The COAM myth
Nowadays almost everyone, without much thought, refers to COAM as if simple commonplace knowledge, wich needs no further discussion, being understood by all. The simple and fundamental fact most people seem to ignore is that to be able to apply the angular momentum conservation law there has to be no net external torque acting on the system. This is referred to as being a closed system. Internal torques, however, don't change the angular momentum of a system.
Usually when explaining COAM reference is made to a figure skater pulling her arms in and increasing her rotational speed. Sometimes one can also see reference to a high-diver who is rotating coming of the board. These two examples have in common to be, for a short time, almost perfect closed systems, i.e., no external torques acting on the system. A golfer's down swing however is definitely not a closed system. Hence comparison of a golfer to skaters or high divers is not correct.
Moreover there is the word 'conservation' in COAM. This implies that there is something to be conserved to start with. But simply looking at a golfer at the top of his back swing it is evident that he has zero angular momentum, hence he cannot conserve any angular momentum since there isn't any when starting the down swing. However a spinning skater or a high diver start their motion with a considerable amount of angular momentum remaining virtually constant during the spin or dive.
To get a feeling for angular momentum I have done some thought experiments using two masses, linked together as shown in Fig1. Mass m1 is pinned and is limited to rotate around a fixed center. Mass m2, connected to m1 via a spring, can slide without friction along a massless slender rod which is solidly connected to mass m1. Hence both masses rotate together, whereas m1 can have additionally a linear motion along the rod. We assume motion on a horizontal plane with zero friction.
We assume prior to time zero the spring to have zero deflection from its neutral state, m2 to be temporarily fixed to the rod, and m1 and m2 to be rotating with the same angular speed around the center, on a horizontal plane without any friction. We then let go of the connection between m2 and the rod at t = 0, and analyze what happens to the system.
We have here typically a closed system, there being no external torques acting on the system. There is however an internal spring force acting between the two masses but it does not affect the conservation of angular momentum of the rotating system of the two masses m1 and m2. The black dot in Fig2 indicates the starting position for mass m2 at t = 0 sec;
Notice that m2 acquires a rather complicated motion around the center. Kind of an ellipse squeezed inward a bit when m2 passes through the vertical axis. There is indeed an intricate tug of war going on between the spring force - providing the centripetal force - and the centrifugal force pulling on the spring.
Fig3 shows the angular momentum for mass1 and mass2 during one complete revolution. Notice the large variations in the respective angular momenta. There is continuously a exchange of angular momentum going on between the two masses but yet their sum remains perfectly constant, illustrating neatly the law of conservation of angular momentum for a closed system, i.e., no external torques acting on the system.
We assume the same initial positions as above at time t = 0 sec. However this time we assume zero initial angular speed and a constant torque acting on the system. This situation is indeed much closer to that of a golfer's down swing.
The resulting motion is shown in Fig4. Indeed a very different type of motion, the mass m2 is gradually spiraling away from the center and mass1 and increasing continuously its angular speed. At each moment there is a temporary equilibrium between spring force and centrifugal force.
The angular momentum, individually and the sum total, is shown in Fig5. Notice a totally different situation. There is no angular momentum to be conserved as there is none to start with at the beginning. In effect it is linearly increasing. This is due to the external torque acting on the system.
To be more precise, internal torques don't effect the total angular momentum as long as they don't provoke a noticeable reaction torque relative to the earth. For Iron Byron this is not possible, having an inner revolute joint, and hence internal torques don't have any influence on the total angular momentum, thus only effecting their redistribution.
Let' s now next look at a golfer's downswing. We assume the golfer to be a Iron Byron golf machine as depicted in Fig6.
Torque1 is an external torque whereas torque2 is to be considered as an internal torque. Torque2 is modeled to constitute a 90 deg dead stop to prevent back knifing of the golf club. Torque1 is very simply taken to be constant during the down swing, see Fig7. The ensuing motion for Iron Byron is depicted in Fig8.
The behavior of the angular momentum is quite similar to that of experiment II. The total angular momentum is initially zero and gradually increases almost linearly with time. Notice in Fig9 that prior to impact there is some redistribution of angular momentum as there is some flow from the arm to the club.
A down swing basically can be thought to have two distinct phases. First phase - from the top both arm and club increase their respective angular momentum. Second phase - when club starts to release half way down the swing there is also some redistribution of angular momentum from arm to club.
The second phase is very much affected by the particular time history of the torque1/2. However to keep things as simple as possible torque2 was modeled to act as a passive 90 deg dead stop and torque1 was assumed to be constant.
Since we have a mathematical model we can readily experiment with various ideas. So let's analyze the case when only an internal torque ('wrist torque') is applied at the joint, as shown in Fig10, to clearly demonstrate that internal torques don't play a role in the conservation of angular momentum in rotating systems.
Imagine club and arm together rotating with a constant angular speed of 10 rad/s, with the joint kept rigid at a 90 deg angle. At t = 0 we let go and free up the joint but instead apply a negative torque (Fig11) for a some limited time such as to keep the joint angle 90 deg preventing release to occur too early.
We assume gravity to be zero as this also creates external torques. Therefore imagine the swing to take place on a horizontal frictionless surface. While it is true that internal torques don't effect the overall angular momentum they do however effect the time history of its distribution between various linked masses.
The resulting motion is shown in Fig12. Notice from Fig13 that the sum total of angular momentum remains perfectly constant. Don't be fooled into taking this for a real golf swing, it being only a thought experiment. However it shows clearly the unrealistic conditions required to have a down swing satisfying the conditions for applying the angular conservation law.
It is clear form above that COAM does not apply to a golf swing -
1) There are external forces acting on a golfer, therefore no closed system, and
2) A golfer starts the downswing with zero angular momentum and hence cannot conserve any.