## Multibody Systems

The time rate of change of kinetic energy of a deformable body is not necessarily equal to the work of external forces and couples acting on the body. This holds true even for a tree of rigid bodies. What is the equation governing the time rate of change of kinetic energy of serially linked rigid bodies?

Consider two rigid objects connected at point A (Fig. 8.8). We want to know whether the forces and moments acting on the objects at point A contribute to the rate of change of kinetic energy of the two objects. Let dT/dt represent the rate of change of kinetic energy of the two bodies as a result of the mechanical power of the forces and moments acting at point A. Then, according to the principle of conservation of kinetic energy:

dT/dt = F • v + (—F) • v + M1-2 • «1 + (-M1-2 • «2) (8.25a)

in which F is the force exerted by body 2 on body 1 at point A, v is the velocity of point A, M1-2 is the moment exerted by body 2 on body 1, and «1 and «2 are, respectively, the angular velocities of the bodies 1 and 2. According to Newton's third law, action is equal to reaction, and thus the

force (moment) exerted by body 1 on body 2 must be equal to minus the force (moment) exerted by body 2 on body 1. Thus, the reaction force at the joint connecting the two bodies cannot contribute to the total kinetic energy.

After algebraic simplifications, the time rate of change of kinetic energy of two bodies connected at a point (because of forces and moments acting at that point) can be put in the following simple form:

This equation illustrates what differentiates a living creature from a tree of bodies that are serially linked using hinges. As a hinge (pin) joint cannot resist a moment, the joint would move in such a way to make the resultant moment acting on the joint equal to zero. Hence, in a serially linked system of bodies, the rate of increase in the kinetic energy of the system from joint reaction forces is equal to zero. In the human body, however, the tendons exert moments of appreciable value to bone joints, and therefore the right-hand side of Eqn. 8.25 is not equal to zero.

Equation 8.25 provides insights into the mechanics of human movement. Consider, for example, the case of a skater spinning about a point on ice. When she draws her arms in, her moment of inertia decreases, resulting in an increase in her angular speed. This increase results from the chemical energy used in producing the movement.

Example 8.4. Transfer of Kinetic Energy During Running. The ground force acting on a runner during an instant in the stance phase was measured to be equal to 2,400 N in magnitude and vertically upward in direction (Fig. 8.9). At that time, the shank had an angular velocity of «2 =

Figure 8.9. The relative motion of the foot and the shank of a runner during an instant in the stance phase. The free-body diagram of the foot is also shown in the figure. Point A identifies the center of rotation of the ankle. The symbol MA denotes the moment generated by the Achilles tendon at point A.

12cm

### 2400N

Figure 8.9. The relative motion of the foot and the shank of a runner during an instant in the stance phase. The free-body diagram of the foot is also shown in the figure. Point A identifies the center of rotation of the ankle. The symbol MA denotes the moment generated by the Achilles tendon at point A.

12cm

### 2400N

—6 rad/s e3 and the foot had an angular velocity = —10.5 rad/s e3. The moment arm h of the ground force with respect to the center of rotation of the ankle joint was 12 cm. Determine the rate of increase of kinetic energy of the shank and the foot at that instant. Determine also the changes in the kinetic energy of the shank and the foot separately.

Solution: Let the foot be represented as body 1 and the shank as body 2. The moment applied by the shank on the foot is equal to the moment of the force exerted by the Achilles tendon with respect to point A. The value of this force is unknown, however, the moment it generates at point A can be computed by considering the free-body diagram of the foot. Because the weight of the foot is small relative to the forces acting on it, we assume that the resultant moment acting on it must be equal to zero. Thus, the resultant moment acting on the foot at the center of rotation of the ankle is given by the following relation:

Ma e3 + 2400 N (0.12 m) e3 = 0 => MA = —288 N-m in which Ma represents the moment generated by the Achilles tendon at A.

The time rate of change of kinetic energy of the foot and shank from the moment acting on joint A can be written as dT/dt = Ma e3-(w1 — œ2) = —288 N-m e3 • ( — 10.5 rad/s + 6 rad/s) e3

Thus, part of the chemical energy used by the calf muscles during running is transformed into kinetic energy at a rate of 1,296 N-m/s.

Next, let us consider the rate of change of kinetic energy of the foot and the shank separately. The force at the joint A balances the ground force Fg. This force is in the vertical direction. Thus, to calculate the power gen erated by this force, we need to determine the vertical component of the velocity of point A. We denote this component as v:

The rate of change of kinetic energy of the foot resulting from the joint forces and moments at A is given by the following relation:

dT(1)/dt = -2,400 N e2 • 1.26 m/s e2 + (-288 N-m) e3 • (-10.5) rad/s e3

The time rate of change in kinetic energy of the shank from the joint forces and moments at A can be written as dT(2)/dt = 2,400 N e2 • 1.26 m/s e2 + (288 N-m) (-6) rad/s e3 dT(2)/dt = 3,024 N-m/s - 1728 N-m/s = 1296 N-m/s

Thus, the kinetic energy of the foot remains constant at the instant considered whereas the kinetic energy of the shank increases at a rate of 1,296 N-m/s.

## Getting Started With Dumbbells

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

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