Chapter 9. Center of Mass and Linear Momentum

Transcription

1 Chapter 9 Center of Mass and Linear Momentum

2 9.2 The Center of Mass Recall we had idealized a body by treating it as a point particle (neglecting its extent). We now start working towards a more complete treatment. We start by defining the 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 and (2) all external forces were applied there. The center of mass (black dot) of a baseball bat flipped into the air follows a parabolic path, but all other points of the bat follow more complicated curved paths.

3 9.2 The Center of Mass: A System of Particles Consider a situation in which n particles are strung out along the x axis. Let the mass of the particles are m 1, m 2,, m n, and let them be located at x 1, x 2,, x n respectively. Then if the total mass is M = m 1 + m m n, then the location of the center of mass, x com, is (definition):

4 The generalization to 3-D is straightforward, the center of mass is given by: Equivalently, in vector notation the center of mass can be expressed as:

5 Sample problem, COM of 3 particles We are given the following data:

6 Sample problem, COM of 3 particles The total mass M of the system is 7.1 kg. The coordinates of the center of mass are therefore: We are given the following data: Note that the z com = 0.

7 9.2 The Center of Mass: Solid Body In the case of a solid body, the particles become differential mass elements dm, the sums become integrals, and the coordinates of the center of mass are defined as where M is the mass of the object. If the object has density, ρ, defined as: Then x com = 1 M xρ dv, y com = 1 M yρ dv, z com = 1 M zρ dv Where V is the volume of the object and the mass is M = ρdv

8 Recall: Consider two extended bodies of mass M1 and M2 with their COMs at x1 and x2. The COM of this system of extended bodies is the same as the COM of two particles of mass M1 and M2 with their COMs at x1 and x2. x com = and substitute in here to obtain 1 M 1 + M 2 R 1 +R 2 xdm= R 1 xdm= M 1 x com,1 = M 1 x 1, x com = 1 M 1 + M 2 1 M 1 + M 2 [M 1 x 1 + M 2 x 2 ] xdm+ R 1 xdm R 2 R 2 xdm= M 2 x com,2 = M 2 x 2

9 Recall: In general the density can change from one place to another within a body. So it is a function, ρ(x, y, z), and the COM can be awfully difficult to calculate. But if the density, ρ, is uniform within the body, finding the COM may be simpler. Uniform density means Then Where V is the volume of the object.

10 sample calculations: COM for rectangular plates While we can guess this by symmetry, it is instructional to calculate y h Thickness of plate t. Infinitesimal volume element: dv = t da = t dx dy x com = 1 xρ dv = 1 x σ M M t (tda) = 1 M w x xσ da Uniform density: = M ta = 1 t σ where σ = M/A is the surface mass density x com = 1 A ρ = M V xda= 1 h wh = 1 hw 0 h 0 w 0 xdxdy ( 1 2 x2 w 0 ) dy = 1 hw 1 2 w2 h or, finally: x com = 1 2 w

11 sample calculations: COM for rectangular plates Not much more difficult to calculate for some non-uniform cases y h For example, if σ(x, y)= C x, then w M = σda= h 0 x w Cx dx dy = 1 2 Chw2 0 and x com = 1 M y com = 1 M xσ da = 1 M yσ da = 1 M h h 0 0 w Cx xdx dy = w Cx ydx dy = Chw3 1 2 Chw2 = 2 3 w Ch 2 w Chw2 = 1 2 h

12 Sample problem, COM

13 Sample problem, COM Calculations: First, put the stamped-out disk (call it disk S) back into the plate P to form the original disk (call it S+P). Because of its circular symmetry, the center of mass x S for the disk S is at the center of S, at xs = R. Similarly, the center of mass x S+P for composite disk S+P is at the center of S+P, at x S+P = 0. The expression for the center of mass x S+P of the combined disk-plate system is Using x S+P = 0, and xs = R we can solve for xp: We only need to know the relative masses:

14 9.3: Newton s 2 nd Law for a System of Particles The vector equation that governs the motion of the center of mass of such a system of particles is: Note that: 1. F net is the net force of all external forces that act on the system. Forces on one part of the system from another part of the system (internal forces) are not included 2. M is the total mass of the system. M remains constant, and the system is said to be closed. 3. a com is the acceleration of the center of mass of the system.

15 9.3: Newton s 2 nd Law for a System of Particles: Proof of final result For a system of n particles, where M is the total mass, and r i are the position vectors of the masses m i. Differentiating, where the v vectors are velocity vectors. This leads to Finally, What remains on the right hand side is the vector sum of all the external forces that act on the system, while the internal forces cancel out by Newton s 3 rd Law.

16 Sample problem: motion of the com of 3 particles Calculations: Applying Newton s second law to the center of mass,

17 9.4: Linear Momentum DEFINITION: in which m is the mass of the particle and v is its velocity. Alternative formulation of Newton s 2nd Law in terms of linear momentum: Prove equivalence by manipulating this equation:

18 9.5: The Linear Momentum of a System of Particles The linear momentum of a system of particles is the sum of the individual linear momenta of the particles in the system. P = p 1 + p p n Taking one derivative and using Newton s 2nd Law for each particle: d P dt = d dt (p 1 + p p n ) = F 1 + F F n = F net which gives Newton s 2nd Law for a system of particles: It is just as if the system was a single particle of momentum P acted on by a single net force Fnet.

19 9.5: The Linear Momentum of a System of Particles In fact, the single particle analogy goes further: The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. Recall (slide 15): Mv com = m 1 v 1 + m 2 v m n v n But the right hand side is just the sum of linear momenta of the particles in the system, that is, the total linear momentum: P = Mv com

20 Collision Isolated system: two particles Initial motion: force free, constant velocity (straight line) Sudden drastic change of motion by (possibly large) internal forces Final motion: force free, constant velocity (straight line)

21 9.6: Collision and Impulse In this case, the collision is brief, and the ball experiences a force that is great enough to slow, stop, or even reverse its motion. The figure depicts the collision at one instant. The ball experiences a force F(t) that varies during the collision and changes the linear momentum of the ball.

22 9.6: Collision and Impulse The change in linear momentum is related to the force by Newton s second law written in the form The right side of the equation is a measure of both the magnitude and the duration of the collision force, and is called the impulse of the collision, J. Define average force during collision: Then we have F avg = 1 t J = p t F avg = 1 t tf t i F (t) dt

23 9.6: Collision and Impulse Instead of the ball, one can focus on the bat. At any instant, Newton s third law says that the force on the bat has the same magnitude but the opposite direction as the force on the ball. That means that the impulse on the bat has the same magnitude but the opposite direction as the impulse on the ball.

24 9.7: Conservation of Linear Momentum If no net external force acts on a system of particles, the total linear momentum, P, of the system cannot change. or P f = P i This is a straightforward consequence of Note also the weaker statement holds: If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

25 Sample problem: 1-D explosion

26 Sample problem: 1-D explosion

27 Sample problem: 2-D explosion

28 Sample problem: 2-D explosion

29 Sample problem: 2-D explosion

30 9.8: Momentum and Kinetic Energy in Collisions In a closed and isolated system, if there are two colliding bodies, and the total kinetic energy is unchanged by the collision, then the kinetic energy of the system is conserved (it is the same before and after the collision). Such a collision is called an elastic collision. If during the collision, some energy is always transferred from kinetic energy to other forms of energy, such as thermal energy or energy of sound, then the kinetic energy of the system is not conserved. Such a collision is called an inelastic collision.

31 9.9: Inelastic collisions in One Dimension Momentum is conserved (always): But energy is not: 1 2 m 1v1i m 2v2i 2 = 1 2 m 1v1f m 2v2f 2 In an inelastic collision some kinetic energy is converted to another type of energy, for example, thermal energy.

32 9.9: Inelastic collisions in 1-D: Velocity of Center of Mass In a completely inelastic collision the velocity of the stuck bodies is the same as the velocity of the COM of the colliding bodies. m 1 v 1i + m 2 v 2i =(m 1 + m 2 )v f v f = m 1v 1i + m 2 v 2i m 1 + m 2 which is the same as the velocity of the COM

33 Sample problem: conservation of momentum

34 Sample problem: conservation of momentum The collision within the bullet block system is brief. Therefore: (1)During the collision, the gravitational force on the block and the force on the block from the cords are still balanced. Thus, during the collision, the net external impulse on the bullet block system is zero. Therefore, the system is isolated and its total linear momentum is conserved. (2) The collision is one-dimensional in the sense that the direction of the bullet and block just after the collision is in the bullet s original direction of motion. As the bullet and block now swing up together, the mechanical energy of the bullet block Earth system is conserved: Combining steps:

35 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. m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 1 2 m 1v 2 1i m 2v 2 2i = 1 2 m 1v 2 1f m 2v 2 2f

36 9.10: Elastic collisions in One Dimension: Stationary Target Newton s cradle: equal masses v 1f =0 v 2f = v 1i

37 9.10: Elastic collisions in One Dimension: Moving Target (see derivation in text)

38 Tennis ball on basket ball For m1 >> m2 these are (set m2 = 0): v 1f = v 1i v 2f =2v 1i v 2i 1 to 2: collision of basketball with floor (1:floor, 2:basketball) basketball bounces up with same speed as before collision 2 to 3: magnitudes of initial v of tennis ball and basket ball equal, but opposite direction. Call this magnitude vi Final velocities (1: basketball, 2: tennis ball): v 1f = v i v 2f =2v i ( v i )=3v i

39 Sample problem: Two pendulums

40 9.11: Collisions in Two Dimensions If elastic, Also, Note there are three equations with four unknowns (the two components of each of the two final velocities). So we need one more piece of information to determine the final velocities completely. Often this is the angle of one of the two final velocities.

41 9.12: Systems with Varying Mass: A Rocket The system here consists of the rocket and the exhaust products released during interval dt. The system is closed and isolated, so the linear momentum of the system must be conserved. Practical information: velocity of rocket relative to products (or vice versa), vrel. So eliminate U left side place the role of a force! dm/dt : rate of fuel consumption vrel: velocity of ejected fuel

42 9.12: Systems with Varying Mass: Finding the velocity recall: in which M i is the initial mass of the rocket and M f its final mass. Evaluating the integrals then gives for the increase in the speed of the rocket during the change in mass from M i to M f.

43 Sample Problem: rocket engine, thrust, acceleration