## Angular Impulse and Angular Momentum

The moment of momentum of a body B with respect to a point O that is fixed on earth was defined by the following integral over the mass of the body:

Ho = /r X v dm in which r and v are, respectively, the position and velocity of mass element dm in the inertial reference frame E (Fig. 7.7). We have shown (Chapter 3) in Eqn. 3.39b that

Ho = m (rc X vc) + Hc in which the superscript c refers to the center of mass. According to the conservation of moment of momentum:

in which 2Mc and 2Mo denote the resultant external moment about the center of mass and the fixed point O, respectively. Integrating Eqn. 7.6 between times ti and tf we obtain:

Integrals in Eqns. 7.7a and 7.7b represent the angular impulses of the external forces about points C and O:

In the presence of impulsive forces, we need only to consider impulsive forces and moments when evaluating Ac or Ac. Because the time duration of an impulsive force is very small, contributions of regular forces to the angular impulse can be neglected. For rigid bodies that are symmetric with respect to the plane of motion, we have already seen in Chapter 4 that

in which a is the angular velocity of the rigid body, e3 denotes the unit vector perpendicular to the plane of motion, and Ic and Io are the mass moment of inertia with respect to the center of mass and point O, respectively. Combining Eqns. 7.8 with Eqns. 7.7, we obtain:

At this point, it is useful to recapture the principal assumptions used in the study of impact and collision.

1. Velocities and angular velocities may change greatly during the brief time interval of impact.

2. Positions of the bodies do not change appreciably during impact.

3. Forces (and moments) that are nonimpulsive (force and moments that do not assume very large values during the time of collision) are neglected.

As examples of negligible forces during impact we cite gravitational force and the force exerted by a stretched or compressed spring. Only the impulsive forces and moments produce sudden changes in velocities and angular velocities.

Example 7.3. A man standing still is struck by an impact force (Fig. 7.8). Assume that the man can be represented by a uniform rod of length L and mass m. The distance between the point of application of the impact force and the center of the rod is d. Determine the velocity of the center of mass and the angular velocity just after the blow. Figure 7.8. A man standing at rest is hit with an impulsive force.

Solution: If u ei and m e3 are the velocity of the center of mass and the angular velocity just after the blow, the conservation of linear and angular momentum leads to the following equations:

m (L2/12) m = d ■ i m = d ■ i/[m (L2/12)] (7.10b)

Consistent with impact analysis, contributions of the body weight and the contact forces to the velocity after the impact were neglected in comparison to the contribution of the impulsive force.

Even when the magnitude of the impulse cannot be measured, it is possible to deduce a mathematical relationship between u and m. Eliminating i from Eqns. 7.10a and 7.10b, we find u = (L2/12) M/d

Suppose the blow is struck at the end of the rod, so that d = L/2; then u = (L/6) m.

Example 7.4. A body of arbitrary planar shape is attached to a fixed axis through A as shown in Fig. 7.9. A blow of impulse i is exerted on the body at an angle 0 with the vertical axis. Determine the angular velocity of the body immediately after the blow and the impulsive reaction at the hinge.

Solution: Let mo and m denote the angular velocity before and after the impulse. The change in angular momentum is related to the angular impulse as follows:

Figure 7.9. The effect of an impulse on the rotation of a rigid body around an axis that passes through point A. in which h is the distance between the center of mass C and the hinge A, k is the radius of gyration of the object B with respect to C, and d is the distance between the point of application of the impulse and the hinge A. Rearranging Eqn. 7.11a, we obtain a = oo + d • f sin 0/[m (h2 + k2)] (7.11b)

To compute the impulse acting on the hinge, we use the principle of conservation of linear momentum (see Eqn. 7.3):

in which A1 and A2 are the components of the impulsive reactive forces the hinge exerts on the object in consequence of the blow. The left-hand side of Eqn. 7.12b is zero because the velocity of the center of mass in the e2 direction is equal to zero.

Example 7.5. Thoracic Injury Potential of Basic Competition Tae Kwon Do Kicks. In a competition, a tae kwon do master hits the other on the chest with a tae kwon do kick. Consider a simple mechanical analysis in which the leg of the tae kwon do master is represented by a rod of mass m and length L (Fig. 7.10). The rod rotates about its center with angular velocity ao, and the end of the rod marked as A collides with the stationary mass M. The velocity of the center of mass of the rod is equal to zero before the collision. After the impact, the points of contact move with the velocity v. At this time the linear velocity of the center of mass of the kicking leg is u and its angular velocity is a. Determine the velocities u, v, and a. Determine the impulse exerted by the tae kwon do master.

Solution: Because points C and A belong to the same rigid body (rod), their velocities are related by the following equation:

before impact after impact before impact after impact Figure 7.10. A tae kwon do master hits his partner on the chest with a tae kwon do kick. In the impact analysis the kicking leg is represented by a rod of mass m and length L. The partner is represented as a sphere with mass M.

The linear momentum of the system before the collision must be equal to the linear momentum after the collision:

Additionally, the moment of momentum about C of the system before the collision must be equal to the corresponding moment of momentum after the collision

Combining Eqns. 7.13a and 7.13b, we find v = -(m/M) u

Using this expression together with Eqn. 7.13b in Eqn. 7.13c, we obtain:

Because the lower limb constitutes about 17% of the body weight, m = 0.17 M, and thus:

These results indicate that right after the kick the center of mass of the kicking leg moves backward, whereas the man who is hit moves forward.

Let us next compute the impulse generated at the collision. The conservation of linear momentum for mass M leads to c = 0.04 M Mo (L/2)

For M = 75 kg, L = 1.02 m, and mo = 22 rad/s, the impulse f turns out to be 33.6 N-s. If the impulsive force of the kick were to last 20 ms, then the average impulsive force would have been 1680 N.

Example 7.6. Impact Force Acting on a Runner's Heel. Runners hit the ground with one leg at a time, then use the heel of the colliding leg as a pivot to push forward (Fig. 7.11). The angular velocity of the leg before hitting the ground is —wo e3. The velocity of runner's trunk before the impact is vo e1. The length of the leg is equal to L. The mass of one leg constitutes 17% of the body mass m. For simplicity, we assume the weight of the leg to be negligible and that the weight of the runner is lumped on the hip. Determine the impulse exerted by the ground on the runner's heel. What is the angular velocity of the leg immediately after the impact?

Solution: When the heel A becomes fixed on the ground, the leg rotates about A. While the impulsive force is acting at A, the rate of change of Figure 7.11a-d. The force of impulse during running (a). The impulsive ground force acting on the heel of a runner is shown in (b). The vertical component of the ground force is plotted as a function of time (c). Stick figures for the first part of the contact phase are shown in (d).

moment of momentum of the runner about A is zero. The moment of momentum A before the impulse (Ho e3) is given by the expression:

where \$ is the angle the leg makes with the vertical axis at the impact.

### And after the impulse

H = —m L2 m in which — m is the angular velocity of the leg after the impact. Because there is no change of moment of momentum about A, we have H = Ho, which gives us m = vo cos \$ /L

To determine the impulse during the impact, we need to know the velocity of the center of mass after the impact. When A becomes fixed and the leg rotates about A, the velocity of the center of mass becomes v = m L cos \$ e1 + m L sin \$ e2 = vo cos2 \$ e1 + vo cos \$ sin \$ e2

This equation shows that the impact alters both the speed and the direction of velocity of the center of mass of the runner. As a result of the impact, the runner gains an upward velocity. Also, the speed of his center of mass is reduced to vo cos \$ during the course of the impact.

To find the components of impulse, we consider the change in linear momentum:

Let L = 1.06 m, \$ = 35°, m = 71 kg, vo = 6 m/s, and mo = 25 rad/s. Then we find:

Assuming the duration of the impact to be 0.1 s, the mean impulsive force in the vertical direction is found to be 2,000 N. The vertical ground reaction force has received the greatest attention of biomechanics researchers because of its magnitude. Particular interest in the first early phase of the ground force (impact force) has been motivated by the concern about the transmission of shock waves upward through the muscu-loskeletal system. Runners who initially contact the ground with their heels tend to elicit a high force of short duration that has been termed as the impact peak. Magnitude of the impact force is considered as a po tential indicator of overuse injuries in running, degenerative changes in joints, and low back pain. At running speed of 6 m/s this force is about three body weights, which is in agreement with our estimate. A typical time history of vertical ground force measured by using a force plate in the study of Bobbert et al. (1992) is shown in Fig. 7.11c. In this graph the impact force peaks at 1850 N. The data were collected on a runner whose steady-state speed was 4.5 m/s. The data indicate that the impact force increases with running speed.

We might ask the question whether the peak ground force measured during running qualifies to be called impulsive force? According to our definition, an impulsive force must be so much greater than any other force acting on a body so that the velocity of the body changes sharply during the brief period the impulsive force acts. Also, the duration of application of impact force must be so short that the position of the body does not change appreciably during impact. Bobbert et al. (1992) filmed the running events as they collected data on the ground reaction force. Stick figures representing a runner's body are shown in Fig. 7.11d for the first part of the contact force in heel-toe running. The leftmost figure shows the segment orientations on the last frame before touchdown. The time duration between two subsequent stick figures is 5 ms. As can be seen in the figure, there is a small change in the configuration of the runner during the first 100 ms of the contact. These observations suggest that the impact must occur in less than 0.1 s. Otherwise, impact force is not an impulsive force in the strictest sense—neither it is too large compared to the body weight nor does the configuration of the runner remains constant during the course of its application.

A note on the foot as a shock observer may be in order. The foot sustains impact forces and reduces potential injury to the body by deforming upon striking the ground. If the foot were a rigid object, the ground reaction force acting on it would be of great magnitude and short duration. The bones of the foot, however, are tied together by flexible ligaments and the movement of these bones relative to each other is also controlled by tendons. The presence of the deformable heel pad reduces the vertical force of impact. As the foot deforms in response to the force of impact, the ligaments and tendons stretch, absorbing much of the shock. As a result, the impulse is caused by a more sustained force of smaller amplitude. 