Why can you balance a broom but not a pencil?
Tonight's idea
Nudge a marble resting in a bowl and it rolls back. Nudge a marble resting on top of a hill and it rolls away. Both marbles were balanced. Only one balance survives a nudge.
A balance is stable if a small push dies away, and unstable if a small push grows.
A broom standing on your palm is a marble on a hilltop: a small lean makes it lean more. Yet you can keep it up, because you watch the lean and slide your hand under it. Using what you see to correct what happens next is called feedback.
Tonight's model. Call the lean \(y\), in millimetres away from upright. A negative \(y\) means leaning the other way. At each tick of a clock, a rule multiplies \(y\) by a fixed number \(k\). This is a simplified model, sensible only for small leans. Real brooms are messier.
Watch one: the same push, two rules
Start both times with a lean of \(y = 1\) mm.
Rule \(y \to y \div 2\) (so \(k = \tfrac12\)): the lean goes \(1,\ 0.5,\ 0.25,\ 0.125,\ \dots\) After 7 ticks it is \(1 \div 128\), less than a hundredth of a millimetre. The push dies away. Stable.
Rule \(y \to 2y\) (so \(k = 2\)): the lean goes \(1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128\). After 7 ticks, a 1 mm lean has become 128 mm. The push grows. Unstable. (Long before 128 mm the lean has stopped being small, so the model has stopped being accurate. Its message is still right: the broom is going over.)
Your turn
- Start with a lean of 8 mm and use the rule \(y \to -y \div 2\). Write the next four leans. Describe in words what the broom is doing. Stable or unstable?
- Start with 10 mm. Write the next three leans for \(k = 1.1\), and then for \(k = 0.9\). Which rule is stable? With the other rule, will the lean ever pass 20 mm?
- A broom with feedback. Left alone, the broom's lean doubles every tick. At the end of every tick your hand moves and removes three quarters of the lean that is there. Start at 8 mm and follow two ticks. What single number is the lean multiplied by each tick? Stable now?
- A slow hand. Same broom, but now your hand only moves at the end of every second tick. Start at 8 mm and follow four ticks. Does the push die away?
- The pencil. For plain sticks let go at the same small lean, the time to fall goes with the square root of the length: 4 times as long takes 2 times as long to fall, 9 times as long takes 3 times as long. The broom is 144 cm long and the pencil 16 cm. How many times faster does the pencil fall? So how many times does its lean double in one broom tick? Your hand still removes three quarters once per broom tick. What is the pencil's lean multiplied by each tick? Stable?
Stretch: exactly which rules are stable?
Try the rule \(y \to ky\) with \(k = 1\), \(k = -1\), \(k = 0\) and \(k = -2\), starting from a lean of 1 mm. Then describe all the numbers \(k\) for which a small push dies away.
Answers
- \(-4,\ 2,\ -1,\ 0.5\). The broom wobbles from one side to the other, and each lean is half the size of the one before. The push dies away, so this is stable. A minus sign in the rule changes the side, not the size.
- \(k = 1.1\): \(11,\ 12.1,\ 13.31\). \(k = 0.9\): \(9,\ 8.1,\ 7.29\). The rule with \(k = 0.9\) is stable. With \(k = 1.1\) the lean grows by a tenth of itself every tick, so each step is bigger than the last (the first is 1 mm), and it does pass 20 mm: after 8 ticks it is about 21.4 mm. A multiplier only a little bigger than 1 still loses in the end.
- Tick one: \(8 \to 16\), then your hand leaves a quarter, \(16 \div 4 = 4\). Tick two: \(4 \to 8 \to 2\). Each tick multiplies the lean by \(2 \times \tfrac14 = \tfrac12\). That is the stable rule from the worked example. Feedback has turned the hilltop into a bowl.
- \(8 \to 16 \to 32\), then the hand leaves a quarter: \(32 \div 4 = 8\). Two more ticks give \(16\), then \(32 \div 4 = 8\) again. Every two ticks the lean is multiplied by \(2 \times 2 \times \tfrac14 = 1\), so it keeps coming back to 8 mm. The push does not die away, so this is not stable. The same hand, moving half as often, no longer does the job.
- \(144 \div 16 = 9\), and \(9 = 3 \times 3\), so the broom takes 3 times as long to fall and the pencil falls 3 times faster. Its lean doubles three times in one broom tick: \(2 \times 2 \times 2 = 8\). Then your hand leaves a quarter: \(8 \times \tfrac14 = 2\). The lean doubles every tick, so the pencil is unstable. The trouble is speed: the pencil falls too fast for your hand. In this model, a hand three times quicker would balance it. Feedback only works when it is fast compared with the fall.
Stretch. \(k = 1\): \(1, 1, 1, \dots\) The lean stays. \(k = -1\): \(-1, 1, -1, \dots\) It swaps sides for ever at the same size. \(k = 0\): \(0, 0, \dots\) Gone in one tick. \(k = -2\): \(-2, 4, -8, \dots\) It swaps sides and grows. Each tick multiplies the size of the lean by the size of \(k\), ignoring the sign. So a small push dies away exactly when \(k\) is between \(-1\) and \(1\), that is \(-1 \lt k \lt 1\). It grows when \(k \gt 1\) or \(k \lt -1\). At \(k = 1\) and \(k = -1\) it neither grows nor dies away, so those two do not count as stable by tonight's definition.