Problems

Problem 8.1. Consider two masses m1 (15 g) and m2 (24 g) that are attached together by a spring with stiffness coefficient k = 120 dyn/mm. The spring is force free at time t = 0 and at that instant these particles are located at the positions r1 = 3 mm e1; r2 = 12 mm e1 + 2 mm e2

Suppose that the particles move with constant velocities:

Determine the work done by the spring on the particle 1 and 2 at t = 2 s. Answer: W = -41 dyn-mm.

Problem 8.2. Show that the kinetic energy of a rigid body rotating around an axis that passes through point O of the body is given by the following equation:

in which Io is the mass moment of inertia with respect to point O and m is the angular speed. Problem 8.3. A Problem Concerning Geometry. Consider a two-rod system in which the rigid links are connected by a contractile cord (Fig. P.8.3). Determine the change in the angle between the two links when the contractile cord shortens by 10%. Answer: (a) 135° ^ 113°

Problem 8.4. Four equal rods, each having mass of 10 kg and length 0.6 m, are hinged together to form a rhombus (Fig. P.8.4). The angle the rods make with the horizontal plane, 6, is 45° at the instant the system is let go from rest. Use the conservation of energy principle for the system of four links to determine the kinetic energy of the system when the angle 6 reduces to 30°. Note that the kinetic energy of the system is the sum of kinetic energy of the four links. What is the work done by the reaction forces at B and C on the part of the system composed of the rod a and b? Answer: T = 48.7 N-m, W = 4.6 N-m.

Figure P.8.3. Two rods connected by a hinge and a contractile cord as a simple model of bending with the use of the abdominal muscle.

Figure P.8.4. A structure consisting of four equal rods that are connected to each other by hinges.

Problem 8.5. The four-rod system shown in Fig. P.8.4 is revised so that elastic springs with spring constant k = 27 N/cm connect the rods at midpoint as shown in Fig. P.8.5. The angle the rods make with the horizontal plane, 6, is 45° at time t = 0. The springs are force free when the system is let go from rest at t = 0. Determine the angle 6 for which the system reaches static equilibrium. Answer: 6 = 12°.

Problem 8.6. Consider the four-rod system shown in Fig. P.8.5. Determine the work done by the springs as 6 decreases from 45° to 40°.

Problem 8.7. When a car or a railway carriage enters upon a circular path of radius r after traveling in a straight road with velocity vo, its velocity decreases (Fig. P.8.7). Determine the velocity at the curved path. Answer: v2 = vo2 r2/(r2 + k2) where k is the radius of gyration of the body.

Problem 8.8. A rod of length L and mass m is held in a vertical position on a rough horizontal surface (Fig. P.8.8). The rod is then slightly displaced from the vertical unstable equilibrium position (Fig. P.8.8a) and let go. The rod will rotate clockwise around point A without slip (Fig.

Figure P.8.5. The four-rod structure with compression springs attached to the rods.

Figure P.8.7. The path of a train car changes from a straight line to a circular arc in the horizontal plane.

P.8.8b), but eventually the slipping must take place before the horizontal position is reached (Fig. P.8.8c). Determine the angular velocity of the rod as a function of its angle of inclination with the vertical axis before the slip takes place. At which value of 0 does the slip takes place? The coefficient of static friction ^ = 0.4.

Hint: Conservation of mechanical energy leads to L(d0/dt)2 = 3g (1 -cos 0). Conservation of angular momentum with respect to point A results in L(d20/dt2) = -(3/2) g sin 0. Answer: 0 = 50°

Problem 8.9. Model the dynamics of a fall of a box containing a computer as a mass of m striking a dashpot with velocity v (Fig. P.8.9). When the box is dropped from a height h, the mass-dashpot system is subjected to an impact. Assume that the resistance force exerted by the dashpot on the computer is given by the equation

Figure P.8.8. Three instances from the time history of the fall of a thin rod released from a vertical upright position.

Figure P.8.9. A computer falling through a viscous damping material after the container hit the ground with velocity vo.

¿xxxXj

m

1111

in which m is the mass of the computer, v is its speed, and t is the dissipation time constant. Initially, before the impact, the vertical distance between the computer and the bottom of the box is d. What is the differential equation governing the motion of the computer after the box hits the ground and before the computer reaches the bottom of the box? What is the resistance force on the computer immediately after the box hits the ground? Use the following parameter values in your calculations: m = 3 kg, h = 2 m, d = 0.3 m, t = 1.2 s.

Answer: F = 15.66 N, (d2x/dt2) + (1/t) (dx/dt) = g where x represents the vertical distance.

Problem 8.10. Back hyperextension is a good exercise for strengthening the back muscle (Fig. P.8.10). Supporting the heels, upper thighs, and pelvis on the padded supports, with arms folded on the chest, lower the upper torso down to the vertical position. Using the back muscle slowly pull up to the horizontal position. Once the upper body is aligned horizontally, its kinetic energy must be equal to zero. Determine the work done by the back muscle in raising the upper body from vertical down-

Figure P.8.10. A schematic diagram of a man performing back extensions.

ward to horizontal alignment. Assume that the mass of the upper body is m and its center of mass is at a distance h from the hip joint.

Problem 8.11. A vaulter running with speed vo = 10 m/s toward the crossbar places the end of the bar onto the ground at time t = 0 and continues running in the same direction for an additional 1.0 m (Fig. P.8.11). The length d of the part of the pole from his grip to the distal end is 7 m. The vaulter weighs 75 kg and his height is 1.91 m. The response of the pole to a thrust is given by the following relation:

in which F is the compressive force acting on the pole, k = 6,000 N/m, and b is the shortest distance between the handgrip and the distal end of the pole as shown in the figure. Determine the speed v of the vaulter at the instant when he has covered a distance of 0.75 m toward the crossbar while keeping the distal end of the pole fixed on the ground. Answer: v = 7.7 m/s.

Problem 8.12. McMahon and Cheng (1990) modeled hopping in place with a mass-spring system constrained to move vertically as it strikes the ground. The mass m represents the body mass, and the spring constant k of the spring represents stiffness properties of the leg (Fig. P.8.12). Let the displacement y of mass m be measured such that y is increasing when the mass is moving upward. The spring is slack, neither stretched nor compressed, when y = 0. The vertical velocity dy/dt

h f ll

0.75m

Figure P.8.11. A vaulter pressing the pole against the ground while running toward the crossbar.

Figure P.8.11. A vaulter pressing the pole against the ground while running toward the crossbar.

Figure P.8.12. The hopping model consisting of a mass-spring system interacting with a horizontal surface.

y g v of the mass at the moment the leg-spring strikes the ground is — v, where v is a positive quantity. Determine the peak force the leg must bear, the time the leg spends in contact with the ground, and the stride frequency as a function of one dimensionless group, (vw/g), where a = (k/m)1/2.

Problem 8.13. McMahon and Cheng (1990) used the mass-spring model to study the mechanics of running (Fig. P.8.13). Zero-force length of the leg was represented by lo. The leg was assumed to have a stiffness of kleg. The parameter y still measures the vertical height of the body mass m, but now y = 0 corresponds to the ground plane. At the beginning of the rebound (the foot strikes the ground), the forward velocity dx/dt of the body mass is u and the vertical velocity dy/dt is — v. During the

Figure P.8.13. A simple mass-spring system used to analyze the dependence of running speed on leg stiffness.

rebound (stance), the angle of the leg with respect to the vertical begins at - 0o and ends at 0o. The x velocity begins and ends with the value u, and the y velocity is reversed by the step, starting with the value - v and ending with v. Determine expressions for the velocity of mass m and the ground force acting on the mass-spring system by assuming that 0o is small (sin 0o = 0o). For a complete solution of the problem, see McMahon and Cheng (1990).

Problem 8.14. In a review article published in August 1998 in the Journal of Bone and Joint Surgery, T. F. Novacheck, M.D., wrote, "Joint power is the product of the net joint moment and the joint angular velocity." How would you transform this sentence into the notation adopted in this book? Is there such a thing as a joint angular velocity?

Problem 8.15. The data presented by Roberts et al. (1997) suggest that the changes in length of the muscle belly itself are relatively slight in running turkeys. Instead, the muscles appear to function as tensioners of the tendons, meaning they act as tension generators. Discuss whether finite angular changes between body segments can occur without significant shortening of the skeletal muscle. Cite examples.

Problem 8.16. In running, both the kinetic energy and the potential energy from gravity are higher at the two ends of the stance phase than at the midstance. Because the ground force performs no work on the feet of the runner, it is clear that the sum of kinetic energy and gravitational potential energy is not constant. Speculate how the elastic (strain) energy stored in the heel pad and the tendons of the leg vary with time during the stance phase.

Problem 8.17. A subject raises the arm from the resting position (0 = 0) to the full arm elevation (0 = such that the angle the arm makes with the horizontal axis changes according to the relation

in which t is time in seconds. Determine the work done on the arm by the muscle-tendon systems and ligaments of the shoulder as a function of arm length L and mass m for a = 1, 2, and 3 and for 0 from 0 to (^/2) and from 0 to

Hint: The sum of kinetic plus potential energy of the arm must be equal to the work W done by forces and moments acting on the arm at the shoulder joint. For 0 = (^/2), the conservation of energy leads to the following equation:

According to this equation, the faster the motion, the greater is the work that must be done by the muscle-tendon systems, in this case, principally the deltoids. Deltoids will have to shorten faster with increasing speed of the motion. Even when a = 1, most of the work done goes to increasing the kinetic energy at 0 = (n/2). In contrast, the resultant work done by joint structures in raising the arm from 0 = 0 to 0 = n is given by the expression:

The work done at 0 = n is smaller than the corresponding work done at 0 = n/2 for the values of a2 > 3g/(Ln4). Thus, in raising the arm from 0 = n/2 to 0 = n, shoulder joint structures must have supplied power in the opposite direction to that supplied by the deltoids.

Problem 8.18. A walking animal pole-vaults over its rigid limb similar to using a rigid pole to clear a hurdle. There is an alternate exchange between kinetic energy of the center of mass of the animal and the gravitational potential energy as the animal rotates over its rigid limb during walking. Experiments on dogs indicate that as much as 55% of the initial kinetic energy might be transformed into gravitational potential energy during rotation over a rigid limb. What should be the angle the limb makes with the vertical axis when it first hits the ground? Represent the animal as a point mass of m attached to one end of a light rigid pole of length d. Assume also the velocity of the center of mass to be in the horizontal direction.

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Getting Started With Dumbbells

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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