### Assumptions

For this investigation, we will model the trapeze and performer as a simple pendulum (i.e. a mass on the end of a light inextensible string). We will assume that the flyer does not try to move their body, and hangs in line with the cable (obviously not the case, but we can approximate to this, in such a simple investigation).

As the mass of the flyer is spread out along their length, we will have to approximate, and model the flyer as a particle, positioned at the flyer's centre of mass, which we will assume to be halfway down their body. The flyer and trapeze is therefore modelled as a mass of 74kg (70kg for flyer, and 4kg for bar), at a distance 4.6m from top of cable (3.7m for cable, and extra 0.9m for half of the flyer's length).
We will also ignore air resistance.

### Energy

There are three types of energy involved in a trapeze swing: kinetic, potential and chemical (in the flyer). As in this investigation we are not allowing the flyer to move themselves, we can ignore the third.

When the flyer starts from the board, they are initially not moving, and so they have their maximum potential energy (PE), given by:
$PE = m \times g \times h$
(m = mass, g = gravitational constant, h = height)

Once they have swung to their lowest point, gravity has accelerated them to their maximum speed and therefore their maximum kinetic energy (KE) which is given by:
$KE = {1}\over{2} \times m \times v^{2}$
(m = mass, v = velocity (in this instance this is equal to speed))

All the kinetic energy that they now have has been converted from potential energy as they swung down to the bottom. At the bottom, they therefore have their lowest amount of potential energy. We will assume the bottom of the swing to be the point of reference for all the energy calculations (i.e. O), as we are only interested in ‘change in height’. Therefore at the bottom he has ZERO potential energy and ALL the PE he had in the first place is now KE.

We can therefore say that:
$PE = KE$

We can then substitute the formulae for PE and KE:
$mgh = {1}\over{2} \times m \times v^{2}$

Cancelling the m and rearranging gives:
$v^{2} = 2gh$

From this expression, we can find their maximum speed, or indeed their speed at any point on their swing, just by knowing h, the distance they have moved VERTICALLY from the board. By symmetry, on the other side of the swing, the KE will turn back to PE and, if no energy is lost from the system, they should end up on the other side at the same height they started at. In practice, they lose energy to heat (the air around them heats up slightly due to friction with the body). There is therefore less energy available to swing them back up to the same height, and, if they do not use some of their own chemical energy in their body to ‘top up’ the energy of the swing (using a more advanced swing technique), they will swing slower and slower and eventually stop.

### Calculating Maximum Speed

Let us now find the maximum speed of a flyer on a trapeze, modelled as above:

$v^{2} = 2gh\\ g = 9.8\ (a\ constant)\\ h = 1.5m\ on\ a\ standard\ rig\\ v^{2} = 2 \times 9.8 \times 1.5\\ v = 5.4ms^{-1}$

### Find time period of swing:

For this we will use Simple Harmonic Motion, however we must first prove than SHM is a valid way to model this problem:

$F = ma$
along the path of pendulum (positive for displacements to right)
$-mg \sin(\theta) = m\ddot{x}\\ But\ x=l\theta,\ so\ \ddot{x} = l\ddot{\theta}\\ -mg\sin(\theta)=m \ddot{\theta}\\ \ddot{\theta}=\frac{-g}{l}\sin(\theta)\\ For\ small\ \theta: \theta\approx \sin(\theta)\\ \frac{\sin(\theta)}{\theta}\rightarrow 1\ (as\ \theta\rightarrow 0)\\ Therefore\ \ddot{\theta}\approx\frac{-g}{l}\theta\\ This\ is\ in\ the\ form\\ \ddot{x}=-\omega^{2}\theta\\ ...where\ in\ this\ case\\ \omega = \sqrt{\frac{g}{l}} which\ is\ the\ condition\ for\ SHM.\\ Therefore\ we\ must\ approximate\\ \sin(\theta)\approx\theta\\ in\ order\ to\ obtain\ the\ expression\\ \ddot{\theta}=-\omega^{2}\theta$
In our case θ is approx. 0.83 radians, so sinθ = 0.73, which is (fairly) close to θ, so we are just about justified in making this approximation.

Therefore time period (time from one point in swing, to same point in next swing) is given by:
$T=\frac{2\pi}{\omega} where\ \omega=\sqrt{\frac{g}{l}}\\ Therefore\ in\ this\ case\\ T=2\pi\sqrt{\frac{l}{g}}\\ (l=4.6m,\ g=9.8)\\ T=2\times\pi\times\sqrt{\frac{4.6}{9.8}}\\ T=4.3s$