Mechanics of a Pole Vault
The mechanics of a pole vault involve the most basic law in physics, Newton’s Second Law – F=ma (The force exerted on an object of mass m experiences an acceleration in the same direction as the force, proportional to the magnitude of the force, and inversely proportional to the object’s mass.). In pole vaulting, this means that the horizontal (x) force applied to the vaulter by the horizontal (x) component of the normal force on the end of the pole is equal to the mass of the vaulter multiplied by their horizontal (x) acceleration. The same theory holds for the force in the vertical (y) direction; except, the additional downward force of gravity is added to the model. From the center of mass diagram below, it can be determined that the acceleration in the x-direction (–> = +) is negative, at a decreasing rate until the pole is completely vertical. The initial x-velocity has the greatest magnitude and decreases to a low constant velocity as the angle of the force exerted on the vaulter gets closer to 90° (assuming 0° to be the [-x]-direction and counted clock-wise). It is a good idea to push the pole back to the ground at the top of the vault because it will 1. add another force in the positive x-direction, thus extending the distance over which the vaulter can clear the bar because of an increase in the final x velocity, and 2. so this won’t happen:
The velocity in the y-direction has its greatest positive magnitude at the top of the vault (where the slope of the center of mass position is steepest). After release of the pole, the vaulter follows a free-fall model, with a constant velocity in the x-direction.
So, the first assumption is that all of the kinetic energy of the vaulter at the plant is converted to potential energy without any loss. This, of course, never happens in real life, as energy is “lost” in the form of sound, heat, pole vibration, improper technique, etc. However, making these assumptions allows for us to create a simple model that only requires inputs of initial height of the vaulter’s center of mass and velocity at the take-off point.
>the mass is inconsequential (beyond its distribution throughout the body – which will affect center of mass) because it will cancel out from both sides of the equation. Leaving:
>Solve for h
>Then, all you have to do is add the calculated h value (from a certain velocity) to the vaulter’s center of mass (≈ 0.55[height of vaulter in m]).
Here is the abstract of a study done on pole vaulting, Energy loss in the pole vault take-off and the advantage of a flexible pole:
A model of pole vaulting with a flexible pole was developed with the aim of predicting the optimum take-off technique and pole characteristics for a typical world-class pole vaulter. The key features of the model are that it includes the interdependence of the take-off angle and the take-off velocity, and that it accounts for the energy losses in the pole plant and take-off phases of the vault. A computer simulation program was used to systematically investigate the effect of different combinations of take-off velocity, take-off angle, grip height and pole stiffness on the performance of a world-class male vaulter. For the highest vault with this model, the vault height and the optimum combination of take-off velocity, take-off angle, grip height and pole stiffness were in good agreement with measured values for world-class vaulters using fiberglass poles. The results from the model were compared with those from a model of vaulting with a rigid pole. There was a clear performance advantage to vaulting with a flexible pole. The flexible pole produced a 90 cm higher vault by allowing a 60 cm higher grip and by giving a 30 cm greater push height. There are two main advantages of a flexible fiberglass pole over a rigid pole made of steel or bamboo. A flexible pole reduces the energy dissipated in the vaulter’s body during the pole plant, and it also lowers the optimum take-off angle so that the athlete loses less kinetic energy when jumping up at take-off.