SKAA 1213 Engineering Mechanics

Transcription

1 SKAA 113 Engineering Mechanic TOPIC 8 KINEMATIC OF PARTICLES Lecturer: Roli Anang Dr. Mohd Yunu Ihak Dr. Tan Cher Siang

2 Outline Introduction Rectilinear Motion Curilinear Motion Problem

3 Introduction General Term & Definition: Mechanic Static equilibrium of a body that i at ret, or the body moe with contant elocity Mechanic Dynamic deal with accelerated motion of a body 1) Kinematic analyi of geometric apect of a motion ) Kinetic analyi of the force that caue the motion

4 Introduction Dynamic: Kinematic of Particle Rectilinear Motion A particle moe in a traight line and doe not rotate about it centre of ma. CircularMotion (Curilinear Motion) A particle moe along a path of a perfect circle. General Plane Motion (Curilinear Motion) A particle moe in a plane, which may follow a path that i neither traight norcircular circular.

5 Rectilinear Motion Rectilinear Kinematic pecifying the particle poition, elocity, and acceleration at any intant (time factor) Factor Symbol Unit Remark Time t econd () Data may be gien in minute or hour (h) Poition meter (m) Data may be gien in millimeter (mm), kilometer (km) Velocity m/ Another common unit i kilometer per hour (km/h) Acceleration a m/

6 Rectilinear Motion Poition: Single coordinate axi, Magnitude of = ditance from origin (O) to current poition (P) Direction: +e = right of origin; e = left of origin t t 1 t Time O Poition 1 Diplacement

7 Rectilinear Motion Diplacement: Change in the particle poition, ector quantity If particle moe from S 1 to S : 1 When i +e / e, particle poition i right / left of it initial poition t t 1 t Time O Poition 1 Diplacement

8 Rectilinear Motion Velocity: The peed of the change of poition. Aerage elocity: 1 t t t t Intantaneou elocity: in lim / t d dt t 1 O 1 Velocity, t 1 t

9 Rectilinear Motion Acceleration: The peed of the change of elocitie. Aerage acceleration: a 1 t t t t Intantaneou acceleration: a a t lim / t d dt d dt 1 O Acceleration, a 1

10 Rectilinear Motion Magnitude and direction Factor +e alue e alue Zero alue Poition, Direction to right Direction to left Velocity, Direction to right Direction to left Particle top moing Acceleration, a Velocity increaed Velocity decreaed Contant elocity

11 Rectilinear Motion Poition, elocity and acceleration a a function of time (t): d dt a d dt ( t) ( t) a( t) Differential ( t) ( t) a( t) dt adt Integration

12 Rectilinear Motion Function of poition, elocity and acceleration without time (t) factor: a d dt d dt dt dt d d a d d a d ad

13 Rectilinear Motion Contant acceleration, a c : d t a c dt a t c t 1 d dt t a ct d a cd a c

14 Rectilinear Motion Summary of Equation: d dt a d dt a d dt When acceleration i contant: a t c d ad 1 t a t o o c ac c

15 Rectilinear Motion A ehicle moe in a traight line uch that for a hort time it elocity i defined by =(.9t +.6t) m/ where t i in econd. When t =, =. Determine it poition () and acceleration (a) when t p = 3.

16 Rectilinear Motion Solution: Poition When = when t =, we hae: d.9t. 6t d dt t When t = 3: 3.9t.6tdt.3t.3t.3t 3 3.3t.3t.3(3).3(3) 1.8m 3.3t t

17 Rectilinear Motion Solution: Acceleration Knowing i a function of time (t), the acceleration can be determined from a = d/dt d dt d dt 1.8 t.6 a.9 t 6.6 t When t = 3: a1.8 t.6 1.8(3). 6 6m/

18 Rectilinear Motion A ball i thrown upward at 75m/ from the top of a 4 m tall building. Determine: a) Maximum height B reached by the ball. A = 75m/ A B B = B b) The peed of the ball jut before it hit the ground. A = 4m C C O

19 Rectilinear Motion Solution: Information gathering: Take origin at O and upward direction i poitie. Acceleration i contant and due to graity: a C = 9.81m/ The ball will reach maximum height at B: = B B = (ball top moing at maximum height) F th ti h From the quetion we hae: t = A = +75m/, A = +4m

20 Rectilinear Motion Solution: At Point B: ac B A 75 a C ( B A ( 9.81)( ) B 4) B B 37 m 19.6

21 Rectilinear Motion Rectilinear Motion Solution: Solution: At Point C: ) ( a 37) 9.81)( ( ) ( C B C C B C a a c ) ( / 8.1 / m m c c ) ( c

22 Rectilinear Motion Erratic Motion: When a particle moe in erratic motion, it can be bet decribed graphically by a erie of cure. A graph i ued to decribe the relationhip with any of the factor: a,,, t Recall kinematic equation: d dt a d dt dt adt

23 Rectilinear Motion Erratic Motion: The t, t and a t Graph When gien the t graph, we can contruct the t graph and a t graph, and ice era: Slope of t graph = ; d d a Slope of t graph = a; dt dt Area under a t graph = Area under t graph = dt adt

24 Rectilinear Motion Erratic Motion: The t, t and a t Graph General behaior of graph: Incline lope poitie Stagnant lope Decline lope negatie Poitie area increae lope Negatie area decreae lope

25 Rectilinear Motion Erratic Motion: Take an example of a bicycle moe along a traight road in the motion with uch that it poition i decribed by: 1 =.t 3 from time t = to t = 1; =.t + t from time t = 1 to t = ; 3 = 1t 1 from time t = to t = 3;

26 Rectilinear Motion Erratic Motion: With the gien information, a t graph can be contructed:

27 Rectilinear Motion Erratic Motion: By uing function = d/dt and a = d/dt: 3 d t 1;.t.6t a. 1t dt d 1 t ;. t t.4 t a. 4 dt d t 3; 1t 1 1 a dt

28 Rectilinear Motion Erratic Motion: The t graph of the bicycle motion:

29 Rectilinear Motion Erratic Motion: The a t graph of the bicycle motion:

30 Rectilinear Motion Erratic Motion: Comparion between t, t and a t graph:

31 Curilinear Motion Introduction: Curilinear occur when a particle i moing along a cured path. Poition i meaured from a fixed point O, by the poition ector r = r(t) Diplacement, r r O Poition, r Path,

32 Curilinear Motion Introduction: Diplacement Δr repreent the change in the particle poition. r' r r Diplacement, r r O Poition, r Path,

33 Curilinear Motion Introduction: Aerage elocity i defined a: ag r t Intantaneou elocity i found when Δt: dr dt Velocity, r O

34 Curilinear Motion Introduction: The aerage and intantaneou acceleration are: a ag t d dt d dt a act tangent to the hodograph and i not tangent to the path, a r a Acceleration Path,

35 Curilinear Motion Projectile Motion Projectile launched at (x, y ) and path i defined in the x y plane where y axi i the ertical axi. Air reitance i neglected The only force exit i the weight downward Projectile acceleration alway act ertically Contant acceleration: a 981m/ c = g = 9.81 a x = ; a y = g = 9.81 m/

36 Curilinear Motion Projectile Motion Horizontal motion: Since a x =, act; 1 ct x x t a ac ( ; ); x ( ) x x x ( x ( ) x ) x t

37 Curilinear Motion Projectile Motion Vertical motion: Poitie y axi i upward, we take a y = g act; 1 ct y y t a ac ( y y ; ); y ( ) gt y 1 y y ( ) yt gt ) g( y y y ( y )

38 Curilinear Motion Projectile Motion A cyclit jump off the 5 o lope track at 5m height, and with the peed m/. Calculate the time (t) that the cyclit i flying in the air, and the ditance (D) from point A when he landed on the ground. y = m/ B A 5 o 5 m h x C D

39 Curilinear Motion Projectile Motion Solution: Vertical motion: y in ()in m / 1 y y ( ) yt gt t (9.81) t 4.95 t 8.45 t 5

40 Curilinear Motion Projectile Motion Solution: Uing mathematicalolution (take poitie anwer): 4.95t 8.45t 5 b t t. 19 b a 4ac ( 8.45) 8.45 (4.95) 4(4.95)( 5)

41 Curilinear Motion Projectile Motion Solution: Horizontal motion: x co ()co m / x C x ) t D xc (18.13)(.19) 39.7 m C A ( Ax

42 Curilinear Motion Planar Circular Motion: Normal & Tangential Component When a particle moe in planar circular motion, the path of motion can be decribed uing n and t coordinate, which h act normal and tangent to the path. O O n u n t u t

43 Curilinear Motion Planar Circular Motion: Normal & Tangential Component Velocity: Particle elocity ha direction that i alway tangent to the path d dt ut

44 Curilinear Motion Planar Circular Motion: Normal & Tangential Component Acceleration: the timerate of change of elocity a a a u t where t a u a t n t u t u n or a d t d and a n R Magnitude: a a t a n

45 Curilinear Motion Planar Circular Motion: Starting from ret, a motor trael around a circular path of R = 3 m at a peed that increae with time: =.5t m/. Find the magnitude of the boat elocity and acceleration at the time t = 3. R = 3m O =.5t

46 Curilinear Motion Planar Circular Motion: Solution: 1) Calculate the elocity at t = 3 The magnitude i gien by: =.5t m/. At t = 3: =.5t =.5(3) =.5 m/

47 Curilinear Motion Planar Circular Motion: Solution: ) Calculatethe the tangential and normalcomponent of acceleration and then the magnitude of the acceleration ector. Tangential Component: d a t.5 t. 5t dt At t = 3: a t =.5(3) = 1.5 m

48 Curilinear Motion Planar Circular Motion: Solution: ) Normal Component: a n At t = 3: a n.5t 3.5(3) 3.169m /

49 Curilinear Motion Planar Circular Motion: Solution: acceleration ector i: a a u t t a n u n u t u n Magnitude of acceleration: a a t a n m /

50 Curilinear Motion Curilinear Motion Angular motion Angular motion Angular motion equation: ; 1 ; t a t y y t a c ; 1 ; t t t ); ( ; y y a t a t y y c c ; ; t t ( ) y y c

51 Problem P1 A poition of a particle i gien by = 1 + 6t t 3, where i ditance in meter and t i time in econd. Determine: a) elocity of the particle when the particle i at 17 m. b) maximum elocity of the particle. c) acceleration of the particle when it top momentary. d) poition of the particle when the elocity i maximum, and e) traelling ditance by the particle in 3 econd.

52 Problem P A particle trael along a traight line with a elocity = (1 3t ) m/, where t i in econd. When t = 1, the particle i located 1 m to the left of the origin. a) Determine the acceleration when t = 4. b) Calculate the diplacement from t = to t = 1. c) Analyze and the ditance the particle trael during thi time period. d) Sketch the moement of the particle.

53 Problem P3 A ball i thrown upward from the ground. The initial elocity i 15 m/. Calculate: The height and itelocity after econd. The maximum height that the ball reached.

54 Problem P4 The t graph of a car while traeling along a road i The t graph of a car while traeling along a road i hown in following Figure. Draw the t and a t graph for the motion.

55 Problem P5 A projectile launched frompoint A with an initial elocity of 15 m/ at an angle of to the horizontal axi. The projectile pae oer the peak of a hill (point B) at a ertical height of 7 m aboe point A, and fall to the ground at point C which i m ertically bl below point ta. When paing point tb the ertical component of the projectile elocity i upward.

56 Problem P5 (cont.) a) Determine the angle,. b) The elocity of the projectile when paing point B. c) The horizontal ditance,.

57 Problem 6 A car race around a horizontal circular track with a A car race around a horizontal circular track with a radiu of 8 m. Starting from ret, the car increae it peed at a contant rate of m/. Find the time (t) needed for it to reach an acceleration of 3 m/. What i it peed at thi intant?

58 The End