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Center-of-Mass ProblemsSample Problems 9-1 and 9-2 provide three strategies for simplifying center-of-mass problems. (1) Make full use of the symmetry of the object, be it about a point, a line, or a plane. (2) If the object can be divided into several parts, treat each of these parts as a particle, located at its own center of mass. (3) Choose your axes wisely: If your system is a group of particles, choose one of the particles as your origin. If your system is a body with a line of symmetry, let that be your x or ó axis. The choice of origin is completely arbitrary; the location of the center of mass is the same regardless of the origin from which it is measured. 9-3 Newton's Second Law for a System of Particles If you roll a cue ball at a second billiard ball that is at rest, you expect that the two-ball system will continue to have some forward motion after impact. You would be surprised, for example, if both balls came back toward you or if both moved to the right or to the left. What continues to move forward, its steady motion completely unaffected by the collision, is the center of mass of the two-ball system. If you focus on this point— which is always halfway between these bodies because they have identical masses— you can easily convince yourself by trial at a billiard table that this is so. No matter whether the collision is glancing, head on, or somewhere in between, the center of mass continues to move forward, as if the collision had never occurred. Let us look into this center-of-mass motion in more detail. To do so, we replace the pair of billiard balls with an assemblage of n particles of (possibly) different masses. We are interested not in the individual motions of these particles but only in the motion of their center of mass. Although the center of mass is just a point, it moves like a particle whose mass is equal to the total mass of the system; we can assign a position, a velocity, and an acceleration to it. We state (and shall prove next) that the (vector) equation that governs the motion of the center of mass of such a system of particles is
This equation is Newton's second law for the motion of the center of mass of a system of particles. Note that it has the same form (Fnet = ma) that holds for the motion of a single particle. However, the three quantities that appear in Eq. 9-14
must be evaluated with some care: 1. 2. 3. Equation 9-14 is equivalent to three equations involving the components of
Now we can go back and examine the behavior of the billiard balls. Once the cue ball has begun to roll, no net external force acts on the (two-ball) system. Thus, because Equation 9-14 applies not only to a system of particles but also to a solid body, such as the bat of Fig. 9-lb. In that case, Figure 9-5 shows another interesting case. Suppose that at a fireworks display, a rocket is launched on a parabolic path. At a certain point, it explodes into fragments. If the explosion had not occurred, the rocket would have continued along the trajectory shown in the figure. The forces of the explosion are internal to the system (first the rocket and then its fragments); that is, they are forces on parts of the system from other parts. If we ignore air drag, the net external force When a ballet dancer leaps across the stage in a grand jete, she raises her arms and stretches her legs out horizontally as soon as her feet leave the stage (Fig. 9-6). These actions shift her center of mass upward through her body. Although the shifting center of mass faithfully follows a parabolic path across the stage, its movement relative to the body decreases the height that is attained by her head and torso, relative to that of a normal jump. The result is that the head and torso follow a nearly horizontal path, giving an illusion that the dancer is floating.
Sample Problem 9-3 The three particles in Fig. 9-1 a are initially at rest. Each experiences an external force due to bodies outside the three-particle system. The directions are indicated, and the magnitudes are SOLUTION: The position of the center of mass, calculated by the method of Sample Problem 9-1, is marked by a dot in the figure. One Key Idea here is that we can treat the center of mass as if it were a real particle, with a mass equal to the system's total mass M = 16 kg. We can also treat the three external forces as if they act at the center of mass (Fig. 9-lb). A second Key Idea is that we can now apply Newton's second law ( or so Equation 9-20 tells us that the acceleration We can evaluate the right side of Eq. 9-21 directly on a vector-capable calculator, or we can rewrite Eq. 9-21 in component form, find the components of
Along the ó axis, we have
From these components, we find that
and the angle (from the positive direction of the x
Date: 2015-01-12; view: 1606
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