pi r\) and a width of \(\mathrm{d}r\), therefore \(\mathrm{d}A = 2\pi r \mathrm{d}r\).

We know that the change in the mass with respect to the surface area \(\frac{\mathrm{d}m}{\mathrm{d}A}\) is \(\frac{M}{A}\) and we also know that \(A=\pi R^2,\) where \(R\) is the radius of the whole disk. We can then use these relations

$$\frac{M}{\textcolor{#00b695}{A}}\\=\frac{\mathrm{d}m}{\textcolor{#56369f}{\mathrm{d}A}}\\$$

$$\frac{M}{\textcolor{#00b695}{\pi R^2}}\\ = \frac{\mathrm{d}m}{\textcolor{#56369f}{2\pi r \mathrm{d}r}}\\$$

isolating \(\mathrm{d}m\):

$$\begin{aligned}\mathrm{d}m &= \frac{2M\pi r \mathrm{d}r}{\pi R^2}\\[8pt] \mathrm{d}m &= \frac{2M r \mathrm{d}r}{ R^2} \end{aligned}$$

Now that we know \(\mathrm{d}m\), we can plug that into our integral equation

$$I=\int r^2 \mathrm{d}m$$

to get

$$I=\int r^2\frac{2M r \mathrm{d}r}{ R^2}\\\mathrm{.}$$

We integrate from \(0\) to \(R\),

$$I=\frac{2M}{R^2}\\ \int_0^R r^3 \mathrm{d}r\mathrm{,}$$

because we want to go from the center of the disk \(r=0\) to the very edge, or the radius of the whole disk \(r=R\). After integrating and evaluating at the corresponding \( r-\text{values} \) we get:

$$I=\frac{2M}{R^2}\\ \frac{R^4}{4}\\ - 0.$$

If we simplify the previous expression, we obtain the equation for the rotational inertia of a disk:

$$I=\frac{1}{2}\\MR^2\mathrm{.}$$

The above derivation shows the usefulness of rotational inertia and its various formulas. Now you are ready to take the world head-on! You are now ready to tackle rotational inertia and thing such as torque and angular motion. If you ever get in an office chair spinning competition, you know how to win, you just need to put your mass closer to the axis of rotation so tuck those arms and legs in!