Chapter 9 System of particles
8- System of particles 1. Two particle system 2. Many particle system 3. Centre of mass of solid objects 4. Linear momentum of a particles 5. Linear momentum of system of particles. 6. Conservation of linear momentum
Learning Objectives After reading this module, you should be able to : Given the positions of several particles along an axis or a plane, determine the location of their center of mass. For a two-dimensional or three-dimensional extended object with a uniform distribution of mass, determine the center of mass by (a) mentally dividing the object into simple geometric figures, each of which can be replaced by a particle at its center and (b) finding the center of mass of those particles
1- Two Particle System. The center of mass: We define the center of mass (com or cm) of a system of particles (such as a person) in order to predict the possible motion of the system. The center of mass of a system of particles is the point that moves as though (1)All of the system s mass were concentrated there and (2) All external forces were applied there.
We consider a two particles of masses m 1 and m 2 separated by distance d. We have arbitrarily chosen the origin of an x axis. We define the position of the center of mass (com or cm) of this two-particle system to be: Or in which M is the total mass of the system. ( M = m 1 + m 2.)
The velocity of the center of mass is found by taking the time derivative of the previous equation:
The acceleration of the center of mass is found by differentiating again 7
2- Many Particle Systems. In this section we generalize the results of the previous section to systems in three dimensions that contain more than two particles. We consider a system consisting of N particles are strung out along the x- axis. Then the total mass is M = m 1 + m 2 +... +m n, location of the center of mass of the system can be defined by 8
If the particles are distributed in three dimensions, the center of mass must be identified by three coordinates We can also define the center of mass with the language of vectors. First recall that the position of a particle at coordinates x i, y i, and z is given by a position vector: Similarly, the position of the center of mass of a system of particles is given by a position vector 9
Sample Problem 1 The Figure shows a system of three initially resting particles of masses m 1 =4.1 kg, m 2 = 8.2 kg, and m 3 = 4.1 kg. The articles are acted on by different external forces, which have magnitudes F 1 = 6 N, F 2 = 12 N, and F3 = 14 N. The directions of the forces are shown in the figure. Where is the center of mass of this system, and what is the acceleration of the center of mass? 10
Solution The X component of the net external force acting on the center of mass is 11
and the y component is 12
3- Centre of mass of solid objects. It is far too tedious to find the center of mass of a solid object by using the previous equation and summing over every atom in the system. Instead we divide the object into tiny elements of mass δm. As these elements become infinitesimally small, the sums of the two previous equations transform into integrals: In vector form, these equations can be written 13
In many cases it is possible to use arguments based on geometry or symmetry to simplify the calculation of the center of mass of solid objects: If an object has spherical symmetry, the center of mass must lie at the geometrical center of the sphere. If a solid has cylindrical symmetry then the center of mass must lie on the axis If its mass is distributed symmetrically about a plane, then the center of mass must be in the plane. Often we encounter solid, irregular objects that can be divided into several parts. We can find the center of mass of each part, and then by treating each part as a particle located at its own center of mass we can find the center of mass of the combination. 14
Sample Problem 3 The figure shows a circular metal plate of radius 2 R from which a disk of radius R has been removed. Let us call it object X. Its center of mass is shown as a dot on the X axis. Locate this point. Solution Figure shows object X, its hole filled with a disk of radius R, which we call object D. Let us label as object C the large uniform composite disk so formed. From symmetry, the center of mass of object C is at the origin of the coordinate system, as shown 15
The position of the center of mass of object C is in which X D and X x are the positions of the centers of mass of objects ٠ and X, respectively. Noting that x c = 0 and solving for X x, we obtain 16
the ratio m D l m x must be the same as the ratio of the areas of objects D and X assuming the plate is of uniform density and thickness). That is 17
4- Linear momentum of a particles. The linear momentum of a particle is a vector defined as quantity that is in which m is the mass of the particle and is its velocity. Newton expressed his second law of motion in terms of momentum: The Kinetic energy can be written in the form of linear momentum and 18
5- Linear momentum of system of particles. Consider a system of n particles, each with its own mass, velocity, and linear momentum. If we compare this equation with the equation of the center of mass, we see that If we take the time derivative of the previous equation, we find 19
6- Conservation of linear momentum.. Suppose that the sum of the external forces acting on a system is zero and that no particles leave or enter the system This result is called the law of conservation of linear momentum. 20
Sample Problem 6 A stream of bullets whose mass m is each 3.8 g is fired horizontally with a speed v of 1100 m/s into a large wooden block of mass M (=12 kg) that is initially at rest on a horizontal table; see Fig. If the block is free to slide without friction across the table, what speed will it acquire after it has absorbed 8 bullets? Solution P = constant is valid only for closed systems, in which no particles leave or enter. Thus our system must include both the block and the 8 bullets, taken together. In the figure we have identified this system by drawing a closed curve around it. Because no (horizontal) external forces act, we can apply the law of conservation of momentum. The initial(horizontal) momentum, measured while the bullets are still in flight and the block is at rest, is
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