Applications to Human Body Dynamics Pole Vaulting

Pole vaulting is an exciting athletic event in which a vaulter clears a crossbar resting on two metal standards placed approximately 4 m apart (Fig. 8.10). On the ground in front of the crossbar is a small wedge-cut hole called the vaulting box that holds the end of the pole during the vault. Behind the standards is a landing pit that is at least 5 m wide. Back in 1877, the first championship was won with a vault of 2.92 m, but today vaulters reach the sky with much longer and flexible poles. We present two examples concerning the mechanical analysis of vaulting.

Example 8.5. Vaulting with a Rigid Pole. A vaulter of mass m grips a rigid vault of length L at a distance d from the front end. With the pole held to his side, the vaulter begins running from a distance of about 30 m from the crossbar (Fig. 8.10). Before he plants his pole in the vaulting box his running velocity is vo and the vertical distance between his center of mass and the floor is h. unlike the vaults presently used in athletic competitions, the one this vaulter uses is stiff; the pole does not bend or change in length in response to the ground force acting on it. Once the distal end of the vault is firmly on the ground, the vaulter pulls his hips forward and then begins rising in the air holding onto the vault. What should be his minimum running speed so that the vaulter can push the pole into a vertical configuration?

Solution: The linear momentum of the athlete and the pole before the pole hits the ground at point O is equal to (mvo). This linear momentum is not conserved during impact because of the impulsive ground forces

Figure 8.10. A vaulter clearing the crossbar using a stiff pole. The schematic diagrams show instances from the pole-vaulting events.

acting at O. However, the angular momentum with respect to O is conserved. Assuming that the mass of the athlete can be lumped at a single point on the weightless pole at a distance d from point O, the conservation of moment of momentum yields the following equation:

in which v1 is the speed of the athlete immediately after the impact. In deriving this equation, we took into consideration the fact that the velocity of the athlete right after the impact is no longer in the horizontal direction, but it is normal to the direction of the pole as shown in Fig. 8.10.

Equation 8.26a shows that the impact of the vault with the ground reduces the speed of the athlete from vo to vo h/d. Thus, the ratio of the kinetic energy after the impact to kinetic energy before the impact is equal to (h/d)2. For h = 0.9 m and d = 2 m, approximately 80% of the kinetic energy of the center of mass of the vaulter is dissipated during the impact.

Let us now consider the conservation of energy between two time points, time t1 right after the impact and time t2 when the pole becomes vertical. Assuming that much of the kinetic energy of the vaulter is associated with his center of mass, we find

Equating the mechanical energies at times t1 and t2, we arrive at the following relationship between the initial speed vo of the vaulter and the vertical distance that his center of mass can be lifted through the use of the vaulting pole:

This is the value of vo necessary to push the pole into the vertical position. Considering that the best of the athletes run 100 m in slightly less than 10 s, it is clear that an upper bound for vo is about 10 m/s. According to Eqn. 8.26d, the center of mass of a pole vaulter of height 2h = 1.8 m would only rise to about 2.0 m for vo = 10.3 m/s. Why is a rigid pole so inefficient in raising a human above a barrier? To answer this question, we need to consider the speed v1 of the vaulter immediately after the impact. For vo = 12 m/s and d = 4 m, v1 = 2.7 m/s. Clearly much of the linear momentum and the kinetic energy of the athlete are wasted when the rigid pole hits the ground. If the athlete were to use a flexible pole, part of the kinetic energy of the vaulter would have been stored as elastic energy in the pole. That additional energy can push the pole into a vertical position, as illustrated in the next example.

Example 8.6. Pole Vaulting with a Flexible Pole. The athlete in the aforementioned example replaces his rigid pole with a flexible one. The new pole is such that under the application of compressive forces at both ends it assumes the shape of an arc (Fig. 8.11). Slender rods subject to thrust buckle when the compressive forces acting on them reach a certain level. If the rod can rotate freely at both ends as shown in Fig. 8.11, it will buckle when the axial force F reaches the value

in which E represents Young's modulus of the circular rod, ro is its radius, and L is its length. This equation shows that the smaller the radius and the longer the rod, the more likely it is to buckle under thrust.

The potential energy stored in a buckled pole is like the potential energy of a spring:

Figure 8.11. Pole vaulting using a flexible pole. The pole buckles in response to the thrust exerted by the vaulter as he runs toward the cross bar while keeping the front end of the pole fixed to the ground.

in which k (n/m) is the stiffness of the pole, and Ax is the difference between the length d of the curved pole between the hand grip and the distal end and the shortest distance between these points (b):

When a vaulter uses a rigid pole, much of his initial kinetic energy is dissipated into heat during the brief period of impact between the pole and the ground. However, in the present case when the vaulter uses a flexible pole, as he hits the pole onto the ground the pole buckles and takes the shape of an arc. The kinetic energy lost before the takeoff is transformed into elastic energy and is stored in the pole.

Because the ground force acting on the distal end of the pole does no work, mechanical energy is conserved from the moment the vaulter places the distal end of the pole onto the ground to the moment when the vaulter lets go of the pole. Assuming again that the vaulter can be represented as a lumped mass of m, the mechanical energy of the vaulter-pole system at time to right before the impact is

where h denotes the vertical distance between the center of mass of the vaulter and the ground.

Initially, during the swing phase of pole vaulting when the front end of the pole is fixed on the ground, the pole will be compressed into a curved shape. The radius of curvature for the pole decreases until it reaches a minimum value. After that, the pole begins to extend, raising the center of mass of the vaulter and supplying him with additional kinetic energy. As the athlete clears the crossbar, all springlike elastic energy stored in the pole has been released (d = b), and the kinetic energy supplied by the pole to the athlete has been used for raising the legs above the crossbar. The mechanical energy at that time point is

Equating the mechanical energy at time to given by Eqn. 8.27d to the energy at time t2 given by Eqn. 8.27e, we arrive at the following relationship:

This equation shows that, using a flexible pole, a 1.8-m-tall athlete (h = 0.9 m) can clear a bar set at 6.0 m with an initial velocity of 9.1 m/s. [For more on the mechanical analysis of pole vaulting, see Ekevad and Lund-berg (1995) and other references at the end of this book.]

Example 8.7. Abdominal Wheel. Late in the 1980s, the abdominal wheel was a popular device for strengthening stomach muscles. An abdominal wheel is composed of a small rubber wheel with holders on both sides. As shown in Fig. 8.12, a person fixed their feet onto the ground and moved the wheel forward and backward in the direction of the body axis. The person begins to contract their abdominal muscles when the angle 0 between the floor and the lower body is 20°. Determine the motion of the person immediately following the contraction of the abdominal muscle. In solving the problem, represent the body with two rods of equal length

Figure 8.12. Schematic drawing of a woman using an abdominal wheel to strengthen abdominal muscle groups. The mechanical model of the person composed of two rigid links plus a contractile cord is shown in the lower part of the figure.

(L) and mass (M) that are hinged together at b. Model the contracting abdominal muscle as a spring connecting the two rods as shown in the figure. The muscle force is zero at 6 = 45°. Assume also that the floor is smooth so that the roller slides along the floor without frictional resistance.

Solution: The vertical forces acting on the rods ab and bc are shown in the figure. For each rod, the principle of conservation of mechanical energy holds: the rate of change of kinetic energy must be equal to the work done on the rod by the external forces. At time t = t1 and 6 = 20°, both rods are stationary and therefore their kinetic energy must be equal to zero. There is potential energy, however, stored in the contracting abdominal muscle. Let us next consider the mechanical energy at a later time t2. The reaction force at point a (feet) acting on rod ab does no work because point a remains stationary. Because the wheel is polished (fric-tionless), the ground force acting at point c is perpendicular to the displacement of c and thus does no work. With these considerations, the conservation of mechanical energy takes the following forms for rods ab and bc:

T2ab + Mg (L/2) sin 6 = Mg (L/2) sin 20° + Wsab + W1-2 (8.28a)

T2bc + Mg (L/2) sin 6 = Mg (L/2) sin 20° + Wsbc - W1-2 (8.28b)

in which Wsab is the work done on rod ab by the spring DE (abdominal muscle) and W1-2 denotes the work done by the reaction force acting on point B on rod ab. Because action equals to reaction, W2-1 = — W1-2. Adding these two equations together, we obtain the following equation for the whole system:

r2ab + T2bc + Mg L sin 6 = Mg L sin 20° + Wsab + Wsbc (8.28c)

The work done by the spring (Wsab + Wsbc) can be expressed as the change in potential energy of the spring, and therefore Eqn. 8.28c reduces to

T2ab + T2bc + Mg L sin 6 + (1/2) k [2d(cos 6 — cos 45°)]2

= Mg L sin 20° + (1/2) k [2d(cos 20° — cos 45°)]2 (8.28d)

in which k denotes the spring stiffness of the abdominal muscle and d is defined in the figure.

Let us now compute the kinetic energy of rods ab and bc. In each case we need to evaluate the component of the kinetic energy due to the velocity of the center of mass and the remaining part due to the rotation of the rod. Each rod has the same angular speed ||d6/dt||. The speeds of the center of masses of ab and bc are obtained from the time derivatives of their respective coordinates:

||vG|| = (d6/dt)(L/2) (cos2 6 + 9 sin2 6)05 (8.28f)

Because the mass moment inertia of a rod with mass M and length L about the center of mass is equal to ML2/12, the kinetic energy of the two rods can be expressed as

T2ab = (1/8) M L2 [(d6/dt)2 + (1/2) (M L2/12) (d6/dt)2 = T2ab = (1/6) M L2 (d6/dt)2

T2bc = (1/8) M L2 (d6/dt)2 (1 + 8 sin2 6) + (1/2) (M L2/12) (d6/dt)2

T2bc = (1/8) M L2 (d6/dt)2 [4/3 + 8 sin2 6] (8.28g)

Substituting Eqn. 8.28g into Eqn. 8.28d, we determine the time rate of change of angle 6 as a function of 6. In particular, we want to know the angle at which the shortening of the abdominal muscle comes to a halt. At that instant kinetic energy must be equal to zero, and thus

Mg L sin 20° + (1/2) k [2d(cos 20° — cos 45°)]2

This equation can be written in the following convenient form:

sin 6 — sin 20° = p{(cos 20° — cos 45°)2 — (cos 6 — cos 45°)2} (8.28h)

in which the dimensionless parameter p is defined by the following equation:

Parameter ¡3 is a measure of the strength of the abdominal muscle relative to the weight of the body. Using the method of trial and error one can show that for ¡3 = 6, the abdominal muscle will contract approximately to the angle 6 = 40°.

Midsection Meltdown

Midsection Meltdown

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