Drill small holes along a wooden metre scale and hang it on a nail through one of them. Through a hole near one end, it swings gently to and fro. Hang it through a hole closer to the middle. Most people expect a shorter swinging part to swing faster, as a shorter simple pendulum does. Yet near the middle the scale swings more and more slowly, and through the centre hole it does not swing at all.
From force to torque
A bob on a string moves along an arc, and a force brings it back. A rigid body hung on a pivot, or twisted on a wire, turns about an axis. What brings it back is a turning effect, a torque. In the chapter Dynamics of Rotation you met the rotational form of Newton's second law, \(\tau = I\alpha\). Here \(\tau\) is the torque about the axis, \(I\) the moment of inertia about it, and \(\alpha = d^2\theta/dt^2\) the angular acceleration.
Suppose the torque points back towards \(\theta = 0\) and is proportional to the angle: \(\tau = -k_\theta\theta\). This is a restoring torque. Then \(I\,d^2\theta/dt^2 = -k_\theta\theta\), which is the law of SHM in the angle, with \(\omega^2 = k_\theta/I\). The torque constant takes the place of \(k\), and the moment of inertia takes the place of \(m\). The two steps are unchanged.
Step 1: turn the body through a small angle \(\theta\) about its axis and find the net torque; write it as \(\tau = -k_\theta\,\theta\). Step 2: find the moment of inertia \(I\) about the same axis. Then \(\omega^2 = k_\theta/I\) and \(T = 2\pi\sqrt{I/k_\theta}\).
A rigid body turns to and fro about a fixed axis under a restoring torque \(\tau = -k_\theta\,\theta\). In the formula for its period, what takes the place of the mass \(m\) of a block on a spring?
A rigid body on a pivot
A physical pendulum, also called a compound pendulum, is a rigid body that swings under its own weight about a fixed horizontal axis. Let the axis pass through a point P at a distance \(d\) from the centre of mass G. At rest, G hangs straight below P. Turn the body through \(\theta\), and G moves aside. The weight \(mg\), acting at G, now has a lever arm \(d\sin\theta\) about P, and its torque turns the body back.
\[ \tau = -mgd\sin\theta \]
Why. the weight acts at G with a lever arm \(d\sin\theta\) about P, turning the body back towards \(\theta = 0\).
\[ \tau \approx -mgd\,\theta \]
Why. for small angles \(\sin\theta \approx \theta\); so step 1 gives \(k_\theta = mgd\).
\[ I\,\dfrac{d^2\theta}{dt^2} = -mgd\,\theta \]
Why. step 2, \(\tau = I\alpha\), with \(I\) the moment of inertia about the pivot axis.
\[ \omega^2 = \dfrac{mgd}{I}, \qquad T = 2\pi\sqrt{\dfrac{I}{mgd}} \]
Read it. the law of SHM in \(\theta\), with \(\omega^2 = k_\theta/I\).
The moment of inertia must be taken about the pivot, not about G. The chapter Rigid Bodies and Moment of Inertia gives it by the parallel axis theorem: \(I = I_G + md^2\). Write \(I_G = mk^2\), where \(k\) is the radius of gyration about G. Then \(I = m(k^2 + d^2)\), the mass cancels, and the period depends only on the shape of the body and the position of the pivot.
The last form compares the body with a simple pendulum. A simple pendulum of length \(L_{\text{eq}} = d + k^2/d\), the equivalent length, has the same period. The point on the line PG at the distance \(L_{\text{eq}}\) from P is the centre of oscillation. For a metre stick hung from one end, \(d = 0.5\,\text{m}\) and \(k^2 = L^2/12 = 0.0833\,\text{m}^2\), so \(L_{\text{eq}} = 0.5 + 0.167 = 0.667\,\text{m}\), two thirds of the stick. Its period is \(2\pi\sqrt{0.667/10} = 1.62\,\text{s}\).
Move the pivot inward from the end. At \(d = 0.5\,\text{m}\) the period is \(1.62\,\text{s}\). It falls as the pivot moves in, reaches its least value \(1.51\,\text{s}\) at \(d = k = 0.289\,\text{m}\), and then rises steeply: \(1.92\,\text{s}\) at \(0.1\,\text{m}\) and \(2.60\,\text{s}\) at \(0.05\,\text{m}\). Close to the centre the weight has almost no lever arm, so the torque is tiny while the moment of inertia is not. Through the centre itself there is no torque at all, and the stick stays where it is put.
The least value comes from the equivalent length \(d + k^2/d\). The two parts pull in opposite directions: \(d\) grows as the pivot moves out, \(k^2/d\) grows as it moves in. Their sum is least when they are equal, at \(d = k\). Then \(L_{\text{eq}} = 2k\) and \(T_{\min} = 2\pi\sqrt{2k/g}\). The same sum also shows that two pivots on one side of G give the same period, at \(d\) and at \(k^2/d\). For the metre stick, the end, \(0.5\,\text{m}\) from G, pairs with the point \(0.167\,\text{m}\) from G: both give \(1.62\,\text{s}\).
A thin ring of radius \(20\,\text{cm}\) hangs on a nail through its rim and swings in its own plane. Take \(g = 10\,\text{m/s}^2\). Find its period.
About its centre the ring has \(I_G = mR^2\). The pivot is on the rim, so \(d = R\), and by the parallel axis theorem \(I = mR^2 + mR^2 = 2mR^2\).
\(T = 2\pi\sqrt{I/mgd} = 2\pi\sqrt{2mR^2/(mgR)} = 2\pi\sqrt{2R/g}\).
With \(R = 0.2\,\text{m}\): \(T = 2\pi\sqrt{0.4/10} = 2\pi \times 0.2 = 1.26\,\text{s}\). The ring swings like a simple pendulum of length \(2R\), its diameter.
A uniform rod \(1.5\,\text{m}\) long swings about a horizontal axis through one end. Its moment of inertia about that axis is \(mL^2/3\). Take \(g = 10\,\text{m/s}^2\). What is its period for small swings, in seconds?
A uniform rod may be hung from a pivot anywhere along it. Where must the pivot be for the swing to be fastest?
A disc twisting on a wire
Hang a disc from the middle by a thin vertical wire, and turn it about the wire through an angle \(\theta\). The twisted wire resists, like a spring wound round its own axis. For moderate twists its restoring torque is proportional to the angle: \(\tau = -C\theta\). The constant \(C\) is the torsion constant of the wire, in N m/rad. A body oscillating like this is a torsional pendulum.
Gravity plays no part here. The weight pulls straight down along the wire and has no lever arm about it. Step 1 gives \(k_\theta = C\) directly, and step 2 gives the moment of inertia \(I\) of the body about the wire. So \(\omega = \sqrt{C/I}\) and \(T = 2\pi\sqrt{I/C}\). A stiff wire gives a fast twist; a large moment of inertia gives a slow one.
For the disc of Fig 5.1, \(\omega = \sqrt{0.01/0.0025} = 2\,\text{rad/s}\), and the period is \(2\pi/2 = \pi = 3.14\,\text{s}\). The graph line has slope \(-C\). At the largest twist, \(60° = 1.05\,\text{rad}\), the wire's torque is \(10.5\,\text{mN m}\). Unlike the simple pendulum, the torsional pendulum stays isochronous even for large twists, as long as the wire obeys \(\tau = -C\theta\).
A disc of moment of inertia \(2.5 \times 10^{-3}\,\text{kg m}^2\) twists on a wire with a period of \(2.0\,\text{s}\). A ring is placed on the disc, centred on the wire, and the period becomes \(3.0\,\text{s}\). Find the moment of inertia of the ring.
The wire is the same, so \(C\) is the same. The ring adds its moment of inertia \(I_r\) to the disc's \(I_0\).
\(T_0 = 2\pi\sqrt{I_0/C}\) and \(T_1 = 2\pi\sqrt{(I_0 + I_r)/C}\), so \((I_0 + I_r)/I_0 = (T_1/T_0)^2\).
\((3.0/2.0)^2 = 2.25\), so \(I_r = 1.25\,I_0 = 1.25 \times 2.5 \times 10^{-3} = 3.1 \times 10^{-3}\,\text{kg m}^2\).
Two timings with a stopwatch measure the ring's moment of inertia, with no need to know the wire.
A torsional pendulum and a simple pendulum both have a period of \(2\,\text{s}\) on the Earth. Both are taken to the Moon, where \(g\) is about one sixth as large. Which statement is true there?
A body of moment of inertia \(4.0 \times 10^{-3}\,\text{kg m}^2\) hangs on a wire of torsion constant \(0.10\,\text{N m/rad}\). What is the period of its twisting oscillation, in seconds?
The metre scale of the opening is a physical pendulum, and \(L_{\text{eq}} = d + k^2/d\) explains it. As the hole moves in from the end, the period first falls slightly, to its least value \(1.51\,\text{s}\) at \(29\,\text{cm}\) from the centre. Closer in, the term \(k^2/d\) takes over and the swing slows sharply. At the centre the weight has no lever arm, there is no restoring torque, and the scale stays wherever it is turned.
In 1798 the English scientist Henry Cavendish hung a light rod by a thin wire, with a small lead ball at each end. Two large lead balls placed nearby pulled the small ones sideways and twisted the wire through a tiny angle. To turn that angle into a torque, Cavendish needed the torsion constant of the wire. He found it from the period of the rod's slow twisting oscillation, about seven minutes, using \(T = 2\pi\sqrt{I/C}\). From the torque he found the density of the Earth.
Hold a uniform rod loosely at one end and strike it. At most points the blow jolts your hand. But a blow at the centre of oscillation, two thirds of the way down, produces no jolt at all. The rod simply starts to turn about your grip. This point is the centre of percussion. The sweet spot of a cricket bat or a tennis racket is near it, and a ball struck there stings the hands least.
- Using \(I\) about the centre of mass in \(T = 2\pi\sqrt{I/mgd}\).The body turns about the pivot, so \(I\) is about the pivot: \(I = I_G + md^2\).
- Taking \(d\) as the length of the body.\(d\) is the distance from the pivot to the centre of mass, not to the far end.
- Expecting a pivot closer to the centre of mass to give a faster swing.The period is least at \(d = k\). Closer in, the lever arm shrinks and the swing slows sharply.