Chapter 12 Kinematics Of A Particle Chapter Outline

Chapter 12 : Kinematics Of A Particle

Chapter Outline • Introduction • Rectilinear Kinematics: Continuous Motion • Rectilinear Kinematics: Erratic Motion • Curvilinear Motion: Rectangular Components • Motion of a Projectile © 2007 Pearson Education South Asia Pte Ltd

Introduction • Mechanics – the state of rest of motion of bodies subjected to the action of forces • Static – equilibrium of a body that is either at rest or moves with constant velocity • Dynamics – deals with accelerated motion of a body 1) Kinematics – treats with geometric aspects of the motion 2) Kinetics – analysis of the forces causing the motion © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Rectilinear Kinematics – specifying at any instant, the particle’s position, velocity, and acceleration • Position 1) Single coordinate axis, s 2) Origin, O 3) Position vector r – specific location of particle P at any instant © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion 4) Algebraic Scalar s in metres Note : - Magnitude of s = Dist from O to P - The sense (arrowhead dir of r) is defined by algebraic sign on s => +ve = right of origin, -ve = left of origin © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Displacement – change in its position, vector quantity © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • If particle moves from P to P’ => is +ve if particle’s position is right of its initial position is -ve if particle’s position is left of its initial position © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Velocity Average velocity, Instantaneous velocity is defined as © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion Representing as an algebraic scalar, Velocity is +ve = particle moving to the right Velocity is –ve = Particle moving to the left Magnitude of velocity is the speed (m/s) © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion Average speed is defined as total distance traveled by a particle, s. T, divided by the elapsed time. The particle travels along the path of length s. T in time => © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Acceleration – velocity of particle is known at points P and P’ during time interval Δt, average acceleration is • Δv represents difference in the velocity during the time interval Δt, ie © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion Instantaneous acceleration at time t is found by taking smaller and smaller values of Δt and corresponding smaller and smaller values of Δv, © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Velocity as a Function of Time Integrate ac = dv/dt, assuming that initially v = v 0 when t = 0. Constant Acceleration © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Position as a Function of Time Integrate v = ds/dt = v 0 + act, assuming that initially s = s 0 when t = 0 Constant Acceleration © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Continuous Motion • Velocity as a Function of Position Integrate v dv = ac ds, assuming that initially v = v 0 at s = s 0 Constant Acceleration © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 1 The car moves in a straight line such that for a short time its velocity is defined by v = (0. 9 t 2 + 0. 6 t) m/s where t is in sec. Determine it position and acceleration when t = 3 s. When t = 0, s = 0. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 1 Solution: Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right. Position. Since v = f(t), the car’s position can be determined from v = ds/dt, since this equation relates v, s and t. Noting that s = 0 when t = 0, we have © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 1 When t = 3 s, s = 10. 8 m © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 1 Acceleration. Knowing v = f(t), the acceleration is determined from a = dv/dt, since this equation relates a, v and t. When t = 3 s, a = 6 m/s 2 © 2007 Pearson Education South Asia Pte Ltd

© 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion • When particle’s motion is erratic, it is best described graphically using a series of curves that can be generated experimentally from computer output. • a graph can be established describing the relationship with any two of the variables, a, v, s, t • using the kinematics equations a = dv/dt, v = ds/dt, a ds = v dv © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Given the s-t Graph, construct the v-t Graph • The s-t graph can be plotted if the position of the particle can be determined experimentally during a period of time t. • To determine the particle’s velocity as a function of time, the v-t Graph, use v = ds/dt • Velocity as any instant is determined by measuring the slope of the s-t graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Slope of s-t graph = velocity © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Given the v-t Graph, construct the a-t Graph • When the particle’s v-t graph is known, the acceleration as a function of time, the a-t graph can be determined using a = dv/dt • Acceleration as any instant is determined by measuring the slope of the v-t graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Slope of v-t graph = acceleration © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion • Since differentiation reduces a polynomial of degree n to that of degree n-1, then if the s-t graph is parabolic (2 nd degree curve), the v-t graph will be sloping line (1 st degree curve), and the a-t graph will be a constant or horizontal line (zero degree curve) © 2007 Pearson Education South Asia Pte Ltd

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 ≤ 30 s. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 6 Solution: v-t Graph. The v-t graph can be determined by differentiating the eqns defining the s-t graph The results are plotted. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 6 We obtain specify values of v by measuring the slope of the s-t graph at a given time instant. a-t Graph. The a-t graph can be determined by differentiating the eqns defining the lines of the v-t graph. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 6 The results are plotted. © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Given the a-t Graph, construct the v-t Graph • When the a-t graph is known, the v-t graph may be constructed using a = dv/dt Change in = Area under velocity a-t graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion • Knowing particle’s initial velocity v 0, and add to this small increments of area (Δv) • Successive points v 1 = v 0 + Δv, for the v-t graph • Each eqn for each segment of the a-t graph may be integrated to yield eqns for corresponding segments of the v-t graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Given the v-t Graph, construct the s-t Graph • When the v-t graph is known, the s-t graph may be constructed using v = ds/dt Displacement = Area under v -t graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion • knowing the initial position s 0, and add to this area increments Δs determined from v-t graph. • describe each of there segments of the v-t graph by a series of eqns, each of these eqns may be integrated to yield eqns that describe corresponding segments of the s -t graph © 2007 Pearson Education South Asia Pte Ltd

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? © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 7 Solution: v-t Graph. The v-t graph can be determined by integrating the straight-line segments of the a-t graph. Using initial condition v = 0 when t = 0, © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 7 When t = 10 s, v = 100 m/s, using this as initial condition for the next time period, we have When t = t’ we require v = 0. This yield t’ = 60 s s-t Graph. Integrating the eqns of the v-t graph yields the corresponding eqns of the s-t graph. Using the initial conditions s = 0 when t = 0, © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 7 When t = 10 s, s = 500 m. Using this initial condition, When t’ = 60 s, the position is s = 3000 m © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Given the a-s Graph, construct the v-s Graph • v-s graph can be determined by using v dv = a ds, integrating this eqn between the limit v = v 0 at s = s 0 and v = v 1 at s = s 1 Area under a-s graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion • determine the eqns which define the segments of the a-s graph • corresponding eqns defining the segments of the v-s graph can be obtained from integration, using vdv = a ds © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion Given the v-s Graph, construct the a-s Graph • v-s graph is known, the acceleration a at any position s can be determined using a ds = v dv Acceleration = velocity times slope of v-s graph © 2007 Pearson Education South Asia Pte Ltd

Rectilinear Kinematics: Erratic Motion • At any point (s, v), the slope dv/ds of the v-s graph is measured • Since v and dv/ds are known, the value of a can be calculated © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 8 The v-s graph describing the motion of a motorcycle is shown in Fig 12 -15 a. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 120 m. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 8 Solution: a-s Graph. Since the eqns for the segments of the v-s graph are given, a-s graph can be determined using a ds = v dv. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 8 Time. The time can be obtained using v-s graph and v = ds/dt. For the first segment of motion, s = 0 at t = 0, At s = 60 m, t = 8. 05 s © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 8 For second segment of motion, At s = 120 m, t = 12. 0 s © 2007 Pearson Education South Asia Pte Ltd

General Curvilinear Motion Curvilinear motion occurs when the particle moves along a curved path Position. The position of the particle, measured from a fixed point O, is designated by the position vector r = r(t). © 2007 Pearson Education South Asia Pte Ltd

General Curvilinear Motion Displacement. Suppose during a small time interval Δt the particle moves a distance Δs along the curve to a new position P`, defined by r` = r + Δr. The displacement Δr represents the change in the particle’s position. © 2007 Pearson Education South Asia Pte Ltd

General Curvilinear Motion Velocity. During the time Δt, the average velocity of the particle is defined as The instantaneous velocity is determined from this equation by letting Δt 0, and consequently the direction of Δr approaches the tangent to the curve at point P. Hence, © 2007 Pearson Education South Asia Pte Ltd

General Curvilinear Motion • Direction of vins is tangent to the curve • Magnitude of vins is the speed, which may be obtained by noting the magnitude of the displacement Δr is the length of the straight line segment from P to P`. © 2007 Pearson Education South Asia Pte Ltd

General Curvilinear Motion Acceleration. If the particle has a velocity v at time t and a velocity v` = v + Δv at time t` = t + Δt. The average acceleration during the time interval Δt is © 2007 Pearson Education South Asia Pte Ltd

General Curvilinear Motion a acts tangent to the hodograph, therefore it is not tangent to the path © 2007 Pearson Education South Asia Pte Ltd

Curvilinear Motion: Rectangular Components Position vector is defined by r = xi + yj + zk The magnitude of r is always positive and defined as The direction of r is specified by the components of the unit vector ur = r/r © 2007 Pearson Education South Asia Pte Ltd

Curvilinear Motion: Rectangular Components Velocity. where The velocity has a magnitude defined as the positive value of and a direction that is specified by the components of the unit vector uv=v/v and is always tangent to the path. © 2007 Pearson Education South Asia Pte Ltd

Curvilinear Motion: Rectangular Components Acceleration. where The acceleration has a magnitude defined as the positive value of © 2007 Pearson Education South Asia Pte Ltd

Curvilinear Motion: Rectangular Components • The acceleration has a direction specified by the components of the unit vector ua = a/a. • Since a represents the time rate of change in velocity, a will not be tangent to the path. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 9 At any instant the horizontal position of the weather balloon is defined by x = (9 t) m, where t is in second. If the equation of the path is y = x 2/30, 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. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 9 Solution: Position. When t = 2 s, x = 9(2) m = 18 m and y = (18)2/30 = 10. 8 m The straight-line distance from A to B is m Velocity. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 9 When t = 2 s, the magnitude of velocity is The direction is tangent to the path, where Acceleration. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 9 The direction of a is © 2007 Pearson Education South Asia Pte Ltd

Motion of a Projectile • Free-flight motion studied in terms of rectangular components since projectile’s acceleration always act vertically • Consider projectile launched at (x 0, y 0) • Path defined in the x-y plane • Air resistance neglected • Only force acting on the projectile is its weight, resulting in constant downwards acceleration • ac = g = 9. 81 m/s 2 © 2007 Pearson Education South Asia Pte Ltd

Motion of a Projectile © 2007 Pearson Education South Asia Pte Ltd

Motion of a Projectile Horizontal Motion Since ax = 0, Horizontal component of velocity remain constant during the motion © 2007 Pearson Education South Asia Pte Ltd

Motion of a Projectile Vertical. ay = - g Positive y axis is directed upward, then © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 12 The chipping machine is designed to eject wood at chips v. O = 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. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 12 Coordinate System. Three unknown h, time of flight, t. OA and the vertical component of velocity (v. B)y. Taking origin at O, for initial velocity of a chip, (v. A)x = (v. O)x = 6. 5 m/s and ay = -9. 81 m/s 2 © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 12. 12 Horizontal Motion. Vertical Motion. Relating t. OA to initial and final elevation of the chips, © 2007 Pearson Education South Asia Pte Ltd