Announcements. Introduction and Rectilinear Kinematics: Continuous Motion - Sections

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1 Announcements Week-of-prayer schedule (10:45-11:30) Introduction and Rectilinear Kinematics: Continuous Motion - Sections Today s Objectives: Students will be able to find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path. In-Class Activities: Reading quiz Applications Relations between s(t), v(t), and a(t) for general rectilinear motion Relations between s(t), v(t), and a(t) when acceleration is constant Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 12 1

2 Reading Quiz 1. In dynamics, a particle is assumed to have. A) both translation and rotational motions B) only a mass C) a mass but the size and shape cannot be neglected D) no mass or size or shape, it is just a point 2. The average speed is defined as. A) r/ t B) s/ t C) s T / t D) None of the above Applications The motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles. Why? If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration? Engr222 Spring 2004 Chapter 12 2

3 Applications - continued A train travels along a straight length of track. Can we treat the train as a particle? If the train accelerates at a constant rate, how can we determine its position and velocity at some instant? 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 Engr222 Spring 2004 Chapter 12 3

4 Position and Displacement Vector form: r = r - r 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. Scalar form: s = s - s The total distance traveled by the particle, s T, is a positive scalar that represents the total length of the path over which the particle travels. 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 v avg = 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: (v sp ) avg = s T / t Engr222 Spring 2004 Chapter 12 4

5 Acceleration Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s 2 or ft/s 2. The instantaneous acceleration is the time derivative of velocity. Vector form: a = dv/dt Scalar form: a = dv/dt = d 2 s/dt 2 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 Summary of Kinematic Relationships: 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. v t dv = vo o Velocity: a dt v or v dv = vo s so a ds Note that s o and v o represent the initial position and velocity of the particle at t = 0. s ds = so Position: t o v dt Engr222 Spring 2004 Chapter 12 5

6 Constant Acceleration The three kinematic equations can be integrated for the special case when acceleration is constant (a = a c ) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, a c = g = 9.81 m/s 2 = 32.2 ft/s 2 downward. These equations are: v v o s s v v o o dv ds v dv t = o t = o a s = s o c dt v dt a c ds yields yields yields v = v o + o a t c s = s + v t + (1/2)a v o 2 = (v ) 2 + o 2a (s - s ) c o t 2 c Example Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s 2. Find: The distance the motorcycle travels before it stops. Plan: Establish the positive coordinate s in the direction the motorcycle is traveling. Since the acceleration is given as a function of time, integrate it once to calculate the velocity and again to calculate the position. Engr222 Spring 2004 Chapter 12 6

7 Solution: Example - continued 1) Integrate acceleration to determine the velocity. v t a = dv / dt => dv = a dt => dv = ( 6t) dt vo o => v v o = -3t 2 => v = -3t 2 + v o 2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use v o = 27 m/s. 0 = -3t => t = 3 s 3) Now calculate the distance traveled in 3s by integrating the velocity using s o = 0: s t v = ds / dt => ds = v dt => ds = ( 3t 2+ vo) dt => s s o = -t 3 so o + v o t => s 0 = -(3) 3 + (27)(3) => s = 54 m Concept Quiz 3 m/s 5 m/s t = 2 s t = 7 s 1. A particle moves along a horizontal path with its velocity varying with time as shown. The average acceleration of the particle is. A) 0.4 m/s 2 B) 0.4 m/s 2 C) 1.6 m/s 2 D) 1.6 m/s 2 2. A particle has an initial velocity of 30 ft/s to the left. If it then passes through the same location 5 seconds later with a velocity of 50 ft/s to the right, the average velocity of the particle during the 5 s time interval is. A) 10 ft/s B) 40 ft/s C) 16 m/s D) 0 ft/s Engr222 Spring 2004 Chapter 12 7

8 Group Problem Solving Given:Ball A is released from rest at a height of 40 ft at the same time that ball B is thrown upward, 5 ft from the ground. The balls pass one another at a height of 20 ft. Find: The speed at which ball B was thrown upward. Plan: Both balls experience a constant downward acceleration of 32.2 ft/s 2. Apply the formulas for constant acceleration, with a c = ft/s 2. Group Problem Solving - continued Solution: 1) First consider ball A. With the origin defined at the ground, ball A is released from rest (v A ) o = 0 at a height of 40 ft (s A ) o = 40 ft. Calculate the time required for ball A to drop to 20 ft (s A = 20 ft) using a position equation. s A = (s A ) o + (v A ) o t + (1/2)a c t 2 20 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t 2 ) => t = s 2) Now consider ball B. It is throw upward from a height of 5 ft (s B ) o = 5 ft. It must reach a height of 20 ft (s B = 20 ft) at the same time ball A reaches this height (t = s). Apply the position equation again to ball B using t = 1.115s. s B = (s B ) o + (v B ) o t + (1/2) a c t 2 20 ft = 5 + (v B ) o (1.115) + (1/2)(-32.2)(1.115) 2 => (v B ) o = 31.4 ft/s Engr222 Spring 2004 Chapter 12 8

9 Attention Quiz 1. A particle has an initial velocity of 3 ft/s to the left at s 0 = 0 ft. Determine its position when t = 3 s if the acceleration is 2 ft/s 2 to the right. A) 0.0 ft B) 6.0 ft C) 18.0 ft D) 9.0 ft 2. A particle is moving with an initial velocity of v = 12 ft/s and constant acceleration of 3.78 ft/s 2 in the same direction as the velocity. Determine the distance the particle has traveled when the velocity reaches 30 ft/s. A) 50 ft B) 100 ft C) 150 ft D) 200 ft Textbook Problem 12- Engr222 Spring 2004 Chapter 12 9

10 Announcements Correction - example 1 (page 13) from last lecture Mid-term #1 will be Tuesday, April 20 more details Friday Rectilinear Kinematics: Erratic Motion Section 12.3 Today s Objectives: Students will be able to determine position, velocity, and acceleration of a particle using graphs. In-Class Activities: Check homework, if any Reading quiz Applications s-t, v-t, a-t, v-s, and a-s diagrams Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 12 10

11 Reading Quiz 1. The slope of a v-t graph at any instant represents instantaneous A) velocity. B) acceleration. C) position. D) jerk. 2. Displacement of a particle in a given time interval equals the area under the graph during that time. A) a-t B) a-s C) v-t C) s-t Applications In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the rocket sled, can we determine its acceleration at position s = 300 meters? How? Engr222 Spring 2004 Chapter 12 11

12 Graphing 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. 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. Engr222 Spring 2004 Chapter 12 12

13 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 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. A-T Graph Given the 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. Engr222 Spring 2004 Chapter 12 13

14 A-S Graph A more complex case is presented by the a-s graph. The area under the acceleration versus position curve represents the change in velocity (recall a ds = v dv ). ½ (v 1 ² v o ²) = s 2 s 1 a ds = area under the a-s graph This equation can be solved for v 1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance. V-S Graph Another complex case is presented by the 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. a = v (dv/ds) Thus, we can obtain a plot of a vs. s from the v-s curve. Engr222 Spring 2004 Chapter 12 14

15 Example Given: v-t graph for a train moving between two stations Find: a-t graph and s-t graph over this time interval Think about your plan of attack for the problem! Solution: Example - continued For the first 30 seconds the slope is constant and is equal to: a 0-30 = dv/dt = 40/30 = 4/3 ft/s 2 Similarly, a = 0 and a = -4/3 ft/s 2 a(ft/s 2 ) 4 3 t(s) -4 3 Engr222 Spring 2004 Chapter 12 15

16 Example - continued s(ft) The area under the v-t graph represents displacement. s 0-30 = ½ (40)(30) = 600 ft s = (60)(40) = 2400 ft 600 t(s) s = ½ (40)(30) = 600 ft Concept Quiz 1. If a particle starts from rest and accelerates according to the graph shown, the particle s velocity at t = 20 s is A) 200 m/s B) 100 m/s C) 0 D) 20 m/s 2. The particle in Problem 1 stops moving at t =. A) 10 s B) 20 s C) 30 s D) 40 s Engr222 Spring 2004 Chapter 12 16

17 Group Problem Solving Given: The v-t graph shown Find: The a-t graph, average speed, and distance traveled for the 30 s interval Plan: Find slopes of the curves and draw the v-t graph. Find the area under the curve--that is the distance traveled. Finally, calculate average speed (using basic definitions!). Group Problem Solving - continued Solution: For 0 t 10 a = dv/dt = 0.8 t ft/s² For 10 t 30 a = dv/dt = 1 ft/s² a(ft/s²) t(s) Engr222 Spring 2004 Chapter 12 17

18 Group Problem Solving - continued s 0-10 = v dt = (1/3) (.4)(10) 3 = 400/3 ft s = v dt = (0.5)(30) (30) 0.5(10) 2 30(10) = 1000 ft s 0-30 = /3 = ft v avg(0-30) = total distance / time = /30 = ft/s Attention Quiz 1. If a car has the velocity curve shown, determine the time t necessary for the car to travel 100 meters. v A) 8 s B) 4 s 75 C) 10 s D) 6 s 6 s 2. Select the correct a-t graph for the velocity curve shown. a a v A) t B) t a a C) t D) t t t Engr222 Spring 2004 Chapter 12 18

19 Textbook Problem 12- Announcements Correction - example 1 (page 13) from last lecture Mid-term #1 will be Tuesday, April 20 more details Friday Section 12-3 will not be covered HW #7 can be turned in on Friday Engr222 Spring 2004 Chapter 12 19

20 Curvilinear Motion: Rectangular Components Sections Today s Objectives: Students will be able to: a) Describe the motion of a particle traveling along a curved path. b) Relate kinematic quantities in terms of the rectangular components of the vectors. In-Class Activities: Reading quiz Applications General curvilinear motion Rectangular components of kinematic vectors Concept quiz Group problem solving Attention quiz Reading Quiz 1. In curvilinear motion, the direction of the instantaneous velocity is always A) tangent to the hodograph. B) perpendicular to the hodograph. C) tangent to the path. D) perpendicular to the path. 2. In curvilinear motion, the direction of the instantaneous acceleration is always A) tangent to the hodograph. B) perpendicular to the hodograph. C) tangent to the path. D) perpendicular to the path. Engr222 Spring 2004 Chapter 12 20

21 Applications The path of motion of each plane in this formation can be tracked with radar and their x, y, and z coordinates (relative to a point on earth) recorded as a function of time. How can we determine the velocity and acceleration of each plane at any instant? Should they be the same for each aircraft? Applications - continued A roller coaster car travels down a fixed, helical path at a constant speed. How can we determine its position and acceleration at any instant? If you are designing the track, why is it important to be able to predict the acceleration of the car? Engr222 Spring 2004 Chapter 12 21

22 Position and Displacement 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 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 v avg = 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 t0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector! Engr222 Spring 2004 Chapter 12 22

23 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: a avg = v/ t = (v - v )/ t The instantaneous acceleration is the timederivative of velocity: a = dv/dt = d 2 r/dt 2 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. Rectangular Components - Position It is often convenient to describe the motion of a particle in terms of its x, y, and 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 = x i + y j + z k The x, y, and 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 = (x 2 + y 2 + z 2 ) 0.5 The direction of r is defined by the unit vector: u r = (1/r)r Engr222 Spring 2004 Chapter 12 23

24 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, and k are constant in magnitude and direction, this equation reduces to v = v x i + v y j + v z k where v x = x = dx/dt, v y = y = dy/dt, v z = z = dz/dt The magnitude of the velocity vector is v = [(v x ) 2 + (v y ) 2 + (v z ) 2 ] 0.5 The direction of v is tangent to the path of motion. Rectangular Components - Acceleration The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): a = dv/dt = d 2 r/dt 2 = a x i + a y j + a z k where a x = = = dv x /dt, a y = = = dv y /dt, v z z a z = = = dv z /dt v x x v y y The magnitude of the acceleration vector is a = [(a x ) 2 + (a y ) 2 + (a z ) 2 ] 0.5 The direction of a is usually not tangent to the path of the particle. Engr222 Spring 2004 Chapter 12 24

25 Example Given: The motion of two particles (A and B) is described by the position vectors: r A = [3t i + 9t(2 t) j] m r B = [3(t 2 2t +2) i + 3(t 2) j] m Find: The point at which the particles collide and their speeds just before the collision. Plan: 1) The particles will collide when their position vectors are equal, or r A = r B. 2) Their speeds can be determined by differentiating the position vectors. Solution: Example - continued 1) The point of collision requires that r A = r B, so x A = x B and y A = y B. r A = [3t i + 9t(2 t) j] m x-components: 3t = 3(t 2 2t + 2) Simplifying: t 2 3t + 2 = 0 Solving: t = {3 ± [3 2 4(1)(2)] 0.5 }/2(1) => t = 2 or 1 s y-components: 9t(2 t) = 3(t 2) Simplifying: 3t 2 5t 2 = 0 Solving: t = {5 ± [5 2 4(3)( 2)] 0.5 }/2(3) => t = 2 or 1/3 s r B = [3(t 2 2t +2) i + 3(t 2) j] m So, the particles collide when t = 2 s. Substituting this value into r A or r B yields x A = x B = 6 m and y A = y B = 0 Engr222 Spring 2004 Chapter 12 25

26 2) Differentiate r A and r B to get the velocity vectors. v A = dr A /dt = Example - continued... i + x A y A At t = 2 s: v A = [3i 18j] m/s Speed is the magnitude of the velocity vector: v A = ( ) 0.5 = 18.2 m/s v B = ( ) 0.5 = 6.71 m/s j = [3i + (18 18t) j] m/s v B = dr B /dt = x B i + y B j = [(6t 6)i + 3j] m/s At t = 2 s: v B = [6i + 3j] m/s r A = [3t i + 9t(2 t) j] m r B = [3(t 2 2t +2) i + 3(t 2) j] m Concept Quiz 1. If the position of a particle is defined by r = [(1.5t 2 + 1) i + (4t 1) j] (m), its speed at t = 1 s is A) 2 m/s B) 3 m/s C) 5 m/s D) 7 m/s 2. The path of a particle is defined by y = 0.5x 2. If the component of its velocity along the x-axis at x = 2 m is v x = 1 m/s, its velocity component along the y-axis at this position is A) 0.25 m/s B) 0.5 m/s C) 1 m/s D) 2 m/s Engr222 Spring 2004 Chapter 12 26

27 Group Problem Solving Given: A particle travels along a path described by the parabola y = 0.5x 2. The x-component of velocity is given by v x = (5t) ft/s. When t = 0, x = y = 0. Find: The particle s distance from the origin and the magnitude of its acceleration when t = 1 s. Plan: Note that v x is given as a function of time. 1) Determine the x-component of position and acceleration by integrating and differentiating v x, respectively. 2) Determine the y-component of position from the parabolic equation and differentiate to get a y. 3) Determine the magnitudes of the position and acceleration vectors. Group Problem Solving - continued Solution: 1) x-components: Velocity: Acceleration: a x = x = v x = d/dt (5t) = 5 ft/s 2 2) y-components: Position: Velocity: v x = x = dx/dt = (5t) ft/s Position: x dx = t 5t dt => x = (5/2)t 2 = (2.5t 2 ) ft 0 0 y = 0.5x 2 = 0.5(2.5t 2 ) 2 = (3.125t 4 ) ft v y = dy/dt = d (3.125t 4 ) /dt = (12.5t 3 ) ft/s Acceleration: a y = v y = d (12.5t 3 ) /dt = (37.5t 2 ) ft/s 2 Engr222 Spring 2004 Chapter 12 27

28 Group Problem Solving - continued 3) The distance from the origin is the magnitude of the position vector: r = x i + y j = [2.5t 2 i t 4 j] ft At t = 1 s, r = (2.5 i j) ft Distance: d = r = ( ) 0.5 = 4.0 ft The magnitude of the acceleration vector is calculated as: Acceleration vector: a = [5 i t 2 j ] ft/s 2 Magnitude: a = ( ) 0.5 = 37.8 ft/s 2 Attention Quiz 1. If a particle has moved from A to B along the circular path in 4s, what is the average velocity of the particle? A) 2.5 i m/s B) 2.5 i +1.25j m/s C) 1.25 π i m/s D) 1.25 π j m/s 2. The position of a particle is given as r = (4t 2 i 2t j) m. Determine the particle s acceleration. A) (4 i +8 j ) m/s 2 B) (8 i -16 j ) m/s 2 C) (8 i) m/s 2 D) (8 j ) m/s 2 A y R=5m B x Engr222 Spring 2004 Chapter 12 28

29 Textbook Problem 12- Announcements Mid-term #1 will be Wednesday, April 21 Closed book. Cheat sheet (review sheet) and calculator allowed. Review sheet on class web page Engr222 Spring 2004 Chapter 12 29

30 Motion of a Projectile - Section 12.6 Today s Objectives: Students will be able to analyze the free-flight motion of a projectile. In-Class Activities: Reading quiz Applications Kinematic equations for projectile motion Concept quiz Group problem solving Attention quiz Reading Quiz 1. The downward acceleration of an object in free-flight motion is A) zero B) increasing with time C) 9.81 m/s 2 D) 9.81 ft/s 2 2. The horizontal component of velocity remains during a free-flight motion. A) zero B) constant C) at 9.81 m/s 2 D) at 32.2 ft/s 2 Engr222 Spring 2004 Chapter 12 30

31 Applications A kicker should know at what angle, θ, and initial velocity, v o, 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? Applications - continued A fireman wishes to know the maximum height on the wall he can project water from the hose. At what angle, θ, should he hold the hose? Engr222 Spring 2004 Chapter 12 31

32 Concept of Projectile Motion 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., 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. Kinematic Equations Horizontal Motion Since a x = 0, the velocity in the horizontal direction remains constant (v x = v ox ) and the position in the x direction can be determined by: x = x o + (v ox )(t) Why is a x equal to zero (assuming movement through the air)? Engr222 Spring 2004 Chapter 12 32

33 Kinematic Equations Vertical Motion Since the positive y-axis is directed upward, a y = -g. Application of the constant acceleration equations yields: v y = v oy g(t) y = y o + (v oy )(t) ½g(t) 2 v y2 = v oy2 2g(y y o ) For any given problem, only two of these three equations can be used. Why? Example Given: v o and Find: The equation that defines y as a function of x. Plan: Eliminate time from the kinematic equations. Solution: Using v x = v o cos and v y = v o sin We can write: x = (v o cos )t By substituting for t: or y = (v o sin )t ½ g(t) 2 t = x v o cos 2 ( ) ( )( ) y = (v o sin ) x g x v o cos 2 v o cos Engr222 Spring 2004 Chapter 12 33

34 Example - continued Simplifying the last equation, we get: y = (x tanθ) g x ( 2 ) (1 + tan2 θ) 2v o 2 The above equation is called the path equation which describes the path of a particle in projectile motion. The equation shows that the path is parabolic. Example Given: Snowmobile is going 15 m/s at point A. Find: The horizontal distance it travels (R) and the time in the air. Solution: First, place the coordinate system at point A. Then write the equation for horizontal motion. x B = x A + v Ax t AB and v Ax = 15 cos 40 m/s Now write a vertical motion equation. Use the distance equation. y B = y A + v Ay t AB 0.5g c t 2 AB v Ay = 15 sin 40 m/s Note that x B = R, x A = 0, y B = -(3/4)R, and y A = 0. Solving the two equations together (two unknowns) yields R = 42.8 m t AB = 3.72 s Engr222 Spring 2004 Chapter 12 34

35 Concept Quiz 1. In a projectile motion problem, what is the maximum number of unknowns that can be solved? A) 1 B) 2 C) 3 D) 4 2. The time of flight of a projectile, fired over level ground with initial velocity V o at angle, is equal to A) (v o sin θ)/g B) (2v o sin θ)/g C) (v o cos θ)/g D) (2v o cos θ)/g Group Problem Solving Given: Skier leaves the ramp at θ A = 25 o and hits the slope at B. Find: The skier s initial speed v A. Plan: Establish a fixed x,y coordinate system (in the solution here, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y- directions. Engr222 Spring 2004 Chapter 12 35

36 Group Problem Solving - continued Solution: Motion in x-direction: Using x B = x A + v ox (t AB ) t AB = (4/5)100 v A (cos 25) = v A Motion in y-direction: Using y B = y A + v oy (t AB ) ½ g(t AB ) 2-64 = 0 + v A (sin 45) 80 v A (cos 25) ½ (9.81) v A 2 v A = m/s Attention Quiz 1. A projectile is given an initial velocity v o at an angle φ above the horizontal. The velocity of the projectile when it hits the slope is the initial velocity v o. A) less than B) equal to C) greater than D) None of the above. 2. A particle has an initial velocity v o at angle θ with respect to the horizontal. The maximum height it can reach is when A) θ = 30 B) θ = 45 C) θ = 60 D) θ = 90 Engr222 Spring 2004 Chapter 12 36

37 Textbook Problem Determine the horizontal velocity v a of a tennis ball at A so that it just clears the net at B. Also, find the distance s where the ball strikes the ground. Announcements Engr222 Spring 2004 Chapter 12 37

38 Curvilinear Motion: Normal and Tangential Components - Section 12.7 Today s Objectives: Students will be able to determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path. In-Class Activities: Reading quiz Applications Normal and tangential components of velocity and acceleration Special cases of motion Concept quiz Group problem solving Attention quiz Reading Quiz 1. If a particle moves along a curve with a constant speed, then its tangential component of acceleration is A) positive. B) negative. C) zero. D) constant. 2. The normal component of acceleration represents A) the time rate of change in the magnitude of the velocity. B) the time rate of change in the direction of the velocity. C) magnitude of the velocity. D) direction of the total acceleration. Engr222 Spring 2004 Chapter 12 38

39 Applications Cars traveling along a clover-leaf interchange experience an acceleration due to a change in speed 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? Applications - continued A motorcycle travels up a hill for which the path can be approximated by a function y = f(x). If the motorcycle starts from rest and increases its speed at a constant rate, how can we determine its velocity and acceleration at the top of the hill? How would you analyze the motorcycle's flight at the top of the hill? Engr222 Spring 2004 Chapter 12 39

40 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. Normal and Tangential Components - continued The positive n and t directions are defined by the unit vectors u n and u t, respectively. The center of curvature, O, always lies on the concave side of the curve. The radius of curvature, ρ, 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. Engr222 Spring 2004 Chapter 12 40

41 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 = vu t where v = s = ds/dt Here v defines the magnitude of the velocity (speed) and u t defines the direction of the velocity vector. Acceleration in the N-T Coordinate System Acceleration is the time rate of change. of. velocity: a = dv/dt = d(vu t )/dt = vu t + vu t. Here v represents the change in. the magnitude of velocity and u t represents the rate of change in the direction of u t. After mathematical manipulation, the acceleration vector can be expressed as:. a = vu t + (v 2 /ρ)u n = a t u t + a n u n. Engr222 Spring 2004 Chapter 12 41

42 Acceleration in the N-T Coordinate System There are two components to the acceleration vector: a = a t u t + a n u n The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity.. a t = v or a t ds = v dv The normal or centripetal component is always directed toward the center of curvature of the curve. a n = v 2 /ρ The magnitude of the acceleration vector is a = [(a t ) 2 + (a n ) 2 ] 0.5 Special Cases of Motion There are some special cases of motion to consider. 1) The particle moves along a straight line. ρ => a n = v 2 /ρ = 0 => a = a t = 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.. a t = v = 0 => a = a n = v 2 /ρ The normal component represents the time rate of change in the direction of the velocity. Engr222 Spring 2004 Chapter 12 42

43 Special Cases of Motion - continued 3) The tangential component of acceleration is constant, a t = (a t ) c. In this case, s = s o + v o t + (1/2)(a t ) c t 2 v = v o + (a t ) c t v 2 = (v o ) 2 + 2(a t ) c (s s o ) As before, s o and v o 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, ρ, at any point on the path can be calculated from ρ = [ 1 + (dy/dx) 2 ]3/2 d 2 y/dx 2 Three-Dimensional Motion If a particle moves along a space curve, the n and t axes are defined as before. At any point, the t-axis is tangent to the path and the n-axis points toward the center of curvature. The plane containing the n and t axes is called the osculating plane. A third axis can be defined, called the binomial axis, b. The binomial unit vector, u b, is directed perpendicular to the osculating plane, and its sense is defined by the cross product u b = u t x u n. There is no motion, thus no velocity or acceleration, in the binomial direction. Engr222 Spring 2004 Chapter 12 43

44 Example Given: Starting from rest, a motorboat travels around a circular path of ρ = 50 m at a speed that increases with time, v = (0.2 t 2 ) m/s. Find: The magnitudes of the boat s velocity and acceleration at the instant t = 3 s. Plan: The boat starts from rest (v = 0 when t = 0). 1) Calculate the velocity at t = 3s using v(t). 2) Calculate the tangential and normal components of acceleration and then the magnitude of the acceleration vector. Solution: Example - continued 1) The velocity vector is v = v u t, where the magnitude is given by v = (0.2t 2 ) m/s. At t = 3s: v = 0.2t 2 = 0.2(3) 2 = 1.8 m/s. 2) The acceleration vector is a = a t u t + a n u n = vu t + (v 2 /ρ)u n.. Tangential component: a t = v = d(.2t 2 )/dt = 0.4t m/s 2 At t = 3s: a t = 0.4t = 0.4(3) = 1.2 m/s 2 Normal component: a n = v 2 /ρ = (0.2t 2 ) 2 /(ρ) m/s 2 At t = 3s: a n = [(0.2)(3 2 )] 2 /(50) = m/s 2 The magnitude of the acceleration is a = [(a t ) 2 + (a n ) 2 ] 0.5 = [(1.2) 2 + (0.0648) 2 ] 0.5 = 1.20 m/s 2 Engr222 Spring 2004 Chapter 12 44

45 Concept Quiz 1. A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing at a constant rate of 4 m/s 2. What is the magnitude of its total acceleration at this instant? A) 3 m/s 2 B) 4 m/s 2 C) 5 m/s 2 D) -5 m/s 2 2. If a particle moving in a circular path of radius 5 m has a velocity function v = 4t 2 m/s, what is the magnitude of its total acceleration at t = 1 s? A) 8 m/s B) 8.6 m/s C) 3.2 m/s D) 11.2 m/s Group Problem Solving Given: A jet plane travels along a vertical parabolic path defined by the equation y = 0.4x 2. At point A, the jet has a speed of 200 m/s, which is increasing at the rate of 0.8 m/s 2. Find: The magnitude of the plane s acceleration when it is at point A. Plan: 1) The change in the speed of the plane (0.8 m/s 2) is the tangential component of the total acceleration. 2) Calculate the radius of curvature of the path at A. 3) Calculate the normal component of acceleration. 4) Determine the magnitude of the acceleration vector. Engr222 Spring 2004 Chapter 12 45

46 Group Problem Solving - continued Solution: 1) The tangential component of acceleration is the rate of. increase of the plane s speed, so a t = v = 0.8 m/s 2. 2) Determine the radius of curvature at point A (x = 5 km): dy/dx = d(0.4x 2 )/dx = 0.8x, d 2 y/dx 2 = d (0.8x)/dx = 0.8 At x =5 km, dy/dx = 0.8(5) = 4, d 2 y/dx 2 = 0.8 [ 1 + (dy/dx) 2 ] 3/2 => ρ = = [1 + (4) 2 ] 3/2 /(0.8) = km d 2 y/dx 2 3) The normal component of acceleration is a n = v 2 /ρ = (200) 2 /(87.62 x 10 3 ) = m/s 2 4) The magnitude of the acceleration vector is a = [(a t ) 2 + (a n ) 2 ] 0.5 = [(0.8) 2 + (0.457) 2 ] 0.5 = m/s 2 Attention Quiz 1. The magnitude of the normal acceleration is A) proportional to radius of curvature. B) inversely proportional to radius of curvature. C) sometimes negative. D) zero when velocity is constant. 2. The directions of the tangential acceleration and velocity are always A) perpendicular to each other. B) collinear. C) in the same direction. D) in opposite directions. Engr222 Spring 2004 Chapter 12 46

47 Textbook Problem 12- Announcements Engr222 Spring 2004 Chapter 12 47

48 Curvilinear Motion: Cylindrical Coordinates Section 12.8 Today s Objectives: Students will be able to determine velocity and acceleration components using cylindrical coordinates. In-Class Activities: Reading quiz Applications Velocity Components Acceleration Components Concept quiz Group problem solving Attention quiz Reading Quiz 1. In a polar coordinate system, the velocity vector can be... written as v = v r u r + v u = ru r + rθu θ. The term θ is called A) transverse velocity. B) radial velocity. C) angular velocity. D) angular acceleration. 2. The speed of a particle in a cylindrical coordinate system is.. A) r B) rθ... C) (rθ) 2 + (r) 2 D) (rθ) 2. + (r) 2. + (z) 2 Engr222 Spring 2004 Chapter 12 48

49 Applications The cylindrical coordinate system is used in cases where the particle moves along a 3-D curve. In the figure shown, the boy slides down the slide at a constant speed of 2 m/s. How fast is his elevation from the ground. changing (i.e., what is z )? Applications - continued A polar coordinate system is a 2-D representation of the cylindrical coordinate system. When the particle moves in a plane (2-D), and the radial distance, r, is not constant, the polar coordinate system can be used to express the path of motion of the particle. Engr222 Spring 2004 Chapter 12 49

50 Position Polar Coordinates We can express the location of P in polar coordinates as r = ru r. Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, θ, is measured counterclockwise (CCW) from the horizontal. Velocity Polar Coordinates The instantaneous velocity is defined as: v = dr/dt = d(ru r )/dt. v = ru r + r du r dt Using the chain rule: du r /dt = (du r /dθ)(dθ/dt). We can prove that du r /dθ = u so du r /dt = θu.. Therefore: v = ru r + rθu. Thus, the velocity vector has two components:. r, called the radial component, and rθ, called the transverse component. The speed of the particle at any given instant is the sum of the squares of both components or.. v = (r θ ) 2 + ( r ) 2 Engr222 Spring 2004 Chapter 12 50

51 Acceleration Polar Coordinates The instantaneous acceleration is defined as:.. a = dv/dt = (d/dt)(ru r + rθu ) After manipulation, the acceleration can be expressed as a = (r rθ 2 )u r + (rθ + 2rθ)u... The term (r rθ 2 ) is the radial acceleration or a r..... The term (rθ + 2rθ) is the transverse acceleration or a θ The magnitude of acceleration is a = (r rθ 2 ) 2 + (rθ + 2rθ) 2 Cylindrical Coordinates If the particle P moves along a space curve, its position can be written as Velocity: r P = ru r + zu z Taking time derivatives and using the chain rule:... v P = ru r + rθu + zu z..... Acceleration: a P = (r rθ )u r + (rθ + 2rθ)u + zu z Engr222 Spring 2004 Chapter 12 51

52 Example Given: r. = 5 cos(2θ) (m) θ = 3t 2 (rad/s) θ o = 0 Find: Velocity and acceleration at θ = Plan: Apply chain rule to determine r and r and evaluate at θ = 30. t t. Solution: θ = θ dt = 3t 2 dt = t 3 t o = 0 π At θ = 30, θ = = t 3. Therefore: t = s. 6. θ = 3t 2 = 3(0.806) 2 = 1.95 rad/s 0 Example - continued.. θ = 6t = 6(0.806) = rad/s 2 r = 5 cos(2θ) = 5 cos(60) = 2.5m r = -10 sin(2θ)θ = -10 sin(60)(1.95) = m/s.. r = -20 cos(2θ)θ 2 10 sin(2θ)θ.. = -20 cos(60)(1.95) sin(60)(4.836).. = -80 m/s 2 Substitute in the equation for velocity.. v = ru r + rθu v = u r + 2.5(1.95)u v = (16.88) 2 + (4.87) 2 = m/s Engr222 Spring 2004 Chapter 12 52

53 Example - continued Substitute in the equation for acceleration:..... a = (r rθ 2.. )u r + (rθ + 2rθ)u a = [ (1.95) 2 ]u r + [2.5(4.836) + 2(-16.88)(1.95)]u a = -89.5u r 53.7u m/s 2 a = (89.5) 2 + (53.7) 2 = m/s 2 Concept Quiz 1. If r. is zero for a particle, the particle is A) not moving. B) moving in a circular path. C) moving on a straight line. D) moving with constant velocity. 2. If a particle moves in a circular path with constant velocity, its radial acceleration is A) zero. B).. r. C) -rθ. 2. D) 2rθ... Engr222 Spring 2004 Chapter 12 53

54 Group Problem Solving Given: The car s speed is constant at 1.5 m/s. Find: The car s acceleration (as a vector). Hint: The tangent to the ramp at any point is at an angle φ = tan ( ) = π(10) Also, what is the relationship between φ and θ? Plan: Use cylindrical coordinates. Since r is constant, all derivatives of r will be zero. Solution: Since r. is constant the velocity only has 2 components:. = rθ = v cosφ and v z = z = v sinφ v θ Group Problem Solving - continued. Therefore: θ = (.. θ = 0 v cosφ r ) = rad/s. v z = z = v sinφ = m/s.. z = 0... r = r = a = (r rθ 2 )u r + (rθ + 2rθ)u + zu z. a = (-rθ 2 )u r = -10(0.147) 2 u r = u r m/s 2 Engr222 Spring 2004 Chapter 12 54

55 Attention Quiz 1. The radial component of velocity of a particle moving in a circular path is always A) zero. B) constant. C) greater than its transverse component. D) less than its transverse component. 2. The radial component of acceleration of a particle moving in a circular path is always A) negative. B) directed toward the center of the path. C) perpendicular to the transverse component of acceleration. D) All of the above. Textbook Problem 12- Engr222 Spring 2004 Chapter 12 55

56 Announcements Absolute Dependent Motion Analysis of Two Particles - Section 12.9 Today s Objectives: Students will be able to relate the positions, velocities, and accelerations of particles undergoing dependent motion. In-Class Activities: Reading quiz Applications Define dependent motion Develop position, velocity, and acceleration relationships Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 12 56

57 Reading Quiz 1. When particles are interconnected by a cable, the motions of the particles are. A) always independent B) always dependent C) dependent, but not always D) None of the above 2. If the motion of one particle is dependent on that of another particle, each coordinate axis of the particles. A) should be directed along the path of motion B) can be directed anywhere C) should have the same origin D) None of the above Applications The cable and pulley system shown here can be used to modify the speed of block B relative to the speed of the motor. It is important to relate the various motions in order to determine the power requirements for the motor and the tension in the cable. If the speed of the cable coming onto the motor pulley is known, how can we determine the speed of block B? Engr222 Spring 2004 Chapter 12 57

58 Applications - continued Rope and pulley arrangements are often used to assist in lifting heavy objects. The total lifting force required from the truck depends on the acceleration of the cabinet. How can we determine the acceleration and velocity of the cabinet if the acceleration of the truck is known? Dependent Motion In many kinematics problems, the motion of one object will depend on the motion of another object. The blocks in this figure are connected by an inextensible cord wrapped around a pulley. If block A moves downward along the inclined plane, block B will move up the other incline. The motion of each block can be related mathematically by defining position coordinates, s A and s B. Each coordinate axis is defined from a fixed point or datum line, measured positive along each plane in the direction of motion of each block. Engr222 Spring 2004 Chapter 12 58

59 Dependent Motion - continued In this example, position coordinates s A and s B can be defined from fixed datum lines extending from the center of the pulley along each incline to blocks A and B. If the cord has a fixed length, the position coordinates s A and s B are related mathematically by the equation s A + l CD + s B = l T Here l T is the total cord length and l CD is the length of cord passing over arc CD on the pulley. Dependent Motion - continued The velocities of blocks A and B can be related by differentiating the position equation. Note that l CD and l T remain constant, so dl CD /dt = dl T /dt = 0 ds A /dt + ds B /dt = 0 => v B = -v A The negative sign indicates that as A moves down the incline (positive s A direction), B moves up the incline (negative s B direction). Accelerations can be found by differentiating the velocity expression. Prove to yourself that a B = -a A. Engr222 Spring 2004 Chapter 12 59

60 Example Consider a more complicated example. Position coordinates (s A and s B ) are defined from fixed datum lines, measured along the direction of motion of each block. Note that s B is only defined to the center of the pulley above block B, since this block moves with the pulley. Also, h is a constant. The red colored segments of the cord remain constant in length during motion of the blocks. Example - continued The position coordinates are related by the equation 2s B + h + s A = l Where l is the total cord length minus the lengths of the red segments. Since l and h remain constant during the motion, the velocities and accelerations can be related by two successive time derivatives: 2v B = -v A and 2a B = -a A When block B moves downward (+s B ), block A moves to the left (-s A ). Remember to be consistent with the sign convention! Engr222 Spring 2004 Chapter 12 60

61 Dependent Motion - Procedures These procedures can be used to relate the dependent motion of particles moving along rectilinear paths (only the magnitudes of velocity and acceleration change, not their line of direction). 1. Define position coordinates from fixed datum lines, along the path of each particle. Different datum lines can be used for each particle. 2. Relate the position coordinates to the cord length. Segments of cord that do not change in length during the motion may be left out. 3. If a system contains more than one cord, relate the position of a point on one cord to a point on another cord. Separate equations are written for each cord. 4. Differentiate the position coordinate equation(s) to relate velocities and accelerations. Keep track of signs! Example Given:In the figure on the left, the cord at A is pulled down with a speed of 8 ft/s. Find: The speed of block B. Plan: There are two cords involved in the motion in this example. The position of a point on one cord must be related to the position of a point on the other cord. There will be two position equations (one for each cord). Engr222 Spring 2004 Chapter 12 61

62 Solution: 1) Define the position coordinates from a fixed datum line. Three coordinates must be defined: one for point A (s A ), one for block B (s B ), and one relating positions on the two cords. Note that pulley C relates the motion of the two cords. DATUM s A s C sb Example - continued Define the datum line through the top pulley (which has a fixed position). s A can be defined to the center of the pulley above point A. s B can be defined to the center of the pulley above B. s C is defined to the center of pulley C. All coordinates are defined as positive down and along the direction of motion of each point/object. DATUM s A s C sb Example - continued 2) Write position/length equations for each cord. Define l 1 as the length of the first cord, minus any segments of constant length. Define l 2 in a similar manner for the second cord: Cord 1: 2s A + 2s C = l 1 Cord 2: s B + (s B s C ) = l 2 3) Eliminating s C between the two equations, we get 2s A + 4s B = l 1 + 2l 2 4) Relate velocities by differentiating this expression. Note that l 1 and l 2 are constant lengths. 2v A + 4v B = 0 => v B = - 0.5v A = - 0.5(8) = - 4 ft/s The velocity of block B is 4 ft/s up (negative s B direction). Engr222 Spring 2004 Chapter 12 62

63 Concept Quiz 1. Determine the speed of block B. A) 1 m/s B) 2 m/s C) 4 m/s D) None of the above. 2. Two blocks are interconnected by a cable. Which of the following is correct? A) v A = - v B B) (v x ) A = - (v x ) B y C) (v y ) A = - (v y ) B D) All of the above. x Group Problem Solving Given:In this pulley system, block A is moving downward with a speed of 4 ft/s while block C is moving up at 2 ft/s. Find: The speed of block B. Plan: All blocks are connected to a single cable, so only one position/length equation will be required. Define position coordinates for each block, write out the position relation, and then differentiate it to relate the velocities. Engr222 Spring 2004 Chapter 12 63

64 Solution: Group Problem Solving - continued 1) A datum line can be drawn through the upper, fixed, pulleys and position coordinates defined from this line to each block (or the pulley above the block). DATUM s A s C sb 2) Defining s A, s B, and s C as shown, the position relation can be written: s A + 2s B + 2s C = l 3) Differentiate to relate velocities: v A + 2v B + 2v C = v B + 2(-2) =0 v B = 0 Attention Quiz 1. Determine the speed of block B when block A is moving down at 6 ft/s while block C is moving down at 18 ft/s. A) 24 ft/s B) 3 ft/s C) 12 ft/s D) 9 ft/s v A =6 ft/s v C =18 ft/s 2. Determine the velocity vector of block A when block B is moving downward with a speed of 10 m/s. A) (8i + 6j) m/s B) (4i + 3j) m/s C) (-8i - 6j) m/s D) (3i + 4j) m/s j i v B =10 m/s Engr222 Spring 2004 Chapter 12 64

65 Textbook Problem 12- Announcements Engr222 Spring 2004 Chapter 12 65

66 Relative Motion Analysis - Section Today s Objectives: Students will be able to: a) Understand translating frames of reference. b) Use translating frames of reference to analyze relative motion. In-Class Activities: Reading quiz Applications Relative position, velocity and acceleration Vector & graphical methods Concept quiz Group problem solving Attention quiz Reading Quiz 1. The velocity of B relative to A is defined as A) v B v A. B) v A v B. C) v B + v A. D) v A + v B. 2. Since vector addition forms a triangle, there can be at most unknowns (either magnitudes and/or directions of the vectors). A) one B) two C) three D) four Engr222 Spring 2004 Chapter 12 66

67 Applications When you try to hit a moving object, the position, velocity, and acceleration of the object must be known. Here, the boy on the ground is at d = 10 ft when the girl in the window throws the ball to him. If the boy on the ground is running at a constant speed of 4 ft/s, how fast should the ball be thrown? Applications - continued When fighter jets take off or land on an aircraft carrier, the velocity of the carrier becomes an issue. If the aircraft carrier travels at a forward velocity of 50 km/hr and plane A takes off at a horizontal air speed of 200 km/hr (measured by someone on the water), how do we find the velocity of the plane relative to the carrier? How would you find the same thing for airplane B? How does the wind impact this sort of situation? Engr222 Spring 2004 Chapter 12 67

68 Relative Position The absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by r A and r B. The position of B relative to A is represented by r B/A = r B r A Therefore, if r B = (10 i + 2 j ) m and r A = (4 i + 5 j ) m, then r B/A = (6 i 3 j ) m. Relative Velocity To determine the relative velocity of B with respect to A, the time derivative of the relative position equation is taken. v B/A = v B v A or v B = v A + v B/A In these equations, v B and v A are called absolute velocities and v B/A is the relative velocity of B with respect to A. Note that v B/A = - v A/B. Engr222 Spring 2004 Chapter 12 68