After a wedding lunch, a boy balances a steel thali on one finger and sets it spinning flat, like a record. His sister holds another thali by two opposite points of its rim and makes it flip over and over, like a tossed coin. Both plates are the same. Which way is easier to spin, and by how much?
Flat bodies and three axes
A lamina is a flat body whose thickness is negligible: a thin plate, a cardboard cut-out, a flat ring of wire. Put it in the x-y plane. Draw two perpendicular axes, x and y, in the plane of the lamina. Draw a third, z, through their meeting point and perpendicular to the lamina.
Every particle of the lamina lies in the plane, at \((x, y, 0)\). Its distance from the x axis is \(|y|\), from the y axis \(|x|\), and from the z axis \(\sqrt{x^2 + y^2}\). The squares of these distances are linked in the simplest possible way, and so are the three moments of inertia.
\[ I_x = \int y^2\,dm, \qquad I_y = \int x^2\,dm \]
Why. the distance of the particle at \((x, y, 0)\) from the x axis is \(|y|\), and from the y axis \(|x|\).
\[ I_z = \int (x^2 + y^2)\,dm \]
Why. its distance from the z axis is \(\sqrt{x^2 + y^2}\); there is no z coordinate, because the body is flat.
\[ I_z = \int x^2\,dm + \int y^2\,dm = I_y + I_x \]
Read it. split the integral. Each part is one of the in-plane moments of inertia.
Take a lamina and two perpendicular axes in its plane. About a third axis, perpendicular to the plane through their meeting point, \(I_z = I_x + I_y\).
The plate in Fig 6.1 checks the theorem against the table. About the x axis each strip of the plate behaves like a rod across it, so \(I_x = Mb^2/12\) with the height \(b = 0.3\,\text{m}\). In the same way \(I_y = Ma^2/12\) with the width \(a = 0.4\,\text{m}\). Their sum, \(M(a^2 + b^2)/12\), is exactly the table's value for a plate about its perpendicular axis. The four masses of lesson 3 obey it too: \(12 = 8 + 4\).
A student applies \(I_z = I_x + I_y\) to a solid sphere, using \(\tfrac25 MR^2\) for each diameter, and finds \(\tfrac45 MR^2\) about a third diameter. The true value is \(\tfrac25 MR^2\). What went wrong?
The diameters of a disc and a ring
The table of lesson 4 promised two entries: a disc and a ring about a diameter. Take x and y along two perpendicular diameters of a uniform disc. By symmetry the disc looks the same about every diameter, so \(I_x = I_y\). The theorem then gives \(I_x + I_y = I_z = \tfrac12 MR^2\). So each diameter has \(I = \tfrac14 MR^2\). The same steps for a ring give \(\tfrac12 MR^2\) about a diameter.
The last two chips in Fig 6.2 need both theorems. A tangent in the plane of the disc is parallel to a diameter, at distance \(R\). So the parallel-axis theorem gives \(I = \tfrac14 MR^2 + MR^2 = \tfrac54 MR^2\). The axis through a rim point perpendicular to the disc is parallel to the central perpendicular axis, again at \(R\): \(I = \tfrac12 MR^2 + MR^2 = \tfrac32 MR^2\).
A thin ring of mass \(0.2\,\text{kg}\) and radius \(0.1\,\text{m}\) hangs from a nail. It can swing in its own plane, or out of it about the tangent at the nail. Find its moment of inertia about each axis.
In its own plane, the ring turns about an axis through the nail perpendicular to the ring. Out of its plane, it turns about the tangent at the nail, which lies in the plane of the ring.
Perpendicular: \(I = MR^2 + MR^2 = 2MR^2\). Tangent in the plane: \(I = \tfrac12 MR^2 + MR^2 = \tfrac32 MR^2\).
\(MR^2 = 0.2 \times 0.01 = 0.002\,\text{kg m}^2\). So \(I = 0.004\,\text{kg m}^2\) in its plane and \(0.003\,\text{kg m}^2\) out of it.
The ring is harder to turn in its own plane, \(0.004\), than about the tangent, \(0.003\,\text{kg m}^2\).
A square plate turns alike about many axes
The theorem also gives a surprise. For a square plate of side \(a\), the table gives \(I_z = \tfrac16 Ma^2\) about the perpendicular axis through the centre. Now take any pair of perpendicular axes in the plane through the centre, at any angle to the sides. The square looks the same after a quarter turn, so the two axes of the pair have equal \(I\). Each is therefore \(\tfrac{1}{12}Ma^2\), whatever the angle: a diagonal has the same \(I\) as a line parallel to a side.
A uniform square plate of mass \(M\) and side \(a\) turns about one of its diagonals. What is its moment of inertia?
A uniform disc has mass \(M\) and radius \(R\). Which of these are correct?
Bodies with holes
Moments of inertia add, so they also subtract. A disc with a hole is what remains when a plug is cut out of a whole disc. Every particle of the whole disc belongs either to what remains or to the plug. So \(I_{\text{whole}} = I_{\text{remaining}} + I_{\text{plug}}\), with every term about the same axis. Rearranged, this gives the moment of inertia of a body with a cavity.
The plug is rarely centred on the axis, so its moment of inertia comes from the parallel-axis theorem. Its mass comes from the same density as the whole body. If a problem gives the mass of the body with the hole, find the whole body's mass first, from the ratio of areas.
A uniform disc of radius \(0.4\,\text{m}\) would have a mass of \(8\,\text{kg}\) if it were whole. A circular hole of radius \(0.2\,\text{m}\) is cut from it, with the centre of the hole \(0.2\,\text{m}\) from the centre O of the disc. Find the moment of inertia of what remains about the axis through O, perpendicular to the disc.
The plug is a disc of radius \(0.2\,\text{m}\), a quarter of the area of the whole disc. So its mass is \(8/4 = 2\,\text{kg}\), and the remaining mass is \(6\,\text{kg}\).
\(I_{\text{whole}} = \tfrac12 MR^2\). The plug about its own centre has \(\tfrac12 mr^2\); moved to O, a distance \(d = 0.2\,\text{m}\), it has \(\tfrac12 mr^2 + md^2\).
\(I_{\text{whole}} = \tfrac12 \times 8 \times 0.16 = 0.64\,\text{kg m}^2\). \(I_{\text{plug}} = \tfrac12 \times 2 \times 0.04 + 2 \times 0.04 = 0.04 + 0.08 = 0.12\,\text{kg m}^2\). So \(I = 0.64 - 0.12 = 0.52\,\text{kg m}^2\).
The disc with the hole has \(I = 0.52\,\text{kg m}^2\) about O. It lost a quarter of its mass but only about a fifth of its moment of inertia, because the plug lay near the axis.
When the hole is centred on the axis, no shift is needed. A washer of outer radius \(R\) and inner radius \(r\) is a disc minus a smaller disc. With \(m\) the washer's own mass, the subtraction gives \(I = \tfrac12 m(R^2 + r^2)\). It lies between the disc's \(\tfrac12 mR^2\) and the ring's \(mR^2\), as it should.
The same subtraction works for solids. A thick spherical shell of outer radius \(R\) and inner radius \(r\) is a solid sphere minus a smaller one. With masses in proportion to \(R^3\) and \(r^3\), it gives \(I = \tfrac25 m\,\dfrac{R^5 - r^5}{R^3 - r^3}\). As \(r\) approaches \(R\), this tends to the thin shell's \(\tfrac23 mR^2\).
A steel washer has an outer radius of \(0.2\,\text{m}\), an inner radius of \(0.1\,\text{m}\) and a mass of \(0.6\,\text{kg}\). Find its moment of inertia about its central axis, perpendicular to it, in kg m².
A uniform disc of mass \(M\) and radius \(R\) has a concentric hole of radius \(R/2\) cut from it. Here \(M\) is the mass before cutting. What is the moment of inertia of what remains about the central perpendicular axis?
In a cavity problem, why must the plug's mass come from the density of the whole body? Why not use the mass of what remains?
The subtraction \(I_{\text{whole}} - I_{\text{plug}}\) works because the whole body is exactly the remaining body plus the plug, made of the same material. So the plug has the same mass per unit area, or per unit volume, as the rest. Its mass is the whole mass times the ratio of the plug's area, or volume, to the whole. Any other mass would describe a different plug.
The chain wheel of a racing bicycle, and many gears in machines, have neat round holes cut through them. Engineers call them lightening holes. They remove mass, but the designers place them carefully. Holes near the rim remove much more moment of inertia than holes near the axle. So the wheel gains and loses speed more readily when the rider pushes or brakes.
Now return to the two thalis. Spinning flat, a thali has \(I = \tfrac12 MR^2\) about its central axis, if we treat it as a disc. Flipped about a diameter it has \(\tfrac14 MR^2\), exactly half. The perpendicular-axis theorem shows why: the flat spin uses both in-plane axes at once, while the flip uses only one of them.
- Using the perpendicular-axis theorem for a sphere, a cylinder or any body with thickness.It holds only for flat bodies, with x and y in the plane and z perpendicular to it.
- Using axes that do not meet at one point.The three axes of \(I_z = I_x + I_y\) must meet at one point of the lamina.
- Taking the plug's moment of inertia about its own centre.Subtract the plug's \(I\) about the same axis as the whole body, using the parallel-axis theorem.
- Taking the given mass of a body with a hole as the whole body's mass.Scale up by the ratio of areas, or volumes, to find the mass the body would have without the hole.