Upcoming SlideShare
×

# slide kinematics of particle - dynamics

7,763 views
7,269 views

Published on

UMP

Published in: Education, Technology
1 Comment
15 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• thank you

Are you sure you want to  Yes  No
Your message goes here
Views
Total views
7,763
On SlideShare
0
From Embeds
0
Number of Embeds
14
Actions
Shares
0
288
1
Likes
15
Embeds 0
No embeds

No notes for slide

### slide kinematics of particle - dynamics

1. 1. BAA1113 Engineering Mechanics Chapter 8 KINEMATICS of a PARTICLE
2. 2. Chapter 8 KINEMATICS of a PARTICLE Topics 8.1 Rectilinear Kinematics: Continuous/Erratic motion 8.2 CO Curvilinear Motion: Normal and Tangential Component In Your Text Book (R.C. Hibbeler) 12.1 – 12.3 *CO3 12.4 – 12.7 * Analyze the kinematics of motion that involves force & acceleration and work & energy principle.
3. 3. Lesson Outcomes • Describe the concept of position, displacement, velocity and acceleration. • Analyze particle motion along a straight line and represent the motion graphically. • Analyze particle motion along a curved path using different coordinate systems.
4. 4. APPLICATIONS A sports car travels along a straight road. Can we treat the car as a particle? If the car accelerates at a constant rate, how can we determine its position and velocity at some instant?
5. 5. An Overview of Mechanics Mechanics: The study of how bodies react to forces acting on them. Statics: The study of bodies in equilibrium. Dynamics: 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics - concerned with the forces causing the motion
6. 6. RECTILINEAR KINEMATICS: CONTINIOUS MOTION A particle travels along a straight-line path defined by the coordinate axis s. The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft). The displacement of the particle is defined as its change in position. Vector form:  r = r’ - r Scalar form:  s = s’ - s The total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.
7. 7. VELOCITY Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s. The average velocity of a particle during a time interval t is vavg = r / t The instantaneous velocity is the time-derivative of position. v = dr / dt Speed is the magnitude of velocity: v = ds / dt Average speed is the total distance traveled divided by elapsed time: (vsp)avg = sT / t
8. 8. ACCELERATION Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2. The instantaneous acceleration is the time derivative of velocity. Vector form: a = dv / dt Scalar form: a = dv / dt = d2s / dt2 Acceleration can be positive (speed increasing) or negative (speed decreasing). As the book indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv
9. 9. SUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds • Integrate acceleration for velocity and position. Position: Velocity: v t v s  dv =  a dt or  v dv =  a ds s t  ds =  v dt vo o vo so so o • Note that so and vo represent the initial position and velocity of the particle at t = 0.
10. 10. CONSTANT ACCELERATION The three kinematic equations can be integrated for the special case when acceleration is constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2 downward. These equations are: v t  dv =  a dt c vo s v yields s = s o + v o t + (1/2) a c t2 yields = (v o) 2 + 2ac (s - so ) v t o s  v dv =  ac ds vo v = v o + ac t o  ds =  v dt so yields so 2
11. 11. Example 12.1 The car moves in a straight line such that for a short time its velocity is defined by v = (0.9t2 + 0.6t) m/s where t is in sec. Determine it position and acceleration when t = 3s. When t = 0, s = 0.
12. 12. Solution Coordinate System Positive is to the right. Position When s = 0 when t = 0, we have  +      ds v=  = 0 .9 t + 0 .6 t dt 2  When t = 3s, s = 10.8m  s 0 ds =  0 . 9 t t 0 2  + 0 . 6 t dt  s s 0  = 0 .3t + 0 .3t 3 2  t = 0 .3t + 0 .3t 3 0 2
13. 13. Solution Acceleration Knowing v = f(t), the acceleration is determined from a = dv/dt, since this equation relates a, v and t. a= dv dt = d dt 0 .9 t When t = 3s, a = 6m/s2 2  + 0 .6 t = 1 .8t + 0 .6
14. 14. Example 12.3 A rocket travel upward at 75m/s. When it is 40m from the ground, the engine fails. Determine max height sB reached by the rocket and its speed just before it hits the ground.
15. 15. Solution Coordinate System Origin O with positive upward. Maximum Height We have: t = 0 vA = +75m/s, s = sB  vB = 0, aC = -9.81m/s2 v = v + 2 a C ( s B - s A )  s B = 327 m 2 B 2 A
16. 16. Solution Velocity The negative root was chosen since the rocket is moving downward. vC = v B + 2 a C ( sC - s B ) 2 2 v C = - 80 . 1 m / s = 80 . 1 m / s 
17. 17. Rectilinear Kinematics: Erratic Motion
18. 18. APPLICATION In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the tank truck, how can we determine its acceleration at position s = 1500 feet?
19. 19. ERRATIC MOTION Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics. The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve.
20. 20. S-T GRAPH Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt). Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph.
21. 21. V-T GRAPH Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv/dt). Therefore, the acceleration vs. time (or a-t) graph can be constructed by finding the slope at various points along the v-t graph. Also, the distance moved (displacement) of the particle is the area under the v-t graph during time t.
22. 22. A-T GRAPH Given the acceleration vs. time or a-t curve, the change in velocity (v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle.
23. 23. A-S GRAPH A more complex case is presented by the acceleration versus position or a-s graph. The area under the a-s curve represents the change in velocity (recall  a ds =  v dv ). ½ (v1² – vo²) =  s2 a ds s1 = area under the a-s graph This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance.
24. 24. V-S GRAPH Another complex case is presented by the velocity vs. distance or v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. Recall the formula a = v (dv/ds). Thus, we can obtain an a-s plot from the v-s curve.
25. 25. Example 12.6 A bicycle moves along a straight road such that it position is described by the graph as shown. Construct the v-t and a-t graphs for 0 ≤ t ≤ 30s.
26. 26. Solution v-t Graph By differentiating the eqns that defining the s-t graph, we have 0  t  10 s ; s = 0 .3t v= 2 ds = 0 .6 t dt 10 s  t  30 s ; s = 6 t - 30 v= ds =6 dt By measuring the slope of the s-t graph at a given time instant, t = 20 s; v= s t = 150 - 30 30 - 10 = 6m / s
27. 27. Solution a-t Graph By differentiating the eqns defining the lines of the v-t graph, 0  t  10 s ; v = 0 .6 t a = dv = 0 .6 dt 10  t  30 s ; v = 6 a= dv dt =0
28. 28. Example 12.7 A test car starts from rest and travels along a straight track such that it accelerates at a constant rate for 10 s and then decelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t’ needed to stop the car. How far has the car traveled?
29. 29. Solution v-t Graph Using initial condition v = 0 when t = 0, 0  t  10 s a = 10 ;  v dv = 0  t 10 dt , v = 10 t 0 When t = 10s, v = 100m/s, 10 s  t  t ; a = - 2;  v dv = 100 When t = t’, v = 0  t’ = 60 s  t 10 - 2 dt , v = - 2 t + 120
30. 30. Solution s-t Graph. Using initial conditions s = 0 when t = 0, 0  t  10 s ; v = 10 t ;  s ds = 0  t 10 t dt , s = 5t 2 0 When t = 10s, s = 500m, 10 s  t  60 s ; v = - 2 t + 120 ;  s ds = 500 s = - t + 120 t - 600 2 When t’ = 60s, s = 3000m   - 2 t + 120 dt t 10
31. 31. Example 12.8 The v-s graph describing the motion of a motorcycle as shown. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 120 m.
32. 32. Solution a-s Graph a-s graph can be determined using a ds = v dv, 0  s  60 m ; a=v v = 0 .2 s + 3 dv = 0 . 04 s + 0 . 6 ds 60 m  s  120 m ; a=v dv ds v = 15 ; =0
33. 33. Solution Time When s = 0 at t = 0, 0  s  60 m ; v = 0 . 2 s + 3; dt = ds = v  dt = o s ds 0 t 0 .2 s + 3  ds 0 .2 + 3  t = 5 ln( 0 . 2 s + 3 ) - 5 ln 3 At s = 60 m, t = 8.05 s For second segment of motion, 60  s  120 m ; v = 15 ; dt = ds v  t 8 . 05 dt = s ds 60 15   t= s 15 At s = 120 m, t = 12.05 s + 4 . 05 = ds 15
34. 34. General Curvilinear Motion
35. 35. APPLICATIONS The path of motion of a plane can be tracked with radar and its x, y, and z coordinates (relative to a point on earth) recorded as a function of time. How can we determine the velocity or acceleration of the plane at any instant?
36. 36. APPLICATIONS A roller coaster car travels down a fixed, helical path at a constant speed. How can we determine its position or acceleration at any instant? If you are designing the track, why is it important to be able to predict the acceleration of the car?
37. 37. GENERAL CURVILINEAR MOTION A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are used to describe the motion. A particle moves along a curve defined by the path function, s. The position of the particle at any instant is designated by the vector r = r(t). Both the magnitude and direction of r may vary with time. If the particle moves a distance s along the curve during time interval t, the displacement is determined by vector subtraction:  r = r’ - r
38. 38. VELOCITY Velocity represents the rate of change in the position of a particle. The average velocity of the particle during the time increment t is vavg = r/t . The instantaneous velocity is the time-derivative of position v = dr/dt . The velocity vector, v, is always tangent to the path of motion. The magnitude of v is called the speed. Since the arc length s approaches the magnitude of r as t→0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector!
39. 39. ACCELERATION Acceleration represents the rate of change in the velocity of a particle. If a particle’s velocity changes from v to v’ over a time increment t, the average acceleration during that increment is: aavg = v/t = (v - v’)/t The instantaneous acceleration is the timederivative of velocity: a = dv/dt = d2r/dt2 A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.
40. 40. CURVILINEAR MOTION: RECTANGULAR COMPONENTS It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference. The position of the particle can be defined at any instant by the position vector r=xi+yj+zk . The x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t) . The magnitude of the position vector is: r = (x2 + y2 + z2)0.5 The direction of r is defined by the unit vector: ur = (1/r)r
41. 41. RECTANGULAR COMPONENTS: VELOCITY The velocity vector is the time derivative of the position vector: v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt Since the unit vectors i, j, k are constant in magnitude and direction, this equation reduces to v = vx i + vy j + vz k • • • x = dx/dt, vy = y = dy/dt, vz = z = dz/dt where vx = The magnitude of the velocity vector is v = [(vx)2 + (vy)2 + (vz)2]0.5 The direction of v is tangent to the path of motion.
42. 42. RECTANGULAR COMPONENTS: ACCELERATION The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): a = dv/dt = d2r/dt2 = ax i + ay j + az k • • •• •• vx = x = dvx /dt, ay = vy = y = dvy /dt, where ax = •• • az = vz = z = dvz /dt The magnitude of the acceleration vector is a = [(ax)2 + (ay)2 + (az)2 ]0.5 The direction of a is usually not tangent to the path of the particle.
43. 43. Example 12.9 At any instant the horizontal position of the weather balloon is defined by x = (9t) m, where t is in second. If the equation of the path is y = x2/30, determine the distance of the balloon from the station at A, the magnitude and direction of the both the velocity and acceleration when t = 2 s.
44. 44. Solution Position When t = 2  x = 9(2)= 18 m, y = (18)2/30 = 10.8m The straight-line distance from A to B is r = 18 2 + 10 . 8 2 = 21 m Velocity We have  vx = x =  vy = y = d dt d dt 9 t  = 9 m / s  x 2  / 30 = 10 . 8 m / s 
45. 45. Solution Velocity When t = 2 s, the magnitude of velocity is v= 9 2 + 10 . 8 2 = 14 . 1m / s The x is tangent to the path is  v = tan Acceleration We have  a x = v x = 0 and thus, a = 2  a y = v y = 5 .4 m / s  0 2 + 5 . 4 2  a = tan -1 5 .4 0 = 90 = 5 .4 m / s  2 -1 vy vx = 50 . 2 
46. 46. Example 12.10 For a short time, the path of the plane in the figure is described by y = (0.001x2)m. If the plane is rising with a constant velocity of 10m/s , determine the magnitudes of the velocity and acceleration of the plane when it is at y= 100 m
47. 47. Solution When y = 100m x = 316.2m When vy = 10m/s  t = 10s Velocity Using the chain rule,  vy = y = d dt 0 .001 x  = 0 .002 xv 2 x 10 = 0 . 002 316 . 2 v x  v x = 15 . 81 m/s The magnitude is v= v x + v y = 18 . 7 m/s 2 2
48. 48. Solution Acceleration Using the chain rule,  2     a = v y = 0 . 002 x v x + 0 . 002 x v x = 0 . 002 v x + xa x  When x = 316.2m, vx = 15.81m/s,  vy = ay = 0   + 316 .2 a  a = - 0 .791 m/s Magnitude of the plane’s acceleration is 0 = 0 . 002 15 . 81 2 a= 2 x x a x + a y = 0 . 791 m/s 2 2 2
49. 49. Motion of a Projectile
50. 50. APPLICATIONS A good kicker instinctively knows at what angle, , and initial velocity, vA, he must kick the ball to make a field goal. For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?
51. 51. APPLICATIONS A basketball is shot at a certain angle. What parameters should the shooter consider in order for the basketball to pass through the basket? Distance, speed, the basket location, … anything else ?
52. 52. APPLICATIONS A firefighter needs to know the maximum height on the wall she can project water from the hose. What parameters would you program into a wrist computer to find the angle, , that she should use to hold the hose?
53. 53. MOTION OF A PROJECTILE Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., from gravity). For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant.
54. 54. KINEMATIC EQUATIONS: HORIZONTAL MOTION Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by: x = xo + (vox) t Why is ax equal to zero (assuming movement through the air)?
55. 55. KINEMATIC EQUATIONS: VERTICAL MOTION Since the positive y-axis is directed upward, ay = – g. Application of the constant acceleration equations yields: vy = voy – g t y = yo + (voy) t – ½ g t2 vy2 = voy2 – 2 g (y – yo) For any given problem, only two of these three equations can be used. Why?
56. 56. Example 12.11 A sack slides off the ramp with a horizontal velocity of 12 m/s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where the sacks begin to pile up.
57. 57. Solution Coordinate System Origin at point A. Initial velocity is (vA)x = 12 m/s and (vA)y = 0m/s Acceleration between A and B is ay = -9.81 m/s2. Since (vB)x = (vA)x = 12 m/s, the 3 unknown are (vB)y, R and tAB
58. 58. Solution Vertical Motion Vertical distance from A to B is known, +   y B = y A + ( v A ) y t AB + 1 2 a c t AB  t AB = 1 . 11 s 2 Horizontal Motion  +     x = x A + ( v A ) x t AB   R = 0 + 12 11 . 1  R = 13 . 3 m
59. 59. Example 12.12 The chipping machine is designed to eject wood at chips vO = 7.5 m/s. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube.
60. 60. Solution Coordinate System 3 unknown h, tOA and (vB)y. Taking origin at O, for initial velocity of a chip, ( v O ) x = ( 7 . 5 cos 30 ) = 6 . 5 m / s   ( v O ) y = ( 7 . 5 sin 30 ) = 3 . 75 m / s   Also, (vA)x = (vO)x = 6.5 m/s and ay = -9.81 m/s2
61. 61. Solution Horizontal Motion x A = x 0 + ( v 0 ) x t OA  t OA = 0 . 9231 s Vertical Motion Relating tOA to initial and final elevation of the chips, y A = h - 2 . 1 = y O + ( v 0 ) y t OA + h = 1 . 38 m 1 2 2 a c t OA
62. 62. Curvilinear Motion: Normal and Tangential Components
63. 63. APPLICATIONS Cars traveling along a clover-leaf interchange experience an acceleration due to a change in velocity as well as due to a change in direction of the velocity. If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration? Why would you care about the total acceleration of the car?
64. 64. APPLICATIONS A roller coaster travels down a hill for which the path can be approximated by a function y = f(x). The roller coaster starts from rest and increases its speed at a constant rate. How can we determine its velocity and acceleration at the bottom? Why would we want to know these values?
65. 65. NORMAL AND TANGENTIAL COMPONENTS When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle). The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion. The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.
66. 66. NORMAL AND TANGENTIAL COMPONENTS The positive n and t directions are defined by the unit vectors un and ut, respectively. The center of curvature, O’, always lies on the concave side of the curve. The radius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point. The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.
67. 67. VELOCITY IN THE n-t COORDINATE SYSTEM The velocity vector is always tangent to the path of motion (t-direction). The magnitude is determined by taking the time derivative of the path function, s(t). v = v ut where v = .s = ds/dt Here v defines the magnitude of the velocity (speed) and ut defines the direction of the velocity vector.
68. 68. ACCELERATION IN THE n-t COORDINATE SYSTEM Acceleration is the time rate of change of velocity: . . a = dv/dt = d(vut)/dt = vut + vut . Here v represents the change in . the magnitude of velocity and ut represents the rate of change in the direction of ut. After mathematical manipulation, the acceleration vector can be expressed as: . a = v ut + (v2/r) un = at ut + an un.
69. 69. ACCELERATION IN THE n-t COORDINATE SYSTEM So, there are two components to the acceleration vector: a = at ut + an un • The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity. . at = v or at ds = v dv • The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r • The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5
70. 70. SPECIAL CASES OF MOTION There are some special cases of motion to consider. 1) The particle moves along a straight line. . 2/r = 0 r  => an = v => a = at = v The tangential component represents the time rate of change in the magnitude of the velocity. 2) The particle moves along a curve at constant speed. . at = v = 0 => a = an = v2/r The normal component represents the time rate of change in the direction of the velocity.
71. 71. SPECIAL CASES OF MOTION 3) The tangential component of acceleration is constant, at = (at)c. In this case, s = so + vo t + (1/2) (at)c t2 v = vo + (at)c t v2 = (vo)2 + 2 (at)c (s – so) As before, so and vo are the initial position and velocity of the particle at t = 0. How are these equations related to projectile motion equations? Why? 4) The particle moves along a path expressed as y = f(x). The radius of curvature, r, at any point on the path can be calculated from [ 1 + (dy/dx)2 ]3/2 r = ________________ d2y/dx 2
72. 72. Example 12.14 When the skier reaches the point A along the parabolic path, he has a speed of 6m/s which is increasing at 2m/s2. Determine the direction of his velocity and the direction and magnitude of this acceleration at this instant. Neglect the size of the skier in the calculation.
73. 73. Solution Coordinate System Establish n, t axes at the fixed point A. Determine the components of v and a. Velocity The velocity is directed tangent to the path where y = 1 2 x , 20 dy dx =1 x = 10 V make an angle of θ = tan-1 = 45° with the x axis vA = 6m / s
74. 74. Solution Acceleration Since 2 d y dx 2 = 1 ,the radius of curvature is 10 2 3/2 r = [1 + ( dy / dx ) ] 2 d y / dx 2 = 28 . 28 m The acceleration becomes 2 aA v  = vut + un  r   2 = { 2 u t + 1 . 273 u n } m / s
75. 75. Solution Acceleration The magnitude and angle is a=  2 2 + 1 . 237 2  = tan -1 2 1 . 237 Thus, 57.5° – 45 ° = 12.5 ° a = 2.37 m/s2 = 2 . 37 m / s = 57 . 5  2
76. 76. Example 12.15 Race car C travels round the horizontal circular track that has a radius of 90 m. If the car increases its speed at a constant rate of 2.1 m/s2, starting from rest, determine the time needed for it to reach an acceleration of 2.4 m/s2. What is its speed at this instant?
77. 77. Solution Coordinate System The origin of the n and t axes is coincident with the car at the instant. Acceleration The magnitude of acceleration can be related to its 2 components using a = a t2 +,a n thus a t = 2.1 m /s The velocity is v = v We have an = v 0 + ( a t ) c t = 2.1t 2 r 2 = 0 . 049 t m / s 2
78. 78. Solution Acceleration The time needed for the acceleration to reach 2.4m/s2 is a= 2.4 = at + a n 2 2  2.4  2 +  0.049 t 2  2  t = 4.87 s The speed at time t = 4.87s is v = 2 . 1t = 10 . 2 m / s
79. 79. END