Biomechanical Basis of Human Movement, 3rd Edition

Section III - Mechanical Analysis of Human Motion

Chapter 11

Angular Kinetics

Objectives

After reading this chapter, the student should be able to:

  1. Define torqueand discuss the characteristics of a torque.
  2. State the angular analogs of Newton's three laws of motion and their impact on human movement.
  3. Discuss the concept of moment of inertia.
  4. Understand the impact of angular momentum on human motion.
  5. Define the concept of center of mass.
  6. Calculate the segment center of mass and the total body center of mass.
  7. Differentiate between the three classes of levers.
  8. Discuss the relationships between torque, angular work, rotational kinetic energy, and angular power.
  9. Define and conduct a static analysis on a single joint motion.
  10. Define stabilityand discuss its effect on human movement.
  11. Define and conduct a dynamic analysis on a single joint motion.
  12. Define the impulse-momentum relationship.
  13. Define the work-energy relationship.

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Chapter 10 discussed the notion that movement does not occur unless an external force is applied. Also discussed were the characteristics of a force, two of which were the line of action and the point of application. If the line of action and the point of application of a force are critical, it would appear that the type of motion produced may depend on these characteristics. For example, a nurse pushing a wheelchair exerts two equal forces, one on each handle. The result is that the lines of action and points of application of the two forces cause the wheelchair to move in a straight line. What happens, though, when the nurse pushes the chair with only one arm, applying a force to only one of the handles? A force is still applied, but the motion is totally different. In fact, the wheelchair will translate and rotate (Fig. 11-1). The situation that has just been described actually represents most of the types of motion that occur when humans move. It is rare that a force or a system of forces cause pure translation. In fact, the majority of force applications in human movement cause simultaneous translation and rotation.

The branch of mechanics that deals with the causes of motion is called kinetics. The branch of mechanics that deals with the causes of angular motion is called angular kinetics.

 

FIGURE 11-1 An overhead view of a wheelchair. A. The wheelchair is translated forward by forces F1 and F2. B. The wheelchair rotates and translates if only one force, F1, is applied.

Torque or Moment of Force

When a force causes a rotation, the rotation occurs about a pivot point, and the line of action of the force must act at a distance from the pivot point. When a force is applied such that it causes a rotation, the product of that force and the perpendicular distance to its line of action is referred to as a torque or a moment of force. These terms are synonymous and are used interchangeably in the literature and will be used interchangeably in this text. A torque is not a force but merely the effect of a force in causing a rotation. A torque is defined, therefore, as the tendency of a force to cause a rotation about a specific axis. In a 2D analysis, the axis about which the torque acts is neither the horizontal nor vertical axis. The torque acts about an axis that is perpendicular to the x-y plane. This axis is called the z-axis. Thus, the torques referred to in this chapter always act about the z-axis (Fig. 11-2).

Characteristics of A Torque

The two important components of a torque are magnitude of the force and the shortest or perpendicular distance from the pivot point to the line of action of the force. Also,

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torque acting about the z-axis any discussion of a torque must be with reference to a specific axis serving as the pivot point. Mathematically, torque is:

 

FIGURE 11-2 The z-axis is perpendicular to the x-y plane. In 2-D, torques act about the z-axis. A positive torque would appear to come out of the page while a negative torque would go into the page.

where T is the torque, F is the applied force in newtons, and r is the perpendicular distance (usually in meters) from the line of action of the force to the pivot point. Because torque is the product of a force, with units of newtons, and a distance, with units of meters, torque has units of newton-meters (N-m). The distance, r, is referred to as the torque arm or the moment arm of the force (these terms may also be used interchangeably). This concept is illustrated in Figure 11-3. If the force acts directly through the pivot point or the axis of rotation, the torque is zero because the moment arm would be zero. Thus, regardless of the magnitude of the force, the torque would be zero (Fig. 11-3A).

In Figure 11-3B, a force of 20 N is applied perpendicular to a lever at a point 1.1 m from the axis. Because the force is not applied through the axis of rotation, the shortest distance to the axis is the perpendicular distance from the point of contact of the force to the axis. The torque generated by this force can be computed as:

Another method of calculating a torque uses trigonometric functions. An example of a 20 N force applied 1.44 m from the axis or rotation at an angle of 50° to the lever is shown inFigure 11-3C. This is a more common situation than the former case. To compute the torque, we can compute the moment arm by taking the product of the sine of the angle and the distance the force acts from

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theaxis of rotation. This product would in essence give us the perpendicular distance of the line of application of the force to the axis of rotation. This computation would be:

 

FIGURE 11-3 When a force is applied, it causes no rotation if it is applied through the pivot point (A), or it generates a torque if the force is applied a distance from the axis (B and C). The moment arm (MA) is the perpendicular distance from the line of action of the force to the axis of rotation Pivot or axis of rotation.

In this case, the perpendicular distance of the moment arm is 1.1 m as we had in the previous example and the torque is the same. The second formula using the sine of the angle for the computation of torque is actually a more general case of the original formula. If the angle between the moment arm and the line of application of the force was 90°, then the sine of 90° is 1, and we simply assume we have a perpendicular distance. In any other case in which the angle is not 90°, taking product of the sine of the angle and the distance from the axis of rotation produces of the perpendicular distance.

When the force is not applied through the pivot point, as in Figure 11-3B and C, a torque is said to result from an eccentric force, literally an off-center force. Although an eccentric force primarily causes rotation, it also causes translation. Examples of torque applications are illustrated in Figure 11-4. Force can be generated by muscles pulling a distance away from the joint (Fig. 11-4A), by the weight of a body segment acting downward away from a joint (Fig. 11-4B), or by a force coming up from the ground acting a distance away from the center of gravity (Fig. 11-4C).

Torque is a vector quantity and thus has magnitude and direction. The magnitude is represented by the quantity of the magnitude of the force times the magnitude of the moment arm perpendicular to the line of force application. The direction is determined by the convention of the right-hand rule. In this convention, the fingers of the right hand are placed along line of the moment arm and curled in the direction of the rotation. The direction in which the thumb points indicates the positive or negative sign of the torque. As noted with angular measurement, a counterclockwise direction is considered positive (23), and a clockwise direction is negative. Thus, a counterclockwise torque is positive, and a clockwise torque is negative. Torques in a system may be evaluated through vector operations as has been described for forces. That is, torques may be composed into a resultant torque.

The concept of torque is prevalent in everyday life, for example, in the use of a wrench to loosen the nut on a bolt. Applying a force to the wrench produces rotation, causing the nut to loosen. Intuitively, the wrench is grasped at the end as in Figure 11-5A. Grasping it in this way maximizes the moment arm and hence the torque. If the wrench were grasped at its midpoint, the torque would be halved, even though the same force magnitude is applied (Fig. 11-5B). To achieve the same torque as in the first case, the force magnitude must be doubled. Thus, increasing the force magnitude, increasing the moment arm, or both, can increase a torque.

The concept of torque is often used in rehabilitation evaluation. For example, if an individual has an injured elbow, the therapist may use a manual resistance technique to evaluate the joint. The therapist would resist the individual's elbow flexion by exerting a force at the mid-forearm position. This creates torque that the patient must overcome. As the individual progresses, the therapist may exert approximately the same force level at the wrist instead of the mid-forearm. By increasing the moment arm while keeping the force constant, the therapist has increased the torque that the patient must overcome. This rather simple technique can be helpful to the therapist in developing a program for the individual's rehabilitation.

 

FIGURE 11-4 How torques are commonly generated by muscle force (A), gravitational force (B), and a ground reaction force (C). MA, movement arm.

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FIGURE 11-5 A wrench with two points of force application. Grasping the wrench at the end (A) generates more torque than a grasp near the point of rotation (B) because the moment arm is greater at A than at B.

Force Couple

A gymnast who wishes to execute a twist about a longitudinal axis applies not one but two parallel forces acting in opposite directions. By applying a backward force with one foot and a forward force with the other, the gymnast creates two torques that produce a rotation about their longitudinal axis (Fig. 11-6). Such a pair of forces is called a force couple. A force couple is two parallel forces that are equal in magnitude and that act in opposite directions.

 

FIGURE 11-6 A torque FR * dR is created by the right foot while another torque, FL * dL, is created by the left foot. Because these two torques are equal and in the same angular direction, the force couple will result in a rotation about the longitudinal axis through the center of mass.

These two forces act at a distance from an axis of rotation and produce rotation about that axis. A force couple can be thought of as two torques or moments of force, each creating a rotation about the longitudinal axis of the gymnast. Torques, however, also cause a translation, but because the translation caused by each torque is in the opposite direction, the translation is canceled out. Thus, a force couple causes a pure rotation about an axis with no translation. By placing the feet slightly farther apart, the gymnast in Figure 11-6 can increase the moment arm and thus cause a great deal more rotation. A force couple is calculated by:

where F is one of the equal and opposite forces and d is the distance between the lines of action of the two forces. Although no true force couples exist in human anatomy, the human body often uses force couples. For example, a force couple is created when one uses the thumb and forefinger to screw open the top on a jar.

Newton's Laws of Motion: Angular Analogs

The linear case of Newton's laws of motion was presented in Chapter 10. These laws can be restated to represent the angular analogs. Each linear quantity has a corresponding angular analog. For example, the angular analog of force is torque, of mass is moment of inertia, and of acceleration is angular acceleration. These analogs can be directly substituted into the linear laws to create the angular analogs.

First Law: Law of Inertia

A rotating body will continue in a state of uniform angular motion unless acted on by an external torque. Stated mathematically as in the linear case:

That is, if the sum of the torques is zero, then the object is either in a state of rest or rotating at a constant angular velocity.

To completely understand this equation, however, we must first discuss the concept of inertia in the angular case. In the linear case, according to Newton's first law of motion, inertia is an object's tendency to resist a change in linear velocity. The measure of an object's inertia is its mass. The angular counterpart to mass is the moment of inertia. It is a quantity that indicates the resistance of an object to a change in angular motion. Unlike its linear counterpart, mass, the moment of inertia of a body is dependent not only on the mass of the object but also on the distribution of mass with respect to an axis of rotation. The moment of inertia also has different values because an object may rotate about many axes. That is, the moment of inertia is not fixed but changeable.

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If a gymnast rotates in the air in a layout body position, the way in which the moment of inertia changes can be illustrated. Suppose the gymnast rotates about the longitudinal axis passing through the center of mass of the total body. The center of mass is the point at which all of the mass appears to be concentrated; the calculations are presented later in this chapter. The mass of the gymnast is distributed along and relatively close to this axis that passes through the center of mass (Fig. 11-7A). If, however, the gymnast rotates about a transverse axis through the center of mass of the total body, the same mass is distributed much farther from the axis of rotation (Fig. 11-7B). Because there is a greater mass distribution rotating about the transverse axis than about the longitudinal axis, the moment of inertia is greater in the latter case. That is, there is a greater resistance to rotation about the transverse axis than about the longitudinal axis. The gymnast may also alter mass distribution about an axis by changing body position, as with assuming a tuck position, bringing more body mass closer to the transverse axis, thus decreasing the moment of inertia. In multiple aerial somersaults, gymnasts assume an extreme tuck position by almost placing the head between the knees in an attempt to reduce the moment of inertia. They do this to provide less resistance to angular acceleration and thus complete the multiple somersaults.

The concept of reducing the moment of inertia to enhance angular motion is also seen in running. During the swing phase, the foot is brought forward from behind the body to the point on the ground where the next foot-ground contact will be made. After the foot leaves the ground, however, the leg flexes considerably at the knee, and the foot is raised up close to the buttocks. The effect of this action is to decrease the moment of inertia of

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the lower extremity relative to a transverse axis through the hip joint. This enables the limb to rotate forward more quickly than would be the case if the lower limb were not flexed. This action is a distinguishing feature of the lower extremity action of sprinters. Figure 11-8 illustrates the change in the moment of inertia of the lower extremity during the recovery action in running.

 

FIGURE 11-7 The mass distribution of an individual about the longitudinal axis through the total body center of mass (A) and about a transverse axis through the total body center of mass (B).

 

FIGURE 11-8 The changes in the leg's moment of inertia during the stride.

The calculation of moment of inertia is not trivial. If all objects are considered to be made up of a number of small particles, each with its own mass and its own distance from the axis of rotation, the moment of inertia can be represented in mathematical terms:

where I is the moment of inertia, n represents the number of particle masses, mi represents the mass of the ith particle, and ri is the distance of the ith particle from the axis of rotation. That is, the moment of inertia equals the sum of the products of the mass and the distance from the axis of rotation squared of all mass particles comprising the object. A dimensional analysis results in units of kilogram-meters squared (kg-m2) for moment of inertia.

Consider the illustration in Figure 11-9. This hypothetical object is composed of five-point masses, each with a mass of 0.5 kg. The distances r1 to r5 represent the distance from the axis of rotation. The point masses are each 0.1 m apart, with the first mass being 0.1 m from the y-y axis. Each point mass is 0.1 m from the x-x axis. The moment of inertia about the y-y axis is:

If the axis of rotation is changed to the x-x axis, the moment of inertia of the object about this axis would be:

The change in the axis of rotation from the y-y axis to the x-x axis thus dramatically reduces the moment of inertia, resulting in less resistance to angular motion about the x-x axis than about the y-y axis. If an axis that passes through the center of mass of the object is used, the mass of point 3 would not influence the moment of inertia because the axis passes directly through this point. Thus:

From these examples, it should now be clear that the moment of inertia changes according to the axis of rotation.

In the human body, the segments are not as simply constructed as in the example. Each segment is made up of different tissue types, such as bone, muscle, and skin, which are not uniformly distributed. The body segments are also irregularly shaped. This means that a segment is not of uniform density, so it would be impractical to determine the moment of inertia of human body segments using the particle-mass method. Values for the moment of inertia of

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each body segment have been determined using a number of methods. Moment of inertia values have been obtained experimentally. The values were generated from cadaver studies (10), mathematical modeling (20,21) and from gamma-scanning techniques (63). Jensen (27) has developed prediction equations specifically for children based on the body mass and height of the child.

 

FIGURE 11-9 A hypothetical five-point mass system.

It is necessary for a high degree of accuracy to calculate a moment of inertia that is unique for a given individual. Most of the techniques that provide values for segment moments of inertia provide information on the segment radius of gyration; from this value, the moment of inertia may be calculated. The radius of gyration denotes the segment's mass distribution about the axis of rotation and is the distance from the axis of rotation to a point at which the mass can be assumed to be concentrated without changing the inertial characteristics of the segment. Thus, a segment's moment of inertia may be calculated by:

where I is the moment of inertia, m is the mass of the segment, l is the segment length, and the Greek letter rho (p) is the radius of gyration of the segment as a proportion of the segment length. For example, consider the leg of an individual with a mass and length of 3.6 kg and 0.4 m, respectively. The proportion of the radius of gyration to segment length is 0.302, based on the data of Dempster (10). This information is sufficient to calculate the moment of inertia of the leg about an axis through the center of mass of the leg. Thus, the moment of inertia is:

Table 11-1 illustrates the radius of gyration as a proportion of the segment length values from Dempster (10).

Using the radius of gyration technique, the moment of inertia about a transverse axis through the proximal and distal ends of the segment may also be calculated. The radius of gyration as a proportion of segment length about the proximal end of the leg in the previous example is 0.528. Therefore, the moment of inertia about the proximal end of the segment is calculated as:

TABLE 11-1 Radii of Gyration as a Proportion of Segment Length About a Transverse Axis

Segment

Center of Mass

Proximal

Distal

Head, neck, trunk

0.503

0.830

0.607

Upper arm

0.322

0.542

0.645

Arm

0.303

0.526

0.647

Hand

0.297

0.587

0.577

Thigh

0.323

0.540

0.653

Leg

0.302

0.528

0.643

Foot

0.475

0.690

0.690

From Dempster, W. T. (1955). Space requirements of the seated operator. WADC Technical Report. Wright-Patterson Air Force Base, 55-159.

At the distal end point of the leg segment, the moment of inertia is:

Using Max TRAQ, import the midstance video file of the woman walking. Digitize the endpoints of the leg (i.e., knee and ankle), convert these values to real units (i.e., meters), and calculate the moment of inertia about transverse axes through the proximal and distal end point of the segment.

The moment of inertia value for any segment is usually given for an axis through the center of mass of the segment. The moment of inertia about an axis through the center of mass is the smallest possible value of any parallel axis through the segment. For example, in Figure 11-10, three parallel transverse axes are drawn through the leg segment. These axes

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are through the proximal end point, through the center of mass, and through the distal end point. Because the mass of the segment is distributed evenly about the center of mass, the moment of inertia about the center of mass axis is small, and the moments of inertia about the other axes are greater but not equal. This was illustrated in the previous moment of inertia calculations. Because the mass of most segments is distributed closer to the proximal end of the segment, the moment of inertia about the proximal axis is less than about a parallel axis through the distal end point.

 

FIGURE 11-10 Parallel transverse axes through the proximal, distal, and center of mass points of the thigh.

The moment of inertia can be calculated about any parallel axis, given the moment of inertia about one axis, the mass of the segment, and the perpendicular distance between the parallel axes. This calculation is known as the parallel axis theorem. Assume that the moment of inertia about a transverse axis through the center of mass of a segment is known and it is necessary to calculate the moment of inertia about a parallel transverse axis through the proximal end point. This theorem would state that:

where Iprox is the moment of inertia about the proximal axis, Icm is the moment of inertia about the center of mass axis, m is the mass of the segment, and r is the perpendicular distance between the two parallel axes. From the calculation in the example of the leg, it was determined that the moment of inertia about an axis through the center of mass was 0.0525 kg-m2. If the center of mass is 43.3% of the length of the segment from the proximal end, the moment of inertia about a parallel axis through the proximal end point of the segment can be calculated. If the length of the segment is 0.4 m, the distance between the proximal end of the segment and the center of mass is:

The moment of inertia about the proximal end point, then, is:

This value is the same as was calculated using the radius of gyration and segment length proportion from Dempster's data (10). Thus, the moment of inertia about an axis through the center of mass is less than the moment of inertia about any other parallel axis through any other point on the segment.

Second Law: Law of Angular Acceleration

An external torque produces an angular acceleration of a body that is proportional to and in the direction of the torque and inversely proportional to the moment of inertia of the body.

This law may be stated algebraically as:

where ΣT is the sum of the external torques acting on an object, I is the moment of inertia of the object, and a is the angular acceleration of the object about the z-axis (on an X-Y plane). For example, if an individual abducts the arm from the body to a horizontal position, the torque at the shoulder results in an angular acceleration of the arm. The greater the moment of inertia of the arm about an axis through the shoulder, the less the angular acceleration of the segment.

The expression of the relationship in Newton's second law is again analogous to the linear case. That is, the sum of the external torques is equal to the time rate of change in angular momentum. That is:

where dIω is the change in angular momentum. Newton's second law can thus be restated:

where H is the angular momentum. That is, torque is equal to the time rate change of angular momentum. To change the angular momentum of an object, external torque must be applied to the object. The angular momentum may increase or decrease, but in either case, external torque is required. Angular momentum is the quantity of angular motion of an object and has units of kg-m2 s-1. Angular momentum is a vector and the right-hand rule determines the direction of the vector. Again, counterclockwise rotations are positive, and clockwise rotations are negative.

As with the linear case, by cross-multiplying the preceding equation, we get:

That is, the product of torque and the time over which it is applied is equal to the change in angular momentum. The quantity T * dt is referred to as the angular impulse.

When gravity is the only external force acting on an object, as in projectile motion, the angular momentum generated at takeoff remains constant for the duration of the flight. This principle is known as the conservation of angular momentum and is derived from Newton's first law, that the angular momentum of a system will remain constant unless an external torque is applied to the system. Angular momentum is conserved during flight because the body weight vector, acting through the total body center of gravity, creates no torque (i.e., the moment arm is zero). No internal movements or torques generated at the segments can influence the angular momentum

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generated at takeoff. This principle enables divers and gymnasts to accomplish aerial maneuvers by manipulating their moments of inertia and angular velocities because their angular momentum is constant.

Consider the angular momentum of a gymnast performing an aerial somersault about a transverse axis through the total body center of mass (Fig. 11-11). The torque applied over time at the point of takeoff determines the quantity of angular momentum. During the flight phase, angular momentum does not change. The gymnast, however, may manipulate the moment of inertia to spin faster or slower about the transverse axis. At takeoff, the gymnast is in a layout position with a relatively large moment of inertia and a relatively small angular velocity or rate of spin. As the gymnast assumes a tuck position, the moment of inertia decreases and the angular velocity increases accordingly because the quantity of angular momentum is constant. Having completed the necessary rotation and in preparation to land, the gymnast opens up, assuming a layout position, increasing the moment of inertia and slowing the rate of spin. If these actions are done successfully, the gymnast will land on his or her feet.

To this point, only angular momentum about a single axis has been discussed. Angular momentum about one axis may be transferred to another axis. This occurs in many activities in which the body is a projectile. Although the total angular momentum is constant, it may be transferred, for example, from a transverse axis through the center of mass to a longitudinal axis through the center of mass. For example, a diver may twist about the longitudinal and initiate actions that produce a somersault about the transverse axis. This dive is known as a full twisting one-and-a-half somersault. Researchers have investigated arm and hip movements to accomplish this change in angular momentum (15,56,64,65). Other activities that use the principles of transferring angular momentum are freestyle skiing and gymnastics.

 

FIGURE 11-11 The angular momentum, moment of inertia, and angular velocity of a diver completing an aerial somersault. Throughout the aerial portion of the dive, the total body angular momentum of the diver is constant. When the diver is in a layout position, the moment of inertia decreases and the angular velocity increases in proportion. During the tuck portion of the dive, the angular velocity increases and the moment of inertia decreases in proportion.

Rotations may be initiated in midair even when the total body angular momentum is zero. These are called zero momentum rotations. A prime example of this is the action of a cat when dropped from an upside-down position. The cat initiates a zero momentum rotation and lands on its feet. As the cat begins to fall, it arches its back, or pikes, to create two body sections, a front and a hind section, and two distinct axes of rotation (Fig. 11-12A). The cat's front legs are brought close to its head, decreasing the moment of inertia of the front section, and the upper trunk is rotated 180° (Fig. 11-12B). The cat extends its hind limbs and rotates the hind section in the opposite direction to counteract the rotation of the front segment. Because the moment of inertia of the hind section is greater than that of the front section, the angular distance that the hind section moves is relatively small. To complete the rotation, the cat brings the hind legs and tail into line with its trunk and rotates the back section about an axis through the hind section (Fig. 11-12C). The reaction of the front portion of the cat to the hind section rotation is small because the cat creates a large moment of inertia by extending its front legs. Finally, the cat has rotated sufficiently to land upright on its four paws (Fig. 11-12D). The use of such actions in sports such as diving and track and field has received considerable attention (12,15).

 

FIGURE 11-12 A cat initiating a rotation in the air in the absence of external torque.

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FIGURE 11-13 An illustration of the local (HL) and remote (HR) angular momenta of the leg segment.

During human movement, multiple segments rotate. When this happens, each individual segment has angular momentum about the segment's center of mass and also about the total body center of mass. The angular momentum of a segment about its own center of mass is referred to as the local angular momentum of the segment. The angular momentum of a segment about the total body center of mass is referred to as the remote angular momentum of the segment. A segment's total angular momentum is made up of both local and remote aspects (Fig. 11-13). Expressed algebraically:

If the total angular momentum of an individual is calculated, the local aspects of each segment and the remote aspects of each segment must be included. Therefore:

where i represents each segment and n is the total number of segments.

Local angular momentum is expressed as:

where Hlocal is the local angular momentum of the segment, I is the moment of inertia about an axis through the segment center of mass, and co is the angular velocity of the segment about an axis through the segment center of mass. The remote aspect of angular momentum is calculated as:

where H remote is the remote angular momentum, m is the mass of the segment, d is the distance from the segment center of mass to the total body center of mass, and co' is the angular velocity of the segment about an axis through the total body center of mass (24). Figure 11-14 illustrates the proportion of local and remote angular momentum of the total angular momentum in a forward two-and-a-half rotation dive. In this instance, the remote angular momentum makes up a greater proportion of the total angular momentum than does the local angular momentum.

This technique for calculating total body angular momentum has been used in a number of biomechanics studies. Diving, for example, has been an area of study concerning the angular momentum requirements of many types of dives. Hamill et al. (19) reported that the angular momentum of tower divers increased as the number of rotations required in the dive increased (Fig. 11-15). Values as great as 70 kg-m2/s have been reported for springboard dives (37). The inclusion of a twisting movement with a multiple rotation dive further increases the angular momentum requirements (49).

 

FIGURE 11-14 The relationship of total body angular momentum, total local angular momentum, and total remote angular momentum of a diver during a forward two-and-a-half rotation dive. The hash mark denotes the instant of takeoff. (Adapted from 

Hamill, J., et al. [1986]. Angular momentum in multiple rotation non-twisting dives.International Journal of Sports Biomechanics, 2:78-87.

)

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Hinrichs (25) analyzed the motion of the upper extremities during running by considering angular momentum. He used a three-dimensional (3D) analysis to determine angular momentum about the three cardinal axes through the total body center of mass. Hinrichs reported that the arms made a meaningful contribution only about the vertical longitudinal axis. The arms generate alternating positive and negative angular momenta and tend to cancel out the opposite angular momentum pattern of the legs. These findings are illustrated in Figure 11-16. The upper portion of the trunk was found to rotate in conjunction with the arms, and the lower portion of the trunk rotates in conjunction with the legs.

 

FIGURE 11-15 Profiles of back dives with half a rotation, one rotation, and two rotations depicting the buildup of angular momentum on the platform. The hash marks denote the instants of takeoff. (Adapted from 

Hamill, J., et al. [1986]. Angular momentum in multiple rotation non-twisting dives. International Journal of Sports Biomechanics, 2:78-87.

)

 

FIGURE 11-16 The vertical component of the angular momentum of the arms, head and trunk, legs, and total body of a runner at a medium running speed. (Adapted from

Hinrichs, R. N. [1987]. Upper extremity function in running. II: Angular momentum considerations. International Journal of Sports Biomechanics, 3:242-263.

)

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Third Law: Law of Action-Reaction

For every torque exerted by one body on another body, there is an equal and opposite torque exerted by the latter body on the former.

This law illustrates exactly the same principal as in the linear case. When two objects interact, the torque exerted by object A on object B is counteracted by a torque equal and opposite exerted by object B on object A. These torques are equal in magnitude but opposite in direction. That is:

Generally, the torque generated by one body part to rotate that part results in a countertorque by another body part. This concept applies in activities such as long jumping. For example, the long jumper swings the legs forward and upward in preparation for landing. To counteract this lower body torque, the remainder of the body moves forward and downward, producing a torque equal and opposite to the lower body torque. Although the torque and countertorques are equal and opposite, the angular acceleration of these two body portions is different because the moments of inertia are different.

When equal torques are applied to different bodies, however, the resulting angular acceleration may not be the same because of the difference in the moments of inertia of the respective bodies. That is, the effect of one body on another may be greater than the latter body on the former. For example, when executing a pivot during a double play, the second baseman jumps into the air and throws the ball to first base. In throwing the ball, the muscles in the second baseman's throwing arm create a torque as the arm follows through. The rest of the body must counter this torque. These torques are equal and opposite but have a much different effect on the respective segments. While the arm undergoes a large angular acceleration, the body angularly accelerates much less because the moment of inertia of the body is greater than that of the arm.

Center of Mass

An individual's body weight is a product of the mass and the acceleration due to gravity. The body weight vector originates at a point referred to as the center of gravity, or the point about which all particles of the body are evenly distributed. The point about which the body's mass is evenly distributed is referred to as the center of mass. The terms center of massand center of gravity are often used synonymously. The center of gravity, however, refers only to the vertical direction because that is the direction in which gravity acts. The more general term is the center of mass.

If the center of mass is the point about which the mass is evenly distributed, it must also be the balancing point of the body. Thus, the center of mass can be further defined as the point about which the sum of the torques equal zero. That is:

Figure 11-17 shows an illustration of two objects with different masses. Object A results in a counterclockwise torque about point C, and object B results in a clockwise torque, also about point C. If these two torques are equal, the objects are balanced and point C may be considered the center of mass. This does not imply that the mass of these two object masses are the same but that the torques created by the masses are equal (Fig. 11-17).

The center of mass is a theoretical point whose location may change from instant to instant during a movement. The change in position of the center of mass results from the rapidly changing positions of the body segments during movement. In fact, the center of mass does not necessarily have to be inside the limits of the object. For example, the center of mass of a doughnut is within the inner hole but outside the physical mass of the doughnut. In the case of a human performer, the positions of the segments can also place the center of mass outside the body. In activities such as high jumping and pole vaulting, in which the body must curl around the bar, the center of mass is certainly outside the limits of the body (23).

 

FIGURE 11-17 A two-point force system balanced at the center of mass.

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Center of Mass Calculation: Segmental Method

Segment Center of Mass Calculation

A number of methods can be used to compute the center of mass of an object using balancing techniques. The most common is to compute the center of mass of individual segments, which are then combined to provide the location of the center of mass of the total system. This approach, called the segmental method, involves knowledge of the masses and the location of the centers of mass of each of the body's segments. Two-dimensional coordinates from digitized data (x, y) and the previously mentioned properties of the segments are used to analyze one segment at a time and then calculate the total body center of mass. The estimation of the total body center of mass position is obtained by applying a model that assumes that the body is a set of rigid segments. The center of mass of the total body is calculated using the inertial parameters of each segment and its position.

Before presenting the computations for this method, the source of the information concerning the body segments must be approached. At least three methods for deriving this information have been utilized. They are measures based on cadaver studies, mathematical geometric modeling, and mass scanning.

Several researchers have presented formulae that estimate the mass and the location of the center of mass of the various segments based on cadaver studies (8,9,10,36). These researchers have generated regression or prediction equations that make it possible to estimate the mass and the location of the center of mass. Table 11-2 presents the prediction equations from Chandler et al. (8). These predicted parameters are based on known parameters, such as the total body weight or the length or circumference of the segment. An example of a regression equation based on Clauser et al. (9) for estimating the segment mass of the leg is as follows:

where all dimensions of length are measured in centimeters. The location of the center of mass of the segment is usually presented as a percentage of the segment length from either the proximal or distal end of the segment.

TABLE 11-2 Segment Weight Prediction Equations and Location of Center of Mass

Segment

Weight (N)

Center of Massa

Head

0.032 BW 1 18.70

66.3

Trunk

0.532 BW - 6.93

52.2

Upper arm

0.022 BW 1 4.76

50.7

Arm

0.013 BW 1 2.41

41.7

Hand

0.005 BW 1 0.75

51.5

Thigh

0.127 BW - 14.82

39.8

Leg

0.044 BW - 1.75

41.3

Foot

0.009 BW 1 2.48

40.0

a Location from proximal end as a percentage.
From Chandler, R. F. et al. (1975). Investigation of Inertial Properties of the Human Body. AMRL Technical Report. Wright-Patterson Air Force Base, 74-137.

Other researchers have used mathematical geometric models to predict the segment masses and locations for the center of mass calculation. This has been done by representing the individual body segments as regular geometric solids (1,20,21,22). Using this method, the body segments are represented as truncated cones (e.g., upper arm, forearm, thigh, leg, and foot), cylinders (e.g., trunk), or elliptical spheres (e.g., head and hand). One such model, the Hanavan model, is presented in Figure 11-18. Regression equations based on the geometry of these solids requires the input of several measurements for each segment. For example, the thigh segment requires measurement of the circumference of the upper and lower thigh, the length of the thigh, and the total body mass to estimate the desired parameters for the segment.

The third method of determining the necessary segment characteristics is the gamma-scanning technique suggested by Zatsiorsky and Seluyanov (66). A measure of

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a gamma-radiation beam is made before and after it has passed through a segment. This allows calculation of the mass per unit surface area and thus generates prediction equations to determine the segment characteristics. The following example equation is based on these data and predicts the mass of the leg segment:

 

FIGURE 11-18 A representation of the human body using geometric solids. (Adapted from 

Miller, D. I., Morrison, W. E. [1975]. Prediction of segmental parameters using the Hanavan human body model. Medicine and Science in Sports, 7[3]:207–212.

)

where y is the leg mass in kilograms. The following equation will predict the location of the center of mass along the longitudinal axis for the same segment:

where y is the location of the center of mass as a percentage of the segment length.

All of these methods have been used in the literature to determine the segment characteristics of the center of mass, and all give reasonable estimates of these parameters. In a typical biomechanical analysis, researchers collect height and weight measurements, collect position data on the segmental end points, and calculate the location of the center of mass and the proportionate weight of the segment using information provided by studies in the literature. Tables 11-3 and 11-4 present estimates of the center of mass location and proportionate weights generated from four studies (8,9,10,41). The thigh segment at toe-off (frame 76) from Appendix B will be used as an example to calculate these parameters. The subject in Appendix B is a female weighing 50 kg. Using Plagenhoef's data shown in Table 11-2 (41), the mass of the thigh of a female is 0.1175 times her total body mass. Thus, for this female subject, the mass of her thigh would be:

According to Plagenhoef's segmental data, the center of mass is at 42.8% of the length of the thigh measured from the proximal end along the long axis of the segment. Consider the segment end point coordinates at toe-off in Figure 11-19. The location of the center of mass in the x- or horizontal direction would be:

where xcm is the location of the center of mass, xp is the location of the hip joint, and xd is the location of the knee joint, all in the horizontal direction. Similarly, for the y- or vertical direction:

TABLE 11-3 COM Location: Percent of Segment Length From Proximal End

Segment

Plagenhoef et al. 1983 (Seven Men, Nine Women)

Clauser, 1969 (13 Male Cadavers)

Dempster, 1955 (Eight Male Cadavers)

Chandler et al., 1975, (Six Male Cadavers)

Head, neck

M 55.0
F 55.0

M 46.4

M 50.0

M 66.3

Trunk

M 44.5
F 39.0

M 43.8

M 45.0

M 52.2

Whole trunk

M 63.0
F 56.9

 

M 60.4

 

Upper arm

M 43.6
F 45.8

M 51.3

M 43.6

M 50.7

Forearm

M 43.0
F 43.4

M 39.0

M 43.0

M 41.7

Hand

M 46.8
F 46.8

M 48.0

M 49.4

M 51.5

Thigh

M 43.3
F 42.8

M 437.2

M 43.3

M 39.8

Lower Leg

M 43.4
F 41.9

M 37.1

M 43.3

M 41.3

Foot

M 50.0
F 50.0

M 44.9

M 42.9

M 40.0

F, Female; M. Male.

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TABLE 11-4 Segment Weight: Percent of Total Body Weight

Segment

Plagenhoef et al. 1983 (37 Men, 100 Women)

Clauser, 1969 (13 Male Cadavers)

Dempster, 1955 (Eight Male Cadavers)

Chandler et al., 1975 (Six Male Cadavers)

Head, neck

M 8.26
F 8.2

M 7.30

M 7.9

M 7.35

Trunk

M 46.8
F 45.22

M 50.7

M 51.1

M 51.66

Whole trunk

M 55.1
F 53.2

 

Upper arm

M 3.25
F 2.90

M 2.60

M 2.70

M 3.26

Forearm

M 1.87
F 1.57

M 1.60

M 1.60

M 1.84

Hand

M 0.65
F 0.50

M 0.70

M 0.60

M 0.67

Thigh

M 10.50
F 11.75

M 10.3

M 9.70

M 9.4

Lower Leg

M 4.75
F 5.35

M 4.30

M 4.50

M 4.01

Foot

M 1.43
F 1.33

M 1.50

M 1.40

M 1.45

F, Female; M. Male.

 

FIGURE 11-19 The thigh segment and the segment end point coordinates at an instant during toe-off in walking (see Appendix C). COM, center of mass.

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where y is the location of the center of mass, y is the location of the hip joint, and yd is the location of the knee joint, all in the vertical direction. Thus, the center of mass of the segment is (852.9, 675.3) in the reference frame. The center of mass must be between the values for the proximal and distal ends of the segments. This procedure must be carried out for each segment using the coordinates of the joint centers to define the segments.

Refer to the walking data in Appendix E. Compute the location of the center of gravity of the leg from touchdown (frame 0) to toe-off (frame 79), alternating every fourth frame. Plot the path of the center of gravity of the lower leg.

Using Max TRAC, digitize the knee and ankle positions in the video of the woman in the midstance position of walking. Calculate the center of mass of the leg. Note: This woman has a body mass of 58 kg.

Total Body Center of Mass Calculation

After the segment center of mass locations have been determined, the total body center of mass can be calculated. Consider the illustration of a hypothetical three-segment model in Figure 11-20. The mass and the location of the center of mass of each segment have been previously determined. To determine the horizontal location of the center of mass of the three segments, the torque about the y-axis is calculated using the concept that the sum of the torques about the total system center of mass is zero. Four torques should be considered, three created by the segment centers of mass and one by the total system center of mass. Thus:

where m is the mass of the respective segments, M is the total system mass, g is the acceleration due to gravity, x is the location of the segment centers of mass, and x is the location of the system center of mass. Because the term g appears in every term in the equation, it can be removed from the equation, resulting in:

 

FIGURE 11-20 Location of the center of mass in a three-point mass system.

After substitution of the values from Figure 11-20 into this equation, only one quantity is unknown, x . Thus:

Therefore, the center of mass is 2 units from the y-axis. Using the same procedure, the vertical location of the system center of mass can be located by determining the torques about the x-axis. Thus, the vertical location of the system center of mass can be calculated by:

The center of mass is 2.3 units from the x-axis. Thus, the center of mass of this total system is (2, 2.3). In the x-direction, the mass is evenly distributed between the left and right sides, and thus the location of the center of mass is basically in the center. This is not the case in the y-direction. The total mass is not proportioned evenly between the top and bottom because most of the mass is nearer the top of the system. As such, ycm is closer to most of the mass of the system. The location of the system center of mass is represented on Figure 11-20 by the intersection of the dashed lines.

The previous example can be used to generalize the procedure for calculating the total body center of mass. In the previous example, for either the horizontal or vertical position, the products of the coordinates of the segment center of mass and the segment mass for each segment were added and then divided by the total body mass. In algebraic terms:

where x is the horizontal location of the total body center of mass, n is the total number of segments, mi is the mass of the ith segment, M is the total body mass, and xi is the horizontal location of the ith segment center of mass. Similarly, for the vertical direction:

where ycm is the vertical location of the total body center of mass, n is the total number of segments, mi is the mass

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of the ith segment, M is the total body mass, and yi is the vertical location of the ith segment center of mass.

This technique for calculating the total body center of mass is used in many studies in biomechanics. The computation is based on the segment characteristics and the coordinates determined from a kinematic analysis. For most situations, the body is thought of as a 14-segment model (head, trunk, and two each of upper arms, lower arms, hands, thighs, legs, and feet). In certain situations in which the actions of the limbs are symmetrical, an eight-segment model (head, trunk, upper arm, lower arm, hand, thigh, leg, and foot) will suffice, although the mass of the segments not digitized must be included.

As an example of this technique, consider the illustration of the jumper in Figure 11-21, which shows the coordinates of the segmental end points of the jumper. Using an eight-segment model (head and neck, trunk, upper arm, forearm, hand, thigh, lower leg, foot), the coordinates of the location of each segment's center of mass is computed using one of the sets of available anthropometric data. For example, if it is assumed that the jumper is a male (body mass = 70 kg), segmental center of gravity locations can be calculated using Dempster's cadaver data (10). The location of the segmental center of mass measured from the proximal end is calculated in for each segment using estimated locations of the center of mass as demonstrated previously (Table 11-5):

In this example, the center of mass of each segment was calculated as a distance from the proximal end point. The x and y coordinate locations of the center of mass are next used to calculate each segmental torque by multiplying the center of mass location times the segment mass proportions. That is:

The xcm can then be calculated by summing the segmental products, and y is similarly calculated using the equations presented previously in this chapter and expressed below.

The total body center of mass coordinates in digitizing units, therefore, are (614,803) and are indicated on Figure 11-21 by the larger solid circle.

Refer to the data in Appendix C. Using Plagenhoef's data for a woman, calculate the location of the center of mass of a three-link system, including the thigh, lower leg, and foot, for frame 15.

Using MaxTRAC, import the video file of the woman at midstance and digitize the right hip, right knee, right ankle, and right fifth metatarsal head. Using Plagenhoef's data for a woman, calculate the location of the center of mass of a three-link system, including the thigh, leg, and foot. Note: The woman's body mass is 58 kg.

 

FIGURE 11-21 Segmental end points, segment centers of mass, and the tota body center of mass for a jumper. The segment centers of mass are designated by an “X” and the total body center of mass by a black dot.

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TABLE 11-5 Calculation of Total Body Center of Mass

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Rotation and Leverage

Definitions

The outcome of a torque is to produce a rotation about an axis. If rotations about a fixed point are considered, the concept of the lever can be discussed. A lever is a rigid rod that is rotated about a fixed point or axis called the fulcrum. A lever consists of a resistance force, an effort force, a barlike structure, and a fulcrum. In addition, two moment or lever arms are designated as the effort arm and the resistance arm. The effort arm is the perpendicular distance from the line of action of the effort force to the fulcrum. The resistance arm is the perpendicular distance from the line of action of the resistance force to the fulcrum. Because both the effort and resistance forces act at a distance from the fulcrum, they create torques about the fulcrum.

An anatomical example, such as the forearm segment, can be used to illustrate a lever (Fig. 11-22). The long bone of the forearm segment is the rigid barlike structure, and the elbow joint is the fulcrum. The resistance force may be the weight of the segment and possibly an added load carried in the hand or at the wrist. The effort force is produced by the tension developed in the muscles to flex the elbow. Figure 11-23 illustrates several examples of simple machines that are in effect different types of levers.

A lever may be evaluated for its mechanical effectiveness by computing its mechanical advantage. Mechanical advantage is defined as the ratio of the effort arm to the resistance arm. That is:

In the construction of a lever, any of three situations may define the function of the lever. The simplest case is when MA = 1, that is, when the effort arm equals the resistance arm. In this case, the function of the lever is to alter the direction of motion or balance the lever but not to magnify either the effort or resistance force. The second case is when the MA is greater than 1, when the effort arm is greater than the resistance arm. In this case, the greater effort arm magnifies the torque created by the effort force. Thus, when MA is greater than 1, the lever is said to magnify the effort force. In the third situation, MA is less than 1, and the effort arm is less than the resistance arm. In this case, a much greater effort force is required to overcome the

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resistance force. The effort force acts over a small distance, however, with the result that the resistance force is moved over a much greater distance in the same amount of time (Fig. 11-24). When MA is less than 1, therefore, velocity or speed of movement is said to be magnified.

 

FIGURE 11-22 An anatomical lever showing the resistance arm, effort arm, and fulcrum (elbow joint).

 

FIGURE 11-23 Levers.

 

FIGURE 11-24 A first-class lever in which the moment arm is less than 1, that is, the effort arm is less than the resistance arm. The lin- Distance moved by ear distance moved by the effort force, how- resistance force ever, is less than that moved by the resistance force in the same time.

Classes of Levers

There are three classes of levers. In a first-class lever, the effort force and the resistance force are on opposite sides of the fulcrum. Everyday examples of this lever configuration are the seesaw, the balance scale, and the crowbar. A

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first-class lever may be configured many ways and may have a mechanical advantage of 1, more than 1, or less than 1. First-class levers exist in the musculoskeletal system of the human body. The agonist and antagonist muscles simultaneously acting on opposite sides of a joint create a first-class lever. In most instances, however, the first-class lever in the human body acts with a mechanical advantage of 1. That is, the lever acts to balance or change the direction of the effort force.

 

FIGURE 11-25 An anatomical first-class lever in which the weight of the head is the resistance force, the splenius muscles provide the effort force, and the fulcrum is the atlanto-occipital joint.

An example of the former is the action of the splenius muscles acting to balance the head on the atlanto-occipital joint (Fig. 11-25). The latter situation, in which the lever changes the direction of the effort force, is seen in the action of many bony prominences called processes. This type of first-class lever is a pulley. One such example is the action of the patella in knee extension, where the angle of pull of the quadriceps muscles is altered by the riding action of the patella on the condylar groove of the femur.

Second-Class Lever

In a second-class lever, the effort force and the resistance force act on the same side of the fulcrum. In this class of lever, the resistance force acts between the fulcrum and the effort force. That is, the resistance force arm is less than the effort arm and thus the mechanical advantage is greater than 1. One example of a second-class lever in everyday situations is the wheelbarrow (Fig. 11-26). Using the wheelbarrow, effort forces can be applied to act against significant resistance forces provided by the load carried in the wheelbarrow. There are few examples of second-class levers in the human body, although the act of rising onto the toes is often proclaimed and also disputed as one such. This action is used in weight training and is known as a calf raise. Because there are so few examples of second-class levers in the human body, it is safe to say that humans are not designed to apply great forces via lever systems.

 

FIGURE 11-26 A wheelbarrow as a second-class lever. The resistance force is between the fulcrum and the effort force. Because the effort arm is greater than the resistance arm, MA>1, and the effort force is magnified.

Third-Class Lever

The effort force and the resistance force are also on the same side of the fulcrum in a third-class lever. In this arrangement, however, the effort force acts between the fulcrum and the line of action of the resistance force. As a result, the effort force arm is less than the resistance force arm and thus the mechanical advantage is less than 1. An example of this type of lever is the shovel when the hand nearest the spade end applies the effort force (Fig. 11-27). Therefore, it would appear that a large effort force must be applied to overcome a moderate resistance force. In a third-class lever, a large effort force is applied to gain the advantage of increased speed of motion. This is the most prominent type of lever arrangement in the human body, with nearly all joints of the extremities acting as third-class

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levers. It is probably safe to conclude that from a design standpoint, greater speed of movement exemplified by third-class levers appears to be emphasized in the musculoskeletal system to the exclusion of greater effort force application ability of the second-class lever. Figure 11-28 illustrates a third-class lever arrangement in the human body.

 

FIGURE 11-27 An individual using a shovel is a third-class lever.

 

FIGURE 11-28 The arm held in flexion at the elbow is an anatomical third-class lever: The resistance force is the weight of the arm, the fulcrum is the elbow joint, and the effort force is provided by the elbow flexor muscles.

 

FIGURE 11-29 The gravitational torques created by the weight of body segments acting at a distance from the joint in movements such as trunk flexion (A) and an arm lateral raise (B) must be countered by muscular torques acting in the opposite direction.

Types of Torque

A torque acting on a body is created by a force acting a distance away from the axis of rotation. Thus, any of the different types of forces discussed in Chapter 10 can produce a torque if applied in a direction that does not go through the axis or pivot point. Gravity, a noncontact force, generates a torque any time the line of gravity does not pass through the hip joint (pivot point). As illustrated in Figure 11-29, gravity acting on the trunk segment produces a clockwise torque about the lumbar vertebrae (Fig. 11-29A), and the weight of the arm and dumbbell produces a clockwise torque about the shoulder joint (Fig. 11-29B). A muscular torque in the clockwise direction to hold the positions statically must counteract both of these gravitational torques.

Contact forces also produce torques if applied correctly. For example, torques are generated about the center of gravity in events such as diving and gymnastics by using the vertical ground reaction force in conjunction with body configurations that move the center of mass in front of or behind the force application. Consider Figure 11-30 A, in which the ground reaction force generated in a backward somersault is applied a distance from the center of mass, causing a clockwise rotation about the center of mass. A muscle force also generates torques about the joint center, as shown in Figure 11-30B.

Representation of Torques Acting on a System

A free body diagram illustrating torques acting on a system is usually combined with linear forces to identify and analyze the causes of motion. Many biomechanical

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analyses start with a free body diagram for each body segment. Known as the rigid link segment model, it can take either a static or dynamic formulation. Consider the model of the dead lift shown in Figure 11-31 showing the lift (Fig. 11-31 A) and the free body diagram for the leg, thigh, trunk, arm, and forearm segments (Fig. 11-31B). If a rigid link segment model is developed, forces acting at the joints (F, F) and the center of mass (W) can be indicated along with moments (M) acting at the joints.

 

FIGURE 11-30 Contact forces such as ground reaction forces (A) and muscular forces (B) create torques because the line of action of the force does not go through the center of mass or joint axis, respectively. COM, center of mass; GRF, ground reaction force; MA, moment arm.

 

FIGURE 11-31 A squat (A) as free body diagram (B) using a linked segment model illustrating forces acting at the joints and the centers of mass, along with the moments acting at the joints.

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Analysis Using Newton's Laws IH of Motion

Chapter 10 presented three variations of Newton's laws that describe the relationship between the kinematics and the kinetics of a movement. An angular analog can be generated for each of the three approaches. In most biomechanical analyses, both the linear and angular relationships are determined together to describe the cause-and-effect relationship in the movement. The linear analyses previously discussed presented three approaches categorized as the effect of a force at an instant in time, the effect of a force over a period of time, and the effect of a force applied over a distance. A thorough analysis also includes the angular counterparts and examines the effect of a torque at an instant in time, the effect of a torque over time, and the effect of a torque applied over a distance. Each approach provides different information and is useful in relation to the specific question asked about torque and angular motion.

Effects of A Torque At An Instant In Time

Newton's second law of motion is considered with the effects of a torque and the resulting angular acceleration. Thus:

When the angular acceleration is zero, a static case is evaluated. A dynamic analysis results when the acceleration is not zero. Note that the torques act about the axis perpendicular to the X-Y plane (i.e., the Z-axis) and the angular acceleration is also about this axis.

The application of Newton's second law for both angular and linear cause-and-effect relationships would consider both linear and angular equations. In linear motion, the effects of a force and the resulting accelerations at an instant in time are determined, but in angular motion, the effects of a torque and the resulting angular accelerations are determined.

Static Analysis

The static case involves systems at rest of moving at a constant velocity. A state of equilibrium exists when the acceleration of the system is zero. As illustrated in Chapter 10, linear equilibrium exists when the sum of the forces acting on the system equal zero. Equilibrium also depends on the balancing of torques acting on the system when forces are not concurrent. Concurrent forces do not coincide at the same point and thus cause rotation about some axis. These rotations all sum to zero, and the resulting system remains at rest or moves at a constant angular velocity. That is, the sum of the moments of force or torques in the system must sum to zero. Stated algebraically, therefore:

Previously, a convention was suggested in which moments or torques causing a counterclockwise rotation were considered to be positive and clockwise torques were considered to be negative. To satisfy this condition of equilibrium, therefore, the sum of the counterclockwise moments must equal the sum of the clockwise moments, and no angular acceleration can occur.

Consider the diagram in Figure 11-32, in which a first-class lever system is described. On the left side of the fulcrum, individual A weighs 670 N and is 2.3 m from the axis of rotation. This individual would cause a counterclockwise or positive moment. Individual B on the right side weighs 541 N, is 2.85 m from the fulcrum, and would cause a clockwise or negative moment. For this system to be in equilibrium, the clockwise torque must equal the counterclockwise torque. Thus:

 

FIGURE 11-32 A first-class lever, a seesaw.

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and

For this system to be in equilibrium:

Because TA is positive, TB is negative, and their magnitudes are equal, these moments cancel each other, and no rotation occurs. With no further external forces, this system will come to rest in a balanced position.

Typically, multiple torques act on a system involving human movement. Figure 11-33A shows the forearm of an individual holding a barbell. To determine the action at the elbow joint, it would be helpful to know the moment caused by the muscles about the elbow joint. If the elbow joint is considered to be the axis of rotation, there are two negative or clockwise torques in this system. One negative torque is a result of the weight of the forearm and hand acting through the center of mass of the forearm-hand system, and the other is the result of the weight of the barbell. The counterclockwise or positive torque is a result of the muscle force acting across the elbow joint. The net moment of the muscles must be equal to the two negative moments for the system to be in equilibrium. Thus:

Consider the free body diagram of this system illustrated in Figure 11-33B. The muscle moment that is necessary to keep the system in equilibrium can be calculated from the information in the free body diagram. The forearm-hand complex weighs 45 N, and the center of mass is 0.15 m from the elbow joint. The weight of the barbell is 420 N, and the center of mass of the barbell is 0.4 m from the elbow joint. The moment due to the weight of the arm and hand is:

and the moment due to the barbell is:

The moment due to the muscle force can then be calculated as:

The muscle must create a torque of 174.75 Nm to counteract the weight of the forearm, hand, and barbell. This muscle torque cannot be directly attributed to any one muscle that crosses the joint. The muscle moment calculated is the net sum of all muscle actions involved. In this case, because the arm is being held in flexion, the net muscle moment is primarily due to the action of the elbow flexors, but it cannot be said exactly which elbow flexors are most involved. In fact, it might be surmised that because this is a static posture, there may be considerable co-contraction.

To determine the muscle action, the net torque about the joint must be considered. The net torque is the sum of all torques acting at the joint, in this case, the elbow. Thus:

The net moment about the elbow joint is zero, indicating that the muscle action must be isometric.

If the moment arm of the elbow flexors was estimated to be 0.05 m from the elbow joint, the muscle force would be:

 

FIGURE 11-33 The forearm during an instant of a biceps curl (A) and a free body diagram of the system (B).

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It can be seen that the muscle force must be considerably greater than the other two forces because the moment arm for the muscle is very small compared with the moment arms for the forearm-hand or the barbell.

Consider Figure 11-34. In this free body diagram, the arm is placed in a posture similar to that in Figure 11-33, but this barbell weighs 100 N, and the measured muscle torque at the elbow is 180 N-m. All other measures in this situation are the same as in the previous example. Thus, evaluating the net moment at the elbow:

Because the net torque at the elbow is positive, the rotation is counterclockwise, and the muscle action therefore is interpreted as a flexor action.

In the previously described situations, the arm was parallel to the ground, and the torques arms from the axis of rotation were simply the distances measured along the segment to the lines of action of the forces. As the arm flexes or extends at the elbow joint, however, the torque arms change. In Figure 11-35, the moment arms for the muscle force are illustrated with the arm in three positions. With the elbow extended (Fig. 11-35A), the moment arm is rather small. As flexion occurs at the elbow (Fig. 11-35B), the moment arm increases until the arm is parallel to the ground. When flexion continues past this point (Fig. 11-35C), the moment arm again becomes smaller. The magnitude of the moment arm of the muscle force, therefore, depends on how much flexion or extension occurs at a joint. The change in moment arm during flexion and extension of the limb is also true for the moment arms for both the weight of the forearm and hand and for anything held in the hand.

 

FIGURE 11-34 Free body diagram of a biceps curl at an instant when the forearm is horizontal.

In Figure 11-36, the arm is held at some angle θ below the horizontal. The value d describes the distance from the axis of rotation to the center of mass of the arm. The moment arm for the weight of the arm, however, is the distance a. The lines a and d form the sides of a right triangle with the line of action of the force. Therefore, with

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the angle θ, the cosine function can be used to calculate the length a. Thus:

 

FIGURE 11-35 The change in magnitude of the moment arm of the biceps muscle throughout the range of motion. As the elbow flexes from an extended position, the moment arm becomes longer. As the arm continues to flex from the horizontal forearm position, the moment arm becomes shorter.

 

FIGURE 11-36 The cosine of the angle of inclination of the forearm is used to calculate the moment arm when the forearm is parallel to the horizontal.

The arm is considered to have a 0° angle when it is parallel to the ground. As the arm extends, the angle increases until it is 90° at full extension. The cosine of 0° is 1, and as the angle becomes greater, the cosine gets smaller, until at 90° the cosine is 0. If the angle 9 becomes greater as the arm is extended, the moment arm should become correspondingly smaller because the distance d does not change. When the angle 9 is 90°, the arm is fully extended and the moment arm is zero because the line of action of the force due to the weight of the arm passes through the axis of rotation.

Consider an individual performing a biceps curl with a weight. The arm is positioned 25° below the horizontal, so that the elbow is slightly extended (Fig. 11-37A). The corresponding free body diagram is presented in Figure 11-37B. The moment arm for the weight of the arm can be calculated from the distance from the axis of rotation to the center of mass and the angle at which the arm is positioned:

Similarly, the moment arm for the weight of the barbell held in the hand can be calculated as:

 

FIGURE 11-37 The forearm at an instant during a biceps curl (A) and the free body diagram of this position (B). The arm is inclined below the horizontal.

The moment arms are less than they would be if the arm were held parallel to the ground.

This problem may be solved for the muscle force using the same principles of a static analysis that were discussed previously. That is:

If, by convention, the moment due to the muscle is considered to be positive and the moments due to the weight of the arm and the weight of the barbell negative, then:

Remember that the torque is the product of a force and its torque arm. Thus, we can substitute all known values into this equation. Rearranging it results in:

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This equation can be solved to determine the muscle force necessary to maintain this position in a static posture. Thus:

The muscle must exert a force much greater than the weight of the arm and the barbell because the moment or torque arm for the muscle is relatively small. It appears that our body is at a great disadvantage when it comes to producing large moments of force about a joint. Most of our joints are arranged as third-class levers, however, indicating that range of motion is magnified. Muscles can, therefore, exert large forces, but only over very short periods of time.

Static Equilibrium: Stability and Balance

The concept of stability is closely related to that of equilibrium. Stability may be defined in much the same way as equilibrium, that is, as the resistance to both linear and angular acceleration. The ability of an individual to assume and maintain a stable position is referred to as balance. Even in a stable or balanced position, an individual may be subject to external forces.

If a body is in a state of static equilibrium and is slightly displaced by a force, the object may experience three conditions: return to its original position, continue to move away from its original position, or stop and assume a new position. If the object is displaced as a result of work done by a force and returns to its original position, it is said to be in a state of stable equilibrium. If the object is displaced and tends to increase its displacement, it is in a state of unstable equilibrium. A state of neutral equilibrium exists if the object is displaced by a force and returns to the position from which it was displaced.

Figure 11-38 illustrates the various states of equilibrium. In Figure 11-38A, a ball on a concave surface exemplifies stable equilibrium. When it is displaced along one side of the surface by some force, it will return to its original position. In Figure 11-38B, an example of unstable equilibrium is presented: A force displaces a ball on a convex surface. The ball will come to rest in a new position, not its original position. Figure 11-38C illustrates neutral equilibrium. In this case, when a ball is placed on a flat surface and a force is applied, the ball will move to a new position.

With its multiple segments, the human body is much more complex than a ball, but it can assume the different states of equilibrium. An individual doing a headstand, for example, is in a position of unstable equilibrium, and a child sitting on a swing is in a state of stable equilibrium.

Several factors determine the stability of an object. The first factor is where the line of gravity falls with respect to the base of support. An object is most stable if the line of gravity is in the geometric center of the base of support. Increasing the area of the base of support generally increases the stability. A body may be stable in one direction, however, but not in another. For example, spreading one's feet apart increases the area of the base of support and makes the individual stable if pushed in the transverse plane. It does not, however, help stability in the anteroposterior plane. Increasing the base of support to allow the line of gravity to fall within the base of support may be illustrated by the example of an individual using a walker, as in Figure 11-39A. The base of support is produced by the positions of the individual's legs and the legs of the walker. The walker increases the base of support, and the individual is positioned so that the line of action of the center of mass is in the geometric center of this base.

 

FIGURE 11-38 Examples of a ball. A. Stable equilibrium. B. Unstable equilibrium. C. Neutral equilibrium.

The stability of an object is also inversely proportional to the height of the center of mass. That is, an object with a low center of mass tends to be more stable than an object with a high center of mass (Fig. 11-39B). If the two objects in Figure 11-39B undergo the same angular displacement as a result of the forces indicated, the line of gravity of the center of mass of the object on the left, the taller object, will move outside the limit of the base of support sooner than that of the object on the right (the shorter object). Therefore, the force dislodging the object on the left could be less than that for the object on the right. In football, for example, defensive linemen crouch in a three-point stance to keep

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their center of mass low. This enhances their stability so that they are less likely to be moved by the offensive linemen.

 

FIGURE 11-39 Factors that influence the stability of an object. A. Increasing the base of support. B. Lowering the center of mass. C. Increasing the mass of the system.

The final factor that influences stability is the mass of the object. According to the equations of motion, the greater the mass of an object, the greater its stability (Fig. 11-39C). Newton's second law states that the force applied to an object is proportional to the mass of the object and its acceleration. Thus, it takes more force to move an object with a greater mass. Moving a piano, for example, is extremely difficult because of its mass. Many sports such as wrestling and judo, in which stability is critical, take body mass into consideration by dividing the contestants into weight divisions because of the disproportionate stability of heavier individuals.

Applications of Statics

It appears that static analyses are limited in their usefulness because they describe situations in which no motion or motion at a constant velocity occurs. A static analysis of muscle forces and moments has been used extensively in ergonomics, however, even though the task may involve some movement. The evaluation of workplace tasks, such as lifting and manual materials handling, has been examined in considerable detail with static analyses. Use of static analysis to determine an individual's static or isometric strength is widely accepted in determining the ability of the person's lifting ability. Many researchers have suggested that this static evaluation should be used as a pre-employment screening evaluation for applicants for manual materials handling tasks (7,29,42). Lind et al. (32) reported that the static endurance in a manual materials handling task is influenced by the posture of the individual performing the task. A model to evaluate static strength evaluations of jobs has been developed by Garg and Chaffin (17). A free body diagram of one such lifting model is presented in Figure 11-40 (6). This model was presented in Chapter 10 to illustrate a free body diagram for the linear forces acting on the system (Fig. 11-40A). Adding the moments of forces at each joint (Fig. 11-40B) allows the total lifting model to be evaluated (Fig. 11-40C).

Static analysis techniques have also been used in clinical rehabilitation. Quite often, no movement of a body or a body segment is desirable, so a static evaluation may be undertaken. For example, placing a patient in traction demands that a static force system be implemented. Many bracing systems for skeletal problems, such as scoliosis and genu valgum (knock-knee), use a static force system to counteract the forces causing the problem (Fig. 11-41). Static analyses have been used in the calculation of muscle forces and have been performed by multiple researchers on many joints (18,39,44).

Dynamic Analysis

As pointed out in Chapter 10, a dynamic analysis should be used when the accelerations are not zero. Newton's second law establishes the basis for the dynamic analysis by examining the force-acceleration relationship. In the linear case, the equations of motion for a two-dimensional case are:

where the linear acceleration is broken down into its horizontal (x) and vertical (y) components. The angular equivalent looking at the torque-angular acceleration relationship is:

in which I is the moment of inertia and a is the angular acceleration. In a two-dimensional system, angular acceleration occurs about the z-axis. If αz =0, the motion is purely linear. If ax =0 and ay = 0, the motion is purely rotational. If αz, ax, and ay = 0, a static case exists. The torques acting on a body are created by a contact or gravitational force acting a distance from the axis or rotation. Iαz is the inertial torque similar to the linear case, ma and ma y.

Angular accelerations and the inertial properties of the body segments resisting these accelerations must be

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considered in the dynamic case. As discussed in Chapter 10, in an inverse dynamics approach, each segment is evaluated from the most distal segment and works systematically up the segment links. Consider Figure 11-42, which illustrates a free body diagram of the foot during the swing phase of the gait cycle. In Chapter 10, a dynamic analysis of the linear forces was conducted by applying the two-dimensional linear kinetic equations. Joint reaction forces at the ankle were determined to be – 1.57 N for Rx and 20.3 N for R y

 

FIGURE 11-40 A free body diagram of a sagittal view static lifting model showing the linear reactive forces (A) and moments of forces (B) at the joints. These are combined to generate the total lifting model (C). (Adapted from 

Chaffin, D. B., Andersson, G. B. J. [1991]. Occupational Biomechanics [2nd ed.]. New York: Wiley.

)

 

FIGURE 11-41 Bracing systems that illustrate static force systems. A. A neck brace. B. A three-point pressure brace to correct genu valgum. C. A brace to correct a foot deformity.

 

FIGURE 11-42 Free body diagram of the foot during the swing phase of a walking stride. Convention dictates that moments and forces at the proximal joint are positive, as indicated in the free body diagram.

Mass of foot 5 1.16 kg, I 5 0.0096 kg-m2

a 5 - 14.66 rad/s2

ax 5 – 1.35 m/s2, a 5 7.65 m/s2

Rx = 1.57 N R = 20.3 N

(See Chapter 10.)

To determine the net moment acting at the ankle joint, all moments acting on the system must be evaluated. If the center of mass of the foot is considered as the axis of rotation, three moments are acting on this system, two as a result of the joint reaction forces and the net ankle moment itself. Because the joint reaction forces and their moment arms, the moment of inertia of the foot about an axis through the center of mass, and the angular acceleration of the foot are known, the net ankle moment can be calculated. Thus:

where M ankle is the net moment at the ankle, MRx is the moment resulting from the horizontal reaction force at the ankle, MR is the moment resulting from the vertical reaction force at the ankle, and Icm αz is the product of the moment of inertia of the foot and the angular acceleration of the foot. In the general equation of moments acting on the foot, the moments MRx and MRy will cause clockwise rotation of the foot about the center of mass of the foot. By convention, clockwise rotations are negative, so moments causing a clockwise rotation are negative. Substituting the appropriate values from Figure 11-42 and those calculated in the previous equations and rearranging the equation:

The net moment at this instant in time is positive, and we conclude that this is a counterclockwise rotation. A counterclockwise rotation of the foot indicates a dorsiflexion action. As with the static case, the exact muscles acting cannot be determined from this type of analysis. Thus, it cannot be stated whether the muscle activity is dorsiflexor concentric or dorsiflexor eccentric.

It was stated that a dynamic analysis usually proceeds from the more distal joints to the more proximal joints. The data from this calculation of the foot segment analysis are then used to calculate the net moment at the knee. The calculation continues to the thigh to calculate the hip moment. The analysis is conducted for each joint at each instant in time of the movement under consideration to create a profile of the net moments for the complete movement.

Applications of Dynamics Dynamic analyses have been used in many biomechanical studies on a number of activities to determine the net moments at various joints. These activities include cycling (43), asymmetrical load carrying (11), weight lifting (31), jumping (57), and throwing (14). Figure 11-43 illustrates the net torques for the hip, knee, and ankle joints in a comparison study of one-legged and two-legged countermovement jumps by 10 volleyball players (57). The joint torques across the push-off phase demonstrated that the one-legged jump generated higher peak joint torques in all three joints and greater mean torques in the hip and ankle joints.

Effects of Torque Applied Over A Period of Time

For angular motion to occur, torques must be applied over a period of time. In the linear case, the product of a force applied over a period of time was referred to as the impulse. For the angular analogue, the application of a torque over time is known as angular impulse. As in the linear case, we can derive this concept from Newton's second law of motion:

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FIGURE 11-43 Hip (A), knee (B), and ankle (C) joint torques produced in the push-off phase of one-legged (solid) and two legged (dotted) coun-termovement jumps. (Modified from 

Van Soest, A. J., et al. [1985]. A comparison of one-legged and two-legged countermovement jumps Medicine and Science in Sports and Exercise, 17:635-639.

)

The left side of the equation represents the angular impulse, and the right side of the equation describes the change in angular momentum. This relationship is known as the impulse-momentum relationship. The equation demonstrates that when a torque is applied over a period of time, a change in the angular momentum occurs. Consider the handspring vault inFigure 11-44. The gymnast comes in contact with the horse after generating a large linear horizontal velocity during the approach. This velocity is converted to some vertical velocity and angular momentum. On contact with the horse, the contact with the horse (called a “blocking action”) generates two torques. Because the gymnast contacts the horse at an angle, the contact with the horse generates both a vertical force (Fy) and a horizontal force (Fx). The vertical force on the horse increases the vertical velocity off the vault but creates a clockwise torque about the center of mass, which is the product of F and d . This angular impulse (T * t) changes the angular momentum and decreases the counterclockwise rotation created by the angular momentum generated at takeoff from the board. The horizontal force on the horse acts in an opposite manner by decreasing horizontal velocity and increasing angular momentum generated on the board. A counterclockwise torque about the center of mass is generated via the impulse generated by F times d, resulting in an increase in the angular momentum from that generated at takeoff. Takei (53) indicated that the loss in angular momentum generated by the application of the vertical force was greater than the corresponding gain generated by the horizontal force, resulting in a loss of angular momentum while on the horse. In elite Olympic gymnasts, the angular momentum at contact averaged 95.3 kg-m2/s for the vaults with the highest scores and 92.4 kg-m2/s in the vaults with the lowest scores. In addition, the angular momentum was reduced by -33 and -27.4 kg-m2/s, respectively, during the horse contact phase as a result of these angular impulses (52).

Angular Work: Effects of Torque Applied Over A Distance

Mechanical angular work is defined as the product of the magnitude of the torque applied against an object and the angular distance that the object rotates in the direction of

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the torqe while the torque is being applied Expessed algebraically:

where T is the torque applied and A0 is the angular distance. Because torque has units of N-m and angular distance has units of radians, the units for angular work are newton-meters (N-m) or joules (J), the same units as in the linear case.

 

FIGURE 11-44 Torques generated on the horse by the vertical (Fy * dx) and anteroposterior force (Fx * dy) generate angular impulses about the center of mass of the vaulter. (Modified from 

Takei, Y. [1992]. Blocking and post flight techniques of male gymnasts performing the compulsory vault at the 1988 Olympics. International Journal of Sport Biomechanics, 8:87-110.

)

To illustrate angular work, if a 40.5 N-m torque is applied over a rotation of 0.79 radians, the work done in rotation is:

Thus, 32 N-m of work is said to be done by the torque, T.

When a muscle contracts and produces tension to move a segment, a torque is produced at the joint, and the segment is moved through some angular displacement. The muscles that rotated the segment do mechanical angular work. To differentiate between the kinds of muscle actions angular work done by muscles is characterized as either positive or negative work. Positive work is associated with concentric muscle actions or actions in which the muscle is shortening as it creates tension. For example, if a weight lifter performs a biceps curl on a barbell, the phase in which the elbows flex to bring the barbell up is the concentric phase (Fig. 11-45A). During this motion, the flexor muscles of the weight lifter do work on the barbell. Negative work, on the other hand, is associated with eccentric muscle actions, or actions in which the muscle is lengthening as it creates tension. In a biceps curl, when the weight lifter is lowering the barbell, resisting the pull of gravity, the flexor muscles are doing negative work (Fig. 11-45B). In this instance, the barbell is doing work on the muscles. Although it has been found that positive work requires a greater metabolic expenditure than negative work, no direct relationship has been reported between mechanical work of muscles and physiological work.

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FIGURE 11-45 Positive muscular work (A) and negative muscular work (B) during a biceps curl of a barbell.

Special Torque Applications

Angular motion has special torque applications comparable to the force applications in the linear case. Many of the angular applications are direct analogs of the linear case and have similar definitions.

Angular Power

In Chapter 10, power was defined in the linear case as the work done per unit time or the product of force and velocity. Angular power may be similarly defined as:

where dW is the angular work done and dt is the time over which the work was done. Angular power may also be defined, as in the linear case, using the angular analogs of force and velocity, torque, and angular velocity. Angular power is the angular work done per unit time and is calculated as the product of torque and angular velocity:

where T is the torque applied in N-m and co is the angular velocity in radians per second. Angular power thus has units of N-m/s, or watts.

The concept of angular power is often used to describe mechanical muscle power. Muscle power is determined by calculating the net torque at the joint and the angular velocity of the joint. The net moment is assumed to describe the net muscle activity across a joint and does not represent any one particular muscle that crosses the joint but the net activity of all muscles. It also does not take into account the situation in which biarticulate muscles may be acting or the fact that there may be co-contraction of these muscles. This net muscle activity at a joint is simply described as flexor or extensor actions, but whether the muscle activity is concentric or eccentric cannot be ascertained directly from the joint moment. The net moment can, however, be used in conjunction with the angular velocity of the joint to determine the concentric or eccentric nature of the muscular action. As discussed previously, concentric actions of muscles are related to the positive work of muscles and eccentric actions as the negative work of muscles. Because the work done by muscles is rarely constant with time, the concept of muscle power can be used. Muscle power is the time rate of change of work and is defined as the product of the net muscle moment and the joint angular velocity. It is expressed algebraically as:

where Pmuscle, is the muscle power in units of watts (W), M.j is the net muscle moment in N-m, and ωj is the joint angular velocity in radians per second.

Muscle power may be either positive or negative. Because power is the time rate of change of work, the area under the power-time curve is the work done. For example, if Mj and ωjare either both positive or both negative, muscle power will be positive. Positive muscle power indicates that the muscle is acting concentrically. If Mj is positive and ωj is negative or Mj is negative and ωj is positive, muscle power will be negative. Negative muscle power indicates a net eccentric action.

Figure 11-46 illustrates the positive and negative work possibilities of the elbow joint. In Figure 11-46 A Mj and v are positive, indicating a flexor moment with the arm moving in a flexor direction. The resulting muscle power is positive, indicating a concentric action of the elbow flexors. In Figure 11-46B, both Mj and v are negative, resulting in a positive muscle power or a concentric action. The arm is extending, indicating the muscle action is a concentric action of the elbow extensors. In Figure 11-46C,

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Mj is positive or a flexor moment, but the arm has a negative v, indicating extension. In this case, an external force is causing the arm to extend while the elbow flexors resist. This results in an eccentric action of the elbow flexors and is verified by the negative muscle power. Figure 11-46D illustrates the case in which an eccentric action of the elbow extensors occurs.

 

FIGURE 11-46 The definition of positive and negative power at the elbow joint: A and B result in positive power; C and D result in negative power.

Figure 11-47 illustrates the angular velocity-time, net moment-time, and muscle power-time profiles during an elbow flexion followed by elbow extension. In the flexion phase of the movement, the net moment initially is positive, but it becomes negative as the arm becomes more flexed. The initial portion, therefore, results in a positive power or a concentric contraction of the flexor muscles. In the latter portion of the flexor phase, the power is negative, indicating an eccentric muscle contraction. The eccentric contraction of the elbow extensor muscles occurs to decelerate the limb. Because the power is negative at this point, however, it does not mean that there is no flexor muscle activity. It simply means that the predominant activity is extensor. In the extension portion of the movement, the situation is reversed. In the initial portion of extension, the muscle activity is concentric extensor activity. In the latter portion, the muscle power profile again indicates flexor concentric activity to decelerate the limb.

The analysis of net moments of force and muscle power has been widely used in research in biomechanics. Winter and Robertson (63) and Robertson and Winter (46) have investigated the power requirements in walking. Robertson (45) described the power functions of the leg muscles during running to identify any common characteristics among a group of runners. Gage (16) reported on the use of the moment of force and the muscle power profiles as a preoperative and postoperative comparison. The analysis of muscle power in the lower extremity during locomotion, therefore, appears to be a powerful research and diagnostic tool.

Energy

In Chapter 10, translational kinetic energy was defined in terms of mass and velocity. Rotational kinetic energy may also be defined similarly using the angular analogs of mass and velocity, moment of inertia, and angular velocity. Thus, rotational kinetic energy is defined algebraically as:

where RKE is the rotational kinetic energy, I is the moment of inertia, and to is the angular velocity. Thus, when the total energy of a system is defined, the rotational kinetic energy must be added to the translational kinetic energy and the potential energy. Total mechanical energy is therefore defined as:

total energy = translational kinetic energy + potential kinetic energy + rotational kinetic energy

In the discussion of angular kinematics, it was noted that single segments undergo large angular velocities during running. Thus, if a single segment is considered, it might be intuitively expected that the rotational kinetic energy might influence the total energy of the segment more significantly than the translational kinetic energy. Winter et al. (62), for example, hypothesized that angular contributions of the leg would be important to the changes in total segment energy in running. Williams (59) offered the following example to illustrate that this is not the case.

Consider a theoretical model of a lower extremity segment undergoing both translational and rotational movements (Fig. 11-48). If the leg segment is considered, the linear velocity of the center of mass of the leg is:

where ω is the angular velocity of the leg and r is the distance from the knee to the center of mass of the leg. If the body segment values of the leg are moment of

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inertia = 0.0393 kg-m2, mass = 3.53 kg, and r = 0.146 m, the magnitude of the rotational kinetic energy is:

 

FIGURE 11-47 Angular velocity (A), moment of force (B), and muscle power (C) profiles during an elbow flexion-extension motion.

The translational kinetic energy of the leg under the same circumstances is:

where m and v are the mass and the linear velocity of the leg. Substituting the expression ωr into this equation for the linear velocity, v, the equation becomes:

Because ω2 has the same magnitude in both RKE and TKE, the two types of energy can be evaluated based on the values of 0.0192 times ω2 for the rotational energy

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and 0.0376 times co2 for the translational energy. It can be seen that the translational kinetic energy is almost double that of the rotational kinetic energy. In fact, Williams (59) stated that the rotational kinetic energy of most segments is much less than the translational kinetic energy. Figure 11-49 illustrates the relationship between the energy components of the foot during a walking stride. In this figure, the magnitude of the total energy is almost completely made up of the translational kinetic energy.

 

FIGURE 11-48 Theoretical representation of the translational and rotational velocities of the thigh and leg during running. (Adapted from 

Williams, K. R. [1980]. A Biomechanical and Physiological Evaluation of Running Efficiency. Unpublished doctoral dissertation, The Pennsylvania State University.

)

Work-Energy Relationship

Mechanical angular work was defined as the product of a torque applied to an object and the distance that the object moved during the torque application. Angular work is said to have been done on an object when a rotation occurs through some angular distance. Rotational energy has also been defined as the capacity to do angular work. Therefore, analogous to the linear case, the work-energy theorem, W = ΔE, also applies. That is, for mechanical angular work to be done, a change in the rotational energy level must occur. The angular work done on an object is:

where Wangular, is the angular work done on the object and ΔRE is the change in rotational kinetic energy about the center of mass. To calculate the total work done on the object, the other forms of energy, such as potential and kinetic energy, must also be considered. With the inclusion of the additional form of energy, the work done on the object becomes:

 

FIGURE 11-49 The relationship between total energy (Total), rotational kinetic energy (RKE), translational kinetic energy (TKE), and potential energy (PE) of the foot during a walking stride.

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where Wobject is the work done on the object, ΔKE Is the change in the linear kinetic energy of the center of mass of the object [Δ(1/2 mv2)], APE is the change in potential energy of the object center of mass [Δ(mgh)], and (RE is the change in rotational energy about the center of mass of the object [Δ(1/2 I to)2]. For example, in baseball batting, the goal is to generate maximum energy at contact so maximum work can be generated in the ball. Both the linear and rotational kinetic energy of the bat are important. Potential energy is also a factor because the bat stores potential energy in the handle that is later transferred as local kinetic energy at impact (13).

Angular Kinetics of Locomotion

The angular kinetics of locomotion, specifically the joint moments of force of the lower extremity, have been widely researched (3,60,61). Winter (60) stated that the resultant moment of force provided powerful diagnostic information when comparing injured gait with uninjured gait.

Another common area of investigation for gait is muscle mechanical power. Muscle power is the product of the net joint moment and the joint angular velocity. Positive power results when there is a concentric muscle action, such as flexor moments accompanying segment movement in the flexion direction. Negative power is associated with eccentric muscle action when the net moment of force occurs in the opposite direction as the segment movement. For example, negative power would result when a net knee extension moment is generated as the knee is moving into flexion. It is common to see power fluctuate between negative and positive multiple times across the cycle in both walking and running.

Figure 11-50 illustrates the joint kinematics; the net muscle moments of force; and the corresponding powers at the hip, knee, and ankle during a walking stride. At the hip joint, there is a net hip extensor moment during the initial loading phase of support continuing through mid-support into late stance. In late stance, there is a power absorption as hip extension is decelerated via hip flexors (40). In preparation for toe-off, the hip flexors shorten to

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produce power for the initiation of the swing phase. Hip flexion continues into swing via power production from a hip flexor moment until it is terminated in late swing by a hip extensor moment.

 

FIGURE 11-50 Angular displacement, moments of force, and power during a single walking stride. A. Hip, B. Knee C. Ankle. The transition from support to swing is indicated at the solid line. (Adapted from 

Ounpuu, S. [1994] The biomechanics of walking and running. Foot and Ankle Injuries, 13:843-863.

)

At the knee joint, the loading response involves knee flexion controlled by a knee extensor moment from touchdown until midsupport. In late stance, there is again a knee flexor moment that moves to a small knee extensor moment. During swing, there is minimal power production until the terminal phase of swing, when the knee flexors act eccentrically to slow knee extension for contact.

The ankle joint exhibits a brief net dorsiflexor moment during the initial loading phase of stance as the foot is lowered to the ground. A transition to a net ankle plan-tarflexion moment first occurs through eccentric plan-tarflexion actions to control the leg's rotation over the foot. This is followed by a continuation of a net plan-tarflexion moment as the plantar flexors concentrically advance the limb into the swing phase. At the actual initiation of the swing phase, plantarflexion continues under the control of eccentric dorsiflexion activity. While in the swing phase, minimal power is produced at the ankle.

The joint moment and power patterns change with the speed of walking. For example, there are three basic moment patterns at the knee joint: a biphasic pattern, a flexor moment resisting an external extensor moment, and an extensor moment after heel strike (26). At slower speeds, there is more use of a flexor moment, with little knee flexion and small knee moments through midstance resulting in negative joint power (26). In a faster walk, there are greater knee flexion and extension moments and more energy generation early in the stance phase followed by energy absorption during early stance.

Figure 11-51 illustrates the joint kinematics; the net muscle moments of force; and the corresponding powers at the hip, knee, and ankle during a running stride (40). The moments of force and powers in running are greater in magnitude than those in walking. The lower extremity moments of force increase in magnitude with increases in locomotor speed. Cavanagh et al. (4) and Mann and Sprague (34) reported considerable variability in the magnitudes of the moments of force between subjects running at the same speed. For slow running, this variability is generally smallest at the ankle and greatest at the hip.

 

FIGURE 11-51 Angular displacement, moments of force, and power during a single running stride. A. Hip, B. Knee. C. Ankle. The transition from support to swing is indicated at the solid line. (Adapted from 

Ounpuu, S. [1994]. The biomechanics of walking and running. Foot and Ankle Injuries, 13:843-863.

)

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Similar to walking, the hip joint extends during both loading and propulsive stages of stance initially via concentric hip extension and later via eccentric hip flexion. Concentric hip flexion continues into the swing phase as the thigh is brought forward. This hip flexion continues into late swing when the hip extensors terminate the hip flexion movement and initiate a hip extension (40).

At the knee joint, the loading response is similar to that of walking, involving flexion controlled by the knee extensors to the point of midsupport. From that point, there is a net knee extensor moment. At the start of the swing phase, a small net knee extensor moment is associated with knee flexion. Later in the swing phase, a net knee flexor moment slows the rapidly extending knee.

Ankle joint net joint moments and powers are also similar to those of walking, depending on the style of running. For runners with a typical heel strike footfall pattern, the ankle joint exhibits a small net dorsiflexor moment during the loading phase followed by a net ankle plan-tarflexion moment for the remainder of the stance phase. At midstance, the net ankle plantarflexion moment controls rapid dorsiflexion, and in later stance, the plantarflexion moment produces rapid plantarflexion. Minimal ankle power is generated in the swing phase.

Angular Kinetics of the Golf Swing

The actual physical dimensions and characteristics of the club influence the angular kinetics of a golf swing. The composition of the materials of both the shaft and the club head influences swing characteristics. Shafts made of graphite or composite materials are usually lighter and stronger, so using this type of club, a golfer can swing a lighter club faster while producing the same amount of angular work. Adding mass to a club increases the joint torque at the shoulder and the trunk in the latter portion of the swing (30). Use of stiffer shafts generally results in straighter shots. Use of flexible shafts generally results in longer shots, but these clubs are difficult to control, thus influencing accuracy (51). The reason the stiffer staff offers better control is because it has less bend and twist and it compresses the ball more, creating a flight that is more representative of the actual angle of the club (35). A steel club with good direction control may be better for a novice golfer than a graphite club that can achieve more distance but is harder to control. The most desirable physical characteristics of the club are a high moment of inertia and a low center of gravity (50).

The physical characteristics of the club head can also influence the performance. An increase in the mass of the club head increases the joint torque at both the shoulder and the trunk for the latter half of the swing (30). Also, if the weight in the club head can be distributed to the periphery of the face, the optimal hitting area (i.e., the “sweet spot”) is increased, offering greater tolerance to off-center hits (50). If the center of gravity of the club head can be lowered, it can produce a higher flight path, and if the center of gravity can be moved toward the heel of the club, a right-to-left spin is promoted (50).

Club lengths influence the magnitude of the torque generated in the shoulder. A longer club produces an increase in the torque. This can open up the shoulder earlier and increases the linear acceleration of the wrist (30). Drivers are 5% longer than they were 10 years ago, and even if they can produce better results, they are more difficult to swing and control (35).

The angle that the shaft makes with the ground is called the “lie angle.” This angle determines how the face of the club is oriented. Lie angles range from approximately 55° for a driver to approximately 63° for the nine-iron. The lie angle can be altered if the club length does not match the physical dimensions of the golfer. Sample club lengths are 110 cm for a one-wood to 90 cm for a nine-iron (51). If a club is too short for a golfer, the drive is usually shorter because of lower swing angular velocities. Likewise, if the club is too long, the golfer is required to stand up more and may choke up on the shaft. Standing more erect changes the lie angle and often reduces the control over the shot (51). With a lower lie angle, the sweet spot of the club face is raised, leading to “topping” the ball.

Altering the length of the club can affect the angular acceleration of the club. For example, at the top of the backswing, the club lever is shortened by increased right elbow flexion. In the downswing, the club lever is lengthened (33). When the club is reversed at the top of the back-swing, the torque applied to the shaft actually causes it to bend where the head trails the line of the shaft (35). Torque applied through the shoulders accelerates the club, and there is a rigid body rotation with the wrist angle held constant. This positions the shaft perpendicular to the arms. Next, the hands allow the club to accelerate about the wrist. In the last portion of the swing, the shaft is rotated 90° at the wrist to bring the shaft face square into impact (35).

The distance a golf ball travels is related to the speed of the club head at contact. This speed is determined by the torque the golfer applies to the arm-club system and the management of the torque on the club by the wrists (28). If the wrists can be unhinged or uncocked later in the downswing, greater club head speed is generated for a given torque. In the follow-through phase, the angular momentum of the swing takes the right arm over the left and stimulates rotation of both the trunk and the head (33).

The torque generated at the wrist about the vertical (y), anteroposterior (z), and mediolateral (x) axes in the swing of a professional golfer is shown in Figure 11-52. In the backswing, a positive torque is generated about the anteroposterior (z) axis as the angle between the club and the arm is maintained. At the initiation of the downswing, a negative torque is generated about the same axis to keep the club back and maintain the same angle between the club and the arm. At the bottom of the downswing, a large

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positive torque is developed as the wrist uncocks and accelerates the club to impact (38). Rotations about the medio-lateral (x) axis are at a maximum as the club is taken back behind the head. Torques about the vertical (y) and anteroposterior (z) axes are zero at this point because the club is parallel to the vertical-anteroposterior plane. Torque about the vertical (y) axis is maximum at the initiation of the downswing as the body rotates around the support.

 

FIGURE 11-52 Joint torques acting at the wrist joint about the vertical (Y), anteroposterior (Z), and mediolatera (X) axes. The solid line indicates impact (Adapted from 

Neal, R. J., Wilson, B. D [1985]. 3D kinematics and kinetics of the golf swing. International Journal of Sports Biomechanics, 1:221–232.

)

Angular Kinetics of Wheelchair Propulsion

Hand rim wheelchair locomotion is strenuous and involves a significant amount of mechanical work production by the upper extremity muscles. The magnitude of the joint moment at the shoulder and elbow joints is influenced by the direction of application of the linear force. A large elbow moment is present when the force application to the rim is perpendicular to a line from the hand to the elbow and is minimized when the force is applied along the line from the hand to the elbow (48). Likewise, propelling forces, which are perpendicular to the line between the hand and the shoulder, result in large shoulder moments.

The shoulder joint torques account for the majority of the external power in wheelchair propulsion (54). The peak torque in the shoulder joint is greater than those generated in the elbow joint (58). A very small torque is generated in the wrist joint, where a braking effect is generated (58). The torques generated at the elbow and wrist joints are one-third and one-fifth of the torques at the shoulder joint (5). Joint torques ranges for the joints approximate 5 to 9 N-m at the wrist joint, 9 to 25 N-m at the elbow joint, and 25 to 50 N-m at the shoulder joint (47,54). Figure 11-53 illustrates a typical net shoulder moment of a paraplegic subject during wheelchair propulsion (55).

At low speeds and resistances, wheelchair propulsion is generated primarily by upper extremity sagittal plane movement. Fifty percent of the shoulder joint work is lost as negative work at the wrist and elbow (5). As the workload and speed increase, a concomitant increase occurs in the direction and magnitudes of the joint torques. Greater extension and abduction at the shoulder coupled with more torque at the elbow joint contribute to the increased output (5). The increase in shoulder abduction is speculated to be a forced abduction because of rotation of the arm in a closed-chain system (54). A sample of the shoulder flexion and extension, shoulder adduction, elbow flexion, and wrist flexion and extension torque patterns for wheelchair propulsion are shown in Figure 11-54. The highest torques are generated via shoulder flexion and shoulder adduction. The shoulder flexion torque peaks just before the peak in elbow extension torque. The shoulder adduction torque controls the shoulder abduction that is created as a result of arm movement in a closed chain. There is a net flexor torque at the elbow at the beginning of the propulsion phase that shifts to an extensor torque, continuing on through the rest of propulsion. The wrist torque is primarily extensor throughout the majority of the propulsion phase. A pronation and a larger

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radial deviation moment act at the wrist over the majority of the propulsive phase (2).

 

FIGURE 11-53 Example of the net shoulder moment (paraplegia subject) during wheelchair propulsion (mean and standard deviation over five trials; time normalized to 100% of a full cycle). (Adapted from 

van Drongelen, S., et al. [2005]. Mechanical load on the upper extremity during wheelchair activities. Archives of Physical Medicine and Rehabilitation, 86:1214-1220.

)

 

FIGURE 11-54 Shoulder flexion torque (top left), shoulder adduction torque (top right), elbow flexion torque (bottom left), and wrist torque in the sagittal plane (bottom right) for the propulsive stage of wheelchair propulsion. BP, beginning propulsion; EP, end propulsion. (Adapted from 

Veeger, H. E. J., et al. [1991]. Load on the upper extremity in manual wheelchair propulsion. Journal of Electromyography and Kinesiology, 1:270-280.

)

Summary

Torque, moment, or the moment of force are terms that can be used synonymously. A torque results when the line of action of a force acts at a distance from an axis of rotation. The magnitude of a torque is the product of the force and the perpendicular distance from the axis of rotation to the line of action. The unit of torque is the newton-meter (N-m). Torques cause rotational motion of an object about an axis. A torque is a vector and thus must be considered in terms of magnitude and direction. The right-hand rule defines whether the torque is positive (counterclockwise rotation) or negative (clockwise rotation).

Newton's laws of motion can be restated from the linear case to their angular analogs. The angular analogs of these laws:

  1. A rotating body will continue in a state of uniform angular motion unless acted on by an external torque.
  2. An external torque will produce an angular acceleration of a body that is proportional to and in the direction of the torque and inversely proportional to the moment of inertia of the body.
  3. For every torque exerted by one body on another body, an equal and opposite torque is exerted by the latter body on the former.

Moment of inertia is the angular analog of mass. The magnitude of the moment of inertia of an object, or the resistance to angular motion, depends on the axis about which the object is rotating. The calculation of the segment moment of inertia can be calculated using the radius of gyration as:

where Icm is the moment of inertia about a transverse axis through the center of mass, m is the mass of the segment, p is a proportion describing the ratio of the radius of gyration about the center of mass to the segment length, and L is the segment length.

Angular momentum is the angular analog of linear momentum and refers to the quantity of rotation of an object. The angular momentum of a multisegment body must be understood in terms of local and remote angular momenta. Local angular momentum is the angular momentum of a segment about its own center of mass. Remote angular momentum is the angular momentum of a segment about the total body center of mass. The total segment angular momentum is the sum of the segment's local and remote aspects. The total body angular momentum is the sum of the local and remote aspects of all segments. The angular analog of Newton's first law of motion is a statement of the law of conservation of angular momentum.

The concept of torque can be used to define the center of mass of an object. The sum of the torques about the center of mass of an object equals zero. That is:

This relationship defines the center of mass as the balancing point of the object.

The center of mass is commonly computed using the segmental method, which requires body segment parameters, such as the location of the center of mass and the segment proportion of the total body mass. The location of a segment center of mass requires the coordinates of both the proximal and distal ends of the segment. It is defined as:

where scm is the location of the segment center of mass, sproximal is the coordinate point of the proximal end of the segment, sdistal is the coordinate point of the distal end of the segment, and %L is the location of the segment center of mass as a proportion of the segment length from the proximal end of the segment. Locating the total body center of mass with the segmental method also uses the concept that the sum of the torques about the center of mass is zero. This computation uses the formula:

is the location of the total body center of mass, n is the number of segments, mi is the mass of the ith segment, si is the location of the segment center of mass, and M is the total body center of mass.

A lever is a simple machine with a balancing point called the fulcrum and two forces, an effort force and a resistance force. The mechanical advantage (MA) of a lever is defined as the ratio of the effort moment arm to the resistance moment arm. Levers can magnify force (MA > 1), magnify speed of rotation (MA < 1), or change the direction of pull (MA = 1). The three classes of levers are based on the relationship of the effort and resistance forces to the fulcrum. In the human body, however, the third-class lever magnifying the speed of movement predominates. Most of the levers in the extremities are third-class levers.

Special torque applications include angular work, rotational kinetic energy, and angular power. These concepts may be developed by substituting the appropriate angular equivalent in the linear case. Angular work is defined as:

where T is the torque applied and Δθ is the angular distance over which the torque is applied. Rotational kinetic energy is the capacity to do angular work and is defined as:

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where I is the moment of inertia of the segment about its center of mass and to is the angular velocity of the segment. Angular power is defined as either the rate of doing angular work:

where W is the mechanical work done and of is the time period over which the work is done or the product of torque and angular velocity:

where T is the torque and Ω is the angular velocity.

Angular motion analysis can be conducted using one of three techniques: the effect of a torque at an instant in time (T = Iα), the effect of a torque applied over time (impulse-momentum relationship), or the effect of a torque applied over a distance (work-energy theorem).

In the first technique, the static 2D case is determined using the following equations:

The dynamic 2D case uses the following equations:

ΣFX = maX for the horizontal component

ΣFY = maY for the vertical component

ΣTZ = IαZ for rotation

The purpose in both cases is to determine the net muscle moment about a joint. The impulse-momentum relationship relates the torque applied over time to the change in momentum:

The left-hand side of the equation, T * dt, is the angular impulse, and the rieht-hand side, IΩfinal – IΩinitial describes the change in angular momentum. In the third type of analysis, mechanical angular work is calculated via the change in mechanical energy. That is:

where RKE is the rotational kinetic energy.

Equation Review for Angular Kinetics

 
 

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Review Questions

True or False

  1. —A torque that causes rotation acts through the pivot point.
  2. —By convention, counterclockwise torques are considered to be positive.
  3. —A third-class lever is favorable for force production
  4. —A first-class lever always favors range of motion
  5. —Reducing the moment of inertia about an axis through the shoulder joint requires more muscle force to move the limb
  6. —A skater who is spinning can reduce the angular momentum by extending the arms and legs
  7. —The moment arm for muscle force changes throughout the range of motion
  8. —A ground reaction force applied in front of the center of mass creates a clockwise turning effect.
  9. —A negative net moment can indicate an eccentric muscle action
  10. —The sum of torques about the center of mass of an object always equals zero
  11. —The mechanical advantage of a second-class lever is always less than 1
  12. —Remote angular momentum is the angular momentum of a segment about another segment's center of mass
  13. —The only torque acting on an arm held out to the side is the torque generated by the muscle force
  14. —Angular power is the product of net moment and the joint angular velocity.
  15. —The moment arm is always measured from the point of application of the force to the axis of rotation
  16. —A higher center of gravity in a golf club can result in desirable flight paths
  17. —A force couple produces rotation and translation
  18. —Total body angular momentum can be increased during a dive by forcefully swinging the arms downward
  19. —The moment of inertia about an axis through the center of mass is less than the moment of inertia about any other perpendicular axis running through any other point on the segment.
  20. —The final acceleration in the golf club during the downswing is attributed to the wrists uncocking
  21. —Using action-reaction in the air, the legs can be brought up by swinging up the arms
  22. —The angular acceleration of a rotating body is not influenced by the moment of inertia
  23. —During the stance phase of running, positive power is produced at the hip, knee, and ankle joints.
  24. —The shoulder joint torques contribute more power in wheelchair propulsion than the elbow joint torques
  25. —The moment of inertia about the center of mass for the longitudinal and the transverse axes are equal

Multiple choice

  1. A force of 213 N is exerted 0.25 m from the axis of rotation What is the resulting moment of force?
  2. 53.25 N-m
  3. 532.50 N-m
  4. 852.00 N-m
  5. 85.20 N-m
  6. An object has a moment of inertia of 150 kg-m2. A torque of 72 N-m is applied to the object. What is the angular acceleration?
  7. 2.08 r/s2
  8. 10,800 r/s2
  9. 0.48 r/s2
  10. 983 r/s2
  11. A 70-kg gymnast applies a vertical ground reaction force of 1300 N at 0.2 m behind the center of mass. What is the torque generated about the center of mass?
  12. 122.7 N-m
  13. -122.7 N-m
  14. 260 N-m
  15. -260 N-m
  16. If a force of 120 N acting 0.29 m from the axis of rotation is balanced by another force of 92 N, what is the moment arm of the second force?
  17. 2.64 m
  18. 0.51 m
  19. 8.12 m
  20. 0.15 m
  21. What is the torque generated at the elbow by a 90-N force pulling on the forearm at an angle of 110° from the horizon at a point 18 cm from the elbow's axis of rotation. The forearm is positioned horizontal to the ground
  22. 16.2 N-m
  23. 15.22 N-m
  24. 5.54 N-m
  25. 19.8 N-m
  26. How much torque must be generated by the deltoid muscle to hold a 60-N dumbbell straight out at a 90° arm position? The dumbbell is 0.6 m from the shoulder joint. The center of mass of the arm, weighing 30 N, is 0.25 m from the shoulder joint. The moment arm for the deltoid muscle is 0.05 m
  27. 870 N-m
  28. _870 N-m
  29. 43.5 N-m
  30. _43.5 N-m
  31. Calculate the rotational energy of a segment, given mass of the segment 4.59 kg, moment of inertia 0.057 kg-m2, and s 1.90 rad/s.
  32. 0.10 J
  33. 0.47 J
  34. 0.21 J
  35. 0.05 J
  36. A force of 200 N is applied at a point 1.3 m from the axis of rotation, causing a revolving door to accelerate at 6.2 rad/s2What is the moment of inertia of the door from its axis of rotation?
  37. 953 kg-m2
  38. 1612 kg-m2

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  1. 24.81 kg-m2
  2. 41.94 kg-m2
  3. If the net moment at a joint is 25.15 N-m and the angular velocity at the same instant in time is 1.64 rad/s, what is the angular power at that joint?
  4. 4.13 W
  5. 41.25 W
  6. 1.53 W
  7. 15.34 W
  8. The center of mass of the following three-point system with masses of 8, 5.4, and 9 kg at the coordinates (4,7), (9.2, 5.6), and (7,4), respectively, is—.
  9. 4.74, 4.82
  10. 6.46, 5.46
  11. 4.9, 7.47
  12. 5.10, 5.98
  13. What is the angular momentum if the force is 54 N, the lever arm is 5.4 m, and the time is 2.7 s?
  14. 27.0 kg-m2/s
  15. 108 kg-m2/s
  16. 787.32 kg-m2/s
  17. 145.8 kg-m2/s
  18. What are the coordinates of the center of mass of a segment with the proximal end (5.7, 3.2) and the distal end (7.9, 4.1) if the center of mass is 48.6% from the proximal end of the segment?
  19. 4.63, 2.77
  20. 2.77, 4.63
  21. 6.77, 3.64
  22. 3.64, 6.77
  23. Calculate the moment of inertia of a baseball bat about its proximal end if its mass is 2 kg and has a radius of gyration of 0.55 and a length of 0.864 m
  24. 0.452 kg-m2
  25. 0.950 kg-m2
  26. 0.903 kg-m2
  27. 0.542 kg-m2
  28. A torque of 68 Nm results in a rotation through 0.57 rad in 0.3 s. How much angular work was done?
  29. 38.8 J
  30. 129.2 J
  31. 11.63 J
  32. 119.30 J
  33. How much power was developed in the example presented in Question 11?
  34. 38.8 J
  35. 129.2 J
  36. 11.6 J
  37. 119.3 J
  38. Consider the following free body diagram. Using static analysis, solve for the muscle torque that will place this system in equilibrium, given mass of the leg and foot 5.3 kg; distance from the knee joint to the center of mass of the leg-foot system 0.152 m; weight of the barbell 150 N; and distance from the knee joint to the center of mass of the barbell 0.521 m
  39. 78.96 N-m
  40. 8.77 N-m
  41. 15.86 N-m
  42. 86.05 N-m
  43. If the leg is at an angle of 15° below the horizontal, calculate the moment arm of the torque caused by the weight of the leg, given that the distance to the center of mass of the leg 0.17 m
  44. 0.17 m
  45. 0.16 m
  46. 0.15 m
  47. 0.14 m
  48. Consider the following free body diagram. Using static analysis, solve for Achilles tendon force that will place this system in equilibrium if d1 = 0.045 m, d2 = 0.045 m, and d3 = 0.13 m
  49. 5500.10 N
  50. 5477.70 N
  51. 11.10 N
  52. 11.14 N
  53. An object has a moment of inertia of 164 kg-m2. A torque of 89.3 N-m is applied to the object in 5 s. What is the angular acceleration?
  54. 2.72 rad/s2
  55. 0.272 rad/s2
  56. 5.55 rad/s2
  57. 0.55 rad/s2
  58. Consider the following free body diagram. Using static analysis, solve for the moment at the elbow if d1 = 0.04 m, d2 = 0.14 m, and d3 = 0.46 m. What is the net muscle force?
  59. 2945.0 N
  60. 2872.4 N
  61. 2367.5 N
  62. 2250.5 N
  63. Consider the following diagram of the biceps brachii acting on the radius in two joint positions. The angle of pull of the biceps brachii force changes from 15° to 30°. If the muscle force is 900 N and the attachment site of the muscle is 0.04 m from the joint axis, what is the change in joint torque applied by the biceps brachii from 15° to 30°?
  64. 3.59 N-m
  65. 7.19 N-m
  66. 9.0 N-m
  67. 18.0 N-m
  68. What is total energy of the segment, given that mass = 4.2 kg; g = 9.81 m/s2; vx= 0.18 m/s; vY = 3.45 m/s; moment of nertia = 0.07234 kg-m2; angular velocity = 6.3 rad/s2; and height of center of mass = 0.39 m
  69. 42.57 J
  70. 26.43 J
  71. 42.43 J
  72. 27.01 J
  73. What is the work done on the segment given the following information?

 

TKE

RKE

PE

Frame 2

18.7

0.98

18.0

Frame 100

9.2

0.2

7.5

  1. 2.078 J
  2. 20.78 J
  3. 10.28 J
  4. 1.028 J

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  1. During a knee extension exercise, the knee extensor muscle group is applying a torque of 230 N-m in an isometric contraction against the machine pad. If the knee joint angle is being held at an angle of 40° below the horizontal and the machine pad is 0.35 m from the knee joint, how much force is being applied at the pad?
  2. 61.67 N
  3. 657.14 N
  4. 80.50 N
  5. 858.21 N
  6. While exercising, an athlete holds a position with 90° of flexion at the hip joint and 90° of flexion at the knee joint. Considering only the iliopsoas muscle group, determine the amount of force necessary to maintain this position, given the moment arm of iliopsoas = 10.5 cm from hip joint, mass of the lower extremity = 11 kg, and moment arm of CM of the lower extremity = 38 cm from the hip joint.
  7. 3.981 N
  8. 39.81 N
  9. 390.53 N
  10. 3905.30 N
  11. During walking, the knee joint generates 50 N-m of extensor force during the same interval of the stance phase when the knee joint moved from 0.14 rad of flexion to 0.2 rad of flexion in 0.02 s. Determine the power of the knee joint muscles
  12. 50 W
  13. _50 W
  14. 150 W
  15. _150 W

References

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Glossary

Glossary

Angular Kinetics

The branch of mechanics that deals with the causes of rotations.

Angular Momentum

The quantity of angular motion determined by the product of the object's angular velocity and its moment of inertia.

Angular Power

The time rate of change of angular work determined by the product of the torque and the angular velocity.

Angular Work

The product of torque applied to an object and the angular distance over which the torque is applied.

Balance

The ability of an individual to assume and maintain a stable position.

Center of Gravity

The point at which all of the body's mass seems to be concentrated; the balance point of the body; the point about which the sum of the torques equals zero.

Center of Mass

A balance point of a body; the point about which all of the mass particles of the body are evenly distributed.

Conservation of Angular Momentum

The concept that angular momentum is constant unless the object is acted on by an external force.

Dynamic Analysis

A calculation of the forces and moments when there are significant linear or angular accelerations.

Dynamics

The branch of mechanics in which the system being studied undergoes acceleration.

Eccentric Force

A force that is not applied through the center of mass of an object.

Effort Arm

See Moment Arm.

Effort Force

A force applied to a lever, causing movement of the lever.

Equilibrium

The state of a system whose acceleration is unchanged.

External Work

The work done by a body on another body.

First-Class Lever

A lever in which the fulcrum is between the effort force and the resistance force.

Force Couple

Two forces that are equal in magnitude, act in opposite directions at a distance from an axis of rotation, and produce rotation with no translation.

Fulcrum

The axis of rotation of a lever.

Impulse

The product of the magnitude of a torque and its time of application.

Impulse-Momentum Relationship

The relationship stating that the impulse is equal to the change in momentum.

Internal Work

The total work done resulting from the motion of all of the body's segments.

Inverse Dynamics

An analytical approach calculating forces and moments based on the accelerations of the object.

Lever

A mechanism for doing work that consists of a fulcrum and two eccentric forces.

Local Angular Momentum

The angular momentum of a body segment about its own center of mass.

Mechanical Advantage

The ratio of the effort arm to the resistance arm of a lever.

Moment

See Torque.

Moment Arm

The perpendicular distance of the line of action of a force to the axis of rotation.

Moment of Force

See Torque.

Moment of Inertia

The resistance of a body to angular acceleration.

Muscle Power

The product of the net muscle moment and the angular velocity of the joint.

Negative Work

The work done on a system when the loading torque is greater than the torque exerted by the muscle.

Neutral Equilibrium

The state of a body in which the body will remain in a location if displaced from another location.

Parallel Axis Theorem

A theorem stating the relationship between the moment of inertia about an axis through the body's center of mass (Icm) and any other parallel axis (Icm) Icm = Icm + mr2 where m is the mass of the body and r is the perpendicular distance between the axes.

Positive Work

The work done by a system when the torque exerted by a muscle is greater than the torque of the external load.

Radius of Gyration

A measure of the distribution of a body's mass about an axis of rotation.

Remote Angular Momentum

The angular momentum of a segment about the total body center of mass.

Resistance Arm

See Moment Arm.

Resistance Force

A force that resists the effort force in a lever.

Rotational Kinetic Energy

The capacity to do angular work; the product of half the moment of inertia and the angular velocity squared.

Second-Class Lever

A lever in which the resistance force acts between the fulcrum and the effort force.

Segmental Method

A method of calculating the total body center of mass of a multisegment body by summing the product of the locations of the centers of mass of the segments and the mass of the respective segment and dividing by the total body mass.

Stability

Resistance to a disturbance in the body's equilibrium.

Stable Equilibrium

The state of a body in which the body will return to its original location if it is displaced.

Statics

The branch of mechanics in which the system being studied undergoes no acceleration.

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Third-Class Lever

A lever in which the effort force acts between the fulcrum and the line of action of the resistance force.

Torque

The product of the magnitude of a force and the perpendicular distance from the line of action of the force to the axis of rotation.

Unstable Equilibrium

The state of a body in which the body continues to increase its displacement if it is displaced.

Work-Energy Theorem

The relationship between work and energy stating that the work done is equal to the change in energy.