At the Dussehra mela a tall pole carries a spinning top, and chairs hang from its rim on long chains. When the ride starts, the chains hang straight down. As the top spins faster, the chairs swing outwards and upwards, and the chains lean further from the vertical. The heaviest uncle and the lightest child ride side by side. Look closely: an empty chair, a child's chair and the uncle's chair all lean at the same angle.
A bob that goes round
Tie a small bob to a string of length \(L\) and hang it from a fixed point. Set it moving so that it goes round a horizontal circle at a steady speed. The string sweeps out the surface of a cone, so the arrangement is called a conical pendulum. Let the string make an angle \(\theta\) with the vertical. The bob then moves on a circle of radius \(r = L\sin\theta\), whose centre lies directly below the point of support.
Only two real forces act on the bob: the weight \(mg\) down and the tension \(T\) along the string, towards the support. Use the method of Lesson 1. The centre of the circle is level with the bob, so take one axis horizontal, towards the centre, and one vertical. Resolve the tension into \(T\cos\theta\) upwards and \(T\sin\theta\) towards the centre.
The bob does not rise or fall, so the vertical forces balance. The horizontal part of the tension is the only force towards the centre, so it does the centripetal job:
\[ T\cos\theta = mg, \qquad T\sin\theta = m\omega^2 r = m\omega^2 L\sin\theta \]
\[ T\sin\theta = m\omega^2 L\sin\theta \]
Why. the horizontal part of the tension is the centripetal force, and \(r = L\sin\theta\).
\[ T = m\omega^2 L \]
Why. divide both sides by \(\sin\theta\), which is not zero while the bob goes round.
\[ m\omega^2 L\cos\theta = mg \]
Why. put \(T = m\omega^2 L\) into the vertical balance \(T\cos\theta = mg\).
\[ \omega = \sqrt{\dfrac{g}{L\cos\theta}} \]
Read it. the mass cancels. Solve for \(\omega\).
The length \(h = L\cos\theta\) is the depth of the circle below the point of support. So \(\omega = \sqrt{g/h}\), and the time for one turn is \(2\pi\sqrt{h/g}\). The mass has cancelled, just as it did for the car on the hump. A heavier bob needs a larger inward force, and it gets a larger tension to supply it. Dividing the two equations also gives \(\tan\theta = \omega^2 r/g\). This is the pendulum in the accelerating car again, with the car's acceleration replaced by \(\omega^2 r\).
Move the slider in the sim from \(3.8\) to \(6\,\text{rad/s}\). The string opens from about \(46°\) to about \(74°\), and the tension grows from \(3.6\) to \(9\,\text{N}\). Yet the vertical part of the tension never changes: it is always the weight, \(2.5\,\text{N}\). As the string leans more, the tension must grow so that this vertical part still holds the bob up. The horizontal part, which turns the bob, grows much faster.
A string can never be exactly horizontal. Its vertical part, \(T\cos\theta\), must hold up the weight, so \(\cos\theta\) cannot be zero however fast the bob goes. Also, below \(\omega = \sqrt{g/L}\) no cone is possible at all, because \(\cos\theta = g/(\omega^2 L)\) would be more than 1. The bob then just hangs, or swings to and fro.
A \(0.2\,\text{kg}\) bob on a \(1\,\text{m}\) string goes round as a conical pendulum with the string at \(60°\) to the vertical. Take \(g = 10\,\text{m/s}^2\). Find (a) the tension, (b) the angular speed, (c) the time for one revolution.
(a) \(T = mg/\cos 60° = 2/0.5 = 4\,\text{N}\), twice the weight.
(b) The depth of the circle is \(h = L\cos 60° = 0.5\,\text{m}\), so \(\omega = \sqrt{g/h} = \sqrt{20} \approx 4.47\,\text{rad/s}\).
(c) \(t_{\text{rev}} = 2\pi/\omega = 6.283/4.47 \approx 1.4\,\text{s}\). The radius of the circle is \(L\sin 60° \approx 0.87\,\text{m}\) and the speed about \(3.9\,\text{m/s}\).
Now the swing ride. Each chair hangs on a chain from the rim of the spinning top. So the radius of its circle is the rim's radius plus \(L\sin\theta\). The same two equations still hold: \(T\cos\theta = mg\) and \(T\sin\theta = m\omega^2 r\). Divide them, and \(\tan\theta = \omega^2 r/g\) has no mass in it. So every chair, empty or full, leans at the same angle, as you saw at the mela.
In a swing ride, each chain is \(5\,\text{m}\) long and hangs from the rim of a top of radius \(3\,\text{m}\). At full speed the chains make \(37°\) with the vertical. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\tan 37° = 0.75\). Find the time for one revolution, in seconds.
Two conical pendulums have strings of the same length at the same angle. One bob is twice as heavy as the other. How do their times of revolution compare?
The rotor
A rotor is a large drum that spins about its vertical axis. Riders stand with their backs to the wall. When the drum spins fast enough, the floor is lowered, and the riders stay stuck to the wall with their feet in the air. What holds them up?
Draw the forces on one rider of mass \(m\), on a wall of radius \(R\). The wall pushes on the rider with a normal force \(N\), perpendicular to the wall, which is towards the axis. That is the only horizontal force, so it is the centripetal force: \(N = m\omega^2 R\). The weight \(mg\) pulls the rider down. The only upward force is static friction \(f\) from the wall, along the wall. The rider does not slide down, so \(f = mg\).
Static friction can be at most its limiting value, \(\mu_s N\). So the rider stays up only if \(mg \le \mu_s m\omega^2 R\). The faster the drum spins, the harder the wall presses, and the more friction it can give. Below a certain angular speed the limit falls short of the weight, and the rider slides down the wall.
Notice in the figure what changes and what does not. The friction needed is always \(600\,\text{N}\), the weight. Friction does not grow with speed. Only its limit grows, because the normal force grows as \(\omega^2\). The least angular speed here is \(\sqrt{g/(\mu_s R)} = \sqrt{10} \approx 3.16\,\text{rad/s}\), about one turn every two seconds. The mass has cancelled once more, so a child and an adult need the same spin.
A rotor has a radius of \(4\,\text{m}\), and \(\mu_s = 0.4\) between the riders and the wall. Take \(g = 10\,\text{m/s}^2\). What is the least angular speed, in rad/s, at which the floor can be lowered?
The wall of a rotor is relined so that \(\mu_s\) doubles. What happens to the least angular speed?
The well of death
At a village fair, the maut ka kuan, or well of death, is a wooden cylinder about ten metres across. A motorcyclist rides round and round on its vertical wall, with nothing below him. The physics is the rotor's. The wall's normal force on the tyres, towards the centre, is the centripetal force. Static friction between the tyres and the wooden wall, upwards, holds up the weight of the rider and the motorcycle.
So the rider must never ride slower than \(v_{\min} = \sqrt{gR/\mu_s}\). For a well of radius \(5\,\text{m}\) with \(\mu_s = 0.5\), that is \(\sqrt{100} = 10\,\text{m/s}\), or \(36\,\text{km/h}\). Riders start on a sloping ramp at the bottom and move up the wall only once they are fast enough.
In the well of death, a motorcyclist who slows down loses grip in a way he cannot feel coming. The friction he needs stays equal to his weight, but the most the wall can give falls as the square of the speed. Just below \(v_{\min}\) the tyres slide down the wall all at once. So the riders keep the throttle open, and they come down to the ramp before they slow.
- Taking the radius of a conical pendulum's circle as \(L\).The circle has radius \(r = L\sin\theta\). Its depth below the support is \(L\cos\theta\).
- Saying the normal force of the rotor's wall holds the rider up.The wall's normal force is horizontal. It is the centripetal force. Friction, along the wall, holds up the weight.
- Setting the rotor's friction equal to \(\mu_s N\) at every speed.Static friction equals \(mg\) while the rider stays put. \(\mu_s N\) is only its limit, reached at \(\omega_{\min}\).