## Problems

GSI = 0f a2'5 dt where a is the instantaneous acceleration of the head and t is the duration of the pulse. If the integrated value exceeds 1,000, severe injury is predicted to result. Consider a fall where the head of a person hits the ground with a vertical velocity of 8 m/s. The impact lasts 0.05 s and the impact force is uniform during the course of impact. Determine the value of GSI for this head impact. Answer: GSI = 16191.

Problem 7.4. If a person were not wearing a seat belt in a car when the car hit a wall or a large tree, the overall effect is that of a person hitting a massive wall with the velocity of the car before collision. In that sense a collision may be considered equivalent to falling from a height h onto a concrete sidewalk. Suppose the speed of the car before the collision is 90 km/h. Determine the height of a free fall that would give the same velocity before collision. Answer: H = 31.8 m.

Problem 7.5. In a study of hip fracture etiology, young healthy athletes weighing 70 kg performed voluntary sideways falls on a thick foam mattress. The mean value for the vertical impact velocity of the center of mass of a falling athlete was 2.75 m/s. Assuming that there was no rebound immediately after the impact, compute the vertical impulse due to the fall. Answer: £ = 192.5 N-s.

Problem 7.6. A uniform rod of mass m and length L can turn freely around point A (Fig. P.7.6). It is held in its highest position and is then allowed to fall. On reaching its lowest position, it encounters a fixed object. The object remains stationary while the rod rebounds. The coefficient of restitution e between the rod and the object is equal to 0.4. Find the impulse exerted on the rod. What is the rebound velocity at the site of the collision? Figure P.7.6. The rotation of a uniform rod around - !

point A and the resulting impact with a stationary M '"'B object. _

Hint: To compute the velocity of the rod before collision, derive a differential equation for angular speed using conservation of angular momentum.

Problem 7.7. Consider sideways fall of an individual whose weight is 60 kg and height is 1.60 m. Represent the falling person as composed of two equal uniform rods AB (lower body) and BD (upper body), hinged together at B (hip joint) as shown in Fig. P.7.7. Both rods have mass m = 30 kg and length L = 0.8 m. Immediately before the impact, the point A (feet) was moving in the -ei direction with speed equal to 1.2 m/s. The angular velocities of the lower body and the trunk immediately before the impact were measured as -7.5 rad/s e3 and -4.8 rad/s e3. The angles the lower body and the trunk made with the e1 direction were 0 and ^/6, respectively. The lower body remained at rest on the ground immediately after the impact, whereas the upper body began rotating in the counterclockwise direction with angular velocity equal to 3.2 rad/s e3. Determine the impulse of the impact acting on the individual. Answer: £ = 24 e1 + 353 e2 (N-s).

Problem 7.8. A man hits a ball of radius R and mass m with a cylindrical rod of length L and mass M (Fig. P.7.8). Before the impact the ball had an initial velocity vo e2 and angular velocity Mo e3 as shown in the figure. The rod, on the other hand, was rotating around the stationary point A with angular velocity -Oo e3. The distance between the point of impact and point A is denoted as d. Assume the force of impulse to be perpendicular to the contact area. Determine the angular velocity O of the rod and velocity of the center of mass v of the ball immediately after the impact. use the following parameter values in your computations: M = 10 kg, L = 1 m, d = 0.8 m, Oo = 5 rad/s, m = 5 kg, R = 0.1 m, vo = 10 m/s, and the coefficient of restitution e = 0.8.

Figure P.7.7. A two-bar model of a sideways fall. The uniform rods AB and BD represent the upper and lower body, respectively. Figure P.7.7. A two-bar model of a sideways fall. The uniform rods AB and BD represent the upper and lower body, respectively.

Figure P.7.8. The collision of a ball with a cylindrical rod that is free to rotate around point A.

 •«=— M B —=► Hint: Because the impulse passes through the center of the spherical ball, it cannot change the ball's angular velocity wo. Also, the moment of momentum of the system (the rod and the ball) about point A (Ho) should not change as a result of the impact: The equation for the coefficient of restitution yields e = (d O - v)/(vo + d Oo) Problem 7.9. A large mass m falls on a spring with downward velocity vo. The spring constant is denoted as k. Determine the force exerted by the spring on the mass m as a function of time. What is the impulse of this force? Assume the downward direction to be positive. Answer: F = -(km)1/2 vo sin [(k/m)1/2 t] - mg (cos [(k/m)1/2 t] - 1) Problem 7.10. The horizontal velocity of the center of mass of a 70-kg runner immediately before he placed a heel on a flat surface was found to be 2.5 m/s. The horizontal ground force acting on the heel of the runner followed the relation: Fi ei = -3,000 sin (^t/0.2) ei in which F1 has the unit of N and time t has the unit of seconds (s). Determine the horizontal velocity component of the center of mass of the runner at 0.05, 0.1, and 0.15 s after the heel strike. Note that experimental values of the horizontal ground force during running is predominantly biphasic. During the initial phase (termed braking), the direction of the horizontal ground force opposes forward motion. During the latter phase (termed propulsion), its direction leads to forward acceleration. The relative magnitudes of the braking and propulsive impulses for a given trial can serve as an objective measure for verifying whether a runner satisfies the so-called constant velocity criterion. In constant velocity running, the forward and backward impulses exerted by the ground must be equal in magnitude. The ground force equation presented above satisfies this condition. Answer: v = 1.7 m/s at t = 0.05 s, v = -0.2 m/s at t = 0, and v = -2.2 m/s at t = 0.15 s. Problem 7.11. The risk of head injury from striking an automobile dashboard is often correlated with the maximum linear acceleration of the head during the collision. To better understand the mechanics of collision, a team of researchers dropped rigid balls of different masses from a height of h onto an elastic surface with spring constant k. They found that the maximal displacement of the surface during the collision was given by the following relationship: Determine a relationship between the mass of the fallen object and its acceleration at maximal displacement. Hint: Write down the equation of motion of the object in the vertical direction and substitute k A for the spring force. Your result should predict that the smaller the mass of the object, the greater the peak acceleration during impact. Based on this observation, some researchers argued that children may be at greater risks than adults when striking a padding surface. Answer: a = (2ghk/m)1/2 - g. Problem 7.12. An individual is running on a treadmill whose speed is set to vo (Fig. P.7.12). The velocity of the center of mass at an instant im- Figure P.7.12. A runner on a treadmill. mediately before one of the legs hit the tank-treading rubber platform is measured to be zero. The angular velocity of the leg at that very instant is — ao. Once the foot hits the platform, it begins to move with it. Assuming that the leg can be represented as a weightless uniform rod of length L with the lumped mass m of the body attached to it at the hip, determine the impulse exerted by the treadmill on the runner. Determine the velocity of the hip immediately after the impact. Does the impact force increase with increasing vo? 