Introductory Physics. Week 2015/06/26

Transcription

1 2015/06/26

2 Part I Multi-particle system

3 multi-particle system So far, we have been limiting ourselfs to the problems of the motion of single particle (except a brief introduction of coupled oscillation). Now we consider the cases when there are multiple particles. Special attention will be paid to the case when there are only two particles.

4 keywords generalized coordinate degrees of freedom the center of mass external force and internal force

5 Coordinate system in 3-dim space When a particle moves freely in 3-dim space, how many variables do we need to describe its position?

6 Coordinate system in 3-dim space When a particle moves freely in 3-dim space, how many variables do we need to describe its position? In case of Cartesian coordinate system, we need x, y, z.

7 Coordinate system in 3-dim space We can think of another coordinate system to describe the position in 3-dim space. This is called Cylindrical coordinate system. We need ξ, φ, z.

8 Coordinate system in 3-dim space We can think of yet another coordinate system to describe the position in 3-dim space. This is called Spherical coordinate system. We need r, θ, φ.

9 generalized coordinates When a particle moves freely in 3-dim space, we needs three scalar variables to describe its position. x, y, z (in Cartesian coordinate system) ξ, φ, z (in Cylindrical coordinate system) r, θ, φ (in Spherical coordinate system) Note that the scalar variables are not necessary to have the dimention of length. In order to clearly specify that point, these variables are called generalized coordinates.

10 the degrees of freedom When a particle moves freely in 3-dim space, we needs three scalar variables to describe its position. x, y, z (in Cartesian coordinate system) ξ, φ, z (in Cylindrical coordinate system) r, θ, φ (in Spherical coordinate system) Note that we always need three independent variables to describe the position. In this case, we say the degrees of freedom of this system is three.

11 the degrees of freedom the degrees of freedom The degrees of freedom is the number of independent generalized coordinates that are needed to describe the system. In order to determine the motion of the system, we need the same number of equations as the degrees of freedom.

12 example: two free particles in 3-dim space We have two particles moving freely in 3-dim space. We need (x 1, y 1, z 1 ) for one particle, (x 2, y 2, z 2 ) for the other particle. The number of variable necessary to describe the system is six. The degree of freedom of this system is six.

13 example: a bead on a wire A bead is running on a smooth wire.

14 example: a bead on a wire A bead is running on a smooth wire. We need only one variable to describe the position of the bead - the displacement along the wire from Origin. The degrees of freedom of this system is one. Even if the wire (the path of the bead) is in 3-dim space, one equation will be enough to determine the motion of the bead.

15 example: two particle connected by a bar Two particles are connected by a solid bar with a fixed length. How much the degree of freedom of this system is? z y x

16 example: two particles connected by a bar Two particles are connected by a solid bar with a fixed length. We need three variables to describe the position of one of the particle. z y x

17 example: two particles connected by a bar Two particles are connected by a solid bar with a fixed length. We need three variables to describe the position of one of the particle. We need extra two variables to describe the direction of the other particle s position. z q j y x

18 example: two particles connected by a bar Two particles are connected by a solid bar with a fixed length. We need three variables to describe the position of one of the particle. We need extra two variables to describe the direction of the other particle s position. z q j y x The degrees of freedom is five.

19 example: two particles connected by a bar Another way to count the degrees of freedom. If there is no constrain, the degrees of freedom is 6. But we have a relation between 6 coordinates reflecting the constrain, two particles are connected by a bar with a fixed length. That relation (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 = l 2 reduces the degrees of freedom by one. The degrees of freedom is 6 1 = 5. z y x

20 center of mass When we have particles P 1, P 2,... P N with masses m 1, m 2,.. m N, and their position is r 1, r 2,... r N, the center of mass is the point whose position is R = m 1 r 1 + m 2 r m N r 3 m 1 + m m N

21 external force and internal force When we have particles P 1, P 2,... P N with masses m 1, m 2,.. m N, and their position are r 1, r 2,... r N, External forces are forces not originating from either particles. Internal forces are forces particles exert on each other. G ji P j F j O r j r i r ij F i P i G ij internal forces external forces G ij = -G ji

22 mechanical energy conservation energy conservation for an unconstrained system When the system is unconstrained, and both the external and internal forces acting on the system are conservative, the sum of kinetic and potential energies remains constant. energy conservation for a constrained system When both the external and internal forces acting on the system are conservative, and the total work done by constarint forces is zero, the sum of kinetic and potential energies remains constant.

23 Constraint forces which do no work We have considered the list of external constraint forces which do no work. A particle connected to a fixed point by a light inextensible string: the string tension does no work. A particle sliding over a smooth fixed surface: the reaction force of the surface does no work. A particle sliding on a smooth fixed wire: the reaction force of the wire does no work. We add another important constraint force. the distance between a pair of particles remains fixed: the total work done by internal constraint forces between the two particles is zero.

24 Energy conservation for a rigid object Any rigid object contains billions and billions of atoms in it. But the distances between any two atoms remains fixed. The internal constraint forces which keep the distance fixed do no work. When the external forces acting on the object are conservative, the sum of the kinetic and potential energies of the system remains constant.

25 proof If the distance between P i and P j remains fixed, the rate of total work by constraint force G ij and G ji is G ji P = G ij v i + G ji v j = G ij ( v i v j ). On the other hand, ( r i r j ) ( r i r j ) = const. Differentiating with t gives, ( r i r j ) ( v i v j ) = 0 P j F j because G ij ( r i r j ), P = 0 r j r i r ij F i P i G ij internal forces external forces G ij = -G ji O

26 Part II Linear momentum

27 introduction By using the concept of mechanical energy and its conservation, we were able to understand an essence of motion of particle(s). We will learn another quantity linear momentum which also helps us to understand the system s behavior.

28 linear momentum A particle P of mass m is under a force F. The equation of motion is F = m d v dt = d dt (m v) taking integral over time t2 t 1 F dt = t2 t 1 d dt (m v)dt = m v 2 m v 1

29 linear momentum Definition When a particle of mass m has velocity of v, its (linear) momentum is defined to be p m v the equation of motion can be written in terms of linear momentum. d p dt = F

30 linear momentum of a multi-particle system For a multi-particle system: When there are particles of masses m 1, m 2,, m N with velocities v 1, v 2,, v N, the total linear momentum is N N P = p i = m i v i i=1 This is called the linear momentum of the multi-particle system. i=1

31 the center of mass and linear momentum The position vector R of the center of mass is R = i m i r i Take the time derivative i m i V = i m i v i P = M V, i m i = P M where V is the velocity of the center of mass, and M is the total mass of the system. The linear momentum of a system is the same as if all its mass was concentrated at the center of mass.

32 The linear momentum principle Linear momentum principle In any motion of a system, the rate of increase of its linear momentum is equal to the total external force acting on the system. d P dt = F

33 proof: the linear momentum principle When external and internal forces are acting on a system, the equation of motion of i-th particle is d v i m i dt = F N i + G ij G ji P j F j j=1 O r j r i r ij F i P i G ij internal forces external forces G ij = -G ji

34 proof: the linear momentum principle Then, dp ( dt = d N ) m i v i = dt i=1 N N N = F i + = = i=1 N F i + i=1 i=1 j=1 ( i 1 N i=1 N F i = F i=1 G ij N i=1 m i d v i dt ) ( G ij + G ji ) j=1

35 Motion of the center of mass The linear momentum principle d P dt = F can be written as M d V dt = F This means that the center of mass of a system moves as if it were a particle whose mass is the total mass of the system, and all external forces are acting on it

36 Conservation of linear momentum This is a special case of the linear momentum principle. If the system is an isolated system, which means there is no external forces acting on it, the linear momentum principle shows d P dt = 0 Conservation of linear momentum In any motion of an isolated system, the total linear momentum is conserved.

37 Conservation of linear momentum The linear momentum principle is a vector equation. Each component of the equation holds true. Especially, Conservation of a component of linear momentum If the total force acting on a system has zero component in a fixed direction, then the component of the total linear momentum in that direction is conserved.

38 Example: rocket A rocket (mass M) ejects a feul (mass m 0 ) backwards at once with speed u relative to the rocket. Initially the rocket (and the fuel) are at rest. There are no external forces acting on the system.

39 Since there is no external forces acting on the system, the linear momentum is concerved. Mv + m 0 (v u) = 0 v = m 0 M + m 0 u

40 Example: more realistic rocket A rocket (mass M) burns a fuel (total mass m 0 ) gradually. Initially the rocket (and the fuel) are at rest.

41 The linear momentum is concerved. (M + m)v = (M + m + dm)(v + dv) dm(v + dv u) = (M + m)v + (M + m)dv + vdm + dmdv vdm dmdv + udm = (M + m)v + (M + m)dv + udm dv = u M + m dm dv dm = u M + m m(t) u v(t) = M + m dm 0 = u log(m + m(t)) + C

42 from initial condition, v(0) = 0 = u log(m + m 0 ) + C C = u log(m + m 0 ) Hense, v(t) = u log(m + m(t)) + u log(m + m 0 ) ( ) M + m0 = u log M + m(t) after burning all fuel, m(t) = 0 ( ) M + m0 v f = u log M

43 comparison: at once or gradually? ( ) m0 Eject at once: v f = u M + m ( 0 ) M + m0 Eject gradually: v f = u log M Interesting observations. If you eject the fuel at once, the resultant velocity of rocket never exceeds the ejection speed u. But if you gradually burns the fuel, the velocity of rocket can exceeds u.

44 How much fuel do we need? ( ) M + m0 v f = u log M How much fuel do we need on the rocket? v f /u = log((m + m 0 )/M) m 0 /M The amount of fuel required increases exponentially as the final speed increases.

45 How to obtain fast speed? In order to get faster speed, we d be better to find a fuel which realize larger u. Most of rocket engines use exothermic chemical reaction to generate heat. A part of heat is converted to the kinetic energy of the molecules of exhaust gas. So the conditions for the ideal fuel for rocket would be the fuel should generate as large energy as possible per its weight the mass of the exhaust gas should be as small as possible The liquid hydrogen is widely used as the fuel for rocket, (partly) because it satisfies both conditions.

46 Collision process We consider an isolated system of two particles, where the internal force between two particles tends to zero as the distance between two particle tends to infinity. If the particles are initially at great distance, each must be moving with constant velocity. If the particles approach each other, the internal force disturbs their straight line motion. Then after the particles retreat to a great distance, they will move with constant velocity again. This is called collision process.

47 Collision process

48 Conservation laws in collision process Since there are no external forces acting on two particles, their total momentum is conserved. But generally, the total kinetic energy is not conserved. If the total kinetic energy is conserved, the collision is said to be elastic. If not, the collision is said to be inelastic. If the force between two particles is conservative, the collision is elastic.

49 Elastic head-on collision We consider head-on collision (one-dimentional motion) between two particles.

50 Elastic head-on collision Kinetic energy conservation m 1 v1i 2 + m 2 v2i 2 = m 1 v1f 2 + m 2 v2f 2 (1) Momentum conservation m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (2) From (1), m 1 (v1i 2 v1f) 2 = m 2 (v2f 2 v2i) 2 m 1 (v 1i v 1f )(v 1i + v 1f ) = m 2 (v 2f v 2i )(v 2f + v 2i ) From (2), m 1 (v 1i v 1f ) = m 2 (v 2f v 2i ) Thus, v 1i + v 1f = v 2f + v 2i (v 2i v 1i ) = v 2f v 1f (3)

51 Coefficient of restitution Coefficient of restitution C R is defined as C R v 2f v 1f v 2i v 1i. C R is a real number 0 C R 1. For elastic collision, C R = 1.

52 Elastic head-on collision Momentum conservation m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i In elastic collision v 1f + v 2f = v 1i v 2i By solving these equations v 1f = (m 1 m 2 )v 1i + 2m 2 v 2i m 1 + m 2 v 2f = 2m 1v 1i + (m 2 m 1 )v 2i m 1 + m 2

53 Elastic head-on collision v 1f = (m 1 m 2 )v 1i + 2m 2 v 2i m 1 + m 2 v 2f = 2m 1v 1i + (m 2 m 1 )v 2i m 1 + m 2 When m 2 m 1, v 1f v 1i + 2v 2f v 2f v 2i

54 Elastic collision We consider an elastic collision, in which one of the particle ( the target ) is initially at rest.

55 Elastic collision Conservation of linear momentum m 1 v 1i = m 1 v 1f + m 2 v 2f the collision is elastic 1 2 m 1v1i 2 = 1 2 m 1v1f m 2v2f 2 Can we extract useful information from these two formulas?

56 From the first equation, m 1 v 1i = m 1 v 1f + m 2 v 2f m 1 v 1i m 1 v 1i = (m 1 v 1f + m 2 v 2f ) (m 1 v 1f + m 2 v 2f ) m 2 1v 2 1i = m 2 1v 2 1f + 2m 1 m 2 v 1f v 2f + m 2 2v 2 2f From the second equation, m 2 1v 2 1i = m 2 1v 2 1f + m 1 m 2 v 2 2f Thus, 0 = 2m 1 m 2 v 1f v 2f + (m 2 m 1 )m 2 v 2 2f 2m 1 v 1f v 2f = (m 1 m 2 )v 2 2f cos θ = (m 1 m 2 )v 2f 2m 1 v 1f, where θ is the angle between two particles velocities after collision.

57 a tip for billiard players The formula cos θ = (m 1 m 2 )v 2f 2m 1 v 1f means that in an elastic collision between particles of equal mass, the particles depart in directions at right angles. A tip for billiard players! Collisions between pool balls are very close to elastic, and the masses of the balls are equal. This means...

58 a tip for billard players

59 a tip for billard players

60 a tip for billard players

61 a tip for billard players

62 a tip for billard players

63 a tip for billard players

64 a tip for billard players In an elastic collision between particles of equal mass, the particles depart in directions at right angles. If the angle is wrong, you need to spin the cue ball (we are disregarding the rotation of the balls when we consider the collision between balls as the collision between particles), or use a bank shot.

65 a tip for buillard players : bank shot

66 Two-body problem When we consider only two particles, interacting each other but without any external force, it is called two-body problem. The movement of a satelite around the Earch (neglecting influence from the Moon, the Sun, and other planets). The movement of the Earch and the Sun with the gravitational interaction The hydrogen atom (an electron and a nucleus)

67 Two-body problem When the magnitude of the interaction force depends only on r, the distance between two particle, the equations of the motion for particle 1 and 2 will be m 1 r1 = F (r) e r m 2 r2 = F (r) e r

68 Two-body problem The relative motion is described by r 1 r 2 = F (r) e r + F (r) e r = m 1 + m 2 F (r) e r m 1 m 2 m 1 m 2 This means the the motion of particle 1 relative to particle 2 is the same as if particle 2 is fixed and particle 1 had the mass m 1 m 2 m 1 + m 2. The quantity m 1 m 2 m 1 + m 2 is called reduced mass.

69 Example: the Sun and the Earth The mass of the Sun is kg, where as that of the Earth is kg. The average distance between the Sun and the Earch is m. The radius of the Sun is m When we consider the motion of the Earth and the Sun, we often assume that the position of the Sun is fixed, and the Earth is moving around the Sun. This is OK because the reduced mass of the Earth is almost same as the mass of the Earth the center of the mass is very close to the center of the Sun