## B

Figure 4.11a-c. A gymnast rotating downward from a straight horizontal configuration while the positioning of the feet remains constant (a). The gymnast is modeled as two rods linked to each other (b). The free-body diagrams of each of the rods are shown in (c).

First, let us express the accelerations of points D and E in terms of the angular accelerations of rods OA and AB. We begin with the determination of the acceleration of D by using Eqn. 4.20b for points O and D. The acceleration at point O is zero because this point is fixed in space. Furthermore, the rods under consideration are released from rest, and hence their angular velocities are zero. Therefore, Eqn. 4.20b dictates that aD = «1 es X (L/2) ei = «1 (L/2) e2

in which «1 e3 is the angular acceleration of rod AB. Let us use the same equation to determine the acceleration of E:

aE = aA + «2 e3 X (L/2) e1 = «1 L e2 + «2 e3 X (L/2) e1 = [«1 L + «2 (L/2)] e2 in which «2 e3 is the angular acceleration of rod AB.

Let us next consider the equations of motion for the center of mass of each of the rods. These equations can be written in the e1 direction as follows:

Thus, the horizontal reaction forces acting on joints O and A are equal to zero. The equations of motion for the rods OA and AB in the e2 direction are

Note that the reacting force —R2 creates a counterclockwise (positive) moment with respect to point E.

These two equations involve four unknowns: F2, R2, a1, and «2. We need two more equations to solve for them. The principle of conservation of angular momentum provides two scalar equations. First, the conservation of angular momentum of bar OA about the fixed point O:

Note that ¡o = (mL2/3) is the moment of inertia of OA about point O. The conservation of angular momentum of the bar AB about its mass center E requires that

We can eliminate R2 from Eqns. 4.36a and 4.36b:

We can similarly eliminate R2 from Eqns. 4.35b and 4.36a:

Solving these last two equations for «1 and «2 we find that:

Thus, bar OA (legs) will begin rotating clockwise whereas the bar AB (trunk) will begin rotating counterclockwise. The gymnast will have to use abdominal muscles to remain aligned along a straight line.

Example 4.9. Body Curls. In body curls, one uses a specially designed bench to support the heels, upper thighs, and pelvis on padded supports (Fig. 4.12a). Lower the upper torso slowly down to vertical position. Then slowly pull up to the horizontal position and repeat the cycle. Assuming that the back muscle, the erector spinae, is the only muscle involved in this exercise, develop a procedure to evaluate muscle tension as a function of the angle the body trunk makes with the vertical axis.

Solution: Figure 4.12b shows a free-body diagram for the conceptual analysis of the back extension exercise. The upper body is represented by Figure 4.12a,b. A man performing back curls to strengthen the back muscle, erector spinae (a). The free-body diagram of the trunk of the man during a back curl (b).

rod AB of length L and mass m. The B end of the rod represents the head. It is free to move in the plane of the motion. The A end represents the pelvic girdle, which remains stationary during the course of the exercise. Gravity pulls the upper trunk down, tending to turn it in the clockwise direction. On the other hand, the back muscle generates a moment in the counterclockwise direction.

The angular momentum of the upper trunk with respect to the fixed point A and its time derivative can be written as

in which Q is the angle the trunk makes with the vertical axis, and the symbol t denotes the time. As usual, e3 is the unit vector perpendicular to the plane of the motion.

The conservation of moment of momentum dictates that (dHA/dt) is equal to the total external moment acting on the upper body with respect to point A. Thus follows the next equation:

(mL2/3) (d2Q/dt2) = — mg (L/2) sin Q + Tbm h (4.38)

where g denotes the gravitational acceleration, Tbm is the force generated by the back muscle, and h denotes the distance from point A to the line of action of the back muscle force. Using Eqn. 4.38, Tbm can be expressed as

Equation 4.39 shows that the force exerted by the erector spinae depends on the angle of inclination of the upper body with the vertical axis as well as the angular acceleration of the trunk. Assuming m = 34 kg, L = 0.9 m, h = 0.06 m, Q = 45°, and d2Q/dt2 = 2 rad/s, we find for the force exerted by the erector muscle T = 2074 N. Thus, the force exerted by the erector spinae is much greater than the weight of the athlete (667 N). About one-quarter of the force results from the upward acceleration of the trunk. However, the dominant reason why erector spinae has to exert such a high level of force during the back curl is because its moment arm with respect to the center of rotation at the hip is small, 0.06 m, as opposed to the moment arm of the weight of the upper body, 0.32 m.

### 4.8 Instantaneous Center of Rotation

The human knee is a multiple joint in which the femur of the thigh interacts with the tibia of the lower leg and the patella, the kneecap (Fig. 4.13). The geometry of the opposing surfaces in the femorotibial joint is complex, with the radius of curvature changing signs within the region of interaction. Where exactly is the center of rotation of this joint?

The answer to this question lies in the definition of instantaneous center of rotation. In planar motion, continuous motion of a rigid body may be considered as a succession of infinitely small displacements. Each small displacement can be brought about by rotation about some fixed axis. Thus, the motion can be regarded as a series of rotations about a moving axis. The axis of rotation at any instant is called the instantaneous axis of rotation. The point where the instantaneous axis meets the plane of motion is called the instantaneous center of rotation. How do we determine the position of the instantaneous center of rotation?

For locating this center, one needs to know the velocities of two points of a rigid body at an instant of time. Consider a rigid object B undergoing planar rotation as shown in Fig. 4.13a. Let P and Q be two points of   Figure 4.13a-c. The instantaneous center of rotation of an arbitrary rigid body in planar motion (a). The instantaneous center is marked with the symbol IC. The instantaneous center of a sphere rolling down an inclined plane is the point of contact with the plane (b). The instantaneous center of the knee is illustrated in (c).

Figure 4.13a-c. The instantaneous center of rotation of an arbitrary rigid body in planar motion (a). The instantaneous center is marked with the symbol IC. The instantaneous center of a sphere rolling down an inclined plane is the point of contact with the plane (b). The instantaneous center of the knee is illustrated in (c).

the object on a plane parallel to the plane of the motion. The velocities of P and Q are represented by vP and vQ, respectively. The center of rotation is then the intersection of the line that passes through P and is perpendicular to vP with the corresponding line passing through point Q (Fig. 4.13a). The velocity of the instantaneous center is zero. The velocities of points in the rigid body are proportional to their distances from the instantaneous center. The instantaneous center of a sphere rolling down an inclined plane without slip is the point of the sphere in contact with the plane at time t (Fig. 4.13b).

In the human knee, the instantaneous center coincides with the intersection of the cruciate ligaments (Fig. 4.13c). Because these ligaments are very stiff, they hardly change length, and therefore the velocities of points of origin and insertion of these ligaments remain perpendicular to the ligament itself. The instantaneous center, thus defined for the knee, may not correspond to a material point belonging to the head of the femur. The arcs corresponding to the successive instant centers of rotation of the tibia rotating in relation to the femur and that of the femur moving in relation to the tibia are called polodes. These two curves give a picture of the movement between the femur and tibia projected on the sagittal plane. In the unstable knee, the instantaneous center may vary significantly from time to time, and doctors have used polodes to detect instability associated with the knee. In the normal knee, polodes trace a compact curve around a point. The length scale of this curve is small in comparison to the lengths of the interacting bones. The center of the polodes is for practical purposes the center of rotation of the joint. For more information on the geometry of articulating surfaces of human joints and their instantaneous centers of rotation, the reader is referred to the article by Kento R. Kaufman and Kai-Nan An, Joint-Articulating Surface Motion, that appeared in Bronzino (1995).

### 4.9 Summary

A rigid body is a solid object such that the distance between any two of its material points remains constant during resting state or in motion. Various body segments of the human body such as head, thighs, and forearms can be reasonably assumed as rigid in the analysis of movement and motion. In planar motion parallel to the (e1, e2) plane, the angular velocity o of a rigid object B with respect to reference frame E is defined as the time rate of change of angle between a straight line fixed in E and another straight line in the rotating body B in the (e1, e2) plane, taken counterclockwise.

in which 6 is the aforementioned angle and e3 is the unit vector that is perpendicular to the plane of motion.

Angular acceleration a is defined by the following relation:

When angular acceleration is in the positive e3 direction, then the rate of rotation increases in the counterclockwise direction.

Velocity vectors of any two points in a rigid object are related by the following equation:

in which vQ and vP denote the velocities of points Q and P, and rQ/P is the position vector connecting point P to point Q.

Acceleration vectors of any two points in a rigid body obey the following relation:

in which aQ and aP denote the acceleration vectors of Q and P, respectively.

The moment of momentum of a rigid object is called angular momentum. For rigid objects that are undergoing planar motion in a plane of symmetry of the object, angular momentum with respect to the center of mass is given as

in which Hc denotes the angular momentum with respect to the center of mass and Ic is the mass moment of inertia of the object with respect to the center of mass. It is a measure of resistance of the object to the changes in rate of rotation.

If a point of the object, say point O, is fixed on earth and the object rotates around O, the angular momentum with respect to point O is given by the relation

The parameter Io, the mass moment of inertia with respect to point O that is fixed in E, is related to Ic by the following equation:

Io = mr2 + Ic in which r is the distance between the center of mass of the object and point O. The conservation of angular momentum dicates that

The right-hand side of these equations refers to the resultant external moment acting on the object with respect to the fixed point O and the center of mass, respectively. The principle of conservation of angular momentum relates the changes in rate of rotation to the resultant moment acting on an object.

### 4.10 Problems

Problem 4.1. Provide an estimate of the mass moment of inertia of your forearm about the three principal axes that pass through its center of mass. Assume the forearm to be a circular cylinder. Measure its length and girth. Assume that the mass density p is equal to 1.0 g/cm3.

Problem 4.2. The moment of inertia of an athlete with respect to his center of mass along an axis from posterior to anterior was experimentally determined to be equal to Ic = 13.6 kg-m2. The height of the man is 1.83 m and his weight is 78.6 kg. Represent the athlete as a slender rigid rod and determine an approximate value (I*) for his moment of inertia. Does the slender rod assumption overestimate the moment of inertia? What would be the effective length h* of the rod that would correctly predict the moment of inertia of the athlete? Answer: I* = 21.2 kg-m2; h* = 1.44 m.

Problem 4.3. Using the oscillating table method, Matsuo et al. (1995) proposed the following equations for determining the mass moment of inertia of adolescent boys between the ages of 13 and 18:

in which I33 and I11 are the mass moments of inertia with respect to the center of mass around the posterior-to-anterior and right-to-left axis, respectively (see Fig. P.4.3). The parameter H is the height of the adolescent, measured in meters, and W is his mass, measured in kilograms. To check whether this formula could also be applicable to adult men, a group of Air Force researchers measured the mass moment inertia of a select group of Air Force men. Following are the data obtained for three men in the group:

 Age Height (m) Mass (kg) I33 (kg-m2) I11 (kg-m2) 29 1.83 78 14.9 13.4 22 1.71 66 10.8 9.2 20 1.75 63 11.4 10.3 Figure P.4.3. Coordinate system attached to an individual in horizontal position.

How applicable are the formulas of Matsuo et al. (1995) to the data on the three men? Compare the model predictions with the data. How far off would be the predictions of these mass moment of inertia components if one represented each individual with a slender rod whose length and mass are equal to that of the individual?

Problem 4.4. Investigate the applicability of the equations developed by Matsuo et al. (1995) for other population groups such as adolescent girls, adult women, elderly men, and elderly women. Conduct a literature search and find relevant data. Determine if there are phenome-nological equations already developed for these subpopulations. If not, how would you go about coming up with your own set of empirical equations?

Problem 4.5. Discuss the various ways of determining the mass moment of inertia of body parts. Provide examples from the literature.

Problem 4.6. Provide an estimate of the spatial location of the center of mass C of the dancer leaping in air as shown in Fig. P.4.6. Compute the location of C using the data given below. Specify in detail any additional assumptions you had to make to arrive at your results. Note that you need to establish a reference frame to compute and specify the lo-

 cation of the center of mass. Assume that the reference frame is cen- tered at the top of the head. Body segment Weight (kg) Length (cm) Trunk 24 51 Head 3.4 27.5 Thigh 4. 46 Shank 2.9 43 Foot 0.8 22 Upper arm 1.7 27 Forearm 1.3 24 Hand 0.4 19 Figure P.4.6. Dancer leaping on stage. Figure P.4.7a,b. Schematic diagram of an individual doing push-ups (a). In this exercise, the man is represented as a rod with uniform distribution of mass (b). The arms are modeled as weightless links (b).

Problem 4.7. Determine the vertical ground forces acting on a man at the feet (FF e2) and hands (FH e2) while performing push-ups as shown in Fig. P.4.7. The man weighs 71 kg and has a height of 1.76 m. At the instant considered (t = 0), the angle his body makes with the horizontal plane (0) is 20°. Again at t = 0, d0/dt = 4 rad/s and (d20/dt2) = 7 rad/s2. The distance from the bottom of his feet to his center of gravity = 1.02 m. The distance between his head and shoulders is 0.32 m. The body is aligned straight and rotates around the fixed point O as shown in the figure. Assume that the arms can be represented by weightless rods hinged at the middle. Arms transmit vertical forces only. Answer: FH = 872 N, Ff = -95.5 N. The fact that Ff is negative implies that somebody must have been pressing at the ankles of the man doing the push-ups.

Problem 4.8. The rods B1 and B2 shown in Fig. P.4.8 each have mass m and length L. They are hinged together and in the resting position are aligned on a straight line. The rods are released while making 30° with the vertical axis. The rod B1 slides on the smooth, frictionless surface of the floor and the center of mass of the system moves parallel to the floor. Determine the reaction force F2 and the angular accelerations of B1 and B2 right after the release. Take m = 28 kg, and L = 0.78 m. Note that this two-rod system might capture some of the essential features of sideway falls. Among the elderly population, a sideway fall is a most frequent cause of hip fracture. The answer to this problem may provide information about the nature of shape change during such a fall Figure P.4.8. Sideway fall of a person onto a floor (left) and its representation using a two-link model (right).

in the absence of a strong reactive contraction of the back and abdominal muscles.

Answer: F2 = 550 N, d20B1/dt2 = 0, d2dBi/dt2 = -75.0 rad/s2.

Problem 4.9. A diver is airborne in full extended position rotating with clockwise angular velocity a = -5 rad/s. At time t = 0 he begins to pull his legs toward his chest at a rate of —w rad/s (Fig. P.4.9). Determine the angular velocity of his trunk and that of the lower extremities. Assume that the diver is composed of two slender rods each weighing 32 kg and 0.86 m long. Hint: Use Eqn. 4.27b and show atrunk = —3.43 rad/s and aleg = —6.57 rad/s. Figure P.4.9. A diver pulling his legs toward his body while airborne. 