## Center of Mass and Its Motion

The center of mass of a body B, living or nonliving, is defined by the following equation:

2m1 is the total mass in B, rc is the position vector for the center of mass of B, mi is the mass of the ith element in B, and r' is its position vector. These entities are shown in Fig. 3.1. Note that the center of mass is also commonly known as the center of gravity.

Let us now perceive center of mass as if it were a particle in space. In reality, the center of mass may not correspond to any point of the object B. The position of center of mass may be occupied by different particles of B at different times during motion. The time rate of change of position vector rc (derivative of rc with respect to time t) is equal to the velocity of the center of mass, which we denote by vc. The acceleration of the center of mass ac is the time derivative of vc:

Using Eqn. 3.8b, the linear momentum L of a system of particles can be written as

Substituting Eqn. 3.8c into Eqn. 3.6, we obtain an equation governing the motion of center of mass of an object:

According to this equation, the net force acting on a system of particles is equal to the mass of the system times the acceleration of the center of mass. For a sphere of uniform mass density, the center of mass is positioned at the center of the sphere. As in the case of an L-shaped body, the center of mass may lie outside the body. The tables at the end of the text provide information about the position of the center of mass for solid bodies of a variety of geometric shapes as well as human body configurations (see Appendix 2).

Example 3.1. The Center of Mass of a Human Body as Represented by Two Rods. Consider a body consisting of two slender rods ab and bc of length L and mass m that are connected with a pin at b. The bar ab is tilted 45° from the horizontal axis. Determine the center of mass of B.

Solution: We draw a reference frame whose origin O is at the pin b connecting the two rods (Fig. 3.2). In a uniform slender rod, the center of mass occupies the midpoint along the axis of the rod. Thus, the centers of mass of the two rods ab and bc are located at the following positions:

x1 = —(L/2) cos 45°; x2 = (L/2) sin 45°; x3 = 0 for the rod ab x1 = (L/2), x2 = 0, x3 = 0 for the rod bc a — L -» Figure 3.2. The location of the center of mass of a system composed of two slender rods that are linked together at point b. Next, we insert these values into Eqn. 3.8a and determine the center of mass of the serially linked rods: (2m) x1c = -m (L/2) cos 45° + m (L/2) x1c = 0.07 L Example 3.2. The Raising of Arms Changes the Position of the Center of Mass. An athlete weighs 163 lb and has a height of 69 in. His center of mass lies 38 in. above the ground during the standard standing position shown in Fig. 3.3. Determine the change in the elevation of his center of mass when he raises his arms from the sides of his body i80° over his head. His arms weigh 7 lb each and the center mass of his arms moves upward 25 in. when he raises them over his head. Solution: We will consider the athlete to be composed of three parts: two arms each having a mass m and the rest of his body (M). The location of the center of mass in the vertical x2 direction before and after raising the arms is given by the following equations: where d denotes the distance between the origin of the coordinate system and the center of mass of the body without the arms, and h is the distance between the origin and the center of mass of the arms when they lie on the sides. The symbol y denotes the unknown additional elevation of the center of mass. When Eqn. 3.10a is subtracted from Eqn. 3.10b, one finds that y = 2.15 in. Note that one can easily estimate the mass of a body part by immersing it in a container full of water and measuring the amount of water displaced. The mass of the body part, say a foot or an arm, is equal to the volume of water displaced times the average specific density of human body, which is about 0.9 g/cm3. Example 3.3. Chin-ups. A 75-kg man is doing chin-ups to work out his latissimus muscle (Fig. 3.4). If during the rising phase his acceleration reaches a peak of 4 m/s2, determine the contact force exerted by the horizontal bar on the athlete at that instant. Solution: The forces acting on the athlete are the gravitational force (—mge2) pointing downward and the contact forces F exerted by the bar on the athlete, pointing upward. The resultant of these forces must be equal to the mass times the acceleration of center of mass. This can be written in the vector notation in the following form: Note that the contact force exerted by the bar is equivalent to a weight of (2 X 518/9.81) = 105.6 kg. This example shows that when the man accelerates upward, the force the bar exerts on the man is an upward force that is greater than the man's body weight. Because action equals to reaction, this is also the force the athlete exerts on the bar as he pulls on the bar. Example 3.4. Vertical Jumping. Animals and humans jump to reach higher places, to leap over obstacles, or for competition. Volleyball and basketball players must react instantaneously to the ball and jump within Figure 3.3. A man raises his arms from the side of his body to over his head. The center of mass (C) of the man also moves up, albeit a much smaller amount. Figure 3.4. A man doing chin-ups to exercise his latissimus muscle. The diagram shows the external forces acting on the man. |mg a fraction of a second. Videotapes of jumping events show that the entire duration of the propulsive stage of the human vertical jump, from backward rotation of the trunk to toe-off, lasts about 0.3 s. In this brief period of time, the angles between various segments of the lower body (feet, shank, thigh, and hips) change to lift the upper body (70% of the body weight) vertically. The mechanics of vertical jumping can be captured with reasonable accuracy by using a four-segment model of a human body composed of foot, shank, thigh, and the upper body (Fig. 3.5a). Here we consider a much simpler model in which a mass M (representing the upper body) is attached to two slender bars of length L (weightless legs) as shown in Fig. 3.5b. The bars ab and bc must be connected by a muscle-tendon system that enables the bars to change the angle between them. If there was no such mechanism, the two bars would collapse onto the floor under the weight of mass M. Our aim is to determine the conditions of jumping for this two-segment model. Solution: The geometry of the assumed structure dictates that the spatial position of the point mass M at any time t is given by the equation: Differentiating r with respect to time twice, we determine the velocity and acceleration of mass M as a function of time: Substituting Eqn. 3.11c into the law of motion for the center of mass r = 2L sin d e2 (Eqn. 3.9), we arrive at the following vectorial equation for the force P exerted by the ground on the two-segment jumper: P - Mg e2 = M [-2L sin 0 (dO/dt)2 + 2L cos 0 (d20/dt2)] e2 (3.12a) P = M [g-2L sin 0 (dO/dt)2 + 2L cos 0 (d20/dt2)] e2 (3.12b) The ground can only exert an upward force on the jumper; it has no capability to pull the jumper toward earth. At the instant the jumper leaves the ground, the ground reaction force P must be equal to zero. This gives us g = 2L sin 0 (d0/dt)2 - 2L cos 0 (d20/dt2) (3.13) Let us explore the physical implications of this equation. Let the lower limbs of the jumper be represented by two linked rods of length L = 0.5 m. Furthermore, let us assume that at the time of the takeoff, 0 = 60° and (d20/dt2) = 0. The rate of rotation (d0/dt) can then be found by using Eqn. 3.13 to be equal to 3.25 rad/s. This means that, for the ground force to diminish to zero, the angle between the shank and thigh must increase at a rate greater than 180°/s. If the legs straighten at lesser speed, jumping cannot occur because the ground will continue to exert some finite level of upward force. Figure 3.5 a,b. Schematic representation of a four-segment model of an athlete performing a vertical jump (a). The athlete performing the jump is represented by even a simpler model in (b): a mass representing the weight of the upper body is connected to two slender rods. The external forces acting on the athlete are also shown in (b). Figure 3.5 a,b. Schematic representation of a four-segment model of an athlete performing a vertical jump (a). The athlete performing the jump is represented by even a simpler model in (b): a mass representing the weight of the upper body is connected to two slender rods. The external forces acting on the athlete are also shown in (b). This simple analysis implies that action of hip, thigh, and calf muscles on the lower limbs must be fast enough for jumping to occur. Additionally, the analysis presented here shows that the upward velocity increases with increasing length of the slender bars ab and bc. According to the model, athletes with long legs reach higher velocities during vertical jumping than athletes with shorter legs, when both the short and the tall athletes have the same capacity for quick rotation of the thigh over the lower leg. Once the two-segment model lifts itself off the ground, how far will it travel in air? The only force acting on the airborne jumper is gravity, and therefore its center of mass will follow the equations of free fall presented in the previous chapter. To determine how far up the center of mass will reach, we need to determine its velocity at the time of takeoff. The vertical velocity at the time of the takeoff can be obtained by substituting L = 0.5 m, 0 = 60°, and (dd/dt) = 3.25 rad/s in Eqn. 3.11b. This initial velocity is found to be 1.625 m/s. After the takeoff, the path of the center of mass is determined by the following equations: The center of mass will reach its highest point when v becomes equal to zero. Eqn. 3.14b shows that t = 1.625/g at this point. Substituting this value of t into Eqn. 3.14c, we find that the maximum distance h that the center of mass of the athlete rises during the jump must be equal to (1.625)2/2g or 13 cm. In general, vertical distance traveled in air during jumping is equal to h = Vo2/2g where Vo is the vertical velocity at the time of takeoff.

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