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King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 2

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Objective l To introduce the concepts of position, displacement, velocity, and acceleration. l To study particle motion along a straight line.

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Rectilinear Kinematics Section 12.2 l Rectilinear : Straight line motion l Kinematics : Study the geometry of the motion dealing with s, v, a. l Rectilinear Kinematics : To identify at any given instant, the particle’s position, velocity, and acceleration. (All objects such as rockets, projectiles, or vehicles will be considered as particles “has negligible size and shape” particles : has mass but negligible size and shape

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Position l Position : Location of a particle at any given instant with respect to the origin r : Displacement ( Vector ) s : Distance ( Scalar )

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Distance & Displacement l Displacement : defined as the change in position. l r : Displacement ( 3 km ) l s : Distance ( 8 km ) Total length l For straight-line Distance = Displacement s = r s r Vector is direction oriented r positive (left ) r negative (right)

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Velocity & Speed l Velocity : Displacement per unit time l Average velocity : V = r t l Speed : Distance per unit time l Average speed : l sp s T t ( Always positive scalar ) l Speed refers to the magnitude of velocity l Average velocity : avg = s / t

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Velocity (con.) l Instantaneous velocity : For straight-line r = s

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Problem l A particle moves along a straight line such that its position is defined by s = (t 3 – 3 t ) m. Determine the velocity of the particle when t = 4 s. At t = 4 s, the velocity = 3 (4) 2 – 6(4) = 24 m/s

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Acceleration l Acceleration : The rate of change in velocity {(m/s)/s} l Average acceleration : l Instantaneous acceleration : l If v ‘ > v “ Acceleration “ l If v ‘ < v “ Deceleration”

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Problem l A particle moves along a straight line such that its position is defined by s = (t 3 – 3 t ) m. Determine the acceleration of the particle when t = 4 s. l At t = 4 a(4) = 6(4) - 6 = 18 m/s 2

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Problem l A particle moves along a straight line such that its position is defined by s = (t 3 – 12 t t -20 ) cm. Describe the motion of P during the time interval [0,9] t02469 s v a Total time = 9 seconds Total distance = ( )= 145 meter Displacement = form -20 to 61 = 81 meter Average Velocity = 81/9= 9 m/s to the right Speed = 9 m/s Average speed = 145/9 = 16.1 m/s Average acceleration = 27/9= 3 m/s 2 to the right

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

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Problem l A car starts from rest and moves along a straight line with an acceleration of a = ( 3 s -1/3 ) m/s 2. where s is in meters. Determine the car’s acceleration when t = 4 s. Rest t = 0, v = 0

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For constant acceleration a = a c

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Velocity as a Function of Time

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Position as a Function of Time

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velocity as a Function of Position

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Free Fall l Ali and Omar are standing at the top of a cliff of height H. Both throw a ball with initial speed v 0, Ali straight down and Omar straight up. The speed of the balls when they hit the ground are v A and v O respectively. Which of the following is true: (a) v A v O v0v0v0v0 v0v0v0v0 OmarAli H vAvAvAvA vOvOvOvO

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Free fall… Free fall… l Since the motion up and back down is symmetric, intuition should tell you that v = v 0 çWe can prove that your intuition is correct: v0v0v0v0 Omar H v = v 0 Equation: This looks just like Omar threw the ball down with speed v 0, so the speed at the bottom should be the same as Ali’s ball. y = 0

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Free fall… l We can also just use the equation directly: Ali : v0v0v0v0 v0v0v0v0 AliOmar y = 0 Omar: same !! y = H

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Summary l Time dependent acceleration l Constant acceleration This applies to a freely falling object:

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Gravity facts: l g does not depend on the nature of the material! çGalileo ( ) figured this out without fancy clocks & rulers! l Nominally, g = 9.81 m/s 2 çAt the equatorg = 9.78 m/s 2 çAt the North poleg = 9.83 m/s 2 l More on gravity in a few lectures!

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