At some amusement parks a swinging boat hangs from rigid steel arms instead of ropes. Each swing goes higher, until the boat goes right over the top. It crawls through the top, with the riders upside down, and then swoops down again. A bucket on a rope moving that slowly at the top would spill everything.
A rod can push
Replace the string by a light rigid rod, pivoted at O. The radial equation of Lesson 1 still holds for the force \(F\) of the rod on the body, counted positive when it pulls towards O: \(F = mv^2/r + mg\cos\theta\). The difference is in what \(F\) may do. A string must have \(F \ge 0\). A rod may also push, so \(F\) may be negative.
At the top, \(F = mv^2/r - mg\). If \(v\) is more than \(\sqrt{gr}\), the rod pulls. At exactly \(\sqrt{gr}\) the rod carries no force. Below \(\sqrt{gr}\) the weight is more than the circle needs, and the rod pushes the body outwards, upwards at the top. Even at a speed near zero the rod holds the body, pushing up with nearly \(mg\).
So a rod sets only one condition: the body must reach the top. Energy gives \(v_{\text{top}}^2 = u^2 - 4gr\), and this must not be negative. So \(u^2 \ge 4gr\), and the least speed at the bottom is \(\sqrt{4gr} = 2\sqrt{gr}\). At exactly that speed the body arrives at the top with zero speed and balances there, so in practice it needs a little more.
The two chips of Fig 4.1 agree exactly up to \(138°\). Both the string and the rod obey the same equation there, so they pull with the same force. At \(138°\) that force reaches zero. Beyond it the equation asks for a push. The string cannot give one and goes slack, while the rod simply pushes.
A small ball is fixed to the end of a light rod \(0.9\,\text{m}\) long, pivoted at the other end. Take \(g = 10\,\text{m/s}^2\). What is the least speed, in m/s, at the lowest point for the ball to go all the way round?
A \(0.5\,\text{kg}\) ball on a light rod \(1\,\text{m}\) long passes the top of its circle at \(2\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). What is the size, in newtons, of the force of the rod on the ball there?
A ball on a light rod passes the top at exactly \(\sqrt{gr}\). What force does the rod exert on it there?
Beads on rings, balls in tubes
A bead on a ring has the ring passing through a hole in it. The ring can push the bead towards the centre or away from it, depending on which side of the hole presses. A ball inside a thin curved tube works the same way. The outer wall pushes it inwards, and the inner wall pushes it outwards. So both are two-way supports, like a rod, and both need only enough energy to reach the top.
A \(0.2\,\text{kg}\) ball moves inside a smooth circular tube of radius \(0.5\,\text{m}\) in a vertical plane. It is launched at \(4.5\,\text{m/s}\) from the lowest point. Take \(g = 10\,\text{m/s}^2\). Does it go round? Which wall pushes it at the top, and how hard?
The least speed for a tube is \(\sqrt{4gr} = \sqrt{20} \approx 4.47\,\text{m/s}\). The launch speed is just above it, so the ball goes round.
At the top, \(v^2 = u^2 - 4gr = 20.25 - 20 = 0.25\), so \(v = 0.5\,\text{m/s}\). The circle needs \(mv^2/r = 0.2 \times 0.25/0.5 = 0.1\,\text{N}\) towards the centre.
The weight gives \(2\,\text{N}\) towards the centre, far more than needed. So the tube must push outwards with \(2 - 0.1 = 1.9\,\text{N}\). At the top, outwards is upwards, so it is the inner wall that pushes.
A string in its place would have gone slack, because \(4.5\,\text{m/s}\) is well below \(\sqrt{5gr} = 5\,\text{m/s}\).
A track pushes one way
A track is different. A body resting on a track can be pushed by it, but only away from the track's surface. On the inside of a loop, the surface is outside the body, so the push points inwards, towards the centre. That is the string's case: the normal force must stay positive, and the body needs at least \(\sqrt{gr}\) at the top. A car driving round the inside of a loop, or a coaster car, is in this case.
On the outside of a track, such as a hump or a dome, the surface is under the body. So the push points outwards, away from the centre. Now the radial equation reads \(mg\cos\varphi - N = mv^2/R\), with \(\varphi\) measured from the top and \(R\) the radius. The track can only reduce the inward force below the weight's part. So the body stays on only while \(mv^2/R \le mg\cos\varphi\). It leaves when it is too fast, not too slow.
On the outside of a hump, the condition at the top is \(v \le \sqrt{gR}\). This is the car over a crest from the chapter on Dynamics of Circular Motion. A car faster than \(\sqrt{gR}\) at the top of a hump leaves the road. The last lesson follows a body that starts slowly at the top of a smooth dome and asks where it leaves.
A small body must pass the top of a vertical circle of radius \(1\,\text{m}\) at only \(1\,\text{m/s}\), well below \(\sqrt{gr} \approx 3.16\,\text{m/s}\). Which supports can keep it on the circle there?
A stunt car drives round the inside of a vertical loop of radius \(10\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). What must be true at the top?
A toy car track often has a loop of radius about \(15\,\text{cm}\). Treat the car as a small block. At the top it needs \(\sqrt{gr} = \sqrt{1.5} \approx 1.2\,\text{m/s}\). Started from rest on a smooth ramp, it must begin at least \(5r/2 \approx 38\,\text{cm}\) above the bottom of the loop. Real toy cars lose some energy to friction in the wheels, so children soon learn to start them higher still.
- Using \(\sqrt{5gr}\) at the bottom for a rod or a tube.A rod or a tube can push, so the body need only reach the top. The least speed is \(\sqrt{4gr}\).
- Taking the least speed at the top of a rod to be \(\sqrt{gr}\).At \(\sqrt{gr}\) the rod carries no force; below it the rod pushes. The least speed at the top is zero.
- Treating the inside of a track like a rod.A track can only push away from its surface. On the inside of a loop that is inwards, so it behaves like a string.
- Thinking a body on a hump falls off when it is too slow.On the outside of a track the push is outwards, so the body leaves when it is too fast: at the top, above \(\sqrt{gR}\).