A science teacher fills a small bucket with water and swings it in a vertical circle over her head. At the top the bucket is upside down, yet not a drop falls out. Then she slows down a little too much. At the top the water leaves the bucket and splashes on the floor.
A string can pull but never push
A string can only pull. If you push on the end of a string, it goes slack, and it carries no force at all. So the tension can never be negative: \(T \ge 0\). Lesson 1 found the tension at the top, where \(\theta = 180°\) and \(\cos\theta = -1\): \(T = mv^2/r - mg\). This value falls as the speed at the top falls.
At the top, both forces in Fig 2.1 point towards the centre. Their sum must be \(mv^2/r\). If the bob is fast, the weight alone is not enough, and the string adds the rest. As the bob slows, the string's share shrinks. At one particular speed the weight alone is exactly enough, and the tension is zero. Any slower, and the weight is more than the circle needs. The bob then falls inside the circle, and the string goes slack.
Set \(T = 0\) at the top: \(mg = mv^2/r\), so \(v^2 = gr\). This speed is the critical speed at the top. It does not depend on the mass, because the weight and the needed force both grow in proportion to \(m\).
Take a body held on a vertical circle of radius \(r\) by a string, or pressed against the inside of a circular track. It stays on the circle at the top only if its speed there is at least \(\sqrt{gr}\). At exactly \(\sqrt{gr}\) the string or the track exerts no force at the top.
The water in the bucket obeys the same rule. The base of the upside-down bucket pushes the water towards the centre, the way a string pulls a bob. The base can push but cannot pull the water back. So the water stays in only while the bucket moves at least at \(\sqrt{gr}\) at the top. For an arm and bucket of radius \(0.9\,\text{m}\), that is \(\sqrt{9} = 3\,\text{m/s}\). At exactly this speed the water presses on the base with no force at all.
A stone is whirled in a vertical circle on a string \(1.6\,\text{m}\) long. Take \(g = 10\,\text{m/s}^2\). What is the least speed, in m/s, that the stone must have at the top to stay on the circle?
From the top to the bottom
The condition at the top is about forces at one instant. Getting to the top at all is a question of energy. From Lesson 1, the speed at the top is given by \(v_{\text{top}}^2 = u^2 - 4gr\), where \(u\) is the speed at the bottom. The condition \(v_{\text{top}}^2 \ge gr\) becomes \(u^2 - 4gr \ge gr\), that is \(u^2 \ge 5gr\).
Read the result as an energy budget, per unit mass. Rising through \(2r\) to the top costs \(2gr\) out of the kinetic energy \(\tfrac12 u^2\). At the top the bob must still have at least \(\tfrac12 gr\). So \(\tfrac12 u^2 \ge 2gr + \tfrac12 gr\), and \(u^2 \ge 5gr\). Four of the five parts pay for the height, and one part keeps the string tight at the top.
A ball on a string \(0.8\,\text{m}\) long must go all the way round a vertical circle. Take \(g = 10\,\text{m/s}^2\). What is the least speed, in m/s, it needs at the lowest point?
The 6mg rule
Compare the tension at the bottom with the tension at the top. The formula of Lesson 1 gives \(T = mu^2/r - 2mg + 3mg\cos\theta\). The first two terms are the same everywhere on the circle. Only the last term changes, from \(+3mg\) at the bottom to \(-3mg\) at the top. So the difference is \(6mg\), whatever the launch speed.
\[ T_{\text{bottom}} = \dfrac{mu^2}{r} + mg \]
Why. at the bottom the tension points up, towards O, and the weight points down. \(T - mg = mu^2/r\).
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Your turn. which expression gives the tension at the top?
\[ T_{\text{top}} = \dfrac{mu^2}{r} - 5mg \]
Why. open the bracket: \(4mgr/r = 4mg\), and \(4mg + mg = 5mg\).
\[ T_{\text{bottom}} - T_{\text{top}} = mg + 5mg = 6mg \]
Read it. subtract; the term \(mu^2/r\) cancels.
At the least launch speed \(u = \sqrt{5gr}\), the tension at the top is zero. The 6mg rule then gives \(6mg\) at the bottom. So a bob that only just goes round pulls on the string with six times its weight at the bottom.
A \(0.3\,\text{kg}\) stone goes round a vertical circle on a string. The tension at the top is \(4\,\text{N}\). Take \(g = 10\,\text{m/s}^2\). What is the tension, in newtons, at the bottom?
The bucket and its water together have a mass of \(2\,\text{kg}\). They move on a circle of radius \(0.9\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). Find the least speed at the top and the least speed at the bottom. Then find the pull on her arm at the bottom when she swings at these least speeds.
The arm pulls like a string, and the base of the bucket pushes the water inwards. Both fail if the speed at the top is below \(\sqrt{gr}\).
\(v_{\text{top}} = \sqrt{gr}\), \(u = \sqrt{5gr}\), and \(T_{\text{bottom}} = T_{\text{top}} + 6mg\).
\(v_{\text{top}} = \sqrt{9} = 3\,\text{m/s}\). \(u = \sqrt{45} \approx 6.71\,\text{m/s}\). \(T_{\text{bottom}} = 0 + 6 \times 2 \times 10 = 120\,\text{N}\).
She must swing at about \(3\,\text{m/s}\) at the top and \(6.7\,\text{m/s}\) at the bottom. At the bottom her arm pulls with \(120\,\text{N}\), six times the weight of the bucket.
In the "maut ka gola" at a mela, a motorcyclist rides round the inside of a steel mesh globe about \(6\,\text{m}\) across, upside down at the top. Estimate the least speed at the top, in km/h.
A loop fed from a ramp
A loop-the-loop in a toy track or a roller coaster has no string. The body rides on the inside of the track, and the track pushes it towards the centre with a normal force \(N\). A track can push but cannot pull, so \(N \ge 0\). This is the string's condition again, with \(N\) in place of \(T\). The body needs at least \(\sqrt{gr}\) at the top of the loop.
A small block slides without friction down a ramp into a loop of radius \(r\). It starts from rest at a height \(h = 2r\), level with the top of the loop. What happens?
Energy alone is misleading here. A start at \(2r\) gives exactly enough energy to reach the height \(2r\), but only with zero speed at the top. The top of a loop needs speed. The block must arrive there with \(v^2 \ge gr\), so its start must be higher.
Let the block start from rest at a height \(h\) above the bottom of the loop, and take the zero of potential energy there. Energy between the start and the top of the loop gives \(mgh = \tfrac12 mv_{\text{top}}^2 + mg(2r)\). Put in the least value \(v_{\text{top}}^2 = gr\): \(mgh = \tfrac12 mgr + 2mgr = \tfrac52 mgr\). So the least height is \(h = 5r/2\). The ramp's shape does not matter, as long as it is smooth.
The three chips of the figure above show the three cases. From \(2r\) the block leaves the track at \(132°\) from the bottom. From \(2.5r\) it goes round with the push falling to zero at the top. From \(3r\) it goes round with a push of one weight at the top, since \(N/mg = 2h/r - 5 = 1\) there.
A toy car runs without friction down a ramp into a vertical loop of radius \(4\,\text{m}\). Treat it as a small block. What is the least starting height, in metres, above the bottom of the loop?
A block is released from a height of \(2r\) on a smooth track that runs into a loop of radius \(r\). A student says it will just reach the top, because energy is conserved. What is wrong with the argument?
Energy conservation is right, but it only says that the block could reach the height \(2r\) with zero speed. On a loop the block cannot be at the top with zero speed. There the track can only push inwards, so the weight must be at most the needed force \(mv^2/r\), which requires \(v^2 \ge gr\). With no speed to spare, the normal force falls to zero before the top, at \(132°\) from the bottom. The block then leaves the track and falls across the loop.
A ball that rolls instead of sliding needs a higher start. Part of its energy goes into spinning, so less is left for moving along the track. For a small solid ball the least height becomes \(2.7r\) instead of \(2.5r\). The chapter on Rolling Motion shows where the extra \(0.2r\) comes from.
In 1895 the Flip Flap Railway at Coney Island in New York sent riders round a circular loop. To get round the top safely, the cars entered the loop very fast. At the bottom of a circular loop the push on a rider is then at least six times the weight, the 6mg rule of this lesson. Riders complained of sore necks, and the ride did not last. Loops built since the 1970s are shaped like a teardrop. They are tight at the top, where a small radius needs only a small speed. They are wide at the bottom, where a large radius keeps the push gentle.
- Taking the least speed at the top to be zero for a string.A string cannot push, so the bob needs \(v \ge \sqrt{gr}\) at the top. Zero is the answer only for a rod or a tube (Lesson 4).
- Using \(u = \sqrt{4gr}\) at the bottom for a string, from "just enough energy to climb \(2r\)".That leaves no speed at the top. The string needs \(\tfrac12 gr\) of kinetic energy per unit mass there, so \(u = \sqrt{5gr}\).
- Starting a loop-the-loop from a height of \(2r\).The least start is \(5r/2\), because the body must arrive at the top with \(v^2 = gr\).
- Adding the tension and the weight at the bottom.At the bottom they point opposite ways: \(T - mg = mv^2/r\). They add only at the top.