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普通物理 General Physics 9 - Center of Mass and Momentum
郭艷光Yen-Kuang Kuo 國立彰化師大物理系暨光電科技研究所 電子郵件: 網頁:
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Outline 9-1 What Is Physics? 9-2 The Center of Mass 9-3 Newton’s Second Law for a System of Particles 9-4 Linear Momentum 9-5 The Linear Momentum of a System of Particles 9-6 Collision and Impulse 9-7 Conservation of Linear Momentum 9-8 Momentum and Kinetic Energy in Collisions 9-9 Inelastic Collisions in One Dimension 9-10 Elastic Collisions in One Dimension 9-11 Collisions in Two Dimensions 9-12 Systems with Varying Mass: A Rocket 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-1 What Is Physics? We will derive the equation of motion for the center of mass, and discuss the principle of conservation of linear momentum. Finally, we will use the conservation of linear momentum to study collisions in one and two dimensions and derive the equation of motion for rockets. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass Center of Mass: 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. (2) all external forces were applied there. Define the center of mass (com) of particles in order to predict the possible motion of the system. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass Systems of Particles: Define the position of the center of mass (com) of this two- particle system to be 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass The coordinate system has been shifted leftward. The position of the center of mass is now defined as M: total mass of the system 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass For N particles the (vector) position of the CM is M: total mass of the system 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass In three dimensions, the center of mass must be identified by three coordinates. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass Center of Mass of Continuous Bodies: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-2 The Center of Mass Consider the thin rod of mass M and length L. The mass of the element of volume dV is For a disk or cylinder, : kg/m3 ; : linear mass density : areal mass density ( kg/m2 ) 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-1 Three particles of masses m1 = 1.2 kg, m2 = 2.5 kg, and m3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm. Where is the center of mass of this system? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-1 Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-1 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-2 Figure shows a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out (removed) in an assembly line. Using the xy coordinate system shown, locate the center of mass comP of the plate. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-2 Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-2 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-3 A thin rod of length 3L is bent at right angles at a distance L from one end. Locate CM with respect to the corner. Take L = 1.2 m. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-3 Solution: The CM of each arm is at its midpoint. If we take m1 = m, then m2 = 2m. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-3 Newton’s Second Law for a System of Particles
Newton’s second law for the motion of the center of mass of a system of particles: (1) is the net force of all external forces that act on the system. (2) M is total mass of the system. (3) is the acceleration of the center of mass. ( System of particles ) 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-3 Newton’s Second Law for a System of Particles
Proof of equation: O m1 m3 m2 F1 F2 F3 x y z 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-3 Newton’s Second Law for a System of Particles
2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-3 Newton’s Second Law for a System of Particles
In terms of components, we have A fireworks rocket explodes in flight. The center of mass of the fragments would continue to follow the original parabolic path. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-4 The three particles 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 F1 = 6.0 N, F2 = 12 N, and F3 = 14 N. What is the acceleration of the center of mass of the system, and in 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-4 what direction does it move? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-4 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-5 A man of mass m1 = 60 kg is at the rear of a stationary boat of mass m2 = 40 kg and length 3 m, which can move freely on the water; see Fig. (a). The front of the boat is 2 m from the dock. What happens when the man walks to the front? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-5 Solution: In terms of the initial position, Since there is no external force, xCM is fixed. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-4 Linear Momentum Linear Momentum: The SI unit is The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force. m: mass of a particle V: velocity of a particle 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
9-4 Linear Momentum In equation form this becomes Proof of equation: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-5 The Linear Momentum of a System of Particles
particles can by changed only by a net force 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-6 Collision and Impulse
The ball experiences a force that varies during the collision and changes the linear momentum of the ball. In time interval dt, the change in the ball’s momentum 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-6 Collision and Impulse
2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-6 Collision and Impulse
In many situations we do know the average force and the duration of the collision. Thus 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-6 Collision and Impulse
Series of Collisions : If the particle stop after the collision, then If the particle bounce backward, then 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-6 When a male bighorn sheep runs head-first into an other male, the rate at which its speed drops to zero is dramatic. Figure gives a typical graph of the acceleration a versus time t for such a collision, with the acceleration taken as negative to correspond to an initially positive velocity. The peak acceleration has magnitude 34 m/s2 and the duration of the collision is 0.27 s. Assume that the sheep’s mass is 90.0 kg. What are 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-6 the magnitudes of the impulse and average force due to the collision? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-7 (a) Figure is an overhead view of the path taken by a racecar driver as his car collides with the racetrack wall. Just before the collision, he is traveling at speed vi = 70 m/s along a straight line at 30° from the wall. Just after the collision, he is traveling at speed vf = 50 m/s along a straight line at 10° from the wall. His mass m is 80 kg. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-7 (a) What is the impulse on the driver due to the collision? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-7 (a) 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-7 (b) The collision lasts for 14 ms. What is the magnitude of the average force on the driver during the collision? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-7 Conservation of Linear Momentum
Law of Conservation of linear momentum: If the net external force on a system is zero, the total linear momentum is constant. (closed, isolated system) 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-8 One-dimensional explosion: A ballot box with mass m = 6.0 kg slides with speed v = 4.0 m/s across a frictionless floor in the positive direction of an x axis. The box explodes into two pieces. One piece, with mass m1 = 2.0 kg, moves in the positive direction of the x axis at v1 = 8.0 m/s. What is the velocity of the second piece, with mass m2? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-8 Solution: The initial momentum The final momenta of the two pieces are 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-9 One-dimensional explosion: Fig. (a) shows a space hauler and cargo module, of total mass M, traveling along an x axis in deep space. They have an initial velocity of magnitude 2100 km/h relative to the Sun. With a small explosion, the hauler ejects the cargo module, of mass 0.20M (Fig. (b)). The hauler then travels 500 km/h faster than the module along the x axis; that is, the relative 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-9 speed between the hauler and the module is 500 km/h. What then is the velocity of the hauler relative to the Sun? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-9 Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-10 (a) Two-dimensional explosion: A firecracker placed inside a coconut of mass M, initially at rest on a frictionless floor, blows the coconut into three pieces that slide across the floor. An overhead view is shown in Fig. (a). Piece C, with mass 0.30M, 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-10 (a) has final speed vfC = 5.0 m/s. What is the speed of piece B, with mass 0.20M? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-10 (b) What is the speed of piece A? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-8 Momentum and Kinetic Energy in Collisions
Type of collision: (1) A perfectly elastic collision is defined as one in which the total kinetic energy of the particles is also conserved: (2) In an inelastic collision, the total kinetic energy of the particles changes. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-9 Inelastic Collisions in One Dimension
One-dimensional inelastic collision: Two bodies just before and just after they have a one-dimensional collision. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-9 Inelastic Collisions in One Dimension
One-Dimensional Completely Inelastic Collision: Two bodies before and after they have a completely inelastic collision (meaning they stick together). 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-9 Inelastic Collisions in One Dimension
Velocity of the Center of Mass: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-11 The ballistic pendulum was used to measure the speeds of bullets before electronic timing devices were developed. The version shown in figure consists of a large block of wood of mass M = 5.4 kg, hanging from two long cords. A bullet of 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-11 mass m = 9.5 g is fired into the block, coming quickly to rest. The block + bullet then swing upward, their center of mass rising a vertical distance h = 6.3 cm before the pendulum comes momentarily to rest at the end of its arc. What is the speed of the bullet just prior to the collision? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-11 Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-12 (a) A 200-kg Cadillac limousine moving east at 10 m/s collides with a 1000-kg Honda Prelude moving west at 26 m/s. The collision is completely inelastic and takes place on an icy (frictionless) patch of road. Find their common velocity after the collision. m1: Calillac m2: Honda 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-12 (a) Solution: We assume the unknown common velocity V is in the + x direction. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-12 (b) What is the fractional loss in kinetic energy? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-10 Elastic Collisions in One Dimension
In an elastic collision, the kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-10 Elastic Collisions in One Dimension
2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-10 Elastic Collisions in One Dimension
Let us look at a few special situations: 1. Equal masses: If m1 = m2, 2. A massive target: If m2 >> m1, 3. A massive projectile: If m1 >> m2, and : Pool player’s result and and 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-13 Two metal spheres, suspended by vertical cords, initially just touch, Sphere 1, with mass m1 = 30 g, is pulled to the left to height h1 = 8.0 cm, and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2, whose mass m2 = 75 g. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-13 What is the velocity v1f of sphere 1 just after the collision? Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-11 Collisions in Two Dimensions
Total linear momentum must be conserved. If it is also elastic, and then For a glancing collision (it is not head-on): (1) Along the x axis as 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-11 Collisions in Two Dimensions
(2) Along the y axis as (3) 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-12 Systems with Varying Mass: A Rocket
Finding the Acceleration and Velocity: In Fig. (a) and (b), we show the rocket at times t and t + dt. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-12 Systems with Varying Mass: A Rocket
U is the velocity of the ejected gases with respect to the initial reference frame in which we measure the rocket’s speed v. 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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9-12 Systems with Varying Mass: A Rocket
( first rocket equation ) ( second rocket equation ) 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-14 A rocket whose initial mass Mi is 850 kg consumes fuel at the rate R = 2.3 kg/s. The speed vrel of the exhaust gases relative to the rocket engine is 2800 m/s. (a) What thrust does the rocket engine provide? (b) What is the initial acceleration of the rocket? 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
Example 9-14 Solution: 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
End of chapter 9! 2018/12/3 普通物理講義-9 / 國立彰化師範大學物理系/ 郭艷光教授
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