Introductory Physics. Week 2015/06/26

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1 2015/06/26

2 coordinate systems Part I Multi-particle system

3 coordinate systems 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 coordinate systems generalized coordinate degrees of freedom the center of mass external force and internal force

5 coordinate systems 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 systems 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 systems 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 systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems example: a bead on a wire A bead is running on a smooth wire.

14 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 coordinate systems 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 introduction Part II Linear momentum

27 introduction 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 keywords introduction principle conservation of elastic collision two-body motion

29 introduction 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

30 introduction 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

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

32 introduction the center of mass and 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 of a system is the same as if all its mass was concentrated at the center of mass.

33 introduction The 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

34 introduction proof: the 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

35 introduction proof: the 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

36 introduction Motion of the center of mass The 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

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

38 introduction Conservation of The principle is a vector equation. Each component of the equation holds true. Especially, Conservation of a component of 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.

39 introduction 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.

40 introduction Since there is no external forces acting on the system, the is concerved. Mv + m 0 (v u) = (M + m)v i = 0 v = m 0 M + m 0 u

41 introduction 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.

42 introduction The 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

43 introduction 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

44 introduction 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.

45 introduction 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.

46 introduction 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.

Introductory Physics. Week 2015/06/26

Introductory Physics. Week 2015/06/26 2015/06/26 Part I Multi-particle system 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).