Eighth Edition VECTOR MECHANICS FOR ENGINEERS: CHAPTER �� DYNAMICS Ferdinand P. Beer Kinematics of Particles E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University � ������������������������������������������������������� ��
Edition Eighth Vector Mechanics for Engineers: Dynamics Contents Introduction Sample Problem 11.5 Rectilinear Motion: Position, Velocity Graphical Solution of Rectilinear-Motion & Acceleration Problems Determination of the Motion of a Other Graphical Methods Particle Curvilinear Motion: Position, Velocity & Sample Problem 11.2 Acceleration Sample Problem 11.3 Derivatives of Vector Functions Uniform Rectilinear-Motion Rectangular Components of Velocity and Uniformly Accelerated Rectilinear- Acceleration Motion Motion Relative to a Frame in Translation Motion of Several Particles: Relative Tangential and Normal Components Motion Radial and Transverse Components Sample Problem 11.4 Sample Problem 11.10 Motion of Several Particles: Dependent Sample Problem 11.12 Motion � ������������������������������������������������������� �� 11 - 2
Edition Eighth Vector Mechanics for Engineers: Dynamics Introduction • Dynamics includes: - Kinematics : study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion. - Kinetics : study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion. • Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line. • Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions. � ������������������������������������������������������� �� 11 - 3
Edition Eighth Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Particle moving along a straight line is said to be in rectilinear motion . • Position coordinate of a particle is defined by positive or negative distance of particle from a fixed origin on the line. • The motion of a particle is known if the position coordinate for particle is known for every value of time t . Motion of the particle may be expressed in the form of a function, e.g., 2 3 = − x 6 t t or in the form of a graph x vs. t . � ������������������������������������������������������� �� 11 - 4
Edition Eighth Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Consider particle which occupies position P at time t and P’ at t + ∆ t , ∆ x = Average velocity ∆ t ∆ x = = lim v Instantaneous velocity ∆ t ∆ → t 0 • Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed . • From the definition of a derivative, ∆ x dx = = v lim ∆ t dt ∆ → t 0 2 3 = − e.g., x 6 t t dx 2 = = − v 12 t 3 t dt � ������������������������������������������������������� �� 11 - 5
Edition Eighth Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with velocity v at time t and v’ at t + ∆ t , ∆ v Instantaneous acceleration = = lim a ∆ t ∆ → t 0 • Instantaneous acceleration may be: - positive: increasing positive velocity or decreasing negative velocity - negative: decreasing positive velocity or increasing negative velocity. • From the definition of a derivative, 2 ∆ v dv d x = = = a lim 2 ∆ t dt ∆ → t 0 dt 2 = − e.g. v 12 t 3 t dv = = − a 12 6 t dt � ������������������������������������������������������� �� 11 - 6
Edition Eighth Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with motion given by 2 3 = − x 6 t t dx 2 = = − 12 3 v t t dt 2 dv d x = = = − a 12 6 t 2 dt dt • at t = 0, x = 0, v = 0, a = 12 m/s 2 • at t = 2 s, x = 16 m, v = v max = 12 m/s, a = 0 • at t = 4 s, x = x max = 32 m, v = 0, a = -12 m/s 2 • at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s 2 � ������������������������������������������������������� �� 11 - 7
Edition Eighth Vector Mechanics for Engineers: Dynamics Determination of the Motion of a Particle • Recall, motion of a particle is known if position is known for all time t . • Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations. • Three classes of motion may be defined for: - acceleration given as a function of time , a = f ( t ) - acceleration given as a function of position , a = f( x ) - acceleration given as a function of velocity , a = f( v ) � ������������������������������������������������������� �� 11 - 8
Edition Eighth Vector Mechanics for Engineers: Dynamics Determination of the Motion of a Particle • Acceleration given as a function of time , a = f ( t ): ( ) v t t t dv � � � ( ) ( ) ( ) ( ) ( ) = = = = − = a f t dv f t dt dv f t dt v t v f t dt 0 dt v 0 0 0 ( ) x t t t dx � � � ( ) ( ) ( ) ( ) ( ) = = = − = v t dx v t dt dx v t dt x t x v t dt 0 dt x 0 0 0 • Acceleration given as a function of position , a = f ( x ): dx dx dv dv ( ) = = = = = v or dt a or a v f x dt v dt dx ( ) v x x x 2 2 � � � ( ) ( ) 1 ( ) 1 ( ) = = − = v dv f x dx v dv f x dx v x v f x dx 0 2 2 v x x 0 0 0 � ������������������������������������������������������� �� 11 - 9
Edition Eighth Vector Mechanics for Engineers: Dynamics Determination of the Motion of a Particle • Acceleration given as a function of velocity, a = f ( v ): ( ) v t t dv dv dv ( ) � � = = = = a f v dt dt ( ) ( ) dt f v f v v 0 0 ( ) v t dv � = t ( ) f v v 0 ( ) ( ) x t v t dv v dv v dv � � ( ) = = = = v a f v dx dx ( ) ( ) dx f v f v x v 0 0 ( ) v t v dv ( ) � − = x t x 0 ( ) f v v 0 � ������������������������������������������������������� �� 11 - 10
Edition Eighth Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 SOLUTION: • Integrate twice to find v ( t ) and y ( t ). • Solve for t at which velocity equals zero (time for maximum elevation) and evaluate corresponding altitude. • Solve for t at which altitude equals zero Ball tossed with 10 m/s vertical velocity (time for ground impact) and evaluate corresponding velocity. from window 20 m above ground. Determine: • velocity and elevation above ground at time t , • highest elevation reached by ball and corresponding time, and • time when ball will hit the ground and corresponding velocity. � ������������������������������������������������������� �� 11 - 11
Edition Eighth Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 SOLUTION: • Integrate twice to find v ( t ) and y ( t ). dv 2 = = − a 9 . 81 m s dt ( ) v t t � � ( ) = − − = − dv 9 . 81 dt v t v 9 . 81 t 0 v 0 0 � � m m � � ( ) = − v t 10 9 . 81 t � 2 � s s dy = = − v 10 9 . 81 t dt ( ) y t t 2 � � ( ) ( ) 1 = − − = − dy 10 9 . 81 t dt y t y 10 t 9 . 81 t 0 2 y 0 0 � � � � m m 2 ( ) � � � � = + − y t 20 m 10 t 4 . 905 t � � � � 2 s s � ������������������������������������������������������� �� 11 - 12
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