We use the example of a trick from Swinging Trapeze to demonstrate how the length of the pendulum changes the swing.
We have now discussed various ways to change the swing, but have ignored the movement of the flyer. By moving their body or by adopting a position for a trick, the flyer can vary the length of the pendulum and therefore change the period of their swing as shown in the previous investigation.
A good and easy to understand example of this is found on the swinging trapeze (although all the same principles apply to flying). The swinging trapeze is a much simpler version of the flying trapeze. More often than not, there is no safety net and the flyer is secured only with a safety line. The main difference between the two is that on the swinging trapeze, the flyer has to produce the swing themselves, rather than generating swing by stepping off a high platform.
The Standing Seats-Off
This trick, best described by a diagram, involves the flyer first standing on the bar, and ending up hanging underneath it from their feet.
We will examine this trick because it involves a very large change in the length of the pendulum, and also a very large change in speed.
When we first examine this trick, it is easy to think that the (intuitive) change in speed comes as a result of the change in length of the pendulum, as the flyer falls underneath it. We must remember that the equation for speed has no term for length in it (v2=2gh) and therefore it cannot be this that causes the acceleration.
When swinging, standing on the bar, the system follows the original principal of a change from PE to KE, as described in Investigation 1. When the flyer falls back, they convert their PE to KE, which is then added to the KE they would already have had at that point in their swing.
We will assume that the angle of swing (θ) is approx. 0.8 radians (50 degrees), although of course this could be considerably larger for a high level artist.
What is the original maximum speed?
$$v^{2}=2gh$$
$$ v^{2}=2\times9.8\times(2.8-(2.8\times\cos(0.8)))$$
$$v=4.1ms^{-1}$$
$$PE\ at\ top\ is$$
$$mgh = 70 \times 9.8 \times(2.8-(2.8\times\cos(0.8)))$$
$$(flyer\ 70kg,\ ignore\ mass\ of\ bar)$$
$$PE = 582J$$
$$therefore KE\ (at\ bottom) = 582J$$
What is their maximum speed after they fall?
Flyer falls a total distance of 1.8m, therefore they lose how much PE?
$$PE=mgh=70\times9.8\times1.8\\ PE = 1234.8J$$ This is all converted to KE (assume no air resistance)
$$ KE = \frac{1}{2}mv^{2}=1234.8+582=1816.8J\\ v^{2}=51.9ms^{-1}\\ v=7.2ms^{-1}$$ This is a huge increase in speed (the flyer has nearly doubled their original speed). Therefore this shows another way that the flyer can vary the speed of their swing; moving their body to increase or decrease his energy. This is a theme we will return to when we investigate advanced swinging.