Kinematics Of A Particle Chapter 12 RC Hibbler

Kinematics Of A Particle
12 Kinematics Of A Particle Course Instructor: Engr. Sajid Yasin Department of Mechanical Technology MNS UET Multan

Lecture Schedule Lecture Timings Tuesday (Theory) 07:30 PM-:09:30 PM
Room F5 07:30 PM-:09:30 PM Wednesday (Lab) D2 06:30 PM-:09:30 PM

Textbooks “Vector Mechanics for Engineers (Dynamics)”
“Engineering Mechanics: Dynamics” R.C. Hibbeler, 12th Edition “Vector Mechanics for Engineers (Dynamics)” Beer and Johnston (Latest Edition) “Engineering Mechanics (Dynamics)” J. L. Meriam (Latest Edition)

Method of Assessment Type Marks * 75 % Attendance Mandatory
Mid-semester Exam 30 End-semester Exam 40 Quiz 10 Assignments Attendance Total 100 * 75 % Attendance Mandatory

Mechanics Mechanics : A branch of physical science which deals with ( the states of rest or motion of ) bodies under action of forces Mechanics can be divided into 3 branches: - Rigid-body Mechanics - Deformable-body Mechanics - Fluid Mechanics Rigid-body Mechanics deals with - Statics - Dynamics

Mechanics Statics – Equilibrium of bodies At rest
Move with constant velocity Dynamics – Accelerated motion of bodies

Application of Mechanics
Statics Dynamics Mech of Materials Fluid Mechanics Vibration Fracture Mechanics Etc. Structures Automotives Robotics Spacecrafts MEMs Etc. Mechanics MEMs = Micro-Electro-Mechanical System

1.2 Fundamentals Concepts
position, velocity, acceleration Space: Collection of points whose relative positions can be described using “a coordinate system” Time : Relative occurrence of events or Measure of succession of events Mass Measure of inertia of a body (Its resistance to change in velocity)

Force: Vector quantity
An agent which produces or tends to produce motion, destroy or tends to destroy the motion of the body. A force acting on the body may a) Change the motion of the body b) Retard the motion of the body c) Balance the forces already acting d) Give rise to internal stresses To determine the effect of forces acting on the body Magnitude of the force Line of the action of the force Nature of the force i.e. Push or Pull Point of the application of force The SI unit = newton ( N= (kg·m/s2)

Idealizations Concentrated Force: Effect of a loading which is assumed to act at a point (CG) on a body. Provided that, the area over which the load is applied is very small compared to the overall size of the body. Example: Contact Force between a wheel and ground. 40 kN 160 kN

Fundamentals Concepts
Particle: Body of negligible dimensions Rigid body: Body with negligible deformations Non-rigid body: Body which can deform A body with mass, but negligible dimensions Size of earth is insignificant compared to the size of its orbit. Earth can be modeled as a particle when studying its orbital motion In Statics, bodies are considered rigid unless stated otherwise.

Fundamentals Concepts
Point: Exact dimension, No size, Only Position ○ Line: No Thickness, Extends in both directions infinitely Ray: Line with just one end Line Segment: Line with both ends

NEWTON’S LAWS OF MOTION (1st Law)
The study of rigid body mechanics is formulated on the basis of Newton’s laws of motion. First Law: An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction, unless acted upon by an unbalanced force.

NEWTON’S LAWS OF MOTION (2nd Law)
Second Law: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. m

NEWTON’S LAWS OF MOTION (3rd Law)
Third Law: The mutual forces of action and reaction between two particles are equal in magnitude, opposite in direction, and collinear. Forces always occur in pairs – equal and opposite action-reaction force pairs. Point: Isolate the body Confusing? Concept of FBD (Free Body Diagram)

Newton’s Law of Gravitation
M F - M & m are particle masses G is the universal constant of gravitation, 6.673 x m3/kg-s2 - r is the distance between the particles. r m For Gravity on earth (at sea level) Mass of Earth : × 10^24 kg Radius of Earth : 6,371 km m W=mg where - m is the mass of the body in question - g = GM/R2 = m/s2 (32.2 ft/s2) M

Units of Measurement Quantities Dimensional Symbol SI Units Symbols Mass M Kilogram Kg Length L Meter m Time T Second s Force F newton N F  ma 1N = kg.m/s2 1 newton is the force required to give a mass of 1 kg an acceleration of 1 m/s2 W  mg 1N = kg.m/s2

The International System of Units
Prefixes For a very large or small numerical quantity, units can be modified by using a prefix Each represent a multiple or sub-multiple of a unit

Significant Figures In any measurement, the accurately known digits and the first doubtful digit are called significant. Nonzero digits are always significant e.g (4) 283 (3) Zeroes are sometimes significant and sometimes not Zeroes at the beginning: never significant (2) Zeroes between: always (3) Zeroes at the end after decimal: always (3) Zeroes at the end with no decimal may or may not: 8,000 kg (three, four, five) depending on the accuracy of measuring instrument )

Rounding Off Numbers The process of removing insignificant figures from measured value till last significant figure to be retained. If the first digit dropped is less than 5, the last digit retained should remain unchanged. If the first digit dropped is more than 5, the digit to be retained is increased by one. If the digit to be dropped is 5, the previous digit which is increased by one if it is odd retained as such if it is even.

Numerical Calculations
Accuracy obtained would never be better than the accuracy of the problem data Calculators or computers involve more figures in the answer than the number of significant figures in the data Calculated results should always be “rounded off” to an appropriate number of significant figures Calculations Retain a greater number of digits for accuracy Work out computations so that numbers that are approximately equal Round off final answers to three significant figures

General Procedure for Analysis
To solve problems, it is important to present work in a logical and orderly way as suggested: Correlate actual physical situation with theory Draw any diagrams and tabulate the problem data Apply principles in mathematics forms Solve equations which are dimensionally homogenous Report the answer with significance figures Technical judgment and common sense

IPE Approach (Problem Solving Strategy)

Chapter Objectives To introduce the concepts of position, displacement, velocity, and acceleration. To study particle motion along a straight line and represent this motion graphically. To investigate particle motion a long a curved path using different coordinate systems. To present an analysis of dependent motion of two particles. To examine the principles of relative motion of two particles using translating axes.

Chapter Outline Introduction Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematics: Erratic Motion Curvilinear Motion: Rectangular Components Motion of a Projectile Curvilinear Motion: Normal and Tangential Components Curvilinear Motion: Cylindrical Components Absolute Dependent Motion Analysis of Two Particles Relative Motion Analysis of Two Particles Using Translating Axes

Introduction Mechanics – the state of rest or 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 Kinematics – treats with geometric aspects of the motion dealing with s, v, a, & t. Kinetics – analysis of the forces causing the motion

Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematics – To identify at any given instant, the particle’s position, velocity, and acceleration Position Location of a particle at any given instant with respect to the origin r : Displacement ( Vector ) s : Distance ( Scalar ) +ve = right of origin, -ve = left of origin

Displacement – change in its position, vector quantity r : Displacement ( 3 km ) s : Distance ( 8 km ) Total length If particle moves from P to P’ => Vector is direction oriented D s positive D s negative

Velocity Average velocity, Instantaneous velocity is defined as

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)

Average speed is defined as total distance traveled by a particle, sT, divided by the elapsed time The particle travels along the path of length sT in time =>

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, i.e.

Instantaneous Acceleration at time t is found by taking smaller and smaller values of Δt and corresponding smaller and smaller values of Δv,

Particle is slowing down, its speed is decreasing => decelerating => will be negative. Consequently, a will also be negative, therefore it will act to the left, in the opposite sense to v If velocity is constant, acceleration is zero

Velocity as a Function of Time Integrate ac = dv/dt, Assuming that initially v = v0 when t = 0. Constant Acceleration

Position as a Function of Time Integrate v = ds/dt = v0 + act, Assuming that initially s = s0 when t = 0 Constant Acceleration

Relation involving s, v, and s No time t
Position s Velocity Acceleration

Velocity as a Function of Position Integrate v dv = ac ds, Assuming that initially v = v0 at s = s0 Constant Acceleration

PROCEDURE FOR ANALYSIS Coordinate System Establish a position coordinate s along the path and specify its fixed origin and positive direction. The particle’s position, velocity, and acceleration, can be represented as s, v and a respectively and their sense is then determined from their algebraic signs.

2) Kinematic Equation If a relationship is known between any two of the four variables a, v, s and t, then a third variable can be obtained by using one of the three the kinematic equations The positive sense for each scalar can be indicated by an arrow shown alongside each kinematics equation as it is applied

When integration is performed, it is important that position and velocity be known at a given instant in order to evaluate either the constant of integration if an indefinite integral is used, or the limits of integration if a definite integral is used Remember that the three kinematics equations can only be applied to situation where the acceleration of the particle is constant.

EXAMPLE 12.1 The car moves in a straight line such that for a short time its velocity is defined by v = (3t2 + 2t) ft/s where t is in sec. Determine it position and acceleration when t = 3s. When t = 0, s = 0.

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

EXAMPLE 12.1 When t = 3s, s = (3)3 + (3)2= 36ft

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 = 3s, a = 6(3) + 2 = 20ft/s2

EXAMPLE 12.2 A small projectile is forced downward into a
fluid medium with an initial velocity of 60m/s. Due to the resistance of the fluid the projectile experiences a deceleration equal to a = (-0.4v3)m/s2, where v is in m/s2. Determine the projectile’s velocity and position 4s after it is fired.

EXAMPLE 12.2 Solution: Coordinate System. Since the motion is
downward, the position coordinate is downwards positive, with the origin located at O. Velocity. Here a = f(v), velocity is a function of time using a = dv/dt, since this equation relates v, a and t.

EXAMPLE 12.2 When t = 4s, v = m/s

EXAMPLE 12.2 Position. Since v = f(t), the projectile’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

EXAMPLE 12.2 When t = 4s, s = 4.43m

EXAMPLE 12.3 During a test, a rocket travel upward at 75m/s. When it is 40m from the ground, the engine fails. Determine max height sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s2 due to gravity. Neglect the effect of air resistance.

EXAMPLE 12.3 Solution: Coordinate System. Origin O for the position coordinate at ground level with positive upward. Maximum Height. Rocket traveling upward, vA = +75m/s when t = 0. s = sB when vB = 0 at max ht. For entire motion, acceleration aC = -9.81m/s2 (negative since it act opposite sense to positive velocity or positive displacement)

EXAMPLE 12.3 sB = 327 m Velocity. The negative root was chosen since the rocket is moving downward

EXAMPLE 12.4 A metallic particle travels downward through a fluid that extends from plate A and plate B under the influence of magnetic field. If particle is released from rest at midpoint C, s = 100 mm, and acceleration, a = (4s) m/s2, where s in meters, determine velocity when it reaches plate B and time need to travel from C to B

EXAMPLE 12.4 Solution: Coordinate System. It is shown that s is taken positive downward, measured from plate A Velocity. Since a = f(s), velocity as a function of position can be obtained by using v dv = a ds. Realizing v = 0 at s = 100mm = 0.1m

EXAMPLE 12.4 At s = 200mm = 0.2m,

EXAMPLE 12.4 At s = 200mm = 0.2m, t = 0.658s

EXAMPLE 12.5 A particle moves along a horizontal path with a velocity of v = (3t2 – 6t) m/s. if it is initially located at the origin O, Determine the distance traveled in 3.5s and the particle’s average velocity and speed during the time interval.

EXAMPLE 12.5 Solution: Coordinate System. Assuming positive motion to the right, measured from the origin, O Distance traveled. Since v = f(t), the position as a function of time may be found integrating v = ds/dt with t = 0, s = 0.

EXAMPLE 12.5 0 ≤ t < 2 s -> -ve velocity -> the particle is moving to the left, t > 2a -> +ve velocity -> the particle is moving to the right

EXAMPLE 12.5 The distance traveled in 3.5s is
sT = = m Velocity. The displacement from t = 0 to t = 3.5s is Δs = – 0 = 6.125m And so the average velocity is Average speed,

Exercise Problems 12.1 to 12.25 Home-Work Problems
Selected problems will be solved in Class-Room Rest of problems will be submitted as assignment within one-week time.

Problem # 12.4 v2 = v ac(s - s0) v = v0 + act

Problem # 12.6 -h a=-32.2 s = s0 + v0t + 1 2 act2
h = 127 ft v = v0 + act

Problem # 12.10 v = v0 + act t = 13.33 s Distance Covered by A
v2 = v ac (s - s0) 802 = 0 + 2(6)(s1 - 0) s1 = ft s2 = vt(total time-13.33) Car B travels in the opposite direction with a constant velocity of and the distance traveled in is s3 = vt1 = 60t1 S1+S2+S3=6000 s1 + s2 + s3 = 6000 (t ) + 60t1 = 6000 t1 = s It is required that The distance traveled by car A is sA = s1 + s2 = ( ) = 3200 ft Ans.

Problem# 12.19 Till Y ac=0.6 v2 = v0 2 + 2 ac (s - s0)
v2max = 0 + 2(0.6)(y - 0) After Y ac=-0.3 0 = v2max + 2(-0.3)(48 - y)-----2 Putting 1 into 2 Y=16 Vmax=4.382<8ft/s Now time ac=0.6,v=4.38 v = v0 + act 4.382 = t1 Now time ac=-0.3,v0=4.38, v=0 0 = t2 Total Time = t1+t2

Problem# 12.21

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). Example : r = {sin (2t) i + cos (2t) j – 0.5 t k} Path is described in three dimensions Position, velocity, and acceleration are vectors * S is a path function

* S is a path function 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 change in particle’s position.

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, As Delta-t = 0 then Delta-r = Delta-s

Since Δr is tangent to the curve at P, therefore, 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`. Delta-r = Delta-s

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

a acts tangent to the hodograph, therefore it is not tangent to the path a is not tangent to the path of motion a directed toward the inside or concave side Hodograph: A curve the radius vector of which represents in magnitude and direction the velocity of a moving object. The Hodograph is a vector diagram showing how velocity changes with position or time.

Curvilinear Motion: Rectangular Components
Position. 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

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.

Acceleration. where The acceleration has a magnitude defined as the positive value of

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.

PROCEDURE FOR ANALYSIS Coordinate System A rectangular coordinate system can be used to solve problems for which the motion can conveniently be expressed in terms of its x, y and z components.

Kinematic Quantities Since the rectilinear motion occurs along each coordinate axis, the motion of each component is found using v = ds/dt and a = dv/dt, or a ds = v ds Once the x, y, z components of v and a have been determined. The magnitudes of these vectors are found from the Pythagorean theorem and their directions from the components of their unit vectors.

EXAMPLE 12.9 At any instant the horizontal position of the weather balloon is defined by x = (9t) m, where t is in second. If the equation of the path is y = x2/30, Determine the distance of the balloon from the station at A, the magnitude and direction of the both the velocity and acceleration when t = 2 s.

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.

EXAMPLE 12.9 When t = 2 s, the magnitude of velocity is
The direction is tangent to the path, where Acceleration.

EXAMPLE 12.9 The direction of a is

EXAMPLE 12.10 For a short time, the path of the plane is described by y = (0.001x2) m. If the plane is rising with a constant velocity of 10 m/s, Determine the magnitudes of the velocity and acceleration of the plane when it is at y=100 m.

EXAMPLE 12.10 Solution: Position.

EXAMPLE 12.10 Velocity.

EXAMPLE 12.10 Acceleration.

Exercise Problems 12.71 to 12.86 Home-Work Problems
Selected problems will be solved in Class-Room

Problem 12.75 Problem 12.75

Problem 80 Problem 12.80

Kinematics Of A Particle Chapter 12 RC Hibbler

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