Kinematics of Particles. Chapter 2

Transcription

1 Kinematics of Particles Chapter 2

2 Introduction Kinematics: is the branch of dynamics which describes the motion of bodies without reference to the forces that either causes the motion or are generated as a result of the motion. Kinematics is often referred to as the geometry of motion Examples of kinematics problems that engage the attention of engineers. The design of cams, gears, linkages, and other machine elements to control or produce certain desired motions, and The calculation of flight trajectory for aircraft, rockets and spacecraft.

3 If the particle is confined to a specified path, as with a bead sliding along a fixed wire, its motion is said to be Constrained. Example 1. - A small rock tied to the end of a string and whirled in a circle undergoes constrained motion until the string breaks

4 If there are no physical guides, the motion is said to be unconstrained. Example 2. - Airplane, rocket

5 Let s consider a particle and its path of travel The position of particle P at any time t can be described by specifying its: Rectangular coordinates; X,Y,Z Cylindrical coordinates; r,θ,z spherical coordinates; R, θ,ф Also described by measurements along the tangent t and normal n to the curve(path variable). The motion of particles(or rigid bodies) may be described by using coordinates measured from fixed reference axis (absolute motion analysis) or by using coordinates measured from moving reference axis (relative motion analysis).

6 Rectilinear Motion Is a motion in which a particle moving along a straight line(one-dimensional motion) Consider a particle P moving along a straight line. The position of P at any instant of time tis. The position of P at time t + t is s + s s

7 Displacement of P Change in position of the particle during the interval of t. The displacement would be negative if the particle moved in the negative s-direction Distance The total length of the path traced by the particle (it is always positive) Average velocity: ie. ( s+ s) s= s For the time interval Δt, it is defined as the ratio of the displacement Δs to the time interval Δt. s V av = t As Δt becomes smaller and approaches zero in the limit, the average velocity approaches the instantaneous velocity of the particle. s ds V = lim Vav = lim = = S (1) t 0 t 0 t dt

8 Average Acceleration For the time interval Δt, it is defined as the ratio of the change in velocity Δv to the time interval Δt. As Δt becomes smaller and approaches zero in the limit, the average acceleration approaches the instantaneous acceleration of the particle. OR a av = v t v dv a = lim = = v (2a) t 0 t dt 2 a lim v dv d ds d s = = = = = s (2b) t 0 2 t dt dt dt dt Note:-The acceleration is positive or negative depending on whether the velocity increasing or decreasing.

9 Considering equation 1 and 2a, we get the following ds v = dt and v a = t dt = ds v and ds = v dt dv a = dv a by equating the two equations OR ads sds = = vdv sdv (3a) (3b) Equation 1,2 and 3 are the differential equations for the rectilinear motion of a particle. Rectilinear motions are solved by integration of these basic differential relation.

10 Example

11 1. A particle a particle moving in a straight line, and assuming that its position is defined by the equation. 2 3 s = 6t -t Where, t is express in seconds and s is in meters. Determine the velocity and acceleration of the particles at any time t.

12 2. The acceleration of a particle is given by, a = 4t 30 where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as function time. The initial displacement at t=0 is s o =-5m, and the initial velocity is v o =30m/s.

13 Graphical Representation of Relationship Among s, v, and t

14 Graph of s Vs t By constructing tangent to the curve at any time t, we obtain the slope, the slope of the s-t curve at any time instant gives the instantaneous velocity v=ds/dt

15 Graph of v Vs t The slope the s-t curve at any instant gives the instantaneous acceleration. The area under the v-t curves is the net displacement of the particle during the The area under the v-t curve is the net displacement of the particle during the interval from t1 to t2. interval from t 1 to t 2. da = vdt A = t t 1 2 vdt ds v =, ds = vdt dt s2 t2 ds = vdt s 1 1 Equating the two equations gives A s A= s s 2 = ds, s t

16 Graph of a Vs t The area under the a-t curve during time dt is the net change in velocity of the particle between t1 and t2. The area under the a-t curve is the net change in velocity of the particles t 1 to t 2. da = adt t2 A = adt t 1 a dv, dt = adt = dv t t adt = v 2 2 v 1 1 dv Equating the two equations gives A = v v 1 2 dv A= v v 2 1

17 Graph of a Vs s The net area under the a-s curve can be found da = ads A = s s 1 2 ads v vdv v = vdv ads = s 2 2 s 1 1 ads Equating the two equations gives A = v v 1 2 vdv 2 2 V2 V1 A = 2

18 dv since, tanθ = ds 1 Similar Triangle CB tanθ = v dv CB = vdv = CBds, vdv = ads ds v Equating the two equations gives CBds = ads CB = a From the graph distance CB is acceleration

19 The graphical representations described are useful for:- visualizing the relationships among the several motion quantities. approximating results by graphical integration or differentiation. experimental data and motions that involve discontinuous relationship b/n variables are frequently analyzed graphically.

20 Methods for determining the velocity and displacement Function

21 Constant Acceleration At the beginning of the interval t = 0, s = s, v= v v v o o dv a = adt = dv dt t dv = adt v v = at 0 v = v + at o o o

22 using vdv = ads v v o v 2 v 2 vdv v 2 v vo 2 2 o = s s o = as ads s s o ( s ) = a s 2 2 o o v = v + 2a(s s ) 0

23 Using 0 ds v = dt ds = vdt s t o s o 2 ds = ( v + at) dt s s = vt+ o o at 2 1 s = s + v t + at o o 2 2

24 These relations are necessarily restricted to the special case where the acceleration is constant. The integration limits depend on the initial and final conditions and for a given problem may be different from those used here. Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations. Three classes of motion may be defined for: - acceleration given as a function of time, a = f(t) - acceleration given as a function of position, a = f(x) - acceleration given as a function of velocity, a = f(v)

25 Acceleration given as a function of time, a=f(t) dv a f ( t) dt = = dv = f ( t) dt ( ) ds dt = v ds v dt = s t ds = s0 0 v v f s ds Acceleration given as a function of displacement, a = f(s) v dv = ads v dv f ( s) ds s s = 0 = ( ) 0 vdt 0 = = ( ) ( ) v v f s ds s s 0 v v s s v dv f s ds 0 0 s = ( ) s 0 t v v f t dt s s vdt 0 t = v = v + f s ds 0

26 Acceleration given as a function of velocity, a=f(v) dv dv a f ( v) dt = = dt f ( v ) = ( ) v v dv f v = 0 0 t dt vdv = ads vdv = f () v ds v v 0 dv f v = t ( ) ds = v dv f v ( ) s s ds = v v 0 0 v dv f v ( ) s v s = 0 v 0 v dv f v ( )

27 Examples

28 1. The position of a particle which moves along a straight lines is defined by the relation Where S is expressed in m and t in second. Determine: 3 2 S t t t = a) The time which the velocity will be zero. b) The position and distance travelled by the particle at that time. c) The acceleration of the particle at that time. d) The distance travelled by the particle between 4 sec and 6 sec.

29 2. A girl rolls a ball up an incline and allows it to return to her. For the angle θ and ball involved, the acceleration of the ball along the incline is constant at 0.25g, directed down the incline. If the ball is released with a speed of 4 m/s, determine the distance s it moves up the incline before reversing its direction and the total time t required for the ball to return to the child s hand.

30 3.The main elevator A of the CN Tower in Toronto rises about 350 m and for most of its run has a constant speed of 22 km/h. Assume that both the acceleration and deceleration have a constant 1 magnitude of and determine the time duration t 4 g of the elevator run.

31 4. A sprinter reaches his top speed of 10.6m/s in t second from rest with essentially constant acceleration. If he maintains his speed and covers the 100m distance in 10.5 s, find the acceleration interval t and his average starting acceleration a.

32 5. A motorcycle patrolman starts from rest at A two seconds after a car, speeding at the constant rate of 120 km/h, passes point A. If the patrolman accelerates at the rate of 6 m/s 2 until he reaches his maximum permissible speed of 150 km/h, which he maintains, calculate the distance s from point A to the point at which be overtakes the car.

33 6. When the effect of aerodynamic drag is included, the y-acceleration of a baseball moving vertically upward is, while the acceleration when the ball is moving downward is, where k is a positive constant and v is the speed in feet per second. If the ball is thrown upward at 30 m/sec from essentially ground level, compute its maximum height h and its speed upon impact with the ground. Take k to be.0066m -1 and assume that g is constant.

34 Plane Curvilinear Motion

35 The motion of a particle along a curved path that lies on a single plane. Consider the continuous motion of a particle along a plane curve. At time t, the particle is at position A, which is located by the position vector r measured from some convenient fixed origin o. At time t + t, the particle is at A located by the position vector r + r. The vector Δr joining A and A represents the change in the position vector during the time interval Δt (displacement). The distance actually traveled by the particle as it moves along the path from A to A is the scalar s.

36 The average velocity of the particle between A and A defined as: The average speed of the particle between A and A defined as: The instantaneous velocity, v V av av r = t s = t r dr v = lim = t = r 0 t dt As t approaches zero, the direction of r approaches to the tangent of the path. Hence the velocity V is always a vector tangent to the path. ds V = v = = s dt

37 Average acceleration, of the particle between A and A Instantaneous acceleration, a av v = t a v dv a = lim = = v t 0 t dt Note: The direction of the acceleration of a particle in curvilinear motion is neither tangent to the path nor normal to the path.

38 For curvilinear motion of a particle in a plane there are three different coordinate systems. Rectangular coordinate (x, y, z) Normal and tangential coordinate (n-t) Polar coordinates (r-ɵ)

39 Rectangular coordinates (x-y) The resulting curvilinear motion is obtained by a vector combination of the x- and y-components of the position vector, the velocity, and the acceleration. Consider the following figure We may write the vectors in terms of their x and y components. r = xi + yj v = r = xi + yj = v x + vy a = v = r = xi + yj = a x + ay r, v and a

40 The direction of the velocity is always tangent to the path The magnitude of the motion vectors 2 2 r = x + y v= v + v 2 2 x y a = a + a 2 2 x y θ = tan 1 v v y x

41 Projectile motion Neglect the aerodynamic drag, the earth curvature and rotation, The altitude range is so small enough so that the acceleration due to gravity can be considered constant, therefore;

42 ax = 0 a y = g at t=0 ; x=x 0,y=y 0 ; vx=vxo and vy=vy0 Position x = x + 0 vx0t 2 1 y = y 0 + v y0 t gt 2 Velocity v x = ( v x ) 0 v v y 2 y = vy 0 gt 2 = v 2g( y y ) yo o

43 Normal and Tangential Coordinate(n-t) The n and t coordinates are considered to move along the path 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.

44 The coordinate n and t will now be used to describe the velocity v and acceleration a. Similarly to the unit vectors i and j introduced for rectangular coordinate system, unit vectors for t-n coordinate system can be used. For this purpose we introduce unit vector e t in the t-direction e n in the n-direction. e t - directed toward the direction of motion. e n -directed toward the center of curvature of the path.

45 During the differential increment of time dt, the particle moves a differential distance ds along the curve from A to A. With the radius of curvature of the path at this position designated by ρ, we see that Velocity ds = ρdβ, β is in radians ds ρdβ dβ v = v = = = ρ dt dt dt Since it is unnecessary to consider the differential change in ρ between A and A, v = ve = ρβ e...(1) t t

46 Acceleration The acceleration is a vector which reflects both the change in magnitude and the change in direction of v. dv a = = dt d dt ( ) ve t Now differentiate the velocity by applying the ordinary rule (chain rule) for the differentiation of the product of a scalar and a vector. a = a = dv dt v et = + d dt v et ( ve ) t = dv dt e t de + v dt t Where the unit vector e t now has a derivative because its direction changes.

47 To find the derivative of consider the following figure Using vector addition e t = e t + e t Since the magnitude e t = e t = 1 The direction of e t and e t are different det = et dθ = dθ The direction of de t is given by e n de = de e = dθe t t n n de t n dθ = e Or dividing by dt det dt de t dt dθen = dt e t = θe n

48 = v et + v et a a = at + an a t change in direction of velocity a n change in magnitude of velocity a = ve = v θe n t n v θ = = ρθ v ρ v v a = v e = e ρ ρ n n n 2 2 v a = en + ve ρ 2 v 2 an = = ρθ = v θ v= ρθ ρ a = v = s t a= a + a 2 2 n t t

49 Note: a n is always directed towards the center of curvature of the path. a t is directed towards the positive t-direction of the motion if the speed v is increasing and towards the negative t-direction if the speed v is decreasing. At the inflection point in the curve, the 2 v normal acceleration, goes to zero since ρ ρ becomes infinity.

50 Special case of motion Circular motion 2 v a n = but ρ=r and ρ 2 a n = rθ dv d at = = rθ = r dt dt a a t = rθ = rθ e t + rθ 2 e n v = rθ d θ dt

51 The particle moves along a path expressed as y = f(x). The radius of curvature,ρ, at any point on the path can be calculated from ρ xy = dy dx 1 + ( 2 d y 2 ) dx 2 3 2

52 Example

53 1. For a certain interval of motion, the pin P is forced to move in the fixed parabolic slot by the vertical slotted guide, which moves in the x - direction at the constant rate of 20 mm/s. All measurements are in millimeters and seconds. Calculate the magnitudes of the velocity v and acceleration a of pin P when x = 60mm.

54 2. If the tennis player serves the ball with a velocity v of 130km/h at the angle θ=5 0, calculate the vertical clearance h of the center of the ball above the net and the distance s from the net where the ball hits the court surface. Neglect air resistance and the effect of ball spin.

55 3. A projectile is launched with an initial speed of 200 m/s at an angle of 60 0 with respect to the horizontal. Compute the range R as measured up the incline.

56 4. A particle is ejected from the tube at A with a velocity v at an angle θ with the vertical y-axis. A strong horizontal acceleration a in the x-direction. If the particle strikes the ground at a point directly under its released position, determine the height h of point A. The downward y-acceleration may be taken as the constant g.

57 5. The motion of the pin A in the fixed circular slot is controlled by the guide B, which is being elevated by its lead screw with a constant upward velocity v o =2 m/s for an interval of its motion. Calculate both the normal and tangential components of acceleration of pin A as its passes the position for which θ=30 0.

58 6. The pin P is constrained to move in the slotted guides which move at right angles to one another. At the instant represented, A has a velocity to the right of 0.2 m/s which is decreasing at the rate of 0.75 m/s each second. At the same time, B is moving down with a velocity of 0.15 m/s which is decreasing at the rate of 0.5 m/s each second. For this instant determine the radius of curvature ρ of the path followed by P. Is it possible to also determine the time rate of change of ρ?

59 Curvilinear Motion Polar coordinate system (r- ѳ)

60 Polar coordinate(r- ѳ) Where the particle is located by the radial distance r from a fixed pole and by an angular measurement ѳ to the radial line. Polar coordinates are particularly useful when a motion is constrained through the control of a radial distance and an angular position, or when an unconstrained motion is observed by measurements of a radial distance and an angular position.

61 An arbitrary fixed line, such as the x- axis, is used as a reference for the measurement ѳ. Unit vectors e r and e ѳ are established in the positive r and ѳ directions, respectively. The position vector to the particle at A has a magnitude equal to the radial distance r and a direction specified by the unit vector e r. We express the location of the particle at A by the vector r = re r

62 During time dt the coordinate direction rotate through the angle dθ and the unit vectors also rotates through the same ' ' angle from e to e and e to e der = er dθ = dθ der = der e θ der = dθe θ der dθ = e θ dt dt e r = θ e θ r r θ θ deθ = eθ dθ = dθ deθ = deθ. eθ deθ = dθe r deθ dθ = er dt dt e θ = θe r

63 Velocity The velocity is obtained by differentiating the vector r v = dr dt v = re + re d = ( rer ) = re r + re dt r r, but er v= ve + ve v = rer + rθ e θ r r θ θ = θ e θ r vr v θ = r = r 2 2 θ v = v vr + v θ α 1 = tan θ v r

64 Acceleration Differentiating the expression for v to obtain the acceleration a. dv d a = v = = ( re r + r θeθ ) dt dt a = re r + re r + r θe θ + r θe θ + r θe θ But e θ = θer e r = θe θ a = re r + r θe θ + r ( θe θ) + r θe θ + r θ( θe r) 2 a = re r + r θe θ + r θe θ + r θe θ r θ e r 2 a = ( r r θ ) e r + (2 r θ + r θ) e r θ a = ar + a 2 θ i.e ar = r r θ 2 2 aθ = 2r θ + r a a a θ = 1 a r + θ φ = tan θ a r

65 For circular motion For circular motion r is constant v= ve + ve r vr v θ r θ = r = = r θ v = v = r θe θ θ θ 0 N.B This description is the same as that obtained with n and t components, but the positive r-direction is in the negative n-direction Hence a r =-a n for circular motion a = ar + a θ 2 a = ( r rθ ) er + (2 rθ + rθ) er θ 2 a = rθ er + rθer θ

66 Example

67 1. The cam is designed so that the center of the roller A which follows the contour moves on a limaçon defined by r = b ccosθ, where b> c. If the cam does not rotate, determine the magnitude of the total acceleration of A in terms of θ if the slotted arm revolves with a constant counterclockwise angular rate θ = ω.

68 2. The fixed horizontal guide carries a slider and pin P whose motion is controlled by the rotating slotted arm OA. If the arm is revolving about O at the constant rate θ = 2 rad/s for an interval of its designed motion, determine the magnitudes of the velocity and acceleration of the slider in the slot for the instant when θ = 60. Also find the r-components of the velocity and acceleration.

69 3. During a portion of a vertical loop, an airplane flies in an arc of radius ρ = 600m with a constant speed v = 400 km / s. When the airplane is at A, the angle made by v with the horizontal is β = 30, o and radar tracking gives r = 800m and θ = 30 o.calculate v, v, a and θ for this instant. r θ r

70 Relative Motion (Translating Axes) If the motion analysis of a particle using coordinates which are referred to a fixed reference system is said to be absolute motion analysis. If the motion analysis of a particle using coordinates which are referred to a moving reference system is said to be relative motion analysis. Relative motion analysis : is the motion analysis of a particle using moving reference system coordinate in reference to fixed reference system. In this portion we will confine our attention to:- moving reference systems that translate but do not rotate. The relative motion analysis is limited to plane motion.

71 Consider two particles A and B that may have separate curvilinear motion in a given plane or in parallel planes. X,Y : inertial frame of reference x,y : translating coordinate system We will arbitrarily attach the origin of a set of translating (non-rotating) axes x-y to particle B and observe the motion A from our moving position B. The position vector A as measured relative to the frame x-y is rab / = xi + yj Where A/B means A relative to B

72 The absolute position of A is determined by the vector Velocity Acceleration r = r + r A B AB / dra d va= = ( rb+ rab / ) dt dt r = r + r v = v + v A B AB / A B AB / dva d aa= = ( vb+ vab / ) dt dt v = v + v A B AB / a = a + a A B AB /

73 Note:- The observed coordinates system can be Rectangular Normal and tangential Polar Therefore, the choice of any coordinates system depends on the type of given conditions for the problem.

74 Example

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79 Constrained Motion of Connected Particles

80 Sometimes the position of a particle will depend upon the position of another or of several particles. If the particles are connected together by an inextensible ropes, the resulting motion is called constrained motion. Velocity Acceleration The motion of A is twice the vertical motion B. One degree of freedom (only one variable is needed to specify the position of all parts of the system). Circumference of the circle C = 2π r 1/2C = π r 1/4C = π r 2 Total length of the cable is r2 L= x+ π + 2y+ π r1 + b 2 But L,r 2, r 1, and b are all constant d d π r2 ( L) = ( x+ + 2 y+ π r + b) dt dt 2 0= x+ 2y OR 0= VA + 2V V = V x = 2y A 2 B 0= x+ 2 y OR 0= aa + 2a x= 2 y a = 2a A 1 B B B The velocity of A have a sign which opposite to that of the velocity of B.

81 Two degree of freedom L1 = YA + 2YD + Constant L2 = YB + YC + ( YC YD) + Constant d d ( L1 ) = ( YA + 2YD + Constant) dt dt 0= Y + 2Y d dt Eliminating one of the terms A and D 2 YA 0= Y B + 2Y C + YA 0= Y B + 2Y C Y + 2Y + 4Y = 0 Y + 2Y + 4Y = 0 A B C V + 2V + 4V = 0 A B C A D 0= Y A + 2Y D d L2 = YB + YC + ( YC YD) + Constant dt 0= Y + 2Y Y ( ) ( ) B C D 0= Y + 2Y Y B C D Y Y = A B C a + 2a + 4a = 0 A B C YA Y D = 2 The signs in the equations are determined by visualization. For Example:- Both A and B have downward (positive) velocity, but C will have an upward (negative) Velocity.

82 Example

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