Seventh Edition VECTOR MECHANICS FOR ENGINEERS: CHAPTER �� DYNAMICS Ferdinand P. Beer Kinematics of E. Russell Johnston, Jr. Rigid Bodies Lecture Notes: J. Walt Oler Texas Tech University � ������������������������������������������������������� ��
Edition Seventh Vector Mechanics for Engineers: Dynamics Contents Introduction Absolute and Relative Acceleration in Plane Motion Translation Analysis of Plane Motion in Terms of a Parameter Rotation About a Fixed Axis: Velocity Sample Problem 15.6 Rotation About a Fixed Axis: Acceleration Sample Problem 15.7 Rotation About a Fixed Axis: Representative Slab Sample Problem 15.8 Equations Defining the Rotation of a Rigid Rate of Change With Respect to a Rotating Frame Body About a Fixed Axis Coriolis Acceleration Sample Problem 5.1 Sample Problem 15.9 General Plane Motion Sample Problem 15.10 Absolute and Relative Velocity in Plane Motion About a Fixed Point Motion General Motion Sample Problem 15.2 Sample Problem 15.11 Sample Problem 15.3 Three Dimensional Motion. Coriolis Acceleration Instantaneous Center of Rotation in Plane Frame of Reference in General Motion Motion Sample Problem 15.15 Sample Problem 15.4 Sample Problem 15.5 � ������������������������������������������������������� �� 15 - 2
Edition Seventh Vector Mechanics for Engineers: Dynamics Introduction • Kinematics of rigid bodies: relations between time and the positions, velocities, and accelerations of the particles forming a rigid body. • Classification of rigid body motions: - translation: • rectilinear translation • curvilinear translation - rotation about a fixed axis - general plane motion - motion about a fixed point - general motion � ������������������������������������������������������� �� 15 - 3
Edition Seventh Vector Mechanics for Engineers: Dynamics Translation • Consider rigid body in translation: - direction of any straight line inside the body is constant, - all particles forming the body move in parallel lines. • For any two particles in the body, � � � = + r r r B A B A • Differentiating with respect to time, � � � � � � � � = + = r r r r B A B A A � � = v v B A All particles have the same velocity. • Differentiating with respect to time again, � � � � � � � � � � � � = + = r r r r B A B A A � � = a a B A All particles have the same acceleration. � ������������������������������������������������������� �� 15 - 4
Edition Seventh Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Velocity • Consider rotation of rigid body about a fixed axis AA’ � = � v d r dt • Velocity vector of the particle P is v = tangent to the path with magnitude ds dt ( ) ( ) ∆ = ∆ θ = φ ∆ θ s BP r sin ∆ θ ds ( ) � = = φ = θ φ v lim r sin r sin ∆ ∆ → dt t t 0 • The same result is obtained from � � � � d r = = ω × v r dt � � � � ω = ω = θ = k k angular ve locity � ������������������������������������������������������� �� 15 - 5
Edition Seventh Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Acceleration • Differentiating to determine the acceleration, � � � � d v d ( ) = = ω × a r dt dt � � ω � � d d r = × + ω × r dt dt � ω � � � d = × + ω × r v dt � ω � d = α = angular ac celeration • dt � � � � � = α = ω � = θ k k k • Acceleration of P is combination of two vectors, � � � � � � = α × + ω × ω × a r r � � α × = r tangentia l accelerati on component � � � ω × ω × = r radial accelerati on component � ������������������������������������������������������� �� 15 - 6
Edition Seventh Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Representative Slab • Consider the motion of a representative slab in a plane perpendicular to the axis of rotation. • Velocity of any point P of the slab, � � � � � = ω × = ω × v r k r = ω v r • Acceleration of any point P of the slab, � � � � � � = α × + ω × ω × a r r � � � 2 = α × − ω k r r • Resolving the acceleration into tangential and normal components, � � � = α × = α a k r a r t t � � 2 2 = − ω = ω a r a r n n � ������������������������������������������������������� �� 15 - 7
Edition Seventh Vector Mechanics for Engineers: Dynamics Equations Defining the Rotation of a Rigid Body About a Fixed Axis • Motion of a rigid body rotating around a fixed axis is often specified by the type of angular acceleration. θ θ d d ω = = • Recall or dt ω dt 2 ω θ ω d d d α = = = ω θ 2 dt d dt • Uniform Rotation, α = 0: θ = θ + ω t 0 • Uniformly Accelerated Rotation, α = constant: ω = ω + α t 0 2 θ = θ + ω + α 1 t t 0 0 2 ( ) 2 2 ω = ω + α θ − θ 2 0 0 � ������������������������������������������������������� �� 15 - 8
Edition Seventh Vector Mechanics for Engineers: Dynamics Sample Problem 15.1 SOLUTION: • Due to the action of the cable, the tangential velocity and acceleration of D are equal to the velocity and acceleration of C. Calculate the initial angular velocity and acceleration. • Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the Cable C has a constant acceleration of 225 pulley after 2 s. mm/s 2 and an initial velocity of 300 mm/s 2 , both directed to the right. • Evaluate the initial tangential and normal acceleration components of D . Determine (a) the number of revolutions of the pulley in 2 s, (b) the velocity and change in position of the load B after 2 s, and (c) the acceleration of the point D on the rim of the inner pulley at t = 0. � ������������������������������������������������������� �� 15 - 9
Edition Seventh Vector Mechanics for Engineers: Dynamics Sample Problem 15.1 SOLUTION: • The tangential velocity and acceleration of D are equal to the velocity and acceleration of C. � � ( ) = = → � � a a 225 mm s ( ) ( ) = = → v v 300 mm s D C t D C 0 0 ( ) = α ( ) a r = ω v r D t D 0 0 ( ) ( ) a 9 v 12 α = D = = 2 t ω = D = = 3 rad s 0 4 rad s 0 r 3 r 3 • Apply the relations for uniformly accelerated rotation to determine velocity and angular position of pulley after 2 s. ( ) ( ) ω = ω + α = + 2 = t 4 rad s 3 rad s 2 s 10 rad s 0 ( ) ( ( )( ) ) θ = ω + α 2 = + 2 2 1 1 t t 4 rad s 2 s 3 rad s 2 s 0 2 2 = 14 rad � � 1 rev ( ) = = � � = 2 . 23 rev N N 14 rad number of revs � � π 2 rad � ( )( ) = ↑ = ω = v 1250 mm s v r 1 25 mm 10 rad s B B ( )( ) ∆ = ∆ = θ = y 1750 mm y r 1 25 mm 14 rad B B � ������������������������������������������������������� �� 15 - 10
Edition Seventh Vector Mechanics for Engineers: Dynamics Sample Problem 15.1 • Evaluate the initial tangential and normal acceleration components of D . � � ( ) = = 2 → a a 225 mm s D C t ( ) ( )( ) 2 = ω 2 = = 2 a r 7 5 mm 4 rad s 1200 mm s D D 0 n � � ( ) ( ) = → = ↓ 2 2 a 225 mm s a 1200 mm s D D t n Magnitude and direction of the total acceleration, ( ) ( ) = 2 + 2 a a a D D D t n = 2 = 225 + 2 2 a 1220 . 9 mm s 1200 D ( ) a φ = D n tan ( ) a D t 1200 = φ = ° 79 . 4 225 � ������������������������������������������������������� �� 15 - 11
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