Physics 170 Week 10, Lecture 1

Transcription

1 Physics 170 Week 10, Lecture 1 gordonws/170 Physics 170 Week 10, Lecture 1 1

2 Textbook Chapter 14: Section Physics 170 Week 10, Lecture 1 2

3 Learning Goals: We will define the concept of work due to a force. We will use a number of examples to illustrate the concept of work and to show how it can be useful in solving problems. We will study the law of work and energy for a system with more than one particle We will illustrate the law of work and energy with an example By the end of this lecture, the student should know how to compute the work due to a force during a motion by doing a line integral. The student should also be able to apply the principle of work and energy to compute the change in kinetic energy of a particle that is moving under the influence of a force. Physics 170 Week 10, Lecture 1 3

4 Work due to a force Consider a particle which moves from position r to position r = r + d r so that the displacement is infinitesimal d r. Assume that a force F is acting on the particle as it moves. Assume that d r is small enough so that the force F is a constant throughout the motion and the displacement is in a straight line. Physics 170 Week 10, Lecture 1 4

5 Work due to a force cont d. The work done by the force on the particle during the motion is defined as du = F d r Units are force time distance Newton-meters = Joules or foot-pounds. This is the usual definition of work as force times distance the dot product projects F on d r, U = F d r = F d r cos θ Physics 170 Week 10, Lecture 1 5

6 Work due to a force cont d. The work done by the force on the particle during the motion in an infinitesimal dispalcement is du = F d r. To get the work for a finite, rather than infinitesimal displacement, we have to integrate the line integral. U 1 2 = r2 r 1 F d r Physics 170 Week 10, Lecture 1 6

7 How to do the line integral The work is the line integral U 1 2 = r 2 r 1 F d r If you know the equation of the curve over which the displacement occurs, r(s) and you know the value of the force F ( r(s)) when the particle is at a position along the curve, then, if r 1 = r(s 1 ) and r 2 = r(s 2 ), the integral is U 1 2 = r2 r 1 F d r = s2 s 1 F d ( r(s)) r(s) ds ds Physics 170 Week 10, Lecture 1 7

8 Work due to a constant force field: Consider a force which is constant, for example, gravity near the surface of the earth action on a particle with mass m, F = mgˆk. Consider a curve r(s) = x(s) î + y(s) ĵ + z(s) ˆk. The work done in moving between two points is U 1 2 = s2 s 1 F d ds r(s)ds = s2 s 1 ( mgˆk) d ds ( x(s)î + y(s)ĵ + z(s)ˆk ) ds U 1 2 = s2 s 1 ( mg) d ds z(s) ds = mg (z(s 2) z(s 1 )) Work U 1 2 = mg(z 2 z 1 ) is mg times the change in height. Physics 170 Week 10, Lecture 1 8

9 Example: The 10 kg block shown in the figure rests on the smooth incline. If the spring is originally stretched by 0.5 m, determine the total work done by all the forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. Physics 170 Week 10, Lecture 1 9

10 Example cont d: Forces: Gravity: W = mgĵ P: P = 400 î N Reaction: N = N(sin 30 î + cos 30 ĵ) Spring: Fs = (30 N/m)(s + (0.5 m)) ( ) cos 30 î + sin 30 ĵ Physics 170 Week 10, Lecture 1 10

11 Example cont d: Displacement: We ( put the origin at the ) point where the motion starts. r(s) = s cos 30 î + sin 30 ĵ d r(s) = ( ) cos 30 î + sin 30 ĵ ds, s [0, 2 m] Physics 170 Week 10, Lecture 1 11

12 Example cont d: Increments ( of work: find du ) = F d r where d r = cos 30 î + sin 30 ĵ ds and W = mgĵ, P = 400 î N, N = N(sin 30 î + cos 30 ĵ), F s = (30 N/m)(s + (0.5 m)) ( ) cos 30 î + sin 30 ĵ ( ) Gravity: du W = mgĵ cos 30 î + sin 30 ĵ ds ( ) P: du P = 400 N î cos 30 î + sin 30 ĵ ds Reaction: ( ) du N = N(sin 30 î + cos 30 ĵ) cos 30 î + sin 30 ĵ ds = 0 ( ) Spring: du s = (30 N/m)(s + (0.5 m)) cos 30 î + sin 30 ĵ ( ) cos 30 î + sin 30 ĵ ds Now integrate from s = 0 to s = 2m. Physics 170 Week 10, Lecture 1 12

13 Example cont d: Work due to gravity: Gravity du W = mgĵ ( ) cos 30 î + sin 30 ĵ ds U W = mg sin ds = mg sin 30 (2m) = (10)(9.81) sin 30(2) J Physics 170 Week 10, Lecture 1 13

14 Example cont d: Total work of each force Gravity: U W = (10)(9.81)(sin 30)(2) J U W = 98.1 J P: U P = 400 cos ds = (400)(2) cos 30J = J U P = 693 J Reaction: U N = N(sin 30 î + cos 30 ĵ) U N = 0 ( ) 2 cos 30 î + sin 30 ĵ 0 ds = 0 Spring: U s = (30 N/m) 2 ds(s + (0.5 m)) = 0 (30 N/m)( 1 2 s2 + (0.5 m)s) 2 0 = (30)( 1 2 (4) + (0.5)(2)) J U s = 90 J Physics 170 Week 10, Lecture 1 14

15 Principle of Work and Energy Now we consider a displacement where the position of the particle evolves according to Newton s second law F = m a = m d dt v(t) where F is all forces acting on the particle. Multiply each side by the velocity F v(t) = m d dt v(t) v(t) = d dt ( ) 1 2 m v2 (t) Integrate this expression t2 t 1 dt F v(t) = t2 t 1 dt d dt ( ) 1 2 m v2 (t) Physics 170 Week 10, Lecture 1 15

16 Principle of Work and Energy cont d t2 dt t2 F v(t) = dt d ( ) 1 t 1 dt 2 m v2 (t) The right-hand-side can be integrated or t m v2 (t 2 ) 1 2 m v2 (t 1 ) = 1 2 m v m v2 1 = t2 r2 r 1 t 1 dt F v(t) F d r 1 2 m v2 is the kinetic energy The change of kinetic energy of a particle is equal to the work done on it by all forces. Physics 170 Week 10, Lecture 1 16

17 Example: The 20 lb crate has a velocity of v A = 12 ft/s when it is at A. Determine its velocity after it slides s = 6 ft down the plane. The coefficient of kinetic friction between the crate and the plane is µ k = 0.2. Physics 170 Week 10, Lecture 1 17

18 Strategy for solving the problem: Find the forces acting on the box. Find the work done by each of the forces and add them together to find the total work. Equate total work to the change in kinetic energy and use this to compute the final speed. Physics 170 Week 10, Lecture 1 18

19 Free-body Diagram Physics 170 Week 10, Lecture 1 19

20 Forces acting on the box: This is a two-dimensional problem Orient x-axis horizontally and y-axis vertically We find a mathematical expression for force vectors: Gravity: W = mg ĵ ) Reaction: N = N ( 3 5 î ( ĵ ) Friction: Ff 4 = ν k N 5 î ĵ Reaction must cancel normal component of W. N = 4 5 mg Physics 170 Week 10, Lecture 1 20

21 Increment of work due to forces: ( 4 r(s) = s 5 î + 3 ) 5 ĵ, d r = ds ( 4 5 î + 3 ) 5 ĵ Increment of work is du = F d r. We compute this for each force: Gravity: W = mgĵ du W = W d r = 3 5mg ds ) Reaction: N = N ( 3 5 î ĵ N d r = 0 du N ( = 0 ) Friction: Ff 4 = ν k N 5 î ĵ du f = F f d r = µ k N ds Physics 170 Week 10, Lecture 1 21

22 Work due to forces: Now we integrate the increments of work to get the total work done by each force: Gravity: du W = W d r = 3 5mg ds U W = W d r = 3 5 mg 6 ft ds = (20 lb)(6 ft) = 72.0 ft.lb Reaction: U N = 0 Friction: du f = F f d r = µ k N ds U f = Ff d r = µ k N 6 ft ds = (0.2)(20 lb) (6 ft) = ft lb Physics 170 Week 10, Lecture 1 22

23 Implement the principle of work and energy We will now implement the principle that the total work done by all of the forces that are acting on the crate must equal the change of its kinetic energy. Total work due to all forces is is U W + U N + U f = ft lb = 1 2 mv mv2 1 From this equation we find the final velocity: v2 2 = (12 ft/s) 2 + 2(32.2 ft/s2 ) ( lb 5 ft lb) v 2 = 17.7 ft/s Physics 170 Week 10, Lecture 1 23

24 Work due to a force in a system of particles The work done on particle with displacement d r i by the force F j du = F j d r i. The work for a finite displacement is the line integral U 1 2 = r2 r 1 i,j F i d r j Physics 170 Week 10, Lecture 1 24

25 Some forces are internal and f ij = f ji. In this case, particles follow trajectories so that [ dt f ij (t) d r j(t) + f dt ji (t) d r ] i(t) dt [ d rj (t) = f ij (t) d r ] i(t) dt dt dt can cancel and we only need to consider external forces. This is useful when particles are tied together can sometimes be used for pulleys. Physics 170 Week 10, Lecture 1 25

26 Example: A 2-lb block rests on a semi-cylindrical surface. An elastic cord having a stiffness k=2lb/ft is attached to the block at B and to the base of the semi-cylinder at C. If the block is released from rest at A (θ = 0), determine the unstretched length of the cord so that the block begins to leave the semicylinder at the instant that θ = 45 degrees. Neglect the size of the block. Physics 170 Week 10, Lecture 1 26

27 Example: Cylindrical coordinates r = rû r, v = r θû θ, a = r θû θ r θ 2 û r Tangential-normal coordinates (û t = û θ, û n = û r ) v = vû t = r θû t, a = r θu t + (r θ) 2 r û n Physics 170 Week 10, Lecture 1 27

28 Example: Total forces acting: gravity, normal reaction, spring F = mgĵ + Nû r + kr(θ 0 θ)û θ ĵ = sin θû r + cos θû θ F = (N mg sin θ)û r + [kr(θ 0 θ) mg cos θ]û θ Physics 170 Week 10, Lecture 1 28

29 Example: Dybnamics: F = m a where a = r θû θ r θ 2 û r F = (N mg sin θ)û r + [kr(θ 0 θ) mg cos θ]û θ (N mg sin θ)û r + [kr(θ 0 θ) mg cos θ]û θ = m (r θû θ r θ ) 2 û r Physics 170 Week 10, Lecture 1 29

30 Example: (N mg sin θ)û r + [kr(θ 0 θ) mg cos θ]û θ = m (r θû θ r θ ) 2 û r equations for components: N mg sin θ = mr θ 2 kr(θ 0 θ) mg cos θ = mr θ Physics 170 Week 10, Lecture 1 30

31 Example: Integrate second equation: d dt [ kr kr(θ 0 θ) mg cos θ = mr θ kr θ(θ 0 θ) mg θ cos θ = mr θ θ (θθ 0 12 ) ] θ2 mg sin θ = d dt [ 1 2 mr θ 2 ] Physics 170 Week 10, Lecture 1 31

32 Example: t 0 dt d dt [ kr (θθ 0 12 ) θ2 ] mg sin θ = t 0 dt d dt Impose initial conditions: θ(0) = 0, θ(0) = 0 kr (θθ 0 12 ) θ2 mg sin θ = 1 2 mr θ 2 [ ] 1 2 mr θ 2 Physics 170 Week 10, Lecture 1 32

33 Example: Interpretations of the formula t dt d [ kr (θθ 0 12 ) dt θ2 0 ] mg sin θ = t 0 dt d dt [ ] 1 2 mr θ 2 from principle of work and energy: (v = r θ) and = θ mr2 θ2 = 1 2 mv2 = 2 1 F d r = θ 0 F θ (θ )rdθ [kr(θ 0 θ ) mg cos θ ]rdθ = kr 2 ( θθ θ2 ) mgr sin θ Physics 170 Week 10, Lecture 1 33

34 Example: Interpretations of the formula t dt d [ kr (θθ 0 12 ) dt θ2 0 ] mg sin θ = t 0 dt d dt [ ] 1 2 mr θ 2 Conservation of energy kr 2 2 (rπ l 0) 2 = 1 2 mr2 θ2 + k 2 (l l 0) 2 Potential energy stored in a spring is k 2 (s s 0) 2 l = r(π θ), l 0 = r(π θ 0 ) ( kr 2 θθ 0 1 ) 2 θ2 mgr sin θ = 1 2 mr2 θ2 = 1 2 mv2 Physics 170 Week 10, Lecture 1 34

35 Example: kr (θθ 0 12 θ2 ) mg sin θ = 1 2 mr θ 2 mr θ 2 (t) = kr ( 2θ(t)θ 0 θ 2 (t) ) 2mg sin θ(t) Block leaves surface when N = 0 N = mg sin θ mr θ 2 = 0 when mr θ 2 = mg sin θ Physics 170 Week 10, Lecture 1 35

36 Example: mr θ 2 (t) = kr ( 2θ(t)θ 0 θ 2 (t) ) + 2mg sin θ(t) N = mg sin θ mr θ 2 = 0 when mr θ 2 = mg sin θ mg sin θ = kr ( 2θθ 0 θ 2) 2mg sin θ Physics 170 Week 10, Lecture 1 36

37 Example: mg sin θ = kr ( 2θθ 0 θ 2) 2mg sin θ θ 0 = θ 2 + 3mg 2θkr sin θ [ l 0 = r(π θ 0 ) = r π θ 2 3mg ] 2θkr sin θ Physics 170 Week 10, Lecture 1 37

38 Example: when θ = π/4, l 0 = 2.77 ft l 0 = (1.5 ft) l 0 = r [ π π 8 [ π θ 2 3mg ] 2θkr sin θ 3(2 lb) 2(π/4)(2 lb/f t)(1.5 f t) ] 1 2 Physics 170 Week 10, Lecture 1 38

39 For the next lecture, please read Textbook Chapter 14: Section Physics 170 Week 10, Lecture 1 39