From a sum to an integral
A solid rod is not a handful of beads. Its mass is spread along its whole length. To use \(I = \sum m_i r_i^2\), cut the body into very small pieces, each a mass element \(dm\). Each element is so small that all of its mass lies at one distance \(r\) from the axis. It then adds \(r^2\,dm\) to the moment of inertia, and the sum of these pieces becomes an integral.
The whole art lies in choosing the element. For a thin rod the element is a short piece of length \(dx\). Its mass is \(dm = \lambda\,dx\), where \(\lambda = M/L\) is the linear mass density, the mass per metre of a uniform rod of mass \(M\) and length \(L\). Every point of the short piece is at the same distance from an axis across the rod.
\[ dm = \lambda\,dx = \dfrac{M}{L}\,dx \]
Why. the element at distance \(x\) from the axis has length \(dx\), and the rod is uniform.
\[ dI = x^2\,dm = \dfrac{M}{L}\,x^2\,dx \]
Why. all of the element lies at distance \(x\) from an axis through the end, perpendicular to the rod.
\[ I = \dfrac{M}{L}\int_0^L x^2\,dx = \dfrac{M}{L}\cdot\dfrac{L^3}{3} \]
Why. add the elements from the end, \(x = 0\), to the far end, \(x = L\).
\[ I_{\text{end}} = \tfrac13 ML^2 \]
Read it. simplify. One factor of \(L\) cancels.
About the centre the working is the same, and only the limits change. The rod now runs from \(x = -L/2\) to \(x = +L/2\), so \(I = \dfrac{M}{L}\left[\dfrac{x^3}{3}\right]_{-L/2}^{L/2} = \dfrac{M}{L}\cdot\dfrac{L^3}{12} = \dfrac{ML^2}{12}\). This is a quarter of the value about the end. No element of the rod is now farther than \(L/2\) from the axis.
Fig 4.1 shows why the ratio is four. About the end, the far elements sit up to \(1.2\,\text{m}\) away, and the curve of \(r^2\) climbs steeply there. About the centre, no element is more than \(0.6\,\text{m}\) away, and the curve stays low at both ends. The shaded area, and so \(I\), falls to a quarter.
A uniform rod of mass \(3\,\text{kg}\) and length \(2\,\text{m}\) turns about an axis through one end, perpendicular to the rod. Find its moment of inertia, in kg m².
The ring and the disc
A thin ring of mass \(M\) and radius \(R\), about the axis through its centre perpendicular to its plane, needs no integral. Every particle of the ring lies at the same distance \(R\) from the axis. So \(I = \sum m_i R^2 = MR^2\). A thin-walled pipe about its long axis is a stack of such rings, and it has the same \(I = MR^2\).
A uniform disc is a nest of rings. Take the ring between radius \(r\) and \(r + dr\). It is a thin strip of length \(2\pi r\) and width \(dr\), so its area is \(2\pi r\,dr\). The disc has a surface mass density \(\sigma = M/\pi R^2\), its mass per square metre. So the ring's mass is \(dm = \sigma\,2\pi r\,dr\), and every bit of it lies at the same distance \(r\).
\[ dm = \sigma\,2\pi r\,dr, \qquad \sigma = \dfrac{M}{\pi R^2} \]
Why. the thin ring has area \(2\pi r\,dr\), and the disc has mass \(\sigma\) per unit area.
\[ dI = r^2\,dm = \dfrac{2M}{R^2}\,r^3\,dr \]
Why. the whole ring lies at distance \(r\); \(\sigma \times 2\pi = 2M/R^2\).
\[ I = \dfrac{2M}{R^2}\int_0^R r^3\,dr = \dfrac{2M}{R^2}\cdot\dfrac{R^4}{4} \]
Why. add the rings from the centre to the rim.
\[ I_{\text{disc}} = \tfrac12 MR^2 \]
Read it. simplify. Half the value of a ring of the same mass and radius.
Fig 4.2 shows why the disc has half the \(I\) of a ring. The mass inside radius \(r\) grows as \(r^2\), but the moment of inertia inside it grows as \(r^4\). So the inner half of the radius holds \(25\,\%\) of the mass and only about \(6\,\%\) of \(I\). The outer rim does most of the work. A ring puts all its mass there, and a disc puts only part of it.
To find the moment of inertia of a disc about its central perpendicular axis, we cut it into thin rings, not into thin straight strips. Why?
A uniform disc of mass \(M\) and radius \(R\) is cut along a diameter, and one half is thrown away. What is the moment of inertia of the remaining half about the original central perpendicular axis?
Cylinders
A solid cylinder about its long axis is a stack of discs. Each disc of mass \(dm\) adds \(\tfrac12\,dm\,R^2\), and adding them gives \(I = \tfrac12 MR^2\). The length of the cylinder never enters. A coin and a long rod of the same mass and radius have the same \(I\) about their long axis. Stacking discs moves no mass away from the axis.
A solid wooden log of mass \(20\,\text{kg}\) and radius \(0.15\,\text{m}\) turns about its long axis. Find its moment of inertia, in kg m².
The solid sphere and the shell
A solid sphere about a diameter is cut into thin discs across that diameter. The slice at height \(x\) above the centre is a disc of radius \(\sqrt{R^2 - x^2}\) and thickness \(dx\). The disc result gives each slice \(dI = \tfrac12\,dm\,(R^2 - x^2)\). Adding the slices from \(x = -R\) to \(x = R\) is a longer integral, done below.
\[ dm = \rho\,\pi(R^2 - x^2)\,dx, \qquad \rho = \dfrac{3M}{4\pi R^3} \]
Why. the slice is a disc of area \(\pi(R^2 - x^2)\) and thickness \(dx\); \(\rho\) is the mass per unit volume.
\[ dI = \tfrac12\,dm\,(R^2 - x^2) = \tfrac12\rho\pi(R^2 - x^2)^2\,dx \]
Why. a disc about its own central axis has \(\tfrac12\) times its mass times its radius squared.
\[ I = \tfrac12\rho\pi\int_{-R}^{R}(R^2 - x^2)^2\,dx = \tfrac12\rho\pi\cdot\tfrac{16}{15}R^5 \]
Why. expand \((R^2 - x^2)^2 = R^4 - 2R^2x^2 + x^4\) and integrate term by term: \(2R^5 - \tfrac43 R^5 + \tfrac25 R^5 = \tfrac{16}{15}R^5\).
\[ I_{\text{sphere}} = \tfrac25 MR^2 \]
Read it. substitute \(\rho = 3M/4\pi R^3\): \(\tfrac12 \cdot \tfrac{3M}{4\pi R^3} \cdot \pi \cdot \tfrac{16}{15}R^5 = \tfrac25 MR^2\).
A thin spherical shell, such as a table-tennis ball, keeps all its mass at the radius \(R\) from the centre. But not all of it is at \(R\) from a diameter: the parts near the poles lie close to the axis. So its \(I\) lies between the ring's \(MR^2\) and the solid sphere's \(\tfrac25 MR^2\). The result is \(\tfrac23 MR^2\).
For the shell, cut it into rings round the axis. The ring at polar angle \(\theta\) has radius \(R\sin\theta\) and mass \(dm = \tfrac12 M\sin\theta\,d\theta\), since the area of a band grows as \(\sin\theta\). Then \(I = \int_0^\pi R^2\sin^2\theta \cdot \tfrac12 M\sin\theta\,d\theta = \tfrac12 MR^2 \cdot \tfrac43 = \tfrac23 MR^2\).
A cricket ball has a mass of \(0.16\,\text{kg}\) and a radius of \(3.6\,\text{cm}\). Treat it as a uniform solid sphere. Find its moment of inertia about a diameter.
\(I = \tfrac25 MR^2 = \tfrac25 \times 0.16 \times 0.036^2\).
\(0.036^2 = 1.296 \times 10^{-3}\), so \(I = 0.4 \times 0.16 \times 1.296 \times 10^{-3} \approx 8.3 \times 10^{-5}\,\text{kg m}^2\).
The number is tiny because the radius is small and enters squared. That is why a spinner can give the ball many turns a second with a flick of the fingers.
The table of standard bodies
The results above, and two that the last lesson of this chapter will prove, make a small table. Every entry assumes a uniform body. Each is about an axis through the centre of mass except the rod about its end.
Four bodies have the same mass and the same radius. Rank them by moment of inertia about an axis through the centre, largest first. Take the perpendicular axis for the ring and the disc, and a diameter for the spheres.
A solid steel cylinder is made twice as long with the same mass, so it becomes thinner. Does its moment of inertia about its long axis increase, decrease or stay the same? Explain.
About its long axis a solid cylinder has \(I = \tfrac12 MR^2\); its length never enters, because stacking discs moves no mass away from the axis. Here the mass stays the same but the radius becomes smaller, since the volume must be spread over twice the length. So \(R^2\) halves, and \(I\) decreases to half.
Treat the Earth as a uniform solid sphere of mass \(6 \times 10^{24}\,\text{kg}\) and radius \(6400\,\text{km}\). Estimate its moment of inertia about its axis of spin, in units of \(10^{37}\,\text{kg m}^2\).
A cricket bat is about \(0.85\,\text{m}\) long but only \(11\,\text{cm}\) wide. Swung about the shoulder, its mass lies far from the axis, and it feels heavy. Twirled in the fingers about its long axis, the same mass lies within a few centimetres of the axis. The table explains the difference: a long rod has a large \(I\) across its length and a tiny one along it.
- Choosing an element whose points lie at different distances from the axis.Choose the element with all its mass at one distance \(r\). Use a short piece of a rod, a thin ring of a disc, a thin disc of a sphere.
- Writing \(MR^2\) for a disc or \(\tfrac12 MR^2\) for a ring.A ring has all its mass at \(R\), so \(MR^2\). A disc spreads its mass inward, so \(\tfrac12 MR^2\).
- Mixing up the solid sphere and the shell.Solid sphere \(\tfrac25 MR^2\), thin shell \(\tfrac23 MR^2\). The shell is larger because all its mass is at \(R\) from the centre.
- Letting the length of a cylinder into its \(I\) about the long axis.About the long axis only the radius matters: \(\tfrac12 MR^2\) solid, \(MR^2\) thin-walled.