Lect1 Dynamics

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DYNAMICS ( BDA 20103) BDA 20103) LECTURE 1 LECTURE 1 Chapter 1: KINEMATICS Chapter 1: KINEMATICS OF PARTICLES OF PARTICLES 1.1 Rectilinear motion 1.1 Rectilinear motion

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Transcript of Lect1 Dynamics

  • DYNAMICS (BDA 20103) LECTURE 1Chapter 1: KINEMATICS OF PARTICLES1.1 Rectilinear motion

  • *CHAPTER OVERVIEW

  • *Todays Objectives:Students will be able to:Find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path.

  • *An Overview of MechanicsStatics: The study of bodies in equilibrium.Dynamics:1. Kinematics concerned with the geometric aspects of motion2. Kinetics - concerned with the forces causing the motion Mechanics: The study of how bodies react to forces acting on them.

  • DynamicsDynamics consists two distinct parts: kinematics and kinetics.

    Kinematics deals with the study of motion without reference to the force which cause motionsKinetics relate the action of forces on bodies to their resulting motionsFKMP - UTHM*

  • Kinematics: Motion onlyFKMP - UTHM*

  • Kinetics: Interaction force - motionFKMP - UTHM*ForcePath of motion because of the force

  • Particle and rigid bodyFKMP - UTHM*A particle is a point mass. This means the mass is concentrated at a single point and the particle has neither dimensions (height, width, etc) nor orientation (angular position) Under certain conditions a physical body can be modeled as a particle; for example, a.when considering translation of a body, orb. when all forces acting on a body pass through the centre of mass, orc. when the dimensions of a body are very much smaller than those of its path of motion

  • When a body can be replaced by a particleFKMP - UTHM*Replace by a particle (no rotation involved)The body motion cannot be replaced by a particle(rotation involved)

  • Translation - RotationFKMP - UTHM*TranslationCombination Translation and Rotation

  • Kinematic of a particle : Translationstraight, curve and circular pathsFKMP - UTHM*There is no orientation of the bodyCan be simplified as particle

  • *RECTILINEAR KINEMATICS: CONTINUOUS MOTION(Section 12.2) 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.Vector form: r = r - rScalar form: s = s - sThe total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.

  • *VELOCITYVelocity 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 isvavg = r/tThe instantaneous velocity is the time-derivative of position.

    v = dr/dtSpeed is the magnitude of velocity: v = ds/dtAverage speed is the total distance traveled divided by elapsed time: (vsp)avg = sT/ t

  • *ACCELERATIONAcceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2.The instantaneous acceleration is the time derivative of velocity.Vector form: a = dv/dtScalar form: a = dv/dt = d2s/dt2Acceleration can be positive (speed increasing) or negative (speed decreasing).

  • FKMP - UTHM*Particle is slowing down, its speed is decreasing=> decelerating=> v = v v 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

  • *SUMMARY OF KINEMATIC RELATIONS: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.Note that so and vo represent the initial position and velocity of the particle at t = 0.

  • *CONSTANT ACCELERATIONThe three kinematic equations can be integrated for the special case when acceleration is constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2 downward. These equations are:Velocity as a Function of TimePosition as a Function of TimeVelocity as a Function of Position

  • FKMP - UTHM*The car moves in a straight line such that for a short time its velocity is defined by v = (0.9t2 + 0.6t) m/s where t is in sec. Determine it position and acceleration when t = 3s. When t = 0, s = 0.EXAMPLE 12.1

  • FKMP - UTHM*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 cars position can be determined from v = ds/dt, since this equation relates v, s and t. Noting that s = 0 when t = 0, we haveEXAMPLE 12.1

  • FKMP - UTHM*When t = 3s,s = 10.8mEXAMPLE 12.1

  • FKMP - UTHM*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 = 6m/s2EXAMPLE 12.1

  • QUIZ.1.The distance the particle travels is a vector quantity. True False 2. Some objects can be considered as particles provided motion of the body is characterized by motion of its mass center and any rotation of the body can be neglected. True False 3. If the acceleration is zero, the particle cannot move. True False

    4. Kinematics is concerned with the forces that cause the motion. True False FKMP - UTHM*

  • Rectilinear motion at constant velocityFKMP - UTHM*

  • Velocity at constant accelerationFKMP - UTHM*

  • Distance at constant accelerationFKMP - UTHM*

  • Practical situationUsually a particle start moving when the time is set to 0 and the distance goes from 0FKMP - UTHM*

  • Problems The displacement of a mechanical components follows a rule path as a function of time. The function dist (s) = 2t3 -24t + 6 meter a) derive the velocity and acceleration function of time b) calculate the time to reach velocity of 72 m/s. what is the corresponding acceleration at that time? 2)A car starts from rest and reaches a speed of 20m/s after traveling 125m along a straight road. Determine its constant acceleration and the time of travel. FKMP - UTHM*

  • FKMP - UTHM*See U again in.Lecture 2.