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 man does not
try to move his body, and hangs in line with the cable (1)
(obviously not the case, but we can approximate to this,
in such a simple investigation). As the mass of the man
is spread out along his length, we will have to approximate,
and model the man as a particle, positioned at the man's
centre of mass, which we will assume to be halfway down
his body (2). The man and trapeze is therefore modelled
as a mass of 74kg (70kg for man, and 4kg for bar), at a
distance 4.6m from top of cable (3.7m for cable, and extra
0.9m for half man's length).(3). We will also ignore air
resistance.
Energy
There are three types of energy involves in a trapeze swing:
kinetic, potential and chemical (in the man). As in this
investigation we are not allowing the man to move himself,
we can ignore the third. When the man starts from the board,
he is initially not moving, and so he has his maximum potential
energy, given by:
PE = m x g x h
Once he has swung to his lowest point, gravity
has accelerated him to his maximum speed (and therefore
his maximum KE), and lowest PE. If we assume that the bottom
of his swing to be the point of reference for all the energy
calculations (i.e. O), as we are only interested in change
in height, at this point he has no PE and all the original
PE he had has turned to KE.
We can therefore say that since all his original PE is now
KE:
PE = KE
mgh = 1/2 mv2
v = (2gh)1/2
From this expression, we can find his maximum
speed, or indeed his speed at any point on his swing, just
by knowing h, the distance he has moved vertically from
the board. By symmetry, on the other side of his swing,
his KE will turn to PE, and if no energy is lost, he should
end up at the same level he started. In practice, since
air resistance is constantly doing work on his body, he
loses energy to heating the air. There is therefore less
energy available to swing him to the same height, and, if
he does not use some of his own KE to increase his PE (a
more advanced swing), he will swing slower and slower and
eventually stop.
Calculating Maximum Speed:
Let us now find the maximum speed of a man on a trapeze,
modelled as above:
v = (2gh)1/2
v = (2 x 9.8 x 1.5m)1/2
v = 5.4ms-1
Using Simple Harmonic Motion, to find time period of swing
we must first prove than SHM is a valid way to model this
problem:
(Let theta = a) Therefore we must approximate sina = a
in order to obtain expression angular acceleration = -w2
a Since in our case, a is approx. 0.83 radians, sina = 0.73,
which is (fairly) close to a, we can make this approximation.
Therefore time period (time from one point in swing, to
same point in next swing) is given by:
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 man does not
try to move his body, and hangs in line with the cable (1)
(obviously not the case, but we can approximate to this,
in such a simple investigation). As the mass of the man
is spread out along his length, we will have to approximate,
and model the man as a particle, positioned at the man's
centre of mass, which we will assume to be halfway down
his body (2). The man and trapeze is therefore modelled
as a mass of 74kg (70kg for man, and 4kg for bar), at a
distance 4.6m from top of cable (3.7m for cable, and extra
0.9m for half man's length).(3). We will also ignore air
resistance.