DYNAMICS Ferdinand P. Beer Kinematics of E. Russell Johnston, - - PowerPoint PPT Presentation

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Seventh Edition Ferdinand P. Beer

• E. Russell Johnston, Jr.

Lecture Notes:

• J. Walt Oler

Texas Tech University CHAPTER

• Kinematics of

Rigid Bodies

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 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

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:
• general motion
• motion about a fixed point
• general plane motion
• rotation about a fixed axis
• curvilinear translation
• rectilinear translation
• translation:

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,

A B A B

r r r

• +

=

• Differentiating with respect to time,

A B A A B A B

v v r r r r

• =

= + = All particles have the same velocity.

A B A A B A B

a a r r r r

• =

= + =

• Differentiating with respect to time again,

All particles have the same acceleration.

Vector Mechanics for Engineers: Dynamics

Rotation About a Fixed Axis. Velocity

• Consider rotation of rigid body about a fixed

axis AA’

• Velocity vector
• f the particle P is

tangent to the path with magnitude dt r d v

• =

dt ds v =

( ) ( ) ( )

φ θ θ φ θ φ θ sin sin lim sin

• r

t r dt ds v r BP s

t

= ∆ ∆ = = ∆ = ∆ = ∆

→ ∆

locity angular ve k k r dt r d v = = = × = =

• θ

ω ω ω

• The same result is obtained from

Vector Mechanics for Engineers: Dynamics

Rotation About a Fixed Axis. Acceleration

• Differentiating to determine the acceleration,

( )

v r dt d dt r d r dt d r dt d dt v d a

• ×

+ × = × + × = × = = ω ω ω ω ω

• k

k k celeration angular ac dt d

• θ

ω α α ω = = = = = component

• n

accelerati radial component

• n

accelerati l tangentia = × × = × × × + × = r r r r a

• ω

ω α ω ω α

• Acceleration of P is combination of two vectors,

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 ω r v r k r v = × = × =

• Acceleration of any point P of the slab,

r r k r r a

• 2

ω α ω ω α − × = × × + × =

• Resolving the acceleration into tangential and

normal components,

2 2

ω ω α α r a r a r a r k a

n n t t

= − = = × =

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
• ften specified by the type of angular acceleration.

θ ω ω θ ω α ω θ θ ω d d dt d dt d d dt dt d = = = = =

2 2

• r
• Recall
• Uniform Rotation, α = 0:

t ω θ θ + =

• Uniformly Accelerated Rotation, α = constant:

( )

2 2 2 2 1

2 θ θ α ω ω α ω θ θ α ω ω − + = + + = + = t t t

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.1

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. SOLUTION:

• Due to the action of the cable, the

tangential velocity and acceleration of D are equal to the velocity and acceleration

• f C. Calculate the initial angular

velocity and acceleration.

• 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.

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.

( ) ( ) ( ) ( )

s rad 4 3 12 s mm 300 = = = = → = = r v r v v v

ω ω

• (

) ( ) ( )

2

s rad 3 3 9 s mm 225 = = = = → = = r a r a a a

t D t D C t D

α α

• Apply the relations for uniformly accelerated rotation to determine

velocity and angular position of pulley after 2 s.

( )(

)

s rad 10 s 2 s rad 3 s rad 4

= + = + = t α ω ω

( )( )

( )(

)

rad 14 s 2 s rad 3 s 2 s rad 4

2 2 2 1 2 2 1

= + = + = t t α ω θ

( )

revs

• f

number rad 2 rev 1 rad 14 =

• =

π N rev 23 . 2 = N

( )( ) ( )( )

rad 14 mm 25 1 s rad 10 mm 25 1 = = ∆ = = θ ω r y r v

mm 1750 s mm 1250 = ∆ ↑ =

B B

y v

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.1

• Evaluate the initial tangential and normal acceleration

components of D.

( )

→ = =

2

s mm 225

C t D

a a

• (

) ( )( )

2 2 2

s mm 1200 s rad 4 mm 5 7 = = = ω

D n D

r a

( ) ( )

↓ = → =

s mm 1200 s mm 225

a a

• Magnitude and direction of the total acceleration,

( ) ( )

1200 225 + = + =

a a a

2

s mm 9 . 1220 =

D

a

( ) ( )

225 1200 tan = =

a a φ

° = 4 . 79 φ

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
• f a translation and rotation.
• Displacement of particles A and B to A2 and B2 can

be divided into two parts:

• translation to A2 and
• rotation of about A2 to B2

1

B′

1

B′

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.

A B A B

v v v

• +

= ω ω r v r k v

A B A B A B

= × =

• A

B A B

r k v v

• ×

+ = ω

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.

θ θ tan tan

A B A B

v v v v = = θ ω θ ω cos cos l v l v v v

A A A B A

= = =

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.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.2

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. 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 may

be written as Evaluate the velocities of points B and D.

A P A A P A P

r k v v v v

• ×

+ = + = ω

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.2

x y 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). θ π θ π

1

2 2 r x r x

A A

− = − = Differentiate to relate the translational and angular velocities. m 0.150 s m 2 . 1

1 1

− = − = − = r v r v

A A

ω ω

( )k

k

• s

rad 8 − = = ω ω

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.2

• For any point P on the gear,

A P A A P A P

r k v v v v

• ×

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

( ) ( ) ( ) ( ) ( )i

i j k i r k v v v

A B A B R

• s

m 8 . s m 2 . 1 m 10 . s rad 8 s m 2 . 1 + = × + = × + = = ω

( )i

vR

• s

m 2 = Velocity of the point D:

( ) ( ) ( )i

k i r k v v

A D A D

• m

150 . s rad 8 s m 2 . 1 − × + = × + = ω

( ) ( )

s m 697 . 1 s m 2 . 1 s m 2 . 1 = + =

D D

v j i v

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.3

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

• f the piston P.

SOLUTION:

• Will determine the absolute velocity of

point D with

B D B D

v v v

• +

=

• The velocity is obtained from the given

crank rotation data.

B

v

• The directions of the absolute velocity

and the relative velocity are determined from the problem geometry.

D

v

• B

D

v

• The unknowns in the vector expression are

the velocity magnitudes which may be determined from the corresponding vector triangle.

B D D

v v and

• The angular velocity of the connecting rod

is calculated from .

B D

v

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.3

SOLUTION:

• Will determine the absolute velocity of point D with

B D B D

v v v

• +

=

• The velocity is obtained from the crank rotation data.

B

v

• (

) ( )( )

s rad 4 . 209 m 075 . s rad 4 . 209 rev rad 2 s 60 min min rev 2000 = = =

• =

AB v ω π ω

The velocity direction is as shown.

• The direction of the absolute velocity is horizontal. The

direction of the relative velocity is perpendicular to BD. Compute the angle between the horizontal and the connecting rod from the law of sines.

D

v

• B

D

v

• °

= = ° 95 . 13 in. 3 sin in. 8 40 sin β β

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.3

• Determine the velocity magnitudes

from the vector triangle.

B D D

v v and

B D B D

v v v

• +

=

° = ° = ° sin76.05 s m 71 . 15 50 sin 95 . 53 sin

v v s m 40 . 12 s m 09 . 13 s m 09 . 13 = = =

B D D

v v s rad . 62 m .2 s m 40 . 12 = = = = l v l v

ω ω

s m 09 . 13 = =

v v

( )k

BD

• s

rad . 62 = ω

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

• f A.
• The same translational and rotational velocities at A are
• btained 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
• riginally 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.

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 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 vectors are parallel, the instantaneous center
• f rotation is at infinity and the angular velocity is zero.
• If the velocity magnitudes are equal, the instantaneous

center of rotation is at infinity and the angular velocity is zero.

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 . θ ω cos l v AC v

A A =

=

( ) ( )

θ θ θ ω tan cos sin

A A B

v l v l BC v = = =

• 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.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.4

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. 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
• n the given velocity at A.
• Evaluate the velocities at B and D based on

their rotation about C.

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. s rad 8 m 0.15 s m 2 . 1 = = = =

A A A A

r v r v ω ω

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

about C.

( )( )

s rad 8 m 25 . = = = ω

B B R

r v v

( )i

vR

• s

m 2 =

( ) ( )( )

s rad 8 m 2121 . m 2121 . 2 m 15 . = = = = ω

D D D

r v r

( )(

)

s m 2 . 1 2 . 1 s m 697 . 1 j i v v

D D

• +

= =

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.5

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

• f the piston P.

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

• f 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.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.5

SOLUTION:

• From Sample Problem 15.3,

( )(

)

° = = − = 95 . 13 s m 71 . 15 s m 3 . 481 9 . 403 β

v j i v

• The instantaneous center of rotation is at the intersection of

the perpendiculars to the velocities through B and D. ° = − ° = ° = + ° = 05 . 76 90 95 . 53 40 β γ β γ

D B

° = ° = ° sin50 mm 00 2 95 . 53 sin 05 . 76 sin CD BC mm 1 . 211 mm 4 . 253 = = CD BC

• Determine the angular velocity about the center of rotation

based on the velocity at B.

( )

mm 253.4 s m 71 . 15 = = = BC v BC v

ω ω

• Calculate the velocity at D based on its rotation about the

instantaneous center of rotation.

( ) ( )( )

s rad . 62 mm 2111 . = =

CD v ω s m 09 . 13 s m 09 . 13 = = =

D P

v v s rad . 62 =

ω

Vector Mechanics for Engineers: Dynamics

Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the slab,

A B A B

a a a

• +

=

• Relative acceleration associated with rotation about A includes

tangential and normal components,

A B

a

• (

) ( )

A B n A B A B t A B

r a r k a

• 2

ω α − = × =

( ) ( )

2

ω α r a r a

n A B t A B

= =

Vector Mechanics for Engineers: Dynamics

Absolute and Relative Acceleration in Plane Motion

• Given

determine , and

A A

v a

• .

and α

• B

a

( ) ( )t

A B n A B A A B A B

a a a a a a

• +

+ = + =

• Vector result depends on sense of and the relative

magnitudes of

( )n

A B A

a a and

A

a

• Must also know angular velocity ω.

Vector Mechanics for Engineers: Dynamics

Absolute and Relative Acceleration in Plane Motion

+ → x components: θ α θ ω cos sin

2

l l a A − + =

↑ +

y components: θ α θ ω sin cos

2

l l aB − − = −

• Solve for aB and α.
• Write

in terms of the two component equations,

A B A B

a a a

• +

=

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. θ sin l xA = θ cos l yB = θ ω θ θ cos cos l l x v

A A

= = =

• θ

ω θ θ sin sin l l y v

B B

− = − = =

• θ

α θ ω θ θ θ θ cos sin cos sin

2 2

l l l l x a

A A

+ − = + − = =

• θ

α θ ω θ θ θ θ sin cos sin cos

2 2

l l l l y a

B B

− − = − − = =

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.6

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. Determine (a) the angular acceleration of the gear, and (b) the acceleration of points B, C, and D. 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 relative accelerations with respect to the center. The latter includes normal and tangential acceleration components.

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. ω θ θ

1 1 1

r r v r x

A A

− = − = − =

• s

rad 8 m 0.150 s m 2 . 1

1

− = − = − = r vA ω α θ

1 1

r r aA − = − =

• m

150 . s m 3

2 1

− = − = r aA α

( )k

k

• 2

s rad 20 − = = α α

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.6

( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( )j

i i j j k i r r k a a a a a a a

A B A B A n A B t A B A A B A B

• 2

2 2 2 2 2 2

s m 40 . 6 s m 2 s m 3 m 100 . s rad 8 m 100 . s rad 20 s m 3 − + = − − × − = − × + = + + = + = ω α

s m 12 . 8 ) s m 40 . 6 ( ) s m 5 ( = − =

a j i a

• The acceleration of each point is
• btained 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.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.6

( ) ( )

( ) ( ) ( )

( ) ( ) ( )j

i i j j k i r r k a a a a

A C A C A A C A C

• 2

2 2 2 2 2 2

s m 60 . 9 s m 3 s m 3 m 150 . s rad 8 m 150 . s rad 20 s m 3 + − = − − − × − = − × + = + = ω α j ac

• )

s m 60 . 9 (

=

( ) ( )

( ) ( ) ( )

( ) ( ) ( )i

j i i i k i r r k a a a a

A D A D A A D A D

• 2

2 2 2 2 2 2

s m 60 . 9 s m 3 s m 3 m 150 . s rad 8 m 150 . s rad 20 s m 3 + + = − − − × − = − × + = + = ω α

2 2 2

s m 95 . 12 ) s m 3 ( ) s m 6 . 12 ( = + =

D D

a j i a

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.7

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. SOLUTION:

• The angular acceleration of the connecting

rod BD and the acceleration of point D will be determined from

( ) ( )n

B D t B D B B D B D

a a a a a a

• +

+ = + =

• The acceleration of B is determined from

the given rotation speed of AB.

• The directions of the accelerations

are determined from the geometry.

a a a ) ( and , ) ( ,

• Component equations for acceleration of

point D are solved simultaneously for acceleration of D and angular acceleration

• f the connecting rod.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.7

• The acceleration of B is determined from the given rotation

speed of AB. SOLUTION:

• The angular acceleration of the connecting rod BD and the

acceleration of point D will be determined from

a a a a a a ) ( ) (

• +

+ = + =

( )( )

s m 3289 s rad 4 . 209 m 075 . constant s rad 209.4 rpm 2000 = = = = = = =

r a ω α ω

) 40 sin 40 cos ( ) s m 3289 (

2

j i aB

• °

− ° − =

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.7

• The directions of the accelerations

are determined from the geometry.

a a a ) ( and , ) ( ,

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

( )

( ) ( )( )

2 2 2

s m 769 s rad . 62 m 0.2 = = =

BD n B D

BD a ω

( )(

)

j i a

• °

+ ° − = 95 . 13 sin 95 . 13 cos s m 769 ) (

( ) ( )

BD a α α α 2 . m 0.2 ) ( = = = The direction of (aD/B)t is known but the sense is not known,

( )

( )(

)

j i a

BD t B D

• °

± ° ± = 05 . 76 cos 05 . 76 sin 2 . 0 α i a a

D D

• =

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.7

a a a a a a ) ( ) (

• +

+ = + =

• Component equations for acceleration of point D are solved

simultaneously. x components:

° + ° − ° − = − 95 . 13 sin 2 . 95 . 13 cos 769 40 cos 3289

a α ° + ° + ° − = 95 . 13 cos 2 . 95 . 13 sin 769 40 sin 3289

α

y components: i a k

D BD

• )

s m 2787 ( ) s rad 9937 (

2 2

− = = α

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.8

In the position shown, crank AB has a constant angular velocity ω1 = 20 rad/s counterclockwise. Determine the angular velocities and angular accelerations of the connecting rod BD and crank DE. SOLUTION:

• The angular velocities are determined by

simultaneously solving the component equations for

B D B D

v v v

• +

=

• The angular accelerations are determined by

simultaneously solving the component equations for

B D B D

a a a

• +

=

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.8

SOLUTION:

• The angular velocities are determined by simultaneously

solving the component equations for

B D B D

v v v

• +

=

( )

j i j i k r v

• ω

ω ω ω 425 425 425 425 − − = + − × = × =

( )

j i j i k r v

B AB B

• 160

280 350 200 20 + − = + × = × = ω

( )

j i j i k r v

• ω

ω ω ω 300 75 75 300 + − = + × = × =

ω ω 75 280 425 − − = −

x components:

ω ω 200 160 425 + + = −

y components:

( ) ( )k

k

DE BD

• s

rad 29 . 11 s rad 33 . 29 = − = ω ω

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.8

• The angular accelerations are determined by simultaneously

solving the component equations for

B D B D

a a a

• +

=

( ) (

) (

)

j i j i j i j i k r r a

• 3

10 2 . 54 10 2 . 54 425 425 425 425 29 . 11 425 425 × × − + − − = + − − + − × = − × = α α α ω α

( ) (

)

j i j i r r a

• 3

10 140 10 80 350 200 20 × + × − = + − = − × = ω α

( ) (

) (

)

j i j i j i j i k r r a

• 3

10 5 . 64 10 258 300 75 75 300 33 . 29 75 300 × − × − + − = + − + × = − × = α α α ω α

x components:

3

10 2 . 392 75 425 × − = + −

BD DE

α α

y components:

3

10 2 . 150 300 425 × − = − −

BD DE

α α

k k

• )

s rad 809 ( ) s rad 645 (

= − = α α

Vector Mechanics for Engineers: Dynamics

Rate of Change With Respect to a Rotating Frame

• Frame OXYZ is fixed.
• Frame Oxyz rotates about

fixed axis OA with angular velocity Ω

• Vector function varies in

direction and magnitude.

( )

t Q

• ( )

k Q j Q i Q Q

z y x Oxyz

• +

+ =

• With respect to the fixed OXYZ frame,

( )

k Q j Q i Q k Q j Q i Q Q

z y x z y x OXYZ

• +

+ + + + =

• rate of change with

respect to rotating frame.

( )

= = + +

Oxyz z y x

Q k Q j Q i Q

• If were fixed within Oxyz then is

equivalent to velocity of a point in a rigid body attached to Oxyz and

( )OXYZ

Q

• Q

k Q j Q i Q

z y x

• ×

Ω = + + Q

• With respect to the rotating Oxyz frame,

k Q j Q i Q Q

z y x

• +

+ =

• With respect to the fixed OXYZ frame,

( ) ( )

Q Q Q

Oxyz OXYZ

• ×

Ω + =

Vector Mechanics for Engineers: Dynamics

Coriolis Acceleration

• Frame OXY is fixed and frame Oxy rotates with angular

velocity . Ω

• Position vector for the particle P is the same in both

frames but the rate of change depends on the choice of frame.

P

r

• The absolute velocity of the particle P is

( )

( )Oxy

OXY P

r r r v

• +

× Ω = =

• Imagine a rigid slab attached to the rotating frame Oxy or

for short. Let P’ be a point on the slab which corresponds instantaneously to position of particle P.

( )

= =

Oxy P

r v

• velocity of P along its path on the slab

=

' P

v

• absolute velocity of point P’ on the slab
• Absolute velocity for the particle P may be written as
• P

P P

v v v

• +

=

′

Vector Mechanics for Engineers: Dynamics

Coriolis Acceleration

( )

• P

P Oxy P

v v r r v

• +

= + × Ω =

′

• Absolute acceleration for the particle P is

( ) ( )

[ ]

Oxy OXY P

r dt d r r a

• +

× Ω + × Ω =

( )

( ) ( )Oxy

Oxy P

r r r r a

• +

× Ω + × Ω × Ω + × Ω = 2

( ) ( ) ( )

[ ] ( )

( )Oxy

Oxy Oxy Oxy OXY

r r r dt d r r r

• ×

Ω + = + × Ω = but,

( )

( )Oxy

P P

r a r r a

• =

× Ω × Ω + × Ω =

′

• Utilizing the conceptual point P’ on the slab,
• Absolute acceleration for the particle P becomes

( ) ( )

2 2 2 = × Ω = × Ω = + + = × Ω + + =

′ ′

• P

Oxy c c P P Oxy P P P

v r a a a a r a a a

• Coriolis acceleration

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.

c P A P

a a a a

• +

+ =

• Absolute acceleration of the collar is

( )

= =

Oxy P

r a

• u

a v a

c P c

ω 2 2 = × Ω =

• The absolute acceleration consists of the radial and

tangential vectors shown

( )

2

ω r a r r a

A A

= × Ω × Ω + × Ω =

• where

Vector Mechanics for Engineers: Dynamics

Coriolis Acceleration

u v v t t u v v t

A A

′ + = ′ ∆ + + =

′

• ,

at , at

• Change in velocity over ∆t is represented by the sum
• f three vectors

T T T T R R v ′ ′ ′ + ′ ′ + ′ = ∆

( )

2

ω r a r r a

A A

= × Ω × Ω + × Ω =

• recall,
• is due to change in direction of the velocity of

point A on the rod,

A A t t

a r r t v t T T = = = = ′ ′

→ → 2

lim lim ω ωω ∆ θ ∆ ∆

∆ ∆

T T ′ ′

• result from combined effects of

relative motion of P and rotation of the rod T T R R ′ ′ ′ ′ and u u u t r t u t T T t R R

t t

ω ω ω ∆ ∆ ω ∆ θ ∆ ∆ ∆

∆ ∆

2 lim lim = + =

• +

=

• ′

′ ′ + ′

→ →

u a v a

c F P c

ω 2 2 = × Ω =

• recall,

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.9

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. SOLUTION:

• The absolute velocity of the point P may

be written as

s P P P

v v v

• +

=

′

• Magnitude and direction of velocity
• f pin P are calculated from the

radius and angular velocity of disk D.

P

v

• Direction of velocity of point P’ on S

coinciding with P is perpendicular to radius OP.

P

v ′

• Direction of velocity of P with

respect to S is parallel to the slot.

s P

v

• Solve the vector triangle for the angular

velocity of S and relative velocity of P.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.9

SOLUTION:

• The absolute velocity of the point P may be written as

s P P P

v v v

• +

=

′

• Magnitude and direction of absolute velocity of pin P are

calculated from radius and angular velocity of disk D.

( )( )

s mm 500 s rad 10 mm 50 = = =

D P

R v ω

• Direction of velocity of P with respect to S is parallel to slot.

From the law of cosines, mm 1 . 37 551 . 30 cos 2

2 2 2 2

= = ° − + = r R Rl l R r From the law of cosines, ° = ° = ° = 4 . 42 742 . 30 sin sin 30 sin R sin β β β r ° = ° − ° − ° = 6 . 17 30 4 . 42 90 γ The interior angle of the vector triangle is

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,

( )

mm 1 . 37 s mm 2 . 151 s mm 2 . 151 6 . 17 sin s mm 500 sin = = = ° = =

′ s s P P

r v v ω ω γ

( )k

s

• s

rad 08 . 4 − = ω

( )

° = = 6 . 17 cos s m 500 cosγ

P s P

v v

( )(

)

j i v

• °

− ° − = 4 . 42 sin 4 . 42 cos s m 477 s mm 500 =

P

v

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.10

In the Geneva mechanism, disk D rotates with a constant counter- clockwise angular velocity of 10 rad/s. At the instant when ϕ = 150o, determine angular acceleration of disk S. SOLUTION:

• The absolute acceleration of the pin P may be

expressed as

c s P P P

a a a a

• +

+ =

′

• 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.

• Resolve each acceleration term into the

component parallel to the slot. Solve for the angular acceleration of Disk S.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.10

SOLUTION:

• Absolute acceleration of the pin P may be expressed as

c s P P P

a a a a

• +

+ =

′

• From Sample Problem 15.9.

( ) ( )(

)

j i v k

s P S

• °

− ° − = − = ° = 4 . 42 sin 4 . 42 cos s mm 477 s rad 08 . 4 4 . 42 ω β

• Considering each term in the acceleration equation,

( )( )

( )(

)

j i a R a

P D P

• °

− ° = = = = 30 sin 30 cos s mm 5000 s mm 5000 s rad 10 mm 500

2 2 2 2

ω

( ) ( ) ( )

( )(

)

( ) ( )(

)

( ) ( )( )(

)

j i a j i r a j i r a a a a

• °

+ ° − = ° + ° − = ° − ° − = + =

4 . 42 cos 4 . 42 sin mm 1 . 37 4 . 42 cos 4 . 42 sin 4 . 42 sin 4 . 42 cos

α α ω note: αS may be positive or negative

Vector Mechanics for Engineers: Dynamics

• The relative acceleration must be parallel to the

slot.

s P

a

• Sample Problem 15.10

s P

v

• The direction of the Coriolis acceleration is obtained by

rotating the direction of the relative velocity by 90o in the sense of ωS.

( )(

)

( )( )(

)

( )(

)

j i j i j i v a

• 4

. 42 cos 4 . 42 sin s mm 3890 4 . 42 cos 4 . 42 sin s mm 477 s rad 08 . 4 2 4 . 42 cos 4 . 42 sin 2

+ ° − = + ° − = + ° − = ω

• Equating components of the acceleration terms

perpendicular to the slot, s rad 233 7 . 17 cos 5000 3890 1 . 37 − = = ° − +

S S

α α

( )k

S

• s

rad 233 − = α

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

• r

dt r d v

• ×

= = ω and the acceleration of the particle P is

( )

. dt d r r a ω α ω ω α

• =

× × + × =

• Angular velocities have magnitude and direction and obey

parallelogram law of addition. They are vectors.

• As the vector moves within the body and in space, it

generates a body cone and space cone which are tangent along the instantaneous axis of rotation. ω

• The angular acceleration represents the velocity of the

tip of . ω

• α

Vector Mechanics for Engineers: Dynamics

General Motion

• For particles A and B of a rigid body,

A B A B

v v v

• +

=

• 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

A B A B

r v v

• ×

+ = ω

• Similarly, the acceleration of the particle P is

( )

A B A B A A B A B

r r a a a a

• ×

× + × + = + = ω ω α

• 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.

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.11

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.
• Angular acceleration of the boom,

( )

2 1 2 2 2 2 1

ω ω ω Ω ω ω ω ω α

• ×

= × + = = + =

Oxyz

• Velocity of boom tip,

r v

• ×

= ω

• Acceleration of boom tip,

( )

v r r r a

• ×

+ × = × × + × = ω α ω ω α SOLUTION: With

• Angular velocity of the boom,

2 1

ω ω ω

• +

=

( )

j i j i r k j

• 6

39 . 10 30 sin 30 cos 12 50 . 30 .

2 1

+ = ° + ° = = = ω ω

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.11

j i r k j

• 6

39 . 10 50 . 30 .

2 1

+ = = = ω ω SOLUTION:

• Angular velocity of the boom,

2 1

ω ω ω

• +

=

( ) ( )k

j

• s

rad 50 . s rad 30 . + = ω

• Angular acceleration of the boom,

( )

( ) ( )k

j

Oxyz

• s

rad 50 . s rad 30 .

2 1 2 2 2 2 1

× = × = × + = = + = ω ω ω Ω ω ω ω ω α

( )i

• 2

s rad 15 . = α

• Velocity of boom tip,

6 39 . 10 5 . 3 . k j i r v

• =

× = ω

( ) ( ) ( )k

j i v

• s

m 12 . 3 s m 20 . 5 s m 54 . 3 − + − =

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.11

j i r k j

• 6

39 . 10 50 . 30 .

2 1

+ = = = ω ω

• Acceleration of boom tip,

( )

k j i i k k j i k j i a v r r r a

• 90

. 50 . 1 60 . 2 94 . 90 . 12 . 3 20 . 5 3 50 . 30 . 6 39 . 10 15 . + − − − = − − + = × + × = × × + × = ω α ω ω α k j i a

• )

s m 80 . 1 ( ) s m 50 . 1 ( ) s m 54 . 3 (

2 2 2

+ − − =

Vector Mechanics for Engineers: Dynamics

Three-Dimensional Motion. Coriolis Acceleration

• With respect to the fixed frame OXYZ and rotating

frame Oxyz,

( ) ( )

Q Q Q

Oxyz OXYZ

• ×

Ω + =

• Consider motion of particle P relative to a rotating frame

Oxyz or for short. The absolute velocity can be expressed as

( )

• P

P Oxyz P

v v r r v

• +

= + × Ω =

′

• The absolute acceleration can be expressed as

( )

( ) ( ) ( )

• n

accelerati Coriolis 2 2 2 = × Ω = × Ω = + + = + × Ω + × Ω × Ω + × Ω =

′

• P

Oxyz c c P p Oxyz Oxyz P

v r a a a a r r r r a

Vector Mechanics for Engineers: Dynamics

Frame of Reference in General Motion

Consider:

• fixed frame OXYZ,
• translating frame AX’Y’Z’, and
• translating and rotating frame Axyz or

.

• With respect to OXYZ and AX’Y’Z’,

A P A P A P A P A P A P

a a a v v v r r r

• +

= + = + =

• 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.

( )

• P

P Axyz A P A P A P

v v r r v v

• +

= + × Ω + =

′

( ) ( ) ( )

c P P Axyz A P Axyz A P A P A P A P

a a a r r r r a a

• +

+ = + × Ω + × Ω × Ω + × Ω + =

′

• 2

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.15

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. SOLUTION:

• Define a fixed reference frame OXYZ at O

and a moving reference frame Axyz or 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

• P

P P

v v v

• +

=

′

• The acceleration of P is found from

c P P P

a a a a

• +

+ =

′

• The angular velocity and angular

acceleration of the disk are

( )

ω Ω ω α ω Ω ω

• ×

+ = + =

• D

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 attached to the arm at A.

j j R i L r

• 1

ω Ω = + = k j R r

D A P

• 2

ω ω = =

• With P’ of the moving reference frame coinciding with

P, the velocity of the point P is found from

( )

i R j R k r v k L j R i L j r v v v v

A P D P P P P P

• 2

2 1 1

ω ω ω ω ω Ω − = × = × = − = + × = × = + =

′ ′

• k

L i R vP

• 1

2

ω ω − − =

Vector Mechanics for Engineers: Dynamics

Sample Problem 15.15

• The acceleration of P is found from

c P P P

a a a a

• +

+ =

′

• (

)

( )

i L k L j r aP

• 2

1 1 1

ω ω ω Ω Ω − = − × = × × =

′

( )

( )

j R i R k r a

A P D D P

• 2

2 2 2

ω ω ω ω ω − = − × = × × =

• (

)

k R i R j v a

P c

• 2

1 2 1

2 2 2 ω ω ω ω Ω = − × = × =

• k

R j R i L aP

• 2

1 2 2 2 1

2 ω ω ω ω + − − =

• Angular velocity and acceleration of the disk,
• D

ω Ω ω

• +

= k j

• 2

1

ω ω ω + =

( )

( )

k j j

• 2

1 1

ω ω ω ω Ω ω α + × = × + =

• i
• 2

1ω

ω α =