Another example of vector product from the field of classical mechanics is a vector called the moment of momentum. Moment of momentum of a particle i about point O, H°, is defined as

where ri/o is the position vector connecting point O to i, mi is the mass of particle i, and vi is its velocity, as measured with respect to a coordinate system E that is fixed on earth (Fig. 3.9). The moment of momentum of a particle of mass m tracing a circle of radius r with speed v is

where O is the center of the circle and e is a unit vector perpendicular to the circle. This equation shows that moment of momentum about point O remains constant so long as the particle traverses a circle with constant speed. Let us next consider the moment of momentum about O of a particle of mass m moving in a straight line with constant speed v:

where d is the distance from point O to the straight line traversed by the particle and e is the unit normal to the plane created by point O and the particle path. Equations 3.22b and 3.22c are quite similar, and yet the motion with which each is associated is different (circular versus linear mo-

Figure 3.9. Moment of momentum of particles about the origin O of the Cartesian reference frame E.

e tion). Thus, it is difficult to assign a simple physical meaning to the concept of moment of momentum. Nevertheless, this concept plays a central role in the analysis of motion of solids and multibody systems. For a system of particles, the vector Ho is defined as follows:

where m1 is the mass and v1 is the velocity of the particle i, and the summation is over all particles in the system considered (i = 1, n).

Let us illustrate this definition with a simple example. Let A (0, 1, 2), B(—3, 1, 0), and C (6, 7, 3) represent the positions of points A, B, and C at time t = 0 in a Cartesian reference frame E. Let us assume further that each have the same velocity, v = 3 e1 (m/s). Particle A has a mass of 2 kg, B 4 kg, and C 3 kg. The moment of momentum about point O (0, 0, 0) for the system of particles at time t = 0 is

+ 3 (6 e1 + 7e2 + 3e3) X 3 e1 = (39 e2 — 81e3) (kg-m2/s)

In the following, we derive an equation that relates the time rate of change of moment of momentum of a system of particles to the rotational capacity of the external forces acting on the system. First, we take the time derivative of moment of momentum given by Eqn. 3.23:

dHo/dt = 2 (dri/dt) X m1 v1 + 2 r1 X m1 (dvi/dt) (3.24)

where time derivative of velocity v1 (dvi/dt) is the acceleration of particle i. The summation i is over all particles of the system under consideration. The first term on the right-hand side of Eqn. 3.24 is equal to zero because, by definition, dr1/dt is equal to the velocity v1, and the cross product of two identical vectors, by definition of the vector product, is always equal to zero. Thus the rate of change of moment of momentum becomes equal to dHo/dt = 2 r1 X m1 a1 (3.25)

Using Newton's second law, the time rate of change of moment of momentum of a system of particles can be related to the external forces acting on the system:

where F1 is the external force acting on particle i of the system under consideration. According to this equation, the time rate of change of moment of momentum with respect to a point fixed in space is equal to the resultant moment about the same point of external forces and force couples. Thus for a system of particles (bodies), the laws of motion state that:

where Mo is the moment generated by external forces about point O plus the sum of force couples. These two equations can be used to determine the translational as well as the average rotational motions of a system of particles. According to Eqn. 3.27b, the moment of momentum remains constant unless the system is acted on by an external moment.

In the following, we discuss examples of simulation of movement and motion based on the method of lumped masses. In this method, the mass of a body part is lumped and positioned at the center of mass of the body part. For example, a body segment such as an upper arm is represented as a weightless rigid rod with a point mass attached to the center of the rod. Lumped-mass models of human body have been used in simulating car crashes. The method is also commonly used in structural analysis. For example, the earthquake analysis of a four-story plane frame could be reduced to determination of the horizontal displacements of four lumped masses, each representing the mass of a level. As we see in the next chapter, this simplification results in modest errors in the estimates of forces acting on a system. Nevertheless, the lumped-mass analysis provides insights as to how inertial terms interact with the forces acting on a system.

Example 3.7. Gymnast Holds onto a Bar. A gymnast holding onto a bar rotates from a horizontal position under the action of gravitational force while keeping her body aligned in a straight line (Fig. 3.10). Determine the equations governing the motion of the gymnast.

Solution: We model the gymnast as a long and slender rod of length L and mass m. We lump the mass of the gymnast at the center of the rod. We assume that the wrists of the gymnast behave as hinge joints as the gymnast rotates around the bar in the clockwise direction under the influence of gravity. Let 0 be the angle the rod makes with the vertical axis. With respect to the Cartesian coordinate system shown in Fig. 3.10, the center of mass of the rod is given by the following position vector:

The velocity and acceleration of the center of mass are determined by taking the time derivative of the position vector given by Eqn. 3.28:

v = (d0/dt) [(L/2) cos 0 e1 + (L/2) sin 0 e2] (3.29a)

+ (d0/dt)2 [-(L/2) sin 0 e1 + (L/2) cos 0 e2] (3.29b)

Using the free-body diagram shown in Fig. 3.10, we can now write the equation of motion for the mass center as follows:

F1 = m [(d20/dt2) (L/2) cos 0 - (d0/dt)2 (L/2) sin 0] (3.30a) -mg + F2 = m [(d20/dt2) (L/2) sin 0 + (d0/dt)2 (L/2) cos 0] (3.30b)

where F1 and F2 are the horizontal and the vertical components, respectively, of the force exerted on the gymnast by the bar. These equations contain three unknowns that need to be determined as a function of time during swing: reaction forces F1 and F2 and the angle $ = $(t). We need yet another equation to determine these parameters. The moment of momentum for the swinging rod can be expressed as follows:

Taking the time derivative of Eqn. 3.31, we arrive at an equation for the rate of change of moment of momentum:

The external moment acting on the rod must be equal to (dHo/dt). The free-body diagram in Fig. 3.10 shows that only gravitational force creates moment with respect to the hinge O. The forces F1 and F2 act on the hinge at O and therefore create no moment. Thus, conservation of moment of momentum requires that m (d2$/dt2) [(L/2)2] e3 = r X (- mg e2)

This leads to the following differential equation:

Equation 3.33 is identical to the differential equation governing the motion of a simple pendulum. Thus, this simple analysis predicts that the period in which a gymnast rotates from horizontal down to vertical position depends on the height of the gymnast but not on their weight. The taller the gymnast, the longer will be the duration of swing.

Example 3.8. Swinging of Arms. While standing straight a man begins swinging his arms at constant frequency (Fig. 3.11). Compute the moment of momentum about the center of mass of the man in the standard standing configuration.

Solution: Let 6 be the angle that one of the arms, arm 1, makes with the vertical axis. Because of the asymmetry of motion, the other arm (arm 2) will make angle —6 with the vertical axis. The position vector of the center of mass of each arm with respect to the Cartesian coordinates shown in Fig. 3.11 are given by the following equations:

r1 = h e2 + d e1 — (L/2) cos 6 e2 — (L/2) sin 6 e3 (3.34a)

r2 = h e2 — d e1 — (L/2) cos 6 e2 + (L/2) sin 6 e3 (3.34b)

in which h is the vertical distance between the center of mass in the standard standing position and the line connecting the shoulders, d is half the length between the shoulders, and L is the length of an upper limb.

The velocity of the center of mass of each arm can be obtained by taking the time derivative of Eqn. 3.34:

v1 = (dd/dt) L sin d e2 — (dd/dt) L cos d e3 (3.35a)

v2 = (dd/dt) L sin d e2 + (dd/dt) L cos 0 e3 (3.35b)

In estimating the total moment of momentum Ho, we lump the distributed masses of the arms and assign them as point masses to the center of mass of each arm:

When the vector multiplication specified in this equation is carried out, the result can be expressed as

This equation shows that Ho varies as a function of time. The ground force must create enough moment about point O to compensate for the time variation in moment of momentum. In practice, what happens is that the trunk rotates slightly to the left and right during swinging of the arms in such a way to make Ho to be equal to zero at all times.

Example 3.9. Push-ups. Determine the reaction forces exerted on the feet and the hands during push-ups.

Solution: The person performing the push-ups is represented by a slender rod of length L with a lumped mass m (body weight) at its center. The arms are assumed to be composed of two weightless rods that are linked to each other. The shoulders are considered to lie 0.2 L from the free end of the rod that represents the body as shown in Fig. 3.12.

The position, velocity, and acceleration of the lumped mass m can be expressed as r = (L/2) (cos 6 e1 + sin 6 e2)

Let us now write the equations of motion of the center of mass:

2 (Qf + Qh) = m [-(L/2) (d26/dt2) sin 6 - (L/2) (d6/dt)2 cos 6] (3.37b)

In these equations, the symbol V denotes vertical forces whereas Q is used for horizontal friction forces. The subscripts f and h refer to foot and hand, respectively. Arms are assumed to transmit vertical forces only (Qh = 0).

The rate of rotation of the body during the push-ups is governed by the conservation of moment of momentum. Let us take moment of momentum with respect to point O, which lies between the points at which the feet touch the floor in the midplane of motion.

The time rate of change of moment of momentum must be equal to the sum of the external moment acting on the body:

Mo = (L/2) (cos 6 ex + sin 6 e2) X (-mg e2) + (0.8 L cos 6 ex)

X (2Vh e2)= [-mg (L/2) + 1.6 LVh ] cos 6 e3 (3.38b)

Equating 3.38a and 3.38b, we obtain an expression for the vertical ground force exerted on each hand:

According to this equation, the ground force carried by each arm is slightly less than one-third of the body weight when d26/dt2 = 0. This is the force that provides an excellent workout for the triceps muscle. Another major muscle group essential for push-ups is the vertebral column flexors (abdominals). These muscles stabilize the body position along a straight line, preventing vertebral column hyperextension.

An expression for the vertical ground force acting on each foot can be obtained by substituting Eqn. 3.38c into Eqn. 3.37a. As for the horizontal ground forces acting on each foot, we substitute Qh = 0 in Eqn. 3.37b. Without this approximation, it would not be possible to assign unique values for these ground forces by only considering the laws of motion for the whole body. Such force systems are said to be statically indeterminate.

A few more remarks about the mechanical analysis of the push-up are in order. First, the slender rod actually represents the body from head to the ankle. We have not taken into account the fact that the feet remain more or less vertical during push-up rather than aligned with the rest of the body. Second, it is relatively easy to determine the time history of 6 using equipment available in most engineering departments. One could hook up a videocamera to a computer and digitize the videoimage of the push-up events. Appropriate software can then be used to assess the values of 6, d6/dt, and d26/dt2 as a function of time. In this way, it is possible to study the effect the rate of motion has on the ground forces. Note that the ground reaction forces can also be measured directly, but one would need two force plates to quantify these contact forces.

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The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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