When a tall eucalyptus tree is felled at a timber depot, the workers shout and run well clear. The trunk tips slowly at first, then gathers speed. By the time it reaches the ground, its top is moving so fast that the crown crashes down and branches snap off. A stone dropped from the height of the top would land more gently.
The energy method
Lesson 3 left the falling rod unfinished. Its torque changes with its angle, so its \(\alpha\) is not constant, and the constant-\(\alpha\) equations fail. Energy does not need \(\alpha\) at all. It compares the start and the end, and everything in between takes care of itself.
Two forces act on a body turning on a smooth hinge: its weight and the hinge force. The hinge force acts at a point that does not move, so it does no work, whatever its size. Only gravity does work. So mechanical energy is conserved. If the centre of mass falls through a height \(h\), then \(Mgh = \tfrac12 I\omega^2\), with \(I\) about the hinge.
A uniform rod of mass \(2\,\text{kg}\) and length \(1.5\,\text{m}\) is hinged at one end and released from rest in the horizontal position. Take \(g = 10\,\text{m/s}^2\). Find its angular velocity as it swings through the vertical, the speed of its tip, and the force of the hinge at that instant.
The centre of mass falls \(h = L/2 = 0.75\,\text{m}\). About the hinge, \(I = \tfrac13 ML^2 = 1.5\,\text{kg m}^2\). The hinge does no work.
\(Mg\,\tfrac{L}{2} = \tfrac12 \cdot \tfrac13 ML^2\,\omega^2\), which gives \(\omega = \sqrt{3g/L}\). At the vertical, \(N - Mg = Ma_{cm}\), with \(a_{cm} = \omega^2 L/2\) pointing up.
\(\omega = \sqrt{30/1.5} = \sqrt{20} \approx 4.47\,\text{rad/s}\). The tip moves at \(\omega L \approx 6.71\,\text{m/s}\). Then \(a_{cm} = 20 \times 0.75 = 15\,\text{m/s}^2\), and \(N = 2 \times (10 + 15) = 50\,\text{N}\).
The rod passes the vertical at \(4.47\,\text{rad/s}\), its tip at \(6.7\,\text{m/s}\). The hinge then pulls up with \(50\,\text{N}\), which is \(\tfrac52 Mg\).
Fig 5.1 shows why the hinge works so hard at the bottom. The centre of mass moves on a circle of radius \(L/2\) at the speed \(\omega L/2\). A body on a circle needs a force towards the centre, here towards the hinge. So the hinge must hold up the weight and also supply \(Ma_{cm}\): \(N = Mg + Ma_{cm} = \tfrac52 Mg\). In symbols, \(a_{cm} = \omega^2 L/2 = \tfrac32 g\), for a rod of any length.
A uniform rod \(1.2\,\text{m}\) long is hinged at one end and released from rest in the horizontal position. Take \(g = 10\,\text{m/s}^2\). Find its angular velocity as it swings through the vertical, in rad/s.
While the rod swings down about its smooth hinge, what work does the hinge force do on the rod?
A race between the tip and a stone
The tip of the rod falls through the full length \(L\) while the centre falls only \(L/2\). A stone dropped through \(L\) lands at \(\sqrt{2gL}\). The tip passes the bottom at \(\omega L = \sqrt{3gL}\). Decide what you expect before you read further.
A uniform rod \(1.5\,\text{m}\) long, hinged at one end, is held horizontal. A stone is held beside its free tip. Both are released together. Which falls faster at first, and which is moving faster once each has dropped \(1.5\,\text{m}\)?
The dots in the figure above mark equal instants. The tip pulls ahead at once and leads by up to \(0.26\,\text{m}\), at about \(45°\). Near the bottom the tip's motion turns sideways, so its downward speed falls away. The stone draws level at about \(79°\) and reaches the depth of \(1.5\,\text{m}\) first, at \(0.55\,\text{s}\), against \(0.59\,\text{s}\) for the tip. But the tip arrives moving faster: \(6.7\,\text{m/s}\) against \(5.5\,\text{m/s}\).
No energy has been created. The parts of the rod near the hinge barely move, so the energy released by their fall cannot stay with them. The rod is rigid, so it passes that energy outward, and the tip gets more than its own share. Every part of the rod gets \(\tfrac12 mv^2\) with \(v = \omega r\), and the total is exactly the \(MgL/2\) released.
Think of a whip or a cricket bowler's arm. The heavy parts near the shoulder move slowly, and the energy they hold is passed along to the light end, which moves fastest of all. The falling rod does a gentle version of the same thing.
A straight tree trunk \(20\,\text{m}\) tall is cut at its base and topples from the vertical, turning about the base. Treat it as a uniform rod and ignore air drag. Take \(g = 10\,\text{m/s}^2\). With what speed does its top strike the ground, in m/s?
A \(1\,\text{kg}\) block hangs from a light string wound on a pulley, a uniform disc of mass \(2\,\text{kg}\) and radius \(0.2\,\text{m}\). The block falls \(2\,\text{m}\) from rest, and the string does not slip. Take \(g = 10\,\text{m/s}^2\). Find the speed of the block as it lands.
System: block, pulley and Earth. The axle does no work. The string does not slip, so no heat is made, and \(\omega = v/R\).
Energy: \(mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2\), with \(\tfrac12 I\omega^2 = \tfrac12 \cdot \tfrac12 MR^2 \cdot v^2/R^2 = \tfrac14 Mv^2\).
\(1 \times 10 \times 2 = \tfrac12 \times 1 \times v^2 + \tfrac14 \times 2 \times v^2\), so \(20 = v^2\) and \(v \approx 4.47\,\text{m/s}\). Check by forces: lesson 2 gave \(a = mg/(m + M/2) = 5\,\text{m/s}^2\), and \(v = \sqrt{2ah} = \sqrt{20}\).
The block lands at about \(4.5\,\text{m/s}\). Of the \(20\,\text{J}\) released, \(10\,\text{J}\) is in the block and \(10\,\text{J}\) in the pulley's spin.
The energy route never met the tension. The tension does negative work on the block and an equal positive work on the pulley's rim, because the string does not slip. It moves energy from one body to the other and keeps none. The route through forces needed the tension as an unknown and then removed it. Energy steps over it.
Try the three chips of Fig 5.2. With a light pulley the block keeps all \(20\,\text{J}\) and lands at \(6.32\,\text{m/s}\). A \(2\,\text{kg}\) disc takes half, and a \(4\,\text{kg}\) disc takes two thirds, so the block lands at \(3.65\,\text{m/s}\). The pulley's share stands to the block's as \(\tfrac14 Mv^2\) to \(\tfrac12 mv^2\), that is as \(M\) to \(2m\), at every instant of the fall.
With a \(2\,\text{kg}\) disc pulley, the \(1\,\text{kg}\) block lands with only \(10\,\text{J}\) of the \(20\,\text{J}\) released. Where is the rest?
A \(10\,\text{kg}\) bucket falls \(5\,\text{m}\) from rest on a rope wound on a well drum, a solid cylinder of mass \(20\,\text{kg}\). The rope does not slip. Take \(g = 10\,\text{m/s}^2\). Find the speed of the bucket after the fall, in m/s.
The dictionary between translation and rotation
Every law of motion along a line so far has a twin for rotation about a fixed axis. The twin comes from the same substitutions every time: \(x\) to \(\theta\), \(v\) to \(\omega\), \(a\) to \(\alpha\), \(m\) to \(I\), and \(F\) to \(\tau\). The table is the translation-rotation dictionary as it stands after this chapter. The rows on angular momentum belong to the next chapter.
The dictionary translates equations, not every fact. Two warnings. First, \(I\) depends on the axis: a rod has \(\tfrac{1}{12}ML^2\) about its centre and \(\tfrac13 ML^2\) about its end, while its mass is the same everywhere. Second, a torque depends on the point about which it is taken. A force has no torque about a point on its line, but it has a torque about almost any other point. Always name the axis for \(I\) and the point for \(\tau\).
Which of these pairs is not a true pair of the dictionary between translation and rotation?
A flywheel turning at \(20\,\text{rad/s}\) is braked with a steady angular deceleration of \(4\,\text{rad/s}^2\). Through what angle does it turn before it stops, in radians?
Now return to the felled tree. From the vertical to the ground its centre of mass falls half its height, just as the rod's did. So its top lands at \(\sqrt{3gL}\). For a trunk \(20\,\text{m}\) tall that is about \(24.5\,\text{m/s}\), close to \(90\,\text{km/h}\). A stone dropped from the top would land at \(20\,\text{m/s}\). The crown hits harder than a falling stone because the slow lower trunk passes its energy up to the top.
- Using the fall of the tip instead of the fall of the centre of mass.The weight acts at the centre of mass. Use its fall, \(L/2\) for a rod from horizontal to vertical.
- Using \(\tfrac{1}{12}ML^2\) in the energy of a rod on a hinge at its end.The rod turns about the hinge, so \(I = \tfrac13 ML^2\).
- Taking the hinge force at the bottom of the swing to be \(Mg\).The centre moves on a circle, so \(N = Mg + M\omega^2 L/2 = \tfrac52 Mg\) for a rod released from the horizontal.
- Forgetting the pulley's kinetic energy in an energy account.A pulley with mass holds \(\tfrac12 I\omega^2\), which is \(\tfrac14 Mv^2\) for a disc when the string does not slip.