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- 1. Eighth EditionCHAPTER VECTOR MECHANICS FOR ENGINEERS:15 DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Kinematics of Lecture Notes: Rigid Bodies J. Walt Oler Texas Tech University © 2007 The McGraw-Hill Companies, Inc. All rights reserved.
- 2. EditionEighth Vector Mechanics for Engineers: Dynamics Contents Introduction Absolute and Relative Acceleration in Plane Translation Motion Rotation About a Fixed Axis: Velocity Analysis of Plane Motion in Terms of a Rotation About a Fixed Axis: Acceleration Parameter Rotation About a Fixed Axis: Sample Problem 15.6 Representative Slab Sample Problem 15.7 Equations Defining the Rotation of a Rigid Sample Problem 15.8 Body About a Fixed Axis Rate of Change With Respect to a Rotating Sample Problem 5.1 Frame General Plane Motion Coriolis Acceleration Absolute and Relative Velocity in Plane Sample Problem 15.9 Motion Sample Problem 15.10 Sample Problem 15.2 Motion About a Fixed Point Sample Problem 15.3 General Motion Instantaneous Center of Rotation in Plane Sample Problem 15.11 Motion Three Dimensional Motion. Coriolis Sample Problem 15.4 Acceleration Sample Problem 15.5 Frame of Reference in General Motion Sample Problem 15.15 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 2
- 3. EditionEighth 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 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 3
- 4. EditionEighth 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 rB = rA + rB A • Differentiating with respect to time, r& r r & & r & rB = rA + rB A = rA r r vB = v A All particles have the same velocity. • Differentiating with respect to time again, r r r r &&B = &&A + &&B A = &&A r r r r r r aB = a A All particles have the same acceleration. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 4
- 5. EditionEighth Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Velocity • Consider rotation of rigid body about a fixed axis AA’ r r • Velocity vector v = dr dt of the particle P is tangent to the path with magnitude v = ds dt ∆s = ( BP )∆θ = (r sin φ )∆θ ds ∆θ v= = lim (r sin φ ) = rθ& sin φ dt ∆t →0 ∆t • The same result is obtained from r r dr r r v= =ω ×r dt r r r ω = ωk = θ&k = angular velocity © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 5
- 6. EditionEighth Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Acceleration • Differentiating to determine the acceleration, r r dv d v r a= = (ω × r ) dt dt r r dω r r dr = ×r +ω × dt dt r dω r r r = ×r +ω ×v dt r dω r • = α = angular acceleration dt r r r = αk = ωk = θ & &&k • Acceleration of P is combination of two vectors, r r r r r r a = α × r + ω ×ω × r r r α × r = tangential acceleration component r r r ω × ω × r = radial acceleration component © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 6
- 7. EditionEighth 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, r r r r r v = ω × r = ωk × r v = rω • Acceleration of any point P of the slab, r r r r r r a = α × r + ω ×ω × r r r r = α k × r − ω 2r • Resolving the acceleration into tangential and normal components, r r r at = αk × r a t = rα r r an = −ω 2 r a n = rω 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 7
- 8. EditionEighth 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. dθ dθ • Recall ω = or dt = dt ω dω d 2θ dω α= = 2 =ω dt dt dθ • Uniform Rotation, α = 0: θ = θ 0 + ωt • Uniformly Accelerated Rotation, α = constant: ω = ω0 + αt θ = θ 0 + ω 0t + 1 α t 2 2 ω 2 = ω 0 + 2α (θ − θ 0 ) 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 8
- 9. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 5.1 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 9 pulley after 2 s. in/s2 and an initial velocity of 12 in/s, 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 9
- 10. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 5.1 SOLUTION: • The tangential velocity and acceleration of D are equal to the velocity and acceleration of C. r r r r (v ) = (v ) = 12 in. s → D 0 C 0 (aD )t = aC = 9 in. s → (vD )0 = rω0 (aD )t = rα (vD )0 (aD )t 9 ω0 = 12 = = 4 rad s α= = = 3 rad s 2 r 3 r 3 • Apply the relations for uniformly accelerated rotation to determine velocity and angular position of pulley after 2 s. ( ) ω = ω 0 + αt = 4 rad s + 3 rad s 2 (2 s ) = 10 rad s ( ) θ = ω 0t + 1 αt 2 = (4 rad s )(2 s ) + 1 3 rad s 2 (2 s )2 2 2 = 14 rad ⎛ 1 rev ⎞ N = (14 rad )⎜ ⎟ = number of revs N = 2.23 rev ⎝ 2π rad ⎠ r vB = rω = (5 in.)(10 rad s ) vB = 50 in. s ↑ ∆y B = rθ = (5 in.)(14 rad ) ∆y B = 70 in. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 10
- 11. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 5.1 • Evaluate the initial tangential and normal acceleration components of D. r r (aD )t = aC = 9 in. s → (aD )n = rDω0 = (3 in.)(4 rad s )2 = 48 in 2 s2 r r (aD )t = 9 in. s2 → (aD )n = 48 in. s2 ↓ Magnitude and direction of the total acceleration, aD = (aD )t2 + (aD )n 2 = 92 + 482 aD = 48.8 in. s 2 (aD )n tan φ = (aD )t 48 = 9 φ = 79.4° © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 11
- 12. EditionEighth Vector Mechanics for Engineers: Dynamics General Plane Motion • General plane motion is neither a translation nor a rotation. • General plane motion can be considered as the sum of a translation and rotation. • Displacement of particles A and B to A2 and B2 can be divided into two parts: ′ - translation to A2 and B1 ′ - rotation of B1 about A2 to B2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 12
- 13. EditionEighth Vector Mechanics for Engineers: Dynamics Absolute and Relative Velocity in Plane Motion • Any plane motion can be replaced by a translation of an arbitrary reference point A and a simultaneous rotation about A. r r r vB = v A + vB A r r r v B A = ω k × rB A v B A = rω r r r r v B = v A + ω k × rB A © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 13
- 14. EditionEighth Vector Mechanics for Engineers: Dynamics Absolute and Relative Velocity in Plane Motion • Assuming that the velocity vA of end A is known, wish to determine the velocity vB of end B and the angular velocity ω in terms of vA, l, and θ. • The direction of vB and vB/A are known. Complete the velocity diagram. vB vA vA = tan θ = = cosθ vA v B A lω v B = v A tan θ vA ω= l cosθ © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 14
- 15. EditionEighth Vector Mechanics for Engineers: Dynamics Absolute and Relative Velocity in Plane Motion • Selecting point B as the reference point and solving for the velocity vA of end A and the angular velocity ω leads to an equivalent velocity triangle. • vA/B has the same magnitude but opposite sense of vB/A. The sense of the relative velocity is dependent on the choice of reference point. • Angular velocity ω of the rod in its rotation about B is the same as its rotation about A. Angular velocity is not dependent on the choice of reference point. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 15
- 16. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.2 SOLUTION: • The displacement of the gear center in one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities. • The velocity for any point P on the gear The double gear rolls on the may be written as stationary lower rack: the velocity of r r r r r r vP = v A + vP A = v A + ωk × rP A its center is 1.2 m/s. Determine (a) the angular velocity of Evaluate the velocities of points B and D. the gear, and (b) the velocities of the upper rack R and point D of the gear. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 16
- 17. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.2 SOLUTION: • The displacement of the gear center in one revolution is equal to the outer circumference. For xA > 0 (moves to right), ω < 0 (rotates clockwise). xA θ =− x A = − r1θ 2π r 2π y Differentiate to relate the translational and angular velocities. x v A = − r1ω vA 1.2 m s r r r ω =− =− ω = ωk = −(8 rad s )k r1 0.150 m © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 17
- 18. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.2 r r r r r r • For any point P on the gear, vP = v A + vP A = v A + ωk × rP A Velocity of the upper rack is equal to Velocity of the point D: velocity of point B: r r r r r r r r r vR = vB = v A + ωk × rB A vD = v A + ωk × rD A r r r r r r = (1.2 m s )i + (8 rad s )k × (0.10 m ) j = (1.2 m s )i + (8 rad s )k × (− 0.150 m )i r r = (1.2 m s )i + (0.8 m s )i r r r r r vR = (2 m s )i vD = (1.2 m s )i + (1.2 m s ) j vD = 1.697 m s © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 18
- 19. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.3 SOLUTION: • Will determine the absolute velocity of point D with r r r vD = vB + vD B r • The velocity v B is obtained from the given crank rotation data. r The crank AB has a constant clockwise • The directions of the absolute velocity v D r angular velocity of 2000 rpm. and the relative velocity v D B are determined from the problem geometry. For the crank position indicated, determine (a) the angular velocity of • The unknowns in the vector expression the connecting rod BD, and (b) the are the velocity magnitudes v D and v D B velocity of the piston P. which may be determined from the corresponding vector triangle. • The angular velocity of the connecting rod is calculated from v D B . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 19
- 20. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.3 SOLUTION: • Will determine the absolute velocity of point D with r r r vD = vB + vD B r • The velocity vB is obtained from the crank rotation data. ⎛ rev ⎞⎛ min ⎞⎛ 2π rad ⎞ ω AB = ⎜ 2000 ⎜ 60 s ⎟⎜ rev ⎟ = 209.4 rad s ⎟⎜ ⎟ ⎝ min ⎠⎝ ⎠⎝ ⎠ vB = ( AB )ω AB = (3 in.)(209.4 rad s ) The velocity direction is as shown. r • The direction of the absolute velocity vD is horizontal. r The direction of the relative velocity vD B is perpendicular to BD. Compute the angle between the horizontal and the connecting rod from the law of sines. sin 40° sin β = β = 13.95° 8 in. 3 in. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 20
- 21. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.3 • Determine the velocity magnitudes vD and vD B from the vector triangle. vD vD B 628.3 in. s = = sin 53.95° sin 50° sin76.05° vD = 523.4 in. s = 43.6 ft s vP = vD = 43.6 ft s vD B = 495.9 in. s r r r vD B = lω BD vD = vB + vD B vD B 495.9 in. s ω BD = = l 8 in. r r = 62.0 rad s ω BD = (62.0 rad s )k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 21
- 22. EditionEighth Vector Mechanics for Engineers: Dynamics Instantaneous Center of Rotation in Plane Motion • Plane motion of all particles in a slab can always be replaced by the translation of an arbitrary point A and a rotation about A with an angular velocity that is independent of the choice of A. • The same translational and rotational velocities at A are obtained by allowing the slab to rotate with the same angular velocity about the point C on a perpendicular to the velocity at A. • The velocity of all other particles in the slab are the same as originally defined since the angular velocity and translational velocity at A are equivalent. • As far as the velocities are concerned, the slab seems to rotate about the instantaneous center of rotation C. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 22
- 23. EditionEighth Vector Mechanics for Engineers: Dynamics Instantaneous Center of Rotation in Plane Motion • If the velocity at two points A and B are known, the instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B . • If the velocity vectors are parallel, the instantaneous center of rotation is at infinity and the angular velocity is zero. • If the velocity vectors at A and B are perpendicular to the line AB, the instantaneous center of rotation lies at the intersection of the line AB with the line joining the extremities of the velocity vectors at A and B. • If the velocity magnitudes are equal, the instantaneous center of rotation is at infinity and the angular velocity is zero. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 23
- 24. EditionEighth Vector Mechanics for Engineers: Dynamics Instantaneous Center of Rotation in Plane Motion • The instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B . v v v ω= A = A v B = ( BC )ω = (l sin θ ) A AC l cosθ l cosθ = v A tan θ • The velocities of all particles on the rod are as if they were rotated about C. • The particle at the center of rotation has zero velocity. • The particle coinciding with the center of rotation changes with time and the acceleration of the particle at the instantaneous center of rotation is not zero. • The acceleration of the particles in the slab cannot be determined as if the slab were simply rotating about C. • The trace of the locus of the center of rotation on the body is the body centrode and in space is the space centrode. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 24
- 25. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.4 SOLUTION: • The point C is in contact with the stationary lower rack and, instantaneously, has zero velocity. It must be the location of the instantaneous center of rotation. • Determine the angular velocity about C based on the given velocity at A. • Evaluate the velocities at B and D based on The double gear rolls on the their rotation about C. stationary lower rack: the velocity of its center is 1.2 m/s. Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 25
- 26. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.4 SOLUTION: • The point C is in contact with the stationary lower rack and, instantaneously, has zero velocity. It must be the location of the instantaneous center of rotation. • Determine the angular velocity about C based on the given velocity at A. v 1.2 m s v A = rAω ω= A= = 8 rad s rA 0.15 m • Evaluate the velocities at B and D based on their rotation about C. vR = vB = rBω = (0.25 m )(8 rad s ) r r vR = (2 m s )i rD = (0.15 m ) 2 = 0.2121 m vD = rDω = (0.2121 m )(8 rad s ) vD = 1.697 m s r r r vD = (1.2i + 1.2 j )(m s ) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 26
- 27. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.5 SOLUTION: • Determine the velocity at B from the given crank rotation data. • The direction of the velocity vectors at B and D are known. The instantaneous center of rotation is at the intersection of the perpendiculars to the velocities The crank AB has a constant clockwise through B and D. angular velocity of 2000 rpm. • Determine the angular velocity about the For the crank position indicated, center of rotation based on the velocity determine (a) the angular velocity of at B. the connecting rod BD, and (b) the • Calculate the velocity at D based on its velocity of the piston P. rotation about the instantaneous center of rotation. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 27
- 28. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.5 SOLUTION: • From Sample Problem 15.3, r r r vB = (403.9i − 481.3 j )(in. s ) vB = 628.3 in. s β = 13.95° • The instantaneous center of rotation is at the intersection of the perpendiculars to the velocities through B and D. • Determine the angular velocity about the center of rotation based on the velocity at B. γ B = 40° + β = 53.95° γ D = 90° − β = 76.05° vB = (BC )ω BD vB 628.3 in. s ω BD = = ω BD = 62.0 rad s BC CD 8 in. BC 10.14 in. = = sin 76.05° sin 53.95° sin50° • Calculate the velocity at D based on its rotation about the instantaneous center of rotation. BC = 10.14 in. CD = 8.44 in. vD = (CD )ω BD = (8.44 in.)(62.0 rad s ) vP = vD = 523 in. s = 43.6 ft s © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 28
- 29. EditionEighth Vector Mechanics for Engineers: Dynamics Absolute and Relative Acceleration in Plane Motion • Absolute acceleration of a particle of the slab, r r r aB = a A + aB A r • Relative acceleration a B A associated with rotation about A includes tangential and normal components, r ( ) r r a B A = α k × rB A t a B A = rα t ( ) r ( )n r a B A = −ω 2 rB A a B A = rω 2 n ( ) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 29
- 30. EditionEighth Vector Mechanics for Engineers: Dynamics Absolute and Relative Acceleration in Plane Motion r r • Given a A and v A , r r determine a B and α . r r r aB = a A + aB A r ( r )n ( = a A + aB A + aB r A t) r • Vector result depends on sense of a A and the relative magnitudes of a A and (a B A ) n • Must also know angular velocity ω. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 30
- 31. EditionEighth Vector Mechanics for Engineers: Dynamics Absolute and Relative Acceleration in Plane Motion r r r • Write a B = a A + a B A in terms of the two component equations, + → x components: 0 = a A + lω 2 sin θ − lα cosθ + ↑ y components: − a B = −lω 2 cosθ − lα sin θ • Solve for aB and α. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 31
- 32. EditionEighth Vector Mechanics for Engineers: Dynamics Analysis of Plane Motion in Terms of a Parameter • In some cases, it is advantageous to determine the absolute velocity and acceleration of a mechanism directly. x A = l sin θ y B = l cosθ vA = xA & vB = y B & = lθ& cosθ = −lθ& sin θ = lω cosθ = −lω sin θ a A = &&A x a B = &&B y = −lθ& 2 sin θ + lθ& cosθ & = −lθ& 2 cosθ − lθ&sin θ & = −lω 2 sin θ + lα cosθ = −lω 2 cosθ − lα sin θ © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 32
- 33. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.6 SOLUTION: • The expression of the gear position as a function of θ is differentiated twice to define the relationship between the translational and angular accelerations. • The acceleration of each point on the gear is obtained by adding the acceleration of the gear center and the The center of the double gear has a relative accelerations with respect to the velocity and acceleration to the right of center. The latter includes normal and 1.2 m/s and 3 m/s2, respectively. The tangential acceleration components. lower rack is stationary. Determine (a) the angular acceleration of the gear, and (b) the acceleration of points B, C, and D. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 33
- 34. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.6 SOLUTION: • The expression of the gear position as a function of θ is differentiated twice to define the relationship between the translational and angular accelerations. x A = − r1θ v A = − r1θ& = − r1ω vA 1.2 m s ω =− =− = −8 rad s r1 0.150 m a A = −r1θ& = −r1α & aA 3 m s2 α =− =− r1 0.150 m r r ( )r α = α k = − 20 rad s 2 k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 34
- 35. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.6 • The acceleration of each point is obtained by adding the acceleration of the gear center and the relative accelerations with respect to the center. The latter includes normal and tangential acceleration components. r r r r r ( r aB = a A + aB A = a A + aB A + aB A t n ) ( ) r r r r = a A + α k × rB A − ω 2 rB A ( 2 r ) ( r ) r r = 3 m s i − 20 rad s k × (0.100 m ) j − (8 rad s ) (− 0.100 m ) j 2 2 ( 2 r ) ( 2 r = 3 m s i + 2 m s i − 6.40 m s j ) ( 2 r ) r ( r ) ( r aB = 5 m s 2 i − 6.40 m s 2 j ) aB = 8.12 m s 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 35
- 36. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.6 r r r r r r 2r aC = a A + aC A = a A + α k × rC A − ω rC A ( 2 r ) ( r ) r r = 3 m s i − 20 rad s k × (− 0.150 m ) j − (8 rad s )2 (− 0.150 m ) j 2 ( r ) ( r ) ( = 3 m s 2 i − 3 m s 2 i + 9.60 m s 2 j r ) r ac = 9.60 m s j(2 r ) r r r r r r r aD = a A + aD A = a A + α k × rD A − ω 2 rD A ( 2 r ) ( r ) r = 3 m s i − 20 rad s k × (− 0.150 m )i − (8 rad s )2 (− 0.150m )i 2 r ( r ) ( r ) ( = 3 m s 2 i + 3 m s 2 j + 9.60 m s 2 i r ) r ( 2 r aD = 12.6 m s i + 3 m s j) ( 2 r ) aD = 12.95 m s 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 36
- 37. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.7 SOLUTION: • The angular acceleration of the connecting rod BD and the acceleration of point D will be determined from r r r r (r )t ( aD = aB + aD B = aB + aD B + aD B r )n • The acceleration of B is determined from the given rotation speed of AB. Crank AG of the engine system has a • The directions of the accelerations constant clockwise angular velocity of r (r ) (r ) a D , a D B , and a D B are t n 2000 rpm. determined from the geometry. For the crank position shown, • Component equations for acceleration determine the angular acceleration of of point D are solved simultaneously for the connecting rod BD and the acceleration of D and angular acceleration of point D. acceleration of the connecting rod. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 37
- 38. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.7 SOLUTION: • The angular acceleration of the connecting rod BD and the acceleration of point D will be determined from r r r r r ( aD = aB + aD B = aB + aD B + aD B )t ( r )n • The acceleration of B is determined from the given rotation speed of AB. ω AB = 2000 rpm = 209.4 rad s = constant α AB = 0 aB = rω AB = 2 (12 ft )(209.4 rad s)2 = 10,962 ft s2 3 r ( ) r r aB = 10,962 ft s (− cos 40°i − sin 40° j ) 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 38
- 39. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.7 r r • The directions of the accelerations aD , aD ( B )t , and (aD B )n are r determined from the geometry. r r aD = m aD i From Sample Problem 15.3, ωBD = 62.0 rad/s, β = 13.95o. (aD B )n = (BD )ω BD = (12 ft )(62.0 rad s)2 = 2563 ft s2 2 8 (aD B )n = (2563 ft s2 )(− cos13.95°ir + sin13.95° r ) r j (aD B )t = (BD )α BD = (12 ft )α BD = 0.667α BD 8 The direction of (aD/B)t is known but the sense is not known, r r ( r )t aD B = (0.667α BD )(± sin 76.05°i ± cos 76.05° j ) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 39
- 40. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.7 • Component equations for acceleration of point D are solved simultaneously. r r r r r ( )t ( r aD = aB + aD B = aB + aD B + aD B )n x components: − aD = −10,962 cos 40° − 2563 cos13.95° + 0.667α BD sin 13.95° y components: 0 = −10,962 sin 40° + 2563 sin 13.95° + 0.667α BD cos13.95° r ( )r α BD = 9940 rad s 2 k aD = −(9290 ft s ) i 2 r r © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 40
- 41. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.8 SOLUTION: • The angular velocities are determined by simultaneously solving the component equations for r r r vD = vB + vD B • The angular accelerations are determined by simultaneously solving the component In the position shown, crank AB has a equations for constant angular velocity ω1 = 20 rad/s r r r counterclockwise. aD = aB + aD B Determine the angular velocities and angular accelerations of the connecting rod BD and crank DE. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 41
- 42. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.8 SOLUTION: • The angular velocities are determined by simultaneously solving the component equations for r r r vD = vB + vD B r r r r r r vD = ω DE × rD = ω DE k × (− 17i + 17 j ) r r = −17ω DE i − 17ω DE j r r r r r r vB = ω AB × rB = 20k × (8i + 14 j ) r r = −280i + 160 j r r r r r r vD B = ω BD × rD B = ω BD k × (12i + 3 j ) r r = −3ω BD i + 12ω BD j x components: − 17ω DE = −280 − 3ω BD y components: − 17ω DE = +160 + 12ω BD r r r r ω BD = −(29.33 rad s )k ω DE = (11.29 rad s )k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 42
- 43. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.8 • The angular accelerations are determined by simultaneously solving the component equations for r r r aD = aB + aD B r r r 2 r aD = α DE × rD − ω DE rD r r r r r = α DE k × (− 17i + 17 j ) − (11.29 ) (− 17i + 17 j ) 2 r r r r = −17α DE i − 17α DE j + 2170i − 2170 j r r r 2 r 2 r r aB = α AB × rB − ω AB rB = 0 − (20 ) (8i + 14 j ) r r = −3200i + 5600 j r r r 2 r aD B = α BD × rB D − ω BD rB D r r r r r = α B D k × (12i + 3 j ) − (29.33)2 (12i + 3 j ) r r r r = −3α B D i + 12α B D j − 10,320i − 2580 j x components: − 17α DE + 3α BD = −15,690 y components: − 17α DE − 12α BD = −6010 r ( α BD = − 645 rad s k 2 r ) r ( α DE = 809 rad s k 2 r ) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 43
- 44. EditionEighth Vector Mechanics for Engineers: Dynamics Rate of Change With Respect to a Rotating Frame • With respect to the rotating Oxyz frame, r r r r Q = Qx i + Q y j + Qz k r &() Q Oxyz = Q r r r & x i + Q y j + Qz k & & • With respect to the fixed OXYZ frame, r & () Q OXYZ = Q r & r & r r r j r & & x i + Q y j + Qz k + Qx i + Q y & + Qz k & • Frame OXYZ is fixed. • Q r r r () r & & x i + Q y j + Qz k = Q & & Oxyz = rate of change with respect to rotating frame. () • Frame Oxyz rotates about r r & • If Q were fixed within Oxyz then Q OXYZ is fixed axis OA with angular r velocity Ω equivalent to velocity of a point in a rigid r r r r r body r & j & attached to Oxyz and Qx i + Q y & + Qz k = Ω × Q • Vector function Q(t ) varies in direction and magnitude. • With respect to the fixed OXYZ frame, r & Q () = Q OXYZ r & () r r + Ω×Q Oxyz © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 44
- 45. EditionEighth Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Frame OXY is fixed and frame Oxy rotates with angular r velocity Ω . r • Position vector rP for the particle P is the same in both frames but the rate of change depends on the choice of frame. • The absolute velocity of the particle P is r r r r v P = (r OXY = Ω × r + (r )Oxy &) & • Imagine a rigid slab attached to the rotating frame Oxy or F for short. Let P’ be a point on the slab which corresponds instantaneously to position of particle P. r r v P F = (r )Oxy = velocity of P along its path on the slab & r v P = absolute velocity of point P’ on the slab • Absolute velocity for the particle P may be written as r r r v P = v P′ + v P F © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 45
- 46. EditionEighth Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Absolute acceleration for the particle P is r r r r r & × r + Ω × (r ) aP = Ω & OXY + d r dt [ (r )Oxy & ] r r r r but, (r )OXY = Ω × r + (r )Oxy & & d r dt [ r (r Oxy = (r Oxy &) ] r r &&) + Ω × (r )& Oxy r r r r r r r r r r r & × r + Ω × (Ω × r ) + 2Ω × (r ) + (&&) r v P = Ω × r + (r )Oxy & aP = Ω & r Oxy Oxy r r = v P′ + v P F • Utilizing the conceptual point P’ on the slab, r r r r r a P′ = Ω × r + Ω × (Ω × r ) r & r r a P F = (&&) r Oxy • Absolute acceleration for the particle P becomes r r r r r a P = a P′ + a P F + 2Ω × (r )Oxy & r r r = a P′ + a P F + ac r r r r v ac = 2Ω × (r Oxy = 2Ω × v P F = Coriolis acceleration &) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 46
- 47. EditionEighth Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Consider a collar P which is made to slide at constant relative velocity u along rod OB. The rod is rotating at a constant angular velocity ω. The point A on the rod corresponds to the instantaneous position of P. • Absolute acceleration of the collar is r r r r a P = a A + a P F + ac where r r r r r a A = Ω × r + Ω × (Ω × r ) r & a A = rω 2 r r a P F = (&&)Oxy = 0 r r r r a c = 2Ω × v P F ac = 2ωu • The absolute acceleration consists of the radial and tangential vectors shown © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 47
- 48. EditionEighth Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Change in velocity over ∆t is represented by the sum of three vectors r ∆v = RR ′ + TT ′′ + T ′′T ′ • TT ′′ is due to change in direction of the velocity of point A on the rod, TT ′′ ∆θ lim = lim v A = rωω = rω 2 = a A ∆t →0 ∆t ∆t →0 ∆t r r r r r recall, a = Ω × r + Ω × (Ω × r ) r & r r r a = rω 2 at t , v = vA + u A A r r r at t + ∆t , v ′ = v A′ + u ′ • R R ′ and T ′′T ′ result from combined effects of relative motion of P and rotation of the rod ⎛ RR ′ T ′′T ′ ⎞ ∆θ ∆r ⎞ lim ⎜ + ⎟ = lim ⎛ u ⎜ +ω ⎟ ∆t →0⎜ ∆t ⎝ ∆t ⎟ ∆t →0⎝ ∆t ⎠ ∆t ⎠ = uω + ωu = 2ωu r r r recall, ac = 2Ω × v P F ac = 2ωu © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 48
- 49. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.9 SOLUTION: • The absolute velocity of the point P may be written as r r r v P = v P′ + v P s • Magnitude and direction of velocity r v P of pin P are calculated from the radius and angular velocity of disk D. r • Direction of velocity v P′ of point P’ on S coinciding with P is perpendicular to Disk D of the Geneva mechanism rotates radius OP. with constant counterclockwise angular velocity ωD = 10 rad/s. r • Direction of velocity v P s of P with At the instant when φ = 150o, determine respect to S is parallel to the slot. (a) the angular velocity of disk S, and (b) • Solve the vector triangle for the the velocity of pin P relative to disk S. angular velocity of S and relative velocity of P. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 49
- 50. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.9 SOLUTION: • The absolute velocity of the point P may be written as r r r vP = vP ′ + vP s • Magnitude and direction of absolute velocity of pin P are calculated from radius and angular velocity of disk D. vP = Rω D = (50 mm )(10 rad s ) = 500 mm s • Direction of velocity of P with respect to S is parallel to slot. From the law of cosines, r 2 = R 2 + l 2 − 2 Rl cos 30° = 0.551R 2 r = 37.1 mm From the law of cosines, sinβ sin 30° sin 30° = sin β = β = 42.4° R r 0.742 The interior angle of the vector triangle is γ = 90° − 42.4° − 30° = 17.6° © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 50
- 51. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.9 • Direction of velocity of point P’ on S coinciding with P is perpendicular to radius OP. From the velocity triangle, vP′ = vP sin γ = (500 mm s )sin 17.6° = 151.2 mm s 151.2 mm s = rω s ωs = 37.1 mm r r ω s = (− 4.08 rad s )k vP s = vP cos γ = (500 m s )cos17.6° r r r vP s = (477 m s )(− cos 42.4°i − sin 42.4° j ) vP = 500 mm s © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 51
- 52. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.10 SOLUTION: • The absolute acceleration of the pin P may be expressed as r r r r a P = a P′ + a P s + ac • The instantaneous angular velocity of Disk S is determined as in Sample Problem 15.9. • The only unknown involved in the acceleration equation is the instantaneous In the Geneva mechanism, disk D angular acceleration of Disk S. rotates with a constant counter- clockwise angular velocity of 10 • Resolve each acceleration term into the rad/s. At the instant when ϕ = 150o, component parallel to the slot. Solve for determine angular acceleration of the angular acceleration of Disk S. disk S. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 52
- 53. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.10 SOLUTION: • Absolute acceleration of the pin P may be expressed as r r r r aP = aP′ + aP s + ac • From Sample Problem 15.9. r r β = 42.4° ω S = (− 4.08 rad s )k r r r vP s = (477 mm s )(− cos 42.4°i − sin 42.4° j ) • Considering each term in the acceleration equation, aP = Rω D = (500mm )(10 rad s )2 = 5000 mm s 2 2 r ( ) r aP = 5000 mm s (cos 30°i − sin 30° j ) 2 r r r r aP′ = (aP′ )n + (aP′ )t r ( ) r r (aP′ )n = rω S (− cos 42.4°i − sin 42.4° j ) 2 r r r (aP′ )t = (rα S )(− sin 42.4°i + cos 42.4° j ) r r r (aP′ )t = (α S )(37.1mm )(− sin 42.4°i + cos 42.4° j ) note: αS may be positive or negative © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 53
- 54. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.10 • The direction of the Coriolis acceleration is obtained r by rotating the direction of the relative velocity vP s by 90o in the sense of ωS. r r r ( ) ac = 2ω S vP s (− sin 42.4°i + cos 42.4 j ) r r = 2(4.08 rad s )(477 mm s )(− sin 42.4°i + cos 42.4 j ) ( ) r = 3890 mm s 2 (− sin 42.4°i + cos 42.4 j ) r r • The relative acceleration aP s must be parallel to the slot. • Equating components of the acceleration terms perpendicular to the slot, 37.1α S + 3890 − 5000 cos17.7° = 0 α S = −233 rad s r r α S = (− 233 rad s )k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 54
- 55. EditionEighth Vector Mechanics for Engineers: Dynamics Motion About a Fixed Point • The most general displacement of a rigid body with a fixed point O is equivalent to a rotation of the body about an axis through O. • With the instantaneous axis of rotation and angular r velocity ω , the velocity of a particle P of the body is r r dr r r v= =ω ×r dt and the acceleration of the particle P is r r r r r r r r dω a = α × r + ω × (ω × r ) α= . dt r • The angular acceleration α represents the velocity of r the tip of ω . r • As the vector ω moves within the body and in space, it generates a body cone and space cone which are tangent along the instantaneous axis of rotation. • Angular velocities have magnitude and direction and obey parallelogram law of addition. They are vectors. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 55
- 56. EditionEighth Vector Mechanics for Engineers: Dynamics General Motion • For particles A and B of a rigid body, r r r vB = v A + vB A • Particle A is fixed within the body and motion of the body relative to AX’Y’Z’ is the motion of a body with a fixed point r r r r v B = v A + ω × rB A • Similarly, the acceleration of the particle P is r r r aB = a A + aB A r r r ( r r r = a A + α × rB A + ω × ω × rB A ) • Most general motion of a rigid body is equivalent to: - a translation in which all particles have the same velocity and acceleration of a reference particle A, and - of a motion in which particle A is assumed fixed. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 56
- 57. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.11 SOLUTION: r r r r With ω1 = 0.30 j ω 2 = 0.50k r r r r = 12(cos 30°i + sin 30° j ) r r = 10.39i + 6 j • Angular velocity of the boom, r r r The crane rotates with a constant ω = ω1 + ω 2 angular velocity ω1 = 0.30 rad/s and the boom is being raised with a constant • Angular acceleration of the boom, r r r r r r r angular velocity ω2 = 0.50 rad/s. The α =ω &1 + ω 2 = ω 2 = (ω 2 ) & & & Oxyz + Ω × ω 2 length of the boom is l = 12 m. r r = ω1 × ω 2 Determine: • angular velocity of the boom, • Velocity of boom tip, r r r • angular acceleration of the boom, v =ω ×r • velocity of the boom tip, and • Acceleration of boom tip, • acceleration of the boom tip. r r r r r r r r r r a = α × r + ω × (ω × r ) = α × r + ω × v © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 57
- 58. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.11 SOLUTION: • Angular velocity of the boom, r r r ω = ω1 + ω 2 r r r ω = (0.30 rad s ) j + (0.50 rad s )k • Angular acceleration of the boom, r r r r r r r α =ω &1 + ω 2 = ω 2 = (ω 2 ) & & & Oxyz + Ω × ω 2 r r r r = ω1 × ω 2 = (0.30 rad s ) j × (0.50 rad s )k r ( 2 r α = 0.15 rad s i ) • Velocity of boom tip, r r r i j k r r r v =ω ×r = 0 0.3 0.5 10.39 6 0 r r r r r r r r ω1 = 0.30 j ω 2 = 0.50k v = −(3.54 m s )i + (5.20 m s ) j − (3.12 m s )k r r r r = 10.39i + 6 j © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 58
- 59. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.11 • Acceleration of boom tip, r r r r r r r r r r a = α × r + ω × (ω × r ) = α × r + ω × v r r r r r r i j k i j k r a = 0.15 0 0+ 0 0.30 0.50 10.39 6 0 −3 5.20 − 3.12 r r r r r = 0.90k − 0.94i − 2.60i − 1.50 j + 0.90k r ( ) ( 2 r ) ( 2 r 2 a = − 3.54 m s i − 1.50 m s j + 1.80 m s k r ) r r r r ω1 = 0.30 j ω 2 = 0.50k r r r r = 10.39i + 6 j © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 59
- 60. EditionEighth Vector Mechanics for Engineers: Dynamics Three-Dimensional Motion. Coriolis Acceleration • With respect to the fixed frame OXYZ and rotating frame Oxyz, () r & Q = Q OXYZ r & () r r + Ω×Q Oxyz • Consider motion of particle P relative to a rotating frame Oxyz or F for short. The absolute velocity can be expressed as r r r r v P = Ω × r + (r )Oxyz & r r = v P′ + v P F • The absolute acceleration can be expressed as r r r r r r r r & × r + Ω × (Ω × r ) + 2Ω × (r ) r aP = Ω & Oxyz + (&&)Oxyz r r r r = a p′ + a P F + ac r r r r r ac = 2Ω × (r Oxyz = 2Ω × v P F = Coriolis acceleration &) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 60
- 61. EditionEighth Vector Mechanics for Engineers: Dynamics Frame of Reference in General Motion • With respect to OXYZ and AX’Y’Z’, r r r rP = rA + rP A r r r vP = v A + vP A r r r aP = a A + aP A • The velocity and acceleration of P relative to AX’Y’Z’ can be found in terms of the velocity and acceleration of P relative to Axyz. r r r r r &( v P = v A + Ω × rP A + rP A ) Axyz r r = v P′ + v P F Consider: r r r r r r r & × r + Ω × (Ω × r ) aP = a A + Ω P A P A - fixed frame OXYZ, r r + 2Ω × (rP A ) + (&&P A ) & r - translating frame AX’Y’Z’, and r Axyz Axyz - translating and rotating frame Axyz r r r or F. = a P′ + a P F + ac © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 61
- 62. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.15 SOLUTION: • Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F attached to the arm at A. • With P’ of the moving reference frame coinciding with P, the velocity of the point P is found from r r r For the disk mounted on the arm, the v P = v P′ + v P F indicated angular rotation rates are • The acceleration of P is found from constant. r r r r a P = a P′ + a P F + a c Determine: • the velocity of the point P, • The angular velocity and angular • the acceleration of P, and acceleration of the disk are r r r • angular velocity and angular ω = Ω +ωD F acceleration of the disk. r r r r α = (ω F & ) + Ω ×ω © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 62
- 63. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.15 SOLUTION: • Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F attached to the arm at A. r r r r r r = L i + Rj rP A = Rj r r r r Ω = ω1 j ω D F = ω2k • With P’ of the moving reference frame coinciding with P, the velocity of the point P is found from r r r v P = v P′ + v P F r r r r r r r v P′ = Ω × r = ω1 j × (Li + Rj ) = −ω1L k r r r r r r v P F = ω D F × rP A = ω 2 k × Rj = −ω 2 R i r r r v P = −ω 2 R i − ω1L k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 63
- 64. EditionEighth Vector Mechanics for Engineers: Dynamics Sample Problem 15.15 • The acceleration of P is found from r r r r a P = a P′ + a P F + a c r r r r r 2 r a P′ = Ω × (Ω × r ) = ω1 j × (− ω1Lk ) = −ω1 Li r r r ( r a P F = ω D F × ω D F × rP A ) r r r 2 r = ω 2 k × (− ω 2 R i ) = −ω 2 R j r r r a c = 2Ω × v P F r r r = 2ω1 j × (− ω 2 R i ) = 2ω1ω 2 Rk r 2 r 2 r r a P = −ω1 L i − ω 2 Rj + 2ω1ω 2 Rk • Angular velocity and acceleration of the disk, r r r r r r ω = Ω +ωD F ω = ω1 j + ω 2 k r r r r α = (ω F & ) + Ω ×ω r r r r = ω1 j × (ω1 j + ω 2 k ) r α = ω1ω 2 i © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 64

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