Moment of Inertia

Why a hammer is hard to spin one way and easy the other

Hold a hammer by the end of its handle and swing it around in a circle. Now hold it by the head and swing it the same way. The second is noticeably easier. You did not change the hammer, and you did not change its mass. All you changed was where the heavy part sits relative to the axis you are turning about.

In the last lesson you worked out what makes something start to turn: torque, $\tau = rF\sin\theta$. You balanced torques so that nothing moved. Now you want to know what happens when they do not balance, and for that you need the rotational answer to a different question. In straight lines, mass tells you how hard it is to speed something up. In rotation, mass is not enough. You need a number that knows about mass and about where the mass is. That number is the moment of inertia, and this whole lesson is about computing it.

Where the formula comes from

Take a rigid body spinning about a fixed axis at angular velocity $\omega$ (radians per second). Chop it into tiny pieces. Call the mass of one piece $m_i$ and its perpendicular distance from the axis $r_i$. Because the body is rigid, every piece goes around at the same $\omega$, so piece $i$ moves at speed $v_i = \omega r_i$: the pieces far out move fast, the pieces near the axis crawl.

Add up the kinetic energy:

Compare that with $K = \tfrac12 m v^2$. The $\omega$ sits exactly where $v$ used to, and the bracket sits exactly where $m$ used to. So define

$I = \sum_i m_i r_i^2, \qquad K_{\text{rot}} = \tfrac12 I \omega^2.$

$I$ is the moment of inertia. Read the definition in words: every gram of the body contributes its own mass times the square of its distance from the axis. The square is the entire story. Mass twice as far out counts four times as much. That is why a flywheel is built with its metal in the rim, why a tightrope walker carries a long pole, and why the hammer swung by its handle fights you.

Two warnings hide in the definition. First, $r_i$ is the distance from the axis, measured perpendicular to it, not the distance from the centre of the object. Second, $I$ is not a property of a body on its own. It is a property of a body and an axis. Change the axis and the number changes. Always write down which axis you mean before you compute anything.

For a solid body the sum becomes an integral, $I = \int r^2 \, dm$. You will only need a few of these, and the standard results all come out of one line of calculus. A thin rod of mass $M$ and length $L$, axis through the centre, perpendicular to the rod: the mass per unit length is $\lambda = M/L$, so $dm = \lambda\,dx$, and $r = |x|$ with $x$ running from $-L/2$ to $L/2$.

A uniform disk of mass $M$ and radius $R$ about its central axis: slice it into thin rings. A ring at radius $r$ of thickness $dr$ has area $2\pi r\,dr$, so with surface density $\sigma = M/\pi R^2$ its mass is $dm = \sigma 2\pi r\,dr$, and every bit of it sits at the same distance $r$:

These are the two you should be able to rebuild from scratch. The rest are worth knowing by heart.

The short table you actually need

All of these are about an axis through the centre of mass, and all have the form $I = kMR^2$ or $kML^2$:

• Point mass at distance $r$: $I = mr^2$. Everything else is built from this.
• Thin hoop or hollow cylinder, axis through the centre: $I = MR^2$ $(k = 1)$.
• Uniform disk or solid cylinder, central axis: $I = \tfrac12 MR^2$.
• Annulus (a washer) with inner radius $R_1$, outer $R_2$: $I = \tfrac12 M(R_1^2 + R_2^2)$.
• Solid sphere about a diameter: $I = \tfrac25 MR^2$.
• Thin spherical shell about a diameter: $I = \tfrac23 MR^2$.
• Solid cone about its symmetry axis: $I = \tfrac{3}{10}MR^2$.
• Thin rod about its centre, perpendicular: $I = \tfrac{1}{12}ML^2$.
• Thin rod about one end, perpendicular: $I = \tfrac13 ML^2$.

Look at the coefficients as a story about where mass sits. Hoop $1$: all the mass is at the rim. Disk $0.5$: mass spread evenly over the area, so most of it is well inside $R$. Shell $2/3$, solid sphere $2/5$: the same contrast for spheres, and the solid sphere is lowest of all because a sphere's mass crowds towards the middle, near the axis. If you forget a number, you can usually bracket it by asking how much of the body lives far from the axis.

One more quantity lives here. If you want a single length that summarises how far out the mass effectively sits, define the radius of gyration $k$ by $I = Mk^2$, so $k = \sqrt{I/M}$. It is the radius at which you would have to put all the mass, as a single hoop, to get the same $I$. For a body of mass $8.0\ \mathrm{kg}$ with $I = 0.50\ \mathrm{kg\,m^2}$, $k = \sqrt{0.50/8.0} = \sqrt{0.0625} = 0.25\ \mathrm{m}$.

Moments of inertia add, and they subtract

Because $I$ is a sum over mass, two bodies rotating about the same axis have a combined moment of inertia equal to the sum of the separate ones. A disk with a lump of clay on its rim, about the central axis:

$I = \tfrac12 M_d R^2 + m_{\text{clay}} R^2.$

With $M_d = 4.0\ \mathrm{kg}$, $R = 0.30\ \mathrm{m}$ and $m = 1.0\ \mathrm{kg}$: $I = \tfrac12(4.0)(0.09) + (1.0)(0.09) = 0.18 + 0.09 = 0.27\ \mathrm{kg\,m^2}$.

The same logic run backwards handles holes. A disk with a piece drilled out is (full disk) minus (the piece that used to be there), so

$I_{\text{with hole}} = I_{\text{full}} - I_{\text{removed piece}},$

both taken about the same axis. This trick is worth more than any table entry, because it turns an impossible-looking shape into two easy ones.

Worked example 1: a disk with a central hole

A uniform disk of radius $0.40\ \mathrm{m}$ and surface density $\sigma = 20\ \mathrm{kg/m^2}$ has a hole of radius $0.20\ \mathrm{m}$ cut out, centred on the axis. Find $I$ about that axis.

You are given density, not mass, which is the hint that you should compute both masses yourself. Full disk:

$I_{\text{full}} = \tfrac12 (10.053)(0.16) = 0.8042\ \mathrm{kg\,m^2}.$

Removed disk, radius $0.20\ \mathrm{m}$:

Subtract: $I = 0.8042 - 0.0503 = 0.754\ \mathrm{kg\,m^2}$.

Sanity check it against the annulus formula, $I = \tfrac12 M(R_1^2 + R_2^2)$ with $M = 10.053 - 2.513 = 7.540\ \mathrm{kg}$: $\tfrac12 (7.540)(0.04 + 0.16) = \tfrac12(7.540)(0.20) = 0.754$. Same number. The annulus formula is not something extra to memorise; it is this subtraction, done once in symbols.

Moving the axis: the parallel axis theorem

Everything so far was about an axis through the centre of mass. Real problems pivot things at their ends, their rims, a third of the way along. You do not need a new integral for each one.

Put the origin at the centre of mass and let the new axis be parallel to the old one, a distance $d$ away. For piece $i$, its position vector from the centre of mass is $\vec{r}_i$, and its distance from the new axis is $|\vec{r}_i - \vec{d}\,|$. Then

The first term is $I_{\text{cm}}$. The last is $Md^2$. The middle term contains $\sum m_i \vec{r}_i$, which is $M$ times the position of the centre of mass: measured from the centre of mass, so it is zero. That leaves

$I = I_{\text{cm}} + Md^2.$

In words: shifting the axis away from the centre of mass always increases $I$, by exactly the amount a point mass $M$ sitting at the centre of mass would contribute. Check it on the rod: about the end, $d = L/2$, so $I = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{12}ML^2 + \tfrac14 ML^2 = \tfrac13 ML^2$, which is the table entry you were told to memorise.

Two consequences fall straight out. The centre of mass axis gives the smallest possible $I$ among all parallel axes, since $Md^2 \ge 0$. And the growth is quadratic in $d$, so moving the pivot out is punishing fast.

Worked example 2: a rod with masses on its ends

A uniform rod of mass $2.0\ \mathrm{kg}$ and length $1.2\ \mathrm{m}$ carries $0.50\ \mathrm{kg}$ point masses at each end, and rotates about its centre. Find $I$.

Write down the pieces before touching a calculator. Three bodies, one axis, so three terms added.

Rod about its centre: $\tfrac{1}{12}(2.0)(1.2)^2 = \tfrac{1}{12}(2.0)(1.44) = 0.24\ \mathrm{kg\,m^2}$.

Each end mass is a point at $r = 0.60\ \mathrm{m}$: $mr^2 = (0.50)(0.36) = 0.18\ \mathrm{kg\,m^2}$, and there are two of them, so $0.36\ \mathrm{kg\,m^2}$.

Total: $I = 0.24 + 0.36 = 0.60\ \mathrm{kg\,m^2}$. Notice that the two little masses, together only half the rod's mass, contribute more than the rod does. The $r^2$ again.

Worked example 3: an off-centre hole

A uniform disk of mass $9.0\ \mathrm{kg}$ and radius $0.30\ \mathrm{m}$ has a hole of radius $0.10\ \mathrm{m}$ cut out, centred $0.15\ \mathrm{m}$ from the axis. Find $I$ about the original central axis.

This one needs both tools at once, which is why it looks hard and is not.

Full disk: $I_{\text{full}} = \tfrac12 (9.0)(0.30)^2 = \tfrac12(9.0)(0.09) = 0.405\ \mathrm{kg\,m^2}$.

The removed piece: same material, so its mass is the full mass scaled by area ratio,

Its moment of inertia about its own centre is $\tfrac12 (1.0)(0.01) = 0.005\ \mathrm{kg\,m^2}$, but you need it about the disk's central axis, $0.15\ \mathrm{m}$ away. Parallel axis:

Subtract: $I = 0.405 - 0.0275 = 0.3775\ \mathrm{kg\,m^2}$.

Worked example 4: a rod whose density is not uniform

A rod runs from $x = -1.0\ \mathrm{m}$ to $x = +1.0\ \mathrm{m}$, with mass per unit length $\lambda = \lambda_0(1 + x^2/L^2)$, where $\lambda_0 = 3.0\ \mathrm{kg/m}$ and $L = 1.0\ \mathrm{m}$. Find $I$ about the centre.

No table entry helps here, so go back to the definition. Take a slice of width $dx$ at position $x$. Its mass is $dm = \lambda\,dx$ and its distance from the axis is $|x|$, so $r^2\,dm = x^2\lambda\,dx$:

Both integrands are even, so integrate from $0$ to $1$ and double:

So $I = 3.0\left(\tfrac23 + \tfrac25\right) = 3.0 \times \tfrac{16}{15} = 3.2\ \mathrm{kg\,m^2}$.

Worth a sanity check. The total mass is $\int \lambda\,dx = 3.0(2 + \tfrac23) = 8.0\ \mathrm{kg}$, so the radius of gyration is $\sqrt{3.2/8.0} = 0.63\ \mathrm{m}$. A uniform rod of the same length would have $k = L_{\text{tot}}/\sqrt{12} = 2.0/3.46 = 0.58\ \mathrm{m}$. Slightly larger, as it should be: this rod is heavier at its ends.

Worked example 5: where to pivot for the fastest swing

A uniform rod of length $L = 1.0\ \mathrm{m}$ hangs as a pendulum from a pivot a distance $d$ from its centre. The period of small swings is $T = 2\pi\sqrt{I/(Mgd)}$, where $d$ is the pivot-to-centre-of-mass distance. Which $d$ makes $T$ smallest?

First put in the parallel axis theorem, $I = \tfrac{1}{12}ML^2 + Md^2$:

The mass cancelled, which already tells you something. $T$ is smallest when the bracket $f(d) = L^2/(12d) + d$ is smallest. Differentiate and set to zero:

That is the tug-of-war Figure 2 shows. Pivot too close to the centre and gravity has almost no lever arm, so the swing is slow. Pivot at the end and $I$ has grown too large. The best point is neither, and it lands where the two terms in $f(d)$ are equal.

For the record, the plain end-pivot case: $d = 0.50\ \mathrm{m}$, $I = \tfrac13(M)(1.0)^2$, so $T = 2\pi\sqrt{\tfrac13/(9.8 \times 0.50)} = 2\pi\sqrt{0.0680} = 2\pi(0.2608) = 1.64\ \mathrm{s}$.

How to recognise these problems

You are looking at a moment-of-inertia problem when the question involves something rigid turning about an axis and you need a number to plug into $K = \tfrac12 I\omega^2$, $\tau = I\alpha$, $L = I\omega$, or a pendulum period. Once you know that, the shape of the body tells you which tool:

• One named standard shape, axis through the centre. Read off the table. Nothing to do but square the radius.
• Several bodies on one axis. Add. Clay on a rim, masses on a rod, a disk on a shaft: compute each $I$ about the shared axis and sum them.
• A hole, a notch, a missing quarter. Subtract. Compute the whole body, then the removed piece, and take the difference. If the removed piece is off centre, you will need the parallel axis theorem for it.
• The axis is not through the centre of mass. Parallel axis theorem, always starting from $I_{\text{cm}}$. Phrases like "about its end", "about its rim", "a third of the way along" are the giveaway.
• A density that depends on position, or a shape with no table entry. Go to $I = \int r^2\,dm$. Write $dm$ in terms of $dx$ or $dr$ first; that step is the whole problem.
• "Smallest", "fastest", "optimum". Write $I$ as a function of the free variable using the parallel axis theorem, then differentiate.

Two habits will save you most of the marks. Write the axis down before anything else, because $I$ is meaningless without it. And check the coefficient at the end: a sensible answer for a body of size $R$ looks like $kMR^2$ with $k$ under one, unless the axis has been shifted away from the centre of mass.

With $I$ in hand, the rotational version of Newton's second law is waiting. Torque plays the part of force, $I$ plays the part of mass, and angular acceleration plays the part of $a$. That is the next lesson: Rotational Dynamics.