DYNAMICS

Seventh Edition

CHAPTER

15

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes:

Kinematics of Rigid Bodies

J. Walt Oler Texas Tech University

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

Seventh Edition

Vector Mechanics for Engineers: Dynamics Contents Introduction Translation Rotation About a Fixed Axis: Velocity Rotation About a Fixed Axis: Acceleration Rotation About a Fixed Axis: Representative Slab Equations Defining the Rotation of a Rigid Body About a Fixed Axis Sample Problem 5.1 General Plane Motion Absolute and Relative Velocity in Plane Motion Sample Problem 15.2 Sample Problem 15.3 Instantaneous Center of Rotation in Plane Motion Sample Problem 15.4 Sample Problem 15.5 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

Absolute and Relative Acceleration in Plane Motion Analysis of Plane Motion in Terms of a Parameter Sample Problem 15.6 Sample Problem 15.7 Sample Problem 15.8 Rate of Change With Respect to a Rotating Frame Coriolis Acceleration Sample Problem 15.9 Sample Problem 15.10 Motion About a Fixed Point General Motion Sample Problem 15.11 Three Dimensional Motion. Coriolis Acceleration Frame of Reference in General Motion Sample Problem 15.15

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Seventh Edition

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

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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,


rB = rA + rB A • Differentiating with respect to time,



rB = rA + rB A = rA

vB = v A All particles have the same velocity. • Differentiating with respect to time again, r
B = r
A + r
B A = r
A

aB = a A All particles have the same acceleration. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Velocity • Consider rotation of rigid body about a fixed axis AA’



• Velocity vector v = dr dtof the particle P is v = ds dt tangent to the path with magnitude ∆ s = ( BP )∆ θ = (r sin φ )∆ θ v=

ds ∆θ = lim (r sin φ ) = rθ sin φ dt ∆t → 0 ∆t

• The same result is obtained from

dr

v= =ω ×r dt


 ω = ω k = θ k = angular ve locity

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Rotation About a Fixed Axis. Acceleration • Differentiating to determine the acceleration,

dv d 
= (ω × r ) a= dt dt

dω

dr = ×r +ω × dt dt
dω


= ×r +ω ×v dt
dω
= α = angular ac celeration • dt


  = α k = ω k = θ k • Acceleration of P is combination of two vectors,







a = α × r + ω ×ω × r

α × r = tangentia l accelerati on component


ω × ω × r = radial accelerati on component © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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,




v = ω × r = ωk × r

v = rω • Acceleration of any point P of the slab,





a = α × r + ω ×ω × r


= α k × r − ω 2r • Resolving the acceleration into tangential and normal components,


at = αk × r a t = rα

a n = −ω 2 r a n = rω 2 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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 • Recall ω = dt

or

dt =

dθ

ω

dω d ω d 2θ α= = 2 =ω dθ dt dt • Uniform Rotation, α = 0:

θ = θ0 + ωt • Uniformly Accelerated Rotation, α = constant: ω = ω0 +αt

θ = θ 0 + ω 0 t + 12 α t 2 ω 2 = ω 02 + 2α (θ − θ 0 ) © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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.

Cable C has a constant acceleration of 225 mm/s2 and an initial velocity of 300 mm/s2, both directed to the right. 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the pulley after 2 s. • Evaluate the initial tangential and normal acceleration components of D.

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Seventh Edition

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 D t C = 225 mm s → (vD )0 = (vC )0 = 300 mm s →

(vD )0 = rω 0 (vD )0 ω0 =

r

(a D )t

12 = = 4 rad s 3

= rα

α=

(a D )t r

=

9 = 3 rad s 2 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 + 12 αt 2 = (4 rad s)(2 s) + 12 3 rad s 2 (2 s)2 = 14 rad  1 rev  N = (14 rad )   = number of revs  2π rad 

v B = rω = (125 mm)(10 rad s) ∆y B = rθ = (125 mm)(14 rad)

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

N = 2 .23 rev


v B = 1250 mm s ↑ ∆y B = 1750 mm 15 - 10

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.1 • Evaluate the initial tangential and normal acceleration components of D. (a
D )t = a
C = 225 mm s 2 →

(aD )n = rDω 02 = (75 mm)(4 rad s)2 = 1200 mm

(a
D )t

(a
D )n = 1200 mm

= 225 mm s 2 →

s2

s2 ↓

Magnitude and direction of the total acceleration,

(a D )t2 + (a D )n2

aD =

= 225 + 1200 2

2

a D = 1220.9 mm s 2

(a D )n tan φ = (a D )t = © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

1200 225

φ = 79.4° 15 - 11

Seventh Edition

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 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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.




vB = v A + vB A


v B A = ω k × rB A



v B = v A + ω k × rB © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

vB

A

= rω

A 15 - 13

Seventh Edition

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 = tan θ vA v B = v A tan θ © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

vA v = A = cosθ v B A lω

ω=

vA l cosθ 15 - 14

Seventh Edition

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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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 double gear rolls on the 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.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• The velocity for any point P on the gear may be written as


vP = v A + vP

A




= v A + ω k × rP

A

Evaluate the velocities of points B and D.

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Seventh Edition

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 =− 2π r 2π y

x A = − r1θ

Differentiate to relate the translational and angular velocities. x

v A = − r1ω

ω =−

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

vA 1 .2 m s =− r1 0.150 m



ω = ωk = −(8 rad s )k


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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.2 • For any point P on the gear,




vP = v A + vP

Velocity of the upper rack is equal to velocity of point B:




v R = v B = v A + ω k × rB A


= (1 .2 m s )i + (8 rad s )k × (0 .10 m ) j

= (1 .2 m s )i + (0 .8 m s )i

vR = (2 m s )i

A




= v A + ω k × rP

A

Velocity of the point D:





v D = v A + ω k × rD A


= (1 .2 m s )i + (8 rad s )k × (− 0 .150 m )i




vD = (1.2 m s )i + (1.2 m s ) j vD = 1.697 m s

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.3 SOLUTION:

The crank AB has a constant clockwise angular velocity of 2000 rpm. For the crank position indicated, determine (a) the angular velocity of the connecting rod BD, and (b) the velocity of the piston P.

• Will determine the absolute velocity of point D with


vD = vB + vD B
• The velocity v Bis obtained from the given crank rotation data.
• The directions of the absolute velocity v D
and the relative velocity v D Bare determined from the problem geometry. • The unknowns in the vector expression are v D and v D B the velocity magnitudes which may be determined from the corresponding vector triangle. • The angular velocity of the connecting rod vD B . is calculated from

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.3 SOLUTION: • Will determine the absolute velocity of point D with


vD = vB + vD B


• The velocity v Bis obtained from the crank rotation data.

ω AB

rev   min   2π rad   =  2000   = 209.4 rad s     min  60 s  rev 

v B = ( AB )ω AB = (0.075 m)(209.4 rad s) The velocity direction is as shown.


vis • The direction of the absolute velocity horizontal. The
D direction of the relative velocity isvperpendicular to BD. D B Compute the angle between the horizontal and the connecting rod from the law of sines. sin 40 ° sin β = 8 in. 3 in.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

β = 13 .95 °

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.3

• Determine the velocity magnitudes from the vector triangle. vD B vD 15.71m s = = sin 53.95° sin 50° sin76.05°

vD = 13.09 m s = 13.09 m s




v D = vB + v D

v D and v D

B

vP = vD = 13.09 m s

vD B = 12.40 m s vD B = lω BD B

ω BD

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

vD B

12.40 m s = = l 0.2 m = 62.0 rad s




ω BD


= (62.0 rad s )k 15 - 21

Seventh Edition

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.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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 . vA vA vA ω= = v B = ( BC )ω = (l sin θ ) 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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. The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s.

• Evaluate the velocities at B and D based on their rotation about C.

Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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.

v R = v B = rBω = (0 .25 m )(8 rad s )



v R = (2 m s )i

rD = (0.15 m ) 2 = 0.2121 m

v D = rDω = (0 .2121 m )(8 rad s ) v D = 1 .697 m s


v D = (1 .2i + 1 .2 j )(m s ) © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.5 SOLUTION: • Determine the velocity at B from the given crank rotation data.

The crank AB has a constant clockwise angular velocity of 2000 rpm. For the crank position indicated, determine (a) the angular velocity of the connecting rod BD, and (b) the velocity of the piston P.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• 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 through B and D. • Determine the angular velocity about the center of rotation based on the velocity at B. • Calculate the velocity at D based on its rotation about the instantaneous center of rotation.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.5 SOLUTION: • From Sample Problem 15.3,




vB = (403.9i − 481.3 j )(m s)

vB = 15.71m s

β = 13.95° • The instantaneous center of rotation is at the intersection of the perpendiculars to the velocities through B and D.

γ B = 40° + β = 53.95° γ D = 90° − β = 76.05° BC CD 200 mm = = sin 76.05° sin 53.95° sin50°

BC = 253.4 mm CD = 211.1 mm

• Determine the angular velocity about the center of rotation based on the velocity at B.

vB = (BC )ω BD

ω BD =

vB 15.71m s = BC 253.4 mm

ω BD = 62.0 rad s

• Calculate the velocity at D based on its rotation about the instantaneous center of rotation.

vD = (CD )ω BD = (0.2111 mm)(62.0 rad s)

vP = vD = 13.09 m s = 13.09 m s © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 28

Seventh Edition

Vector Mechanics for Engineers: Dynamics Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the slab,


aB = a A + aB A


• Relative acceleration a B Aassociated with rotation about A includes tangential and normal components,


a B A = rα a B A = α k × rB A

( )t (a
B A )n = −ω 2 r
B A

( )t (a B A )n = rω 2

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Absolute and Relative Acceleration in Plane Motion

• Given a A and v A ,

determine a B and α .


aB = a A + aB A


= a A + aB A + aB

(

)n (


• Vector result depends on sense of the relative a and A magnitudes of a A and a B A

(

)n

• Must also know angular velocity ω. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

)

A t

15 - 30

Seventh Edition

Vector Mechanics for Engineers: Dynamics Absolute and Relative Acceleration in Plane Motion

• Write




a B = a A + a B A in terms of the two component equations,

+ →

x components:

+ ↑

y components:

0 = a A + lω 2 sin θ − lα cos θ

− a B = − lω 2 cos θ − lα sin θ

• Solve for aB and α. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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 cos θ

v A = x A = lθ cos θ

v B = y B

= lω cos θ a A = xA

= −lθ sin θ = −lω sin θ a B = yB

= − lθ 2 sin θ + lθ cos θ

= − lθ 2 cos θ − lθ sin θ

= − lω 2 sin θ + lα cos θ

= − lω 2 cos θ − lα sin θ

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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 center of the double gear has a velocity and acceleration to the right of 1.2 m/s and 3 m/s2, respectively. The lower rack is stationary.

• The acceleration of each point on the gear 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.

Determine (a) the angular acceleration of the gear, and (b) the acceleration of points B, C, and D. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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

ω =−

vA 1.2 m s =− = −8 rad s r1 0.150 m

a A = − r1θ = − r1α

aA 3 m s2 α =− =− r1 0.150 m

(

)



2 α = α k = − 20 rad s k


© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

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.

(

) (

)







aB = a A + aB A = a A + aB A + aB A t n



= a A + α k × rB A − ω 2rB A


2 2
2 = 3 m s i − 20 rad s k × (0.100 m ) j − (8 rad s ) (− 0.100 m ) j


= 3 m s 2 i + 2 m s 2 i − 6.40 m s 2 j

( (

) ( ) (

)

) (

)


2
2
aB = (5 m s ) i − (6.40 m s ) j

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

aB = 8.12 m s 2

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.6




2
α ω = a + k × r − rC A A A C A


2 2
2 = 3 m s i − 20 rad s k × (− 0 .150 m ) j − (8 rad s ) (− 0 .150 m ) j 2
2
2
= 3 m s i − 3 m s i + 9 .60 m s j
2
ac = (9.60 m s ) j






a D = a A + a D A = a A + α k × rD A − ω 2 rD A


2 2
2 = 3 m s i − 20 rad s k × (− 0 .150 m ) i − (8 rad s ) (− 0 .150 m ) i 2
2
2
= 3 m s i + 3 m s j + 9.60 m s i


aD = (12.6 m s 2 ) i + (3 m s 2 ) j aD = 12.95 m s 2


aC = a A + aC

( (

) ( ) (

) (

( (

) ( ) (

)

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

)

) (

)

)

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Seventh Edition

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





aD = aB + aD B = aB + aD B + aD B

(

)t (

• The acceleration of B is determined from the given rotation speed of AB. Crank AG of the engine system has a constant clockwise angular velocity of 2000 rpm. For the crank position shown, determine the angular acceleration of the connecting rod BD and the acceleration of point D.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• The directions of the accelerations


a D , ( a D B ) t , and ( a D B ) n are determined from the geometry. • Component equations for acceleration of point D are solved simultaneously for acceleration of D and angular acceleration of the connecting rod. 15 - 37

)n

Seventh Edition

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





aD = a B + a D B = a B + (aD B )t + (aD B ) 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 a B = rω 2AB = (0.075 m)(209.4 rad s) = 3289 m s 2 2




2 aB = (3289 m s ) ( − cos 40°i − sin 40° j )

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.7

• The directions of the accelerations determined from the geometry.

aD = ∓ aD i




a D , ( a D B ) t , and ( a D Bare )n

From Sample Problem 15.3, ωBD = 62.0 rad/s, β = 13.95o.

(a )

D B n

2 = (BD )ω BD = (0.2 m)(62.0 rad s ) = 769 m s 2 2

(

)




( aD B ) n = 769 m s 2 (− cos13.95°i + sin 13.95° j ) ( aD B )t = (BD )α BD = (0.2 m)α BD = 0.2 α BD

The direction of (aD/B)t is known but the sense is not known,


aD B t = (0.2 α BD ) (± sin 76.05°i ± cos 76.05° j )

(

)

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.7

• Component equations for acceleration of point D are solved simultaneously.





aD = aB + aD B = aB + ( aD B )t + ( aD B ) n x components:

− a D = −3289 cos 40° − 769 cos13.95° + 0.2α BD sin 13.95° y components:

0 = −3289 sin 40° + 769 sin 13.95° + 0.2α BD cos13.95°
α BD = (9937 rad s ) k
2
aD = −(2787 m s ) i


© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

2

15 - 40

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.8 SOLUTION: • The angular velocities are determined by simultaneously solving the component equations for


vD = vB + vD B

In the position shown, crank AB has a constant angular velocity ω1 = 20 rad/s counterclockwise.

• The angular accelerations are determined by simultaneously solving the component equations for


aD = aB + aD B

Determine the angular velocities and angular accelerations of the connecting rod BD and crank DE.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 41

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.8 SOLUTION: • The angular velocities are determined by simultaneously solving the component equations for


v D = vB + v D B







v D = ω DE × rD = ω DE k × (− 425i + 425 j )

= −425ω DE i − 425ω DE j





v B = ω AB × rB = 20k × (200i + 350 j )

= −280i + 160 j





v D B = ω BD × rD B = ω BD k × (300i + 75 j )

= −75ω BD i + 300ω BD j x components: − 425ω DE = −280 − 75ω BD y components: − 425ω DE = +160 + 200 ω BD


ω BD © 2003 The McGraw-Hill Companies, Inc. All rights reserved.


= −(29.33 rad s )k




ω DE


= (11.29 rad s )k 15 - 42

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.8 • The angular accelerations are determined by simultaneously solving the component equations for


aD = aB + aD B




2
a D = α DE × rD − ω DE rD




2 = α DE k × (− 425i + 425 j ) − (11.29) (− 425i + 425 j )



3 3 = −425α DE i − 425α DE j + 54.2×10 i − 54.2×10 j




2 2
a B = α AB × rB − ω AB rB = 0 − (20) (200i + 350 j )

3 3 = −80 × 10 i + 140 × 10 j


2
a D B = α BD × rB D − ω BD rB D




2 = α B D k × (300i + 75 j ) − (29.33) (300i + 75 j )



3 3 = −75α B D i + 300α B D j − 258 × 10 i − 64.5 × 10 j x components:

− 425α DE + 75α BD = −392.2 × 10 3

y components: − 425α DE − 300α BD = −150.2 × 10 3



2 2 α BD = −(645 rad s ) k α DE = (809 rad s ) k © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 43

Seventh Edition

Vector Mechanics for Engineers: Dynamics Rate of Change With Respect to a Rotating Frame • With respect to the rotating Oxyz frame,



Q = Qxi + Q y j + Qz k



Q Oxyz = Q x i + Q y j + Q z k

()

• With respect to the fixed OXYZ frame,






    Q OXYZ = Q x i + Q y j + Q z k + Q x i + Q y j + Q z k

()

• Frame OXYZ is fixed. • Frame Oxyz rotates about fixed axis
OA with angular velocity Ω
• Vector function Q (t varies in ) direction and magnitude.

()





   • Q x i + Q y j + Q z k = Q Oxyz =rate of change with respect to rotating frame.

 • If Q were fixed within Oxyz then Q OXYZ is equivalent to velocity of a point in a rigid
body



  attached to Oxyz and Q x i + Q y j + Q z k = Ω × Q

()

• With respect to the fixed OXYZ frame,



Q = Q + Ω×Q

()

OXYZ

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

()

Oxyz

15 - 44

Seventh Edition

Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Frame OXY
is fixed and frame Oxy rotates with angular velocity Ω .
• Position vector rPfor 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



 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.

v P F = (r )Oxy = velocity of P along its path on the slab
v P ' = absolute velocity of point P’ on the slab • Absolute velocity for the particle P may be written as


v P = v P′ + v P F © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 45

Seventh Edition

Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Absolute acceleration for the particle P is



d

 (r )Oxy a P = Ω × r + Ω × (r )OXY + dt
but, (r
 )OXY = Ω × r
+ (r
 )Oxy

[

[




v P = Ω × r + (r )Oxy

= v P′ + v P F

]

]



d

  (r )Oxy = (r )Oxy + Ω × (r )Oxy dt








   a P = Ω × r + Ω × (Ω × r ) + 2Ω × (r )Oxy + (r )Oxy • Utilizing the conceptual point P’ on the slab,





a P ′ = Ω × r + Ω × (Ω × r )

a P F = (r) Oxy

• Absolute acceleration
for the particle P becomes



a P = a P ′ + a P F + 2Ω × (r )Oxy


= a P′ + a P F + ac



 a c = 2Ω × (r )Oxy = 2 Ω × v P F = Coriolis acceleration © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 46

Seventh Edition

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



a P = a A + a P F + ac where





a A = Ω × r + Ω × (Ω × r ) a A = rω 2

  a P F = (r )Oxy = 0


a c = 2Ω × v P F ac = 2ωu • The absolute acceleration consists of the radial and tangential vectors shown

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 47

Seventh Edition

Vector Mechanics for Engineers: Dynamics Coriolis Acceleration • Change in velocity over ∆t is represented by the sum of three vectors
∆ v = R R ′ + T T ′′ + T ′′T ′

at t , at t + ∆ t ,




v = vA + u


v ′ = v A′ + u ′

• T T ′′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 ∆t → 0 ∆ t ∆t → 0





recall, a = Ω × r + Ω × (Ω × r ) a = rω 2 A

A

• R R ′ and T ′′T ′result from combined effects of relative motion of P and rotation of the rod

 R R ′ T ′′T ′  ∆r   ∆θ   + lim  = lim  u +ω   ∆ t  ∆ t → 0 ∆ t ∆t  ∆ t → 0 ∆ t = uω + ω u = 2ω u


ac = 2ωu recall, ac = 2Ω × v P F © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 48

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.9 SOLUTION: • The absolute velocity of the point P may be written as


v P = v P′ + v P s

Disk D of the Geneva mechanism rotates with constant counterclockwise angular velocity ωD = 10 rad/s. At the instant when φ = 150o, determine (a) the angular velocity of disk S, and (b) the velocity of pin P relative to disk S. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• Magnitude and direction of velocity
v P of pin P are calculated from the radius and angular velocity of disk D.
• Direction of velocity v Pof ′ point P’ on S coinciding with P is perpendicular to radius OP.
• Direction of velocity v P of s P with respect to S is parallel to the slot. • Solve the vector triangle for the angular velocity of S and relative velocity of P. 15 - 49

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.9 SOLUTION: • The absolute velocity of the point P may be written as


vP = vP′ + vP s • Magnitude and direction of absolute velocity of pin P are calculated from radius and angular velocity of disk D.

v P = R ω 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 ° = 0.551 R 2 From the law of cosines, sin β sin 30 ° sin 30 ° = sin β = R r 0 .742

r = 37 .1 mm

β = 42 .4°

The interior angle of the vector triangle is γ = 90 ° − 42 .4 ° − 30 ° = 17 .6 ° © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 50

Seventh Edition

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,

v P ′ = v P sin γ = (500 mm s )sin 17 .6 ° = 151 .2 mm s = rω s

vP

s

151 .2 mm s ωs = 37 .1 mm


ω s = (− 4 .08 rad s )k


= v P cos γ = (500 m s ) cos 17 .6 °


v P s = (477 m s ) (− cos 42 .4°i − sin 42 .4 ° j )

vP = 500 mm s

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 51

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.10 SOLUTION: • The absolute acceleration of the pin P may be expressed as



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 angular acceleration of Disk S.

In the Geneva mechanism, disk D rotates with a constant counterclockwise angular velocity of 10 rad/s. • Resolve each acceleration term into the At the instant when ϕ = 150o, determine component parallel to the slot. Solve for the angular acceleration of Disk S. angular acceleration of disk S.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 52

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.10 SOLUTION: • Absolute acceleration of the pin P may be expressed as



aP = aP′ + aP s + ac • From Sample Problem 15.9.

β = 42 .4° ω S = (− 4 .08 rad s )k


v P s = (477 mm s )(− cos 42 .4 °i − sin 42 .4 ° j ) • Considering each term in the acceleration equation,

a P = R ω D2 = (500 mm )(10 rad s )2 = 5000 mm s 2


2 a P = 5000 mm s (cos 30 °i − sin 30 ° j )


a P ′ = (a P ′ )n + (a P ′ )t


2 (a P ′ )n = rω S (− cos 42 .4°i − sin 42 .4° j )


(a P ′ )t = (rα S )(− sin 42 .4°i + cos 42 .4° j )


(a P ′ )t = (α S )(37 .1 mm )(− sin 42 .4°i + cos 42 .4° j )

(

)

( )

note: αS may be positive or negative © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 53

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.10 • The direction of the Coriolis acceleration is obtained
by vP s rotating the direction of the relative velocity by 90o in the sense of ωS.


ac = 2ω S v P s (− sin 42 .4°i + cos 42 .4 j )

= 2(4 .08 rad s )(477 mm s ) (− sin 42 .4 °i + cos 42 .4 j )

2 = 3890 mm s (− sin 42 .4 °i + cos 42 .4 j )

(

(

)

)

• The relative acceleration slot.


a P must be parallel to the s

• Equating components of the acceleration terms perpendicular to the slot, 37 .1α S + 3890 − 5000 cos 17 .7 ° = 0

α S = − 233 rad s


α S = (− 233 rad s )k


© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 54

Seventh Edition

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
velocity ω , the velocity of a particle P of the body is

dr

v= =ω ×r dt and the acceleration of the particle P is







dω a = α × r + ω × (ω × r ) α= . dt
• The angular
acceleration αrepresents the velocity of the ω. tip of
• 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 55

Seventh Edition

Vector Mechanics for Engineers: Dynamics General Motion • For particles A and B of a rigid body,


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



v B = v A + ω × rB A • Similarly, the acceleration of the particle P is


aB = a A + aB A





= 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 56

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.11 SOLUTION:



With ω1 = 0.30 j ω 2 = 0.50 k


r = 12 (cos 30 °i + sin 30 ° j )

= 10 .39 i + 6 j The crane rotates with a constant angular velocity ω1 = 0.30 rad/s and the boom is being raised with a constant angular velocity ω2 = 0.50 rad/s. The length of the boom is l = 12 m. Determine: • angular velocity of the boom, • angular acceleration of the boom, • velocity of the boom tip, and • acceleration of the boom tip. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• Angular velocity of the boom,


ω = ω1 + ω 2 • Angular acceleration of the boom,






 α = ω1 + ω 2 = ω 2 = (ω 2 )Oxyz + Ω × ω 2

= ω1 × ω 2 • Velocity of boom tip,


v =ω×r • Acceleration of boom tip,









a = α × r + ω × (ω × r ) = α × r + ω × v 15 - 57

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.11 SOLUTION: • Angular velocity of the boom,


ω = ω1 + ω 2


ω = (0.30 rad s ) j + (0 .50 rad s )k • Angular acceleration of the boom,






 α = ω1 + ω 2 = ω 2 = (ω 2 )Oxyz + Ω × ω 2



= ω1 × ω 2 = (0 .30 rad s ) j × (0 .50 rad s )k

α = 0.15 rad s 2 i

(

• Velocity of boom tip,
i


v =ω ×r = 0 10 .39


ω1 = 0.30 j ω 2 = 0.50k


r = 10.39i + 6 j


© 2003 The McGraw-Hill Companies, Inc. All rights reserved.


j

)


k

0 .3 0 .5 6 0



v = − (3 .54 m s )i + (5.20 m s ) j − (3 .12 m s )k

15 - 58

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.11 • Acceleration of boom tip,









a = α × r + ω × (ω × r ) = α × r + ω × v





i j k i j k
a = 0 .15 0 0 + 0 0 .30 0.50 10 .39 6 0 −3 5.20 − 3 .12




= 0.90 k − 0.94 i − 2 .60 i − 1.50 j + 0 .90 k



2
2
2 a = −(3.54 m s ) i − (1.50 m s ) j + (1.80 m s ) k




ω1 = 0.30 j ω 2 = 0.50k


r = 10.39i + 6 j


© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 59

Seventh Edition

Vector Mechanics for Engineers: Dynamics Three-Dimensional Motion. Coriolis Acceleration • With respect to the fixed frame OXYZ and rotating frame Oxyz,



Q = Q + Ω×Q

()

OXYZ

()

Oxyz

• Consider motion of particle P relative to a rotating frame Oxyz or F for short. The absolute velocity can be expressed as



v P = Ω × r + (r )Oxyz

= v P′ + v P F • The absolute acceleration can be expressed as








a P = Ω × r + Ω × (Ω × r ) + 2 Ω × (r )Oxyz + (r)Oxyz


= a p′ + a P F + ac




 a c = 2Ω × (r )Oxyz = 2 Ω × v P F = Coriolis accelerati on © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 60

Seventh Edition

Vector Mechanics for Engineers: Dynamics Frame of Reference in General Motion • With respect to OXYZ and AX’Y’Z’,


rP = rA + rP A


vP = v A + vP A


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.

(

Consider: - fixed frame OXYZ, - translating frame AX’Y’Z’, and - translating and rotating frame Axyz or F. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

)






 v P = v A + Ω × rP A + rP A Axyz

= v P′ + v P F






a P = a A + Ω × rP A + Ω × (Ω × rP A )


+ 2Ω × (rP A ) + (rP A ) Axyz Axyz


= a P′ + a P F + ac 15 - 61

Seventh Edition

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.

For the disk mounted on the arm, the indicated angular rotation rates are constant. Determine: • the velocity of the point P, • the acceleration of P, and • angular velocity and angular acceleration of the disk.

• With P’ of the moving reference frame coinciding with P, the velocity of the point P is found from


v P = v P′ + v P F • The acceleration of P is found from



a P = a P′ + a P F + ac • The angular velocity and angular acceleration of the disk are


ω = Ω +ωD F



 α = (ω )F + Ω × ω

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 62

Seventh Edition

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 = L i + Rj

Ω = ω1 j


rP


ωD

A F


= Rj
= ω 2k

• With P’ of the moving reference frame coinciding with P, the velocity of the point P is found from




v P = v P′ + v P F






v P ′ = Ω × r = ω1 j × (L i + R j ) = −ω1 L k





v P F = ω D F × rP A = ω 2 k × R j = −ω 2 R i


v P = −ω 2 R i − ω1 L k

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

15 - 63

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 15.15 • The acceleration of P is found from





a P = a P′ + a P F + ac





2
a P′ = Ω × (Ω × r ) = ω1 j × (− ω1 Lk ) = −ω1 Li



a P F = ω D F × ω D F × rP A

2
= ω 2 k × (− ω 2 R i ) = −ω 2 R j


a c = 2Ω × v P F


= 2ω1 j × (− ω 2 R i ) = 2ω1ω 2 Rk

2
2
a P = −ω1 L i − ω 2 Rj + 2ω1ω 2 Rk

(

)

• Angular velocity and acceleration of the disk,










ω = Ω +ωD



ω = ω1 j + ω 2 k


F




 α = (ω )F + Ω × ω


= ω1 j × (ω1 j + ω 2 k )


© 2003 The McGraw-Hill Companies, Inc. All rights reserved.


α = ω1ω 2 i


15 - 64