Press one end of a metre ruler on the edge of a desk with a finger, so that the ruler sticks out flat. Rest a one-rupee coin on its far end. Now lift the finger and let the ruler swing down about the desk edge. For an instant the coin floats in the air. The ruler drops away from under it, and the coin lands a moment later. The end of the ruler falls faster than a falling coin.
The first instant of the fall
Take a uniform rod of mass \(M\) and length \(L\), hinged at one end on a smooth hinge. Hold it horizontal and release it. Two forces act on it: its weight \(Mg\) at the centre, and the force of the hinge. Take torques about the hinge. The hinge force acts at the hinge and has no moment arm. The weight acts at \(L/2\) from the hinge, at right angles to the rod.
So \(\tau = Mg\,L/2\). About its end the rod has \(I = \tfrac13 ML^2\), from the chapter on Rigid Bodies and Moment of Inertia. Then \[ \alpha = \dfrac{\tau}{I} = \dfrac{MgL/2}{ML^2/3} = \dfrac{3g}{2L}. \] The mass cancels. A long rod starts with a small \(\alpha\), and a short rod with a large one.
At the first instant the rod is at rest, so no point has any centripetal acceleration. A point at distance \(r\) from the hinge starts to move down with acceleration \(a = \alpha r\), growing in proportion to \(r\). The centre starts at \(\alpha L/2 = \tfrac34 g\). The tip starts at \(\alpha L = \tfrac32 g\), faster than a body in free fall. The point at \(r = 2L/3\) starts at exactly \(g\).
A uniform rod of mass \(2\,\text{kg}\) and length \(1.5\,\text{m}\) is hinged at one end, held horizontal and released. Take \(g = 10\,\text{m/s}^2\). Find its angular acceleration and the accelerations of its centre and its tip at the instant of release.
About the hinge, \(\tau = Mg\,L/2 = 2 \times 10 \times 0.75 = 15\,\text{N m}\) and \(I = \tfrac13 ML^2 = \tfrac13 \times 2 \times 2.25 = 1.5\,\text{kg m}^2\).
So \(\alpha = 15/1.5 = 10\,\text{rad/s}^2\). The formula \(3g/2L = 30/3\) gives the same.
The centre starts at \(a_{cm} = \alpha L/2 = 7.5\,\text{m/s}^2\), and the tip at \(\alpha L = 15\,\text{m/s}^2\), one and a half times \(g\).
Switch Fig 3.1 to "accelerations". The tips of the green arrows lie on one straight line from the hinge, because \(a = \alpha r\). The coin of the ruler experiment needs \(g\) to fall. Any point of the rod beyond two thirds of its length starts faster than that, so a coin resting there is left behind. A coin nearer the hinge stays on the rod.
A small coin rests on the free end of a horizontal hinged rod. The rod is released. What happens to the coin at that instant?
A uniform rod \(1.5\,\text{m}\) long is hinged at one end and released from the horizontal. How far from the hinge is the point that starts to fall with an acceleration of exactly \(g\), in metres?
How hard the hinge pushes
The torque equation cannot give the hinge force, because that force has no moment arm about the hinge. It needs a second law. The chapter on Centre of Mass gave a second law for any body. The net external force moves its centre of mass as if all the mass were there: \(\vec F_{ext} = M\vec a_{cm}\). This holds for a turning body too.
At the instant of release, the centre moves straight down at \(a_{cm} = \tfrac34 g\). The hinge pushes up with a force \(N\), and the weight pulls down. So \(Mg - N = Ma_{cm} = \tfrac34 Mg\), which gives \(N = \tfrac14 Mg\). For the rod of the example, \(N = 5\,\text{N}\), a quarter of its \(20\,\text{N}\) weight. The rod has no angular velocity yet, so the centre has no centripetal acceleration and the hinge gives no sideways push.
The hinge force of a moving body is not its value at rest. A horizontal rod held on two supports, one at each end, rests with \(Mg/2\) on each. Remove one support. At that instant the other support's force drops at once to \(Mg/4\), because the centre of mass has begun to accelerate. Always find a hinge force from \(\vec F_{ext} = M\vec a_{cm}\), with the acceleration of that instant.
A uniform plank of weight \(W\) rests horizontally on two supports at its ends. One support is suddenly knocked away. Just after that, what force does the other support exert on the plank?
A uniform rod of mass \(6\,\text{kg}\) and length \(2\,\text{m}\) is hinged at one end and released from rest in the horizontal position. Take \(g = 10\,\text{m/s}^2\). Find the force of the hinge on the rod at the instant of release, in newtons.
The angular acceleration follows the torque
As the rod swings down, the weight's moment arm shrinks. At an angle \(\theta\) below the horizontal, the centre is only \((L/2)\cos\theta\) from the hinge, measured across. So the torque is \(Mg\,(L/2)\cos\theta\), and \[ \alpha = \dfrac{3g}{2L}\cos\theta. \] At \(60°\) below the horizontal, \(\alpha\) has fallen to half its starting value. When the rod hangs straight down, the weight points at the hinge and \(\alpha = 0\).
So the angular acceleration of a falling rod is not constant. The equations for constant \(\alpha\) from Kinematics of Circular Motion do not apply to it. To find how fast the rod is turning when it hangs straight down, a different tool is needed. Lesson 5 finds it in one line, from energy.
Watch Fig 3.2 at the two ends of the swing. There the rod is at rest for an instant, \(\omega = 0\). Yet \(\alpha\) is at its largest, \(10\,\text{rad/s}^2\), because the weight has its longest moment arm. At the bottom the opposite holds: \(\alpha = 0\), while \(\omega\) is largest, \(4.47\,\text{rad/s}\). The law \(\tau = I\alpha\) links \(\alpha\) to the torque of that instant. It says nothing about \(\omega\).
The same thing happens to a ball thrown straight up. At the top it is at rest for an instant, yet its acceleration is still \(g\). A zero velocity does not mean a zero acceleration, and a zero \(\omega\) does not mean a zero \(\alpha\).
A rod pendulum swings out to one end of its swing, where it is at rest for an instant. What is its angular acceleration at that instant?
A uniform rod \(1.5\,\text{m}\) long, hinged at one end, is released from the horizontal. Take \(g = 10\,\text{m/s}^2\). What is its angular acceleration when it has swung to \(60°\) below the horizontal, in rad/s²?
At a railway level crossing, a long striped boom swings down to block the road. It is a rod hinged at one end. Released freely, it would crash down, its tip starting faster than a falling stone, and the hinge would jolt. So the boom carries a heavy counterweight on the short side of its hinge. The counterweight moves the centre of mass close to the hinge. The weight then has almost no torque, and a small motor lowers the boom gently.
Now return to the ruler and the coin. A metre ruler hinged at the desk edge starts with \(\alpha = 3g/2L = 15\,\text{rad/s}^2\). Its far end starts to fall at \(\tfrac32 g\), faster than the coin can follow, so the coin floats for an instant. Put the coin at the \(50\,\text{cm}\) mark instead. That point starts at \(\tfrac34 g\), slower than the coin, and the coin rides down on the ruler.
- Using \(I = \tfrac{1}{12}ML^2\) for a rod turning about its end.About the hinge at its end, \(I = \tfrac13 ML^2\).
- Taking the hinge force of a falling rod to be its value at rest.Find it from \(Mg - N = Ma_{cm}\) at that instant. At release it is \(Mg/4\).
- Using the constant-\(\alpha\) equations for a swinging rod.Its torque changes with the angle, so \(\alpha\) changes. Use energy for speeds.
- Thinking that \(\omega = 0\) means \(\alpha = 0\).\(\alpha\) follows the torque of the instant. At the end of a swing it is largest.