Dynamics_chapter_2_kinematics_of_particles Rectilinear Motion To Normal Tangential Coordinates.pdf

CHAPTER 2 KINEMATICS OF PARTICLES

CHAPTER OUTLINE

2/1 Introduction

2/2 Rectilinear Motion

2/3 Plane Curvilinear Motion

2/4 Rectangular Coordinates (x-y)

2/5 Normal and Tangential Coordinates (n-t)

2/6 Polar Coordinates (r-θ)

2/7 Space Curvilinear Motion

2/8 Relative Motion (Translating Axes)

2/9 Constrained Motion of Connected Particles Sean Cayton/The Image Works

Article 2/1 Introduction • Kinematics is the “geometry of motion.” • Particle Motion • Constrained • Unconstrained

• Choice of Coordinates • • • •

Rectangular (x, y, z) Cylindrical (r, θ, z) Spherical (R, θ, 𝜙) Path (n, t)

• Reference Frame • Absolute Motion • Relative Motion

Article 2/2 Rectilinear Motion • Description • Sign Convention • Position, s • Displacement, Δs

Article 2/2 – Velocity • Average Velocity, vav = Δs/Δt

• Instantaneous Velocity, v = ds/dt =

• Sign Convention

• Speed

Article 2/2 – Acceleration • Average Acceleration, aav = Δv/Δt

• Instantaneous Acceleration, a = dv/dt =

=

• Sign Convention

• Other Differential Relationships: a = v dv/ds or

= d /ds

Article 2/2 – Graphical Interpretations (1 of 2) • Functions of Time • Velocity at time t is the slope of the position curve at time t. • Acceleration at time t is the slope of the velocity curve at time t. • The area under the v-t curve during the interval t1 to t2 is the net displacement of the particle during that time interval. • The area under the a-t curve during the interval t1 to t2 is the net change in velocity of the particle during that time interval.

Article 2/2 – Graphical Interpretations (2 of 2) • Functions of Position • The area under the a-s curve between the positions s1 and s2 is one-half the difference of the squared velocities of the particle at the two positions.

• A line drawn perpendicular to the slope of the v-s curve at a position s, can be extended to the position axis to give the acceleration of the particle at that position. 𝐶𝐵/v = dv/ds or 𝐶𝐵 = v dv/ds = a

Article 2/2 – Analytical Differentiation • If position is given as a function of time, s(t), then… • Differentiate once to obtain velocity as a function of time, v(t) • Differentiate a second time to obtain acceleration as a function of time, a(t) • The functions for position, velocity, and acceleration are easily plotted and evaluated at times of interest to obtain desired information. • If position is not given as a function of time, it must be determined by successive integrations of the acceleration, which is determined by the forces which act on the particle.

Article 2/2 – Analytical Integration (1 of 4) • Case 1: Constant Acceleration, a • At time t = t0 = 0, the particle has velocity v = v0 and is at position s = s0. • Substitute into a dt = dv and integrate once to obtain velocity as a function of time.

• Substitute the previous result into v dt = ds and integrate a second time to obtain position as a function of time.

• Substitute into a ds = v dv/ds and integrate once with respect to position to obtain a different relationship.

Article 2/2 – Analytical Integration (2 of 4) • Case 2: Acceleration as a Function of Time, a = f(t) • At time t = t0 = 0, the particle has velocity v = v0 and is at position s = s0. • Substitute into a dt = dv and integrate once to obtain velocity as a function of time.

• Substitute the previous result into v dt = ds and integrate a second time to obtain position as a function of time.

• If desired, the displacement s can be obtained by a direct solution of a second-order differential equation of the form 𝑠 = f(t), but this is more difficult.

Article 2/2 – Analytical Integration (3 of 4) • Case 3: Acceleration as a Function of Velocity, a = f(v) • At time t = t0 = 0, the particle has velocity v = v0 and is at position s = s0. • Substitute into a = f(v) = dv/dt, separate variables, and integrate once to obtain the time as a function of velocity.

• This function could be inverted to obtain velocity as a function of time, which could then be integrated, as before, to obtain position as a function of time, or differentiated, to obtain acceleration as a function of time. • Substitute into a = f(v) = v dv/ds, separate variables, and integrate once to obtain a relationship between position and velocity.

Article 2/2 – Analytical Integration (4 of 4) • Case 4: Acceleration as a Function of Position, a = f(s) • At time t = t0 = 0, the particle has velocity v = v0 and is at position s = s0. • Substitute into a = f(s) = v dv/ds, separate variables, and integrate once to obtain a relationship between position and velocity.

• Solve for velocity as a function of position v = g(s), substitute into v = ds/dt, separate variables, and integrate once to obtain time as a function of position.

• This function can be inverted to obtain position as a function of time, differentiated once to obtain velocity as a function of time, and differentiated a second time to obtain acceleration as a function of time.

Article 2/2 – Sample Problem 2/1 (1 of 4) • Problem Statement The position coordinate of a particle which is confined to move along a straight line is given by s = 2t3 − 24t + 6, where s is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (b) the acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval from t = 1 s to t = 4 s.

Article 2/2 – Sample Problem 2/1 (2 of 4) • Velocity and Acceleration

• Plots of the Motion • (a) Time when v = 72 m/s

Article 2/2 – Sample Problem 2/1 (3 of 4) • (b) Acceleration when v = 30 m/s

• (c) Net Displacement from t = 1 s to t = 4 s

Article 2/2 – Sample Problem 2/1 (4 of 4) • Comment about the Displacement

Article 2/2 – Sample Problem 2/2 (1 of 3) • Problem Statement A particle moves along the x-axis with an initial velocity vx = 50 ft/sec at the origin when t = 0. For the first 4 seconds it has no acceleration, and thereafter it is acted on by a retarding force which gives it a constant acceleration ax = −10 ft/sec2. Calculate the velocity and the xcoordinate of the particle for the conditions of t = 8 sec and t = 12 sec and find the maximum positive x-coordinate reached by the particle.

Article 2/2 – Sample Problem 2/2 (2 of 2) • Velocity

• Position

Article 2/2 – Sample Problem 2/2 (3 of 3) • Position

• Maximum Positive x-Coordinate

Article 2/2 – Sample Problem 2/3 (1 of 3) • Problem Statement The spring-mounted slider moves in the horizontal guide with negligible friction and has a velocity v0 in the s-direction as it crosses the mid-position where s = 0 and t = 0. The two springs together exert a retarding force to the motion of the slider, which gives it an acceleration proportional to the displacement but oppositely directed and equal to a = −k2s, where k is constant. (The constant is arbitrarily squared for later convenience in the form of the expressions.) Determine the expressions for the displacement s and velocity v as functions of the time t.

Article 2/2 – Sample Problem 2/3 (2 of 3) • Solution I

Article 2/2 – Sample Problem 2/3 (3 of 3) • Solution II

Article 2/2 – Sample Problem 2/4 (1 of 3) • Problem Statement A freighter is moving at a speed of 8 knots when its engines are suddenly stopped. If it takes 10 minutes for the freighter to reduce its speed to 4 knots, determine and plot the distance s in nautical miles moved by the ship and its speed v in knots as functions of the time t during this interval. The deceleration of the ship is proportional to the square of its speed, so that a = −kv2. Note: one knot is the speed of one nautical mile (6076 ft) per hour. Work directly in the units of nautical miles and hours for simplicity.

Article 2/2 – Sample Problem 2/4 (2 of 3) • Velocity

Article 2/2 – Sample Problem 2/4 (3 of 3) • Distance

Article 2/3 Plane Curvilinear Motion • Introduction • Reference Frame • Position, r • Displacement, Δr • Distance, Δs

Article 2/3 – Velocity • Average Velocity, vav = Δr/Δt • Average Speed, vav = Δs/Δt • Instantaneous Velocity, v = dr/dt = • Instantaneous speed, v = ds/dt = • Velocity is always tangent to the path in the direction of motion.

Article 2/3 – Acceleration • Average Acceleration, aav = Δv/Δt • Instantaneous Acceleration, a = dv/dt = • Instantaneous speed, v = ds/dt = • Acceleration, in general, is neither tangent to the path nor normal to the path.

Article 2/3 – Visualization of Motion • General Scenario

• Hodograph • Plot the velocity vectors to scale, emanating from a common point. • Acceleration vectors will be tangent to the path traced out by the tips of the velocity vectors.

Article 2/4 Rectangular Coordinates (x-y) • Introduction • Illustration • Unit Vectors, i and j

• Rectangular coordinates are particularly useful for describing motions where the x- and y-components of acceleration are independently generated or determined.

Article 2/4 – Vector Representation • Vector Relationships of Interest

• Scalar Relationships of Interest

• Note that the time derivatives of the unit vectors i and j are zero because the unit vectors always have the same magnitude and always point in the same directions.

Article 2/4 – Projectile Motion (1 of 2) • Illustration • Assumptions • • • •

Ignore Aerodynamic Drag Ignore Curvature of the Earth Ignore Rotation of the Earth Constant Acceleration: ax = 0 and ay = -g

• Launch Conditions • Launch Speed, v0 • Launch Angle, θ • Launch Coordinates, (x0, y0)

Article 2/4 – Projectile Motion (2 of 2) • Equations of Interest

• Typical Calculations • • • • •

Find the maximum height which a projectile reaches. Find the maximum range or distance covered by a projectile. Find a launch speed and/or launch angle to ensure a particular impact point. Locate a projectile at a particular instant after launch. Determine the velocity components of a projectile at a particular instant after launch.

Article 2/4 – Sample Problem 2/5 (1 of 3) • Problem Statement The curvilinear motion of a particle is defined by vx = 50 − 16t and y = 100 − 4t2, where vx is in meters per second, y is in meters, and t is in seconds. It is also known that x = 0 when t = 0. Plot the path of the particle and determine its velocity and acceleration when the position y = 0 is reached.

Article 2/4 – Sample Problem 2/5 (2 of 3) • Velocity and Acceleration Expressions

• Plot of the Displacement

Article 2/4 – Sample Problem 2/5 (2 of 3) • Velocity and Acceleration Values

Article 2/4 – Sample Problem 2/6 (1 of 3) • Problem Statement A team of engineering students designs a medium-size catapult which launches 8-lb steel spheres. The launch speed is v0 = 80 ft/sec, the launch angle is 𝜃 = 35° above the horizontal, and the launch position is 6 ft above ground level. The students use an athletic field with an adjoining slope topped by an 8-ft fence as shown. Determine: a) the time duration tƒ of the flight b) the x-y coordinates of the point of first impact c) the maximum height h above the horizontal field attained by the ball d) the velocity (expressed as a vector) with which the projectile strikes the ground (or the fence) Repeat part (b) for a launch speed of v0 = 75 ft/sec.

Article 2/4 – Sample Problem 2/6 (2 of 3) • Check y-Displacement at Fence

Article 2/4 – Sample Problem 2/6 (3 of 3) • Find Maximum height

Article 2/5 Normal and Tangential Coordinates (n-t) • Introduction

• Path Coordinates • The t-axis is tangent to the path and is positive in the direction of motion. • The n-axis is perpendicular to the path and is positive toward the center of curvature of the path. It can shift from one side of the path to the other if the curvature changes. • Because the coordinates travel with the particle, there is no utility in writing a position vector. • These coordinates provide a very natural description for curvilinear motion and are often the most direct and convenient coordinates to use.

Article 2/5 – Velocity with n-t Coordinates • Unit Vectors, et and en • Center of Curvature, C • Radius of Curvature, ρ • Differential Angle, dβ • Arclength, ds = ρ dβ • Speed, v = ds/dt = ρ dβ/dt = ρ • Velocity, v = vet = ρ et

Article 2/5 – Acceleration with n-t Coordinates (1 of 4) • Time Derivative of the Velocity

• Time Rate of Change of Speed, • Question: What is

𝑡?

• While the unit vector in the t-direction is constant in magnitude, it does change directions as it follows the particle along the curved path, therefore, its time derivative is not zero.

Article 2/5 – Acceleration with n-t Coordinates (2 of 4) • Time Derivative of et

• Unit vector et will rotate through a differential angle dβ in a differential time dt, to the et′ configuration.

• The vector difference det has a magnitude equal to the arclength et dβ = 1 dβ = dβ, and points along the n-axis. • Thus, the vector difference is written as det = dβ en. • Dividing both sides by the differential time dt yields the following result: det/dt = 𝐞𝑡 = dβ/dt en = 𝛽en. • Now, the term v𝐞𝑡 = v𝛽en = v2/ρen. • Note the preceding derivation assumes dβ is in radians.

Article 2/5 – Acceleration with n-t Coordinates (3 of 4) • Final Expression for Acceleration

• Scalar Expressions of Interest

Article 2/5 – Acceleration with n-t Coordinates (4 of 4) • Comments • The normal component of acceleration is an = v2/ρ and represents the component of acceleration responsible for changing the direction of the velocity vector as the particle moves along the path. • The tangential component of acceleration is at = 𝑣 and represents the component of acceleration responsible for changing the speed of the particle as it moves along the path. For this reason, it is referred to as the time rate of change of speed.

Article 2/5 – Geometric Interpretation (1 of 2) • The change in the velocity vector as the particle moves from A to A′ is dv. This establishes the direction of the acceleration vector a. • The change in velocity has components along the n- and taxes which are dvn and dvt, respectively. • The normal component of acceleration is an = dvn/dt, and this component is always directed toward the center of curvature of the path. • The tangential component of acceleration is at = dvt/dt, and this component is in the direction of the velocity if the particle is speeding up, and opposite the direction of the velocity if the particle is slowing down.

Article 2/5 – Geometric Interpretation (2 of 2) • Acceleration Vectors for General Scenarios

Article 2/5 – Circular Motion in n-t Coordinates (1 of 2) • Illustration

• Special Case of Curvilinear Motion • Radius of Curvature is a Constant Radius, ρ → r • Angle β is replaced by the angle θ, β → θ • New Expressions

Article 2/5 – Circular Motion in n-t Coordinates (2 of 2) • Angular Velocity, • SI Units: rad/s or deg/s • U.S. Units: rad/sec or deg/sec

• Angular Acceleration, • SI Units: rad/s2 or deg/s2 • U.S. Units: rad/sec2 or deg/sec2

• Conversions – Always Work in Radians! • To convert from deg/s to rad/s (or deg/sec to rad/sec) multiply the degrees by π/180° • The same conversion holds for deg/s2 to rad/s2 (or deg/sec2 to rad/sec2). • The conversion is on the degrees, not the unit of time.

Article 2/5 – Sample Problem 2/7 (1 of 4) • Problem Statement To anticipate the dip and hump in the road, the driver of a car applies her brakes to produce a uniform deceleration. Her speed is 100 km/h at the bottom A of the dip and 50 km/h at the top C of the hump, which is 120 m along the road from A. If the passengers experience a total acceleration of 3 m/s2 at A and if the radius of curvature of the hump at C is 150 m, calculate (a) the radius of curvature 𝜌 at A, (b) the acceleration at the inflection point B, and (c) the total acceleration at C.

Article 2/5 – Sample Problem 2/7 (2 of 4) • Velocities and Acceleration

Article 2/5 – Sample Problem 2/7 (3 of 4) • Condition at A

• Condition at B

Article 2/5 – Sample Problem 2/7 (4 of 4) • Condition at C

Article 2/5 – Sample Problem 2/8 (1 of 3) • Problem Statement A certain rocket maintains a horizontal attitude of its axis during the powered phase of its flight at high altitude. The thrust imparts a horizontal component of acceleration of 20 ft/sec2, and the downward acceleration component is the acceleration due to gravity at that altitude, which is g = 30 ft/sec2. At the instant represented, the velocity of the mass center G of the rocket along the 15° direction of its trajectory is 12,000 mi/hr. For this position determine (a) the radius of curvature of the flight trajectory, (b) the rate at which the speed v is increasing, (c) the angular rate 𝛽of the radial line from G to the center of curvature C, and (d) the vector expression for the total acceleration a of the rocket.

Article 2/5 – Sample Problem 2/8 (2 of 3) • Acceleration Components

• Radius of Curvature

Article 2/5 – Sample Problem 2/8 (3 of 3) • Rate of Change of Speed

• Angular Rate of line GC

• Vector Expression for Acceleration