Extended or Composite Systems Systems of Many Particles Deformation

Transcription

1 Extended or Composite Systems Systems of Many Particles Deformation Lana Sheridan De Anza College Nov 15, 2017

2 Overview last center of mass example systems of many particles deforming systems

3 Continuous mass distribution: Another Example Center of mass of a cylinder of uniform density, ρ. z dr r R L (Exam- ) Calculating I e z axis for a solid cylinder.

4 Continuous mass distribution: Another Example Let s choose our axes so that z points along the length of the cylinder, and the origin is right in the center of the cylinder. Probably, you can easily guess where the CM will be.

5 Continuous mass distribution: Another Example Let s choose our axes so that z points along the length of the cylinder, and the origin is right in the center of the cylinder. Probably, you can easily guess where the CM will be. At the origin!

6 Continuous mass distribution: Another Example How could we prove it?

7 Continuous mass distribution: Another Example How could we prove it? Along the z-direction this cylinder is very similar to the thin rod we just considered. Observe that dm = (πr 2 )ρ dz: z CM = 1 z(πr 2 ρ dz) M = (πr2 )ρ πr 2 Lρ L/2 L/2 z dz

8 Continuous mass distribution: Another Example How could we prove it? Along the z-direction this cylinder is very similar to the thin rod we just considered. Observe that dm = (πr 2 )ρ dz: z CM = 1 z(πr 2 ρ dz) M = (πr2 )ρ πr 2 Lρ = 0 L/2 L/2 z dz

9 Continuous mass distribution: Another Example Along the x-direction this cylinder we are looking down on a circle. Let x = R cos θ, y = R sin θ Observe that dm = (2yL)ρ dx: (Looking at just the positive x part of the cylinder, viewed top down.) 1 Or, leave the integral in terms of x and use the substitution u = x 2.

10 Continuous mass distribution: Another Example Along the x-direction this cylinder we are looking down on a circle. Let x = R cos θ, y = R sin θ Observe that dm = (2yL)ρ dx: x CM = 1 x(2yl)ρ dx M = = = 0 Lρ πr 2 L ρ πr 2 0 π R 2 (2 sin θ cos θ)( R sin θ dθ) [ 2 3 R3 sin 3 θ ] 0 π 1 Or, leave the integral in terms of x and use the substitution u = x 2.

11 Continuous mass distribution: Another Example By symmetry the evaluation for y CM must go exactly the same way as for x CM, therefore, y CM = 0. This is what we expected all along, but this same technique can be used in cases where we cannot guess the answer.

12 Systems of Particles Last lecture we talked about the CM point standing in for an extended object to model the object as point-like. Now we justify why that works. Velocity of the center-of-mass: v CM = dr CM dt Multiply both sides by M: = 1 M i m i dr i dt = 1 M m i v i i M v CM = i p i = p tot

13 Systems of Particles Last lecture we talked about the CM point standing in for an extended object to model the object as point-like. Now we justify why that works. Velocity of the center-of-mass: v CM = dr CM dt Multiply both sides by M: = 1 M i m i dr i dt = 1 M m i v i i M v CM = i p i = p tot The momentum of an entire system of many particles is the same as just calculating the momentum from the total mass and the velocity of the center of mass.

14 Systems of Particles M v CM = p tot Implication: we can model linear momentum of the entire system collectively by imagining that the mass of the entire system is at the center of mass point.

15 Systems of Particles We can also consider how a collection of forces that act on different particles in the system affect the motion of the center of mass. a CM = dv CM dt = 1 M i m i dv i dt = 1 M Since F i = m i a i, we can deduce that F i is the net force on particle i. i F i

16 Systems of Particles We can also consider how a collection of forces that act on different particles in the system affect the motion of the center of mass. a CM = dv CM dt = 1 M i m i dv i dt = 1 M Since F i = m i a i, we can deduce that F i is the net force on particle i. Some of the forces on particle i might come from other particles in the system. i F i

17 Systems of Particles Some of the forces on particle i might come from other particles in the system. For example consider a system of two particles. They exert a force on each other, and also experience an external force: F net,1 = F F ext,1 and F net,2 = F F ext,2 = F F ext,2

18 Systems of Particles Some of the forces on particle i might come from other particles in the system. For example consider a system of two particles. They exert a force on each other, and also experience an external force: F net,1 = F F ext,1 and F net,2 = F F ext,2 = F F ext,2 Then The internal forces cancel! F net,1 + F net,2 = F ext,1 + F ext,2

19 Systems of Particles Therefore, a CM = 1 M Simplifies to: F ext,i = M a CM where both sides were multiplied by M. i i F i

20 Systems of Particles Therefore, a CM = 1 M Simplifies to: F ext,i = M a CM where both sides were multiplied by M. i i F i This shows that the center of mass of the system of particles moves just as a point mass would under the influence of a net external force F ext. Newton s second law: F net,ext = M a CM

21 Systems of Particles Lastly, we still have that the impulse is the total change in momentum of the entire system collectively. I = F ext,i dt = i M dv CM dt dt = M dv CM = M v CM

22 Systems of Particles Lastly, we still have that the impulse is the total change in momentum of the entire system collectively. I = F ext,i dt = i M dv CM dt dt = M dv CM = M v CM = p tot

23 Deformable Systems Some systems will change the distribution of their mass during their motion.

24 Deformable Systems Some systems will change the distribution of their mass during their motion. 1 Image from

25 Deformable Systems In a system that is deformed, the positions of the masses of particles may change relative to the center of mass, but we can still study the system by considering what happens to the center of mass.

26 leave the floor? (d) Does it make sense to say that this momentum came from the floor? Explain. (e) With what kinetic energy does the person leave the floor? (f) Does it make sense to say that this energy came from the floor? Explain. Deformable Systems Problem Page 288, # Figure P9.59a shows an overhead view of the initial S configuration of two pucks of mass m on frictionless ice. The pucks are tied together with a string of length, and negligible mass. At time t 5 0, a constant force of magnitude F begins to pull to the right on the center point of the string. At time t, the moving pucks strike each other and stick together. At this time, the force has moved through a distance d, and the pucks have attained a speed v (Fig. P9.59b). (a) What is v in terms of F, d,,, and m? (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy? m (a) It has an exhaust speed dizer is requi design could amount of fue same task? (c (b) is 2.50 tim why the requ factor of A rocket has 330 kg of fu it starts from engine at tim ative speed v 2.50 kg/s. Th M f /k k ing the burn time is given b, m t 0 F d CM d v t t F (b) Make a gr tion of time fo that the accel a b (d) Graph th

27 the person? (c) With what momentum does the person leave the floor? (d) Does it make sense to say that this momentum came from the floor? Explain. (e) With what kinetic energy does the person leave the floor? (f) Does it make sense to say that this energy came from the floor? Explain. Deformable Systems Problem Page 288, # Figure P9.59a shows an overhead view of the initial S configuration of two pucks of mass m on frictionless ice. The pucks are tied together with a string of length, and negligible mass. At time t 5 0, a constant force of magnitude F begins to pull to the right on the center point of the string. At time t, the moving pucks strike each other and stick together. At this time, the force has moved through a distance d, and the pucks have attained a speed v (Fig. P9.59b). (a) What is v in terms of F, d,,, and m? (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy? m en (a ex di de am sa (b wh fa 64. A 33 it en at 2. M in tim d CM v

28 each other and stick together. At this time, the force has moved through a distance d, and the pucks have attained a speed v (Fig. P9.59b). (a) What is v in terms of F, d,,, and m? (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy? Deformable Systems Problem Page 288, #59 m 3 i e a 2 M i t, m t 0 F d CM d v t t F ( t t a b Figure P9.59 Section 9.9 Rocket Propulsion 60. A model rocket engine has an average thrust of 5.26 N. ( (

29 Deformable Systems (a) Speed of pucks, v, after collision?

30 Deformable Systems (a) Speed of pucks, v, after collision? F = M a CM

31 Deformable Systems (a) Speed of pucks, v, after collision? F = M a CM F is constant a CM is constant.

32 Deformable Systems (a) Speed of pucks, v, after collision? F = M a CM F is constant a CM is constant. a CM = F 2m vf 2 = vi 2 + 2a CM x ( ) ( F vf 2 = d l ) 2m 2 2Fd F l v f = 2m

33 Deformable Systems (b) Increase in internal energy? system: pucks + pucks and ice s internal degrees of freedom W = K + E int

34 Deformable Systems (b) Increase in internal energy? system: pucks + pucks and ice s internal degrees of freedom W = K + E int E int = W K f = Fd 1 2 (2m)v 2 = Fd 1 2 (2m)2Fd F l 2m = Fd Fd + F l 2 = 1 2 F l

35 Summary systems of many particles deformations 3rd Collected Homework! due Monday, Nov 20. Next test Monday, Nov 27. (Uncollected) Homework Serway & Jewett, Ch 9, onward from page 288. Probs: 51, 55, 57, 92 Read ahead in Chapter 10. Ch 10, onward from page 288. Probs: 3