Dynamics_Lecture6 [Compatibility Mode]

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Kinematics of Rigid Bodies



Introduction

• rectilinear translation

- translation:

• 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:

- motion about a fixed point

- general plane motion

- rotation about a fixed axis

• curvilinear translation

- general motion



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.





Rotation About a Fixed Axis. Velocity

• Consider rotation of rigid body about a

fixed axis AA’

• Velocity vector of the particle P is

tangent to the path with magnitude

dt r d v =dt dsv =

( ) ( ) θ φ θ sinr BPs ∆=∆=∆

( ) φ θ φ sinsinlim0

r t r dt

sv

t =∆== →∆

locityangular vek k

r dt r d v

===

×==

θ ω ω

ω

• The same result is obtained from



Rotation About a Fixed Axis. Acceleration

• Differentiating to determine the acceleration,

( )

vr dt

d

dt r d r

dt d

r dt

d

dt

vd a

×+×=

×+×=

×==

ω ω

ω ω

ω

•

k k k

celerationangular acdt d

θ ω α

α

===

==

componentonacceleratiradial

componentonacceleratiltangentia

=××

=×

××+×=

r

r

r r a

ω ω

α

ω ω α

• Acceleration of P is combination of twovectors,



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

2ω α

ω α

−×=

××+×=

• Resolving the acceleration into tangential andnormal components,

22 ω ω

α α

r ar a

r ar k a

nn

t t

=−=

=×=



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.

ω θ ω α

ω

θ θ ω

d d d

d dt

dt

d

===

==

2

or• Recall

dt t

• Uniform Rotation, α = 0:

t θ θ += 0

• Uniformly Accelerated Rotation, α = constant:

( )020

2

2

21

00

0

2 θ θ α ω ω

α ω θ θ

α

−+=

++=

+=

t t

t



Sample Problem 15.1

SOLUTION:

• Due to the action of the cable, the

tangential velocity and acceleration of

D are equal to the velocity andacceleration of C. Calculate the initial

angular velocity and acceleration.

• Apply the relations for uniformly

Cable C has a constant acceleration of

225 mm/s2 and an initial velocity of 300

mm/s, 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.

accelerated rotation to determine thevelocity and angular position of the

pulley after 2 s.

• Evaluate the initial tangential and

normal acceleration components of D.



Sample Problem 5.1

SOLUTION:

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

velocity and acceleration of C.

( ) ( )

( )

( )srad4

75

300

smm300

00

00

00

===

=

→==

r

v

r v

vv

D

D

C D

ω

ω

( )

( )( ) 2

srad33

9

sin.9

===

=

→==

r

a

r a

aa

t D

t D

C t D

α

α

• pp y e re a ons or un orm y acce era e ro a on o

determine velocity and angular position of pulley after 2 s.

( ) srad10s2srad3srad4 20 =+=+= t α ω ω

( )( ) ( )

rad14

s2srad3s2srad422

212

21

0

=

+=+= t t α ω θ

( ) revsof numberrad2

rev1rad14 =

=

π N rev23.2= N

( )( )

( )( )rad14mm125

srad10mm125

==∆

==

θ r y

r v

B

B

m75.1

sm25.1

=∆

↑=

B

B

y

v



Sample Problem 5.1

• Evaluate the initial tangential and normal acceleration

components of D.

( ) →==2

smm225C t Daa

( ) ( )( )222

0 smm1200srad4mm57 === ω Dn D r a

( ) ( ) ↓=→= 22 smm1200smm225n Dt D aa

Magnitude and direction of the total acceleration,

( ) ( )

22

22

1200225 +=

+=n Dt D D aaa

2smm1220= Da

( )

( )

225

1200

tan

=

=t D

n D

a

aφ

°= 4.79φ





Absolute and Relative Velocity in Plane Motion

ω ω r vr k v A B A B A B =×=

• 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 vvv

+=

A B A B r k vv

×+= ω




• Assuming that the velocity v A of end A is known, wish to determine the

velocity v B of end B and the angular velocity ω in terms of v A, l, and θ.

• The direction of v B and v B/A are known. Complete the velocity diagram.

θ

θ

tan

tan

A B

A

B

vv

v

v

=

=

θ

ω

θ ω

cos

cos

l

v

l

v

v

v

A

A

A B

A

=

==




• Selecting point B as the reference point and solving for the velocity v A of end A

and the angular velocity ω leads to an equivalent velocity triangle.

• v A/B has the same magnitude but opposite sense of v B/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.



Sample Problem 15.2

SOLUTION:

• The displacement of the gear center in

one revolution is equal to the outer

circumference. Relate the translationaland 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.

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

Evaluate the velocities of points B and D.

AP A AP AP r k vvvv

×+=+= ω



Sample Problem 15.2

SOLUTION:

• The displacement of the gear center in one revolution is

equal to the outer circumference.

For x A > 0 (moves to right), ω < 0 (rotates clockwise).

1

12 2

A A

x x r

r

θ θ

π π = − = −

x

y Differentiate to relate the translational and angularvelocities.

m0.150

sm2.1

1

1

−=−=

−=

r

v

r v

A

A

ω ( )k k

srad8−== ω ω



Sample Problem 15.2

• For any point P on the gear, AP A AP AP r k vvvv

×+=+= ω

Velocity of the upper rack is equal to

velocity of point B:

( ) ( ) ( )( ) ( )ii

jk i

r k vvv A B A B R

sm8.0sm2.1

m10.0srad8sm2.1

+=×+=

×+== ω

( )iv R

sm2=

Velocity of the point D:

( ) ( ) ( )ik i

r k vv A D A D

m150.0srad8sm2.1 −×+=

×+= ω

( ) ( )

sm697.1

sm2.1sm2.1

=

+=

D

D

v

jiv



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

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 .




• 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

• 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.




• The instantaneous center of rotation lies at the intersection of

the perpendiculars to the velocity vectors through A and B .

θ

ω

cosl

v

AC

v A A == ( ) ( )

θ θ

θ ω

tan cos

sin

A

A B

v l

vl 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.



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 theinstantaneous 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.

Determine (a) the angular velocityof the gear, and (b) the velocities of

the upper rack R and point D of the

gear.

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



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 thegiven velocity at A.

srad8m0.15

sm2.1====

A

A A A

r

vr v ω ω

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

about C .

( )( )srad8m25.0=== B B R r vv

( )iv R

sm2=

( )( )( )srad8m2121.0

m2121.02m15.0

====

ω D D

D

r v

r

( )( )sm2.12.1

sm697.1

jiv

v

D

D

+=

=



Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the slab,

A B A B aaa

+=

• Relative acceleration associated with rotation about A includes

tangential and normal components,

A Ba

( ) A Bn A B

A Bt A B

r a

r k a

2ω

α

−=

×=

( ) 2ω

α

r a

r a

n A B

t A B

=

=





Absolute and Relative Acceleration in Plane Motion

+→ x components: θ α θ ω cossin0 2

lla A −+=

↑+ y components: θ α θ ω sincos2lla B −−=−

• Solve for a B and α .

• Write in terms of the two component equations, A B A B aaa

+=



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 sinl x A = θ cosl y B =

xv A A

=

=

yv B B

−=

=

θ cosl= θ sinl−=

θ α θ

θ θ θ θ

cossin

cossin

2

2

ll

ll

xa A A

+−=

+−=

=

θ α θ

θ θ θ θ

sincos

sincos

2

2

ll

ll

ya B B

−−=

−−=

=



Sample Problem 15.6

SOLUTION:

• The expression of the gear position as a

function of θ is differentiated twice to

define the relationship between thetranslational and angular accelerations.

• The acceleration of each point on the

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.

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.



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

11

1

r r v

r x

A

A

−=−=

−=

srad8

sm2.1−=−=−=

v A m0.1501r

α θ 11 r r a A −=−=

m150.0

sm32

1

−=−=

r

a Aα

k k 2

srad20−== α α



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.

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) jii

j jk i

r r k a

aaaaaa

A B A B A

n A Bt A B A A B A B

222

222

2

sm40.6sm2sm3

m100.0srad8m100.0srad20sm3

−+=

−−×−=

−×+=

++=+=

ω α

222sm12.8sm40.6m5 =−= B B a jisa



Sample Problem 15.6

r r k aaaa AC AC A AC AC 2−×+=+= ω α

( ) ( ) ( )( ) ( ) ( ) jii j jk i

222

222

sm60.9sm3sm3m150.0srad8m150.0srad20sm3

+−=

−−−×−=

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )i ji

iik i

r r k aaaa A D A D A A D A D

222

222

2

sm60.9sm3sm3

m150.0srad8m150.0srad20sm3

++=

−−−×−=

−×+=+=ω α

222sm95.12sm3m6.12 =+=

D Da jisa



Sample Problem 15.7

SOLUTION:

• The angular acceleration of the

connecting rod BD and the acceleration

of point D will be determined from

n B Dt B D B B D B D aaaaaa

++=+=

• The acceleration of B is determined from

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.

e g ven ro a on spee o .

• The directions of the accelerations

are

determined from the geometry.n B Dt B D D aaa

and,,

• Component equations for acceleration

of point D are solved simultaneously for

acceleration of D and angular

acceleration of the connecting rod.



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

n B Dt B D B B D B D aaaaaa

++=+=

constantsrad209.4rpm2000 === AB

( )( ) 22

100072

AB

sm3289srad4.209m0

===

=

AB B r a ω α

( )( ) jia B

°−°−= 40sin40cossm3289 2



Sample Problem 15.7

• The directions of the accelerations are

determined from the geometry.n B Dt B D D aaa

and,,

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

( ) ( ) ( )( ) 22

10002002

sm8.768srad0.62m === BDn B D BDa ω

( ) ( )( ) jian

B D

°+°−= 95.13sin95.13cossm8.768 2

( ) ( )BD BD BDt B D BDa α α α 2.0m

1000200 ===

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

( ) ( )( ) jia BDt

B D

°±°±= 05.76cos05.76sin2.0 α

iaa D D

=

S 1



Sample Problem 15.7

• Component equations for acceleration of point D are solved

simultaneously.

n B Dt B D B B D B D

x components:

°+°−°−=− 95.13sin2.095.13cos8.76840cos3289 BD Da α

°+°+°−= 95.13cos2.095.13sin8.76840sin32890 BDα

y components:

( )( ) ia

k

D

BD

2

2

sm2790

srad9940

−=

=α

S l P bl 15 8



Sample Problem 15.8

SOLUTION:

• The angular velocities are determined by

simultaneously solving the component

equations for

B D B D vvv

+=

In the position shown, crank AB has a

constant angular velocity ω 1 = 20 rad/s

counterclockwise.

Determine the angular velocities andangular accelerations of the connecting

rod BD and crank DE .

• e angu ar acce era ons are e erm ne

by simultaneously solving the componentequations for

B D B D aaa

+=

S l P bl 15 8



Sample Problem 15.8

SOLUTION:

• The angular velocities are determined by simultaneously

solving the component equations for

B D B D vvv

+=

( ) ji

jik r v

DE DE

DE D DE D

ω ω

ω ω

1717

340340

−−=

+−×=×=

ik r v

28016020 +×=×=

ji

160280 +−=

( )

ji

jik r v

BD BD

BD B D BD B D

ω ω

ω ω

123

60240

+−=

+×=×=

BD DE 328017 −−=− x components:

BD DE 1216017 ++=− y components:

( ) ( )k k DE BD

srad29.11srad33.29 =−= ω ω

S l P bl 15 8



Sample Problem 15.8

• The angular accelerations are determined by

simultaneously solving the component equations for

B D B D aaa

+=

( ) ( ) ( ) ji ji

ji jik

r r a

DE DE

DE

D DE D DE D

33.4333.4334.034.0

34.034.029.1134.034.02

2

−+−−=

+−−+−×=

−×=

α α

α

ω α

ir r a

28.01620022 +−=−×= α

ji

11264 +−=

( ) ( ) ( )

ji ji

ji jik

r r a

D B D B

D B

D B BD D B BD B D

61.514.20624.006.0

06.024.033.2906.0242

2

−−+=

+−+×=

−×=

α α

α

ω α

x components: 7.31306.034.0 −=+− BD DE α α

y components: 28.12024.034.0 −=−− BD DE α α

k k DE BD

22srad809srad645 =−= α α

Rate of Change With Respect to a Rotating Frame



Rate of Change With Respect to a Rotating Frame

k Q jQiQQ z y xOxyz

++=

• With respect to the fixed OXYZ frame,

k Q jQiQk Q jQiQQ z y x z y xOXYZ

+++++=

• With respect to the rotating Oxyz frame,

k Q jQiQQ z y x

++=

• Frame OXYZ is fixed.

• Frame Oxyz rotates about

fixed axis OA with angular

velocityΩ

• Vector function varies

in direction and magnitude.

( )t Q

• rate of changewith respect to rotating frame.

==++ Oxyz z y x Qk Q jQiQ

• If were fixed within Oxyz then is

equivalent to velocity of a point in a rigid body

attached to Oxyz and

OXYZ Q

Qk Q jQiQ z y x ×Ω=++

Q

• With respect to the fixed OXYZ frame,

QQQ OxyzOXYZ

×Ω+=

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.

Pr

• The absolute velocity of the particle P is

• 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.

( )==

OxyP r v

F

velocity of P along its path on the slab

='Pv

absolute velocity of point P’ on the slab

OxyOXY P r r r v ×==

• Absolute velocity for the particle P may be written as

F

PPP

vvv

+= ′







• 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.

cP AP aaaa

++=F

• Absolute acceleration of the collar is

where

uava cPc 22 =×Ω= F

• The absolute acceleration consists of the radial and

tangential vectors shown

( )2

ω r ar r a A A =×Ω×Ω+×Ω=

( ) 0== OxyP r a F





• Change in velocity over ∆t is represented by the

sum of three vectors

T T T T R Rv ′′′+′′+′=∆

• is due to change in direction of the velocity of point A on the rod,

A A ar r vT T

====′′ 2limlim ω ωω

θ ∆

T T ′′

( ) 2ω r ar r a A A =×Ω×Ω+×Ω= recall,

→→

• result from combined effects of

relative motion of P and rotation of the rod

T T R R ′′′′ and

uuu

t

r

t u

t

T T

t

R R

t t ∆

∆ω

∆

θ ∆

∆∆ ∆∆

2

limlim00

=+=

+=

′′′+′→→

uavacPc

22 =×Ω=F

recall,

uvvt t

uvvt

A

A

′+=′∆+

+=

′

,at

,at

Sample Problem 15 9



Sample Problem 15.9

SOLUTION:

• The absolute velocity of the point P

may be written as

sPPP vvv

+=

′

• Magnitude and direction of velocity

of pin P are calculated from thePv

Disk D of the Geneva mechanism rotates

with constant counterclockwise angular

velocity ω D = 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.

.

• Direction of velocity of point P’ on

S coinciding with P is perpendicular to

radius OP.

Pv ′

• Direction of velocity of P with

respect to S is parallel to the slot.sP

v

• Solve the vector triangle for the

angular velocity of S and relative

velocity of P.

Sample Problem 15.9



Sample Problem 15.9

SOLUTION:

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

sPPP vvv

+= ′

• Magnitude and direction of absolute velocity of pin P arecalculated from radius and angular velocity of disk D.

( )( ) smm500srad10mm50 === DP Rv

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

From the law of cosines,

mm1.37551.030cos22222 ==°−+= r R Rll Rr

From the law of cosines,

°=°

=°

= 4.42742.030sinsin30sin

Rsin β β β

r

°=°−°−°= 6.17304.4290γ

The interior angle of the vector triangle is

Sample Problem 15.9



Sample Problem 15.9

• Direction of velocity of point P’ on S coinciding with P is

perpendicular to radius OP. From the velocity triangle,

( )

mm1.37

smm2.151

smm2.1516.17sinsmm500sin

==

=°==′

ss

PP

r

vv

ω ω

( )k s

srad08.4−=ω

( ) °== 6.17cossm500cosγ PsP vv

( )( ) jiv sP

°−°−= 4.42sin4.42cossm477

smm500=Pv

Sample Problem 15.10




SOLUTION:

• The absolute acceleration of the pin P may

be expressed as

csPPP aaaa

++=

′

• The instantaneous angular velocity of Disk

S is determined as in Sample Problem 15.9.

In the Geneva mechanism, disk D

rotates with a constant counter-

clockwise angular velocity of 10rad/s. At the instant when ϕ = 150o,

determine angular acceleration of

disk S.

• The only unknown involved in the

acceleration equation is the instantaneous

angular acceleration of Disk S.

• Resolve each acceleration term into thecomponent parallel to the slot. Solve for

the angular acceleration of Disk S.






p

• The direction of the Coriolis acceleration is obtained

by rotating the direction of the relative velocity

by 90o in the sense of ω S.

( )( )( )( )

( )( ) ji

ji jiva sPSc

4.42cos4.42sinsmm3890

4.42cos4.42sinsmm477srad08.424.42cos4.42sin2

2 +°−=

+°−=

+°−=

• The relative acceleration must be parallel tothe slot.

sPa

• Equating components of the acceleration terms

perpendicular to the slot,

srad233

07.17cos500038901.37

−=

=°−+

S

S

α

α

( )k S

srad233−=α

Motion About a Fixed Point



• The most general displacement of a rigid body with afixed 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,

r dt

r d v

×== ω

and the acceleration of the article 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.

• The angular acceleration represents the velocity of

the tip of .

α

General Motion



• For particles A and B of a rigid body,

A B A B vvv

+=

• Particle A is fixed within the body and motion of

the body relative to AX’Y’Z’ is the motion of abody with a fixed point

A B A B r vv

×+=

• Similarly, the acceleration of the particle P is

( ) A B A B A

A B A B

r r a

aaa

××+×+=

+=

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.




The crane rotates with a constant

SOLUTION:

With

• Angular velocity of the boom,

=

( )

ji

jir

k j

639.10

30sin30cos12

50.030.0 21

+=

°+°=

== ω ω

angular velocity ω 1 = 0.30 rad/s and the

boom is being raised with a constant

angular velocity ω 2 = 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,

( )

21

22221

ω ω

ω Ω ω ω ω ω α

×=

×+==+= Oxyz

• Velocity of boom tip,r v

×=

• Acceleration of boom tip,

( ) vr r r a

×+×=××+×= α α




SOLUTION:

• Angular velocity of the boom,

21

+=

( ) ( )k j

srad50.0srad30.0 +=ω

• Angular acceleration of the boom,

( )

k

Oxyz

srad50.0srad30.0

22221

×=×=

×+==+= ω Ω ω ω ω ω α

jir

k j

639.10

50.030.0 21

+=

== ω ω

i

2

srad15.0=α

• Velocity of boom tip,

0639.10

5.03.00

k ji

r v

=×= ω

( ) ( ) ( )k jiv

sm12.3sm20.5sm54.3 −+−=




• Acceleration of boom tip,

( )

k jiik

k jik ji

a

vr r r a

90.050.160.294.090.0

12.320.5350.030.00

0639.100015.0

+−−−=

−−+=

×+×=××+×= α α

jir

k j

639.10

50.030.0 21

+=

== ω ω

k jia 222 sm80.1sm50.1sm54.3 +−−=

Three-Dimensional Motion. Coriolis Acceleration



• With respect to the fixed frame OXYZ and rotating

frame Oxyz,

QQQ OxyzOXYZ

×Ω+=

• Consider motion of particle P relative to a rotating

frame Oxyz or F for short. The absolute velocity can

be expressed as

( )F PP

OxyzP

vvr r v

+=

+×Ω=

′

• The absolute acceleration can be expressed as

( ) ( ) ( )

( ) onacceleratiCoriolis22

2

=×Ω=×Ω=

++=+×Ω+×Ω×Ω+×Ω=

′

F

F

POxyzc

cP p

OxyzOxyzP

vr a

aaar r r r a

Frame of Reference in General Motion



• With respect to OXYZ and AX’Y’Z’,

AP AP

AP AP

AP AP

aaa

vvv

r r r

+=

+=

+=

• The velocity and acceleration of P relative to

AX’Y’Z’can be found in terms of the velocity

Consider:

- fixed frame OXYZ ,

- translating frame AX’Y’Z’, and

- translating and rotating frame Axyz

or F .

and acceleration of P relative to Axyz.

F PP

Axyz AP AP AP

vv

r r vv

+=

+×Ω+=

′

( )( ) ( )

cPP

Axyz AP Axyz AP

AP AP AP

aaa

r r

r r aa

++=

+×Ω+

×Ω×Ω+×Ω+=

′ F

2




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

For the disk mounted on the arm, theindicated 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.

F PPPvvv

+=′

• The acceleration of P is found from

cPPP aaaa

++= ′ F

• The angular velocity and angularacceleration of the disk are

( ) ω Ω ω α

ω Ω ω

×+=

+=

F

F D




SOLUTION:

• Define a fixed reference frame OXYZ at O and amoving reference frame Axyz or F attached to the

arm at A.

j

j Ri Lr

1ω Ω =+=

k

j Rr

D

AP

2ω ω =

=

F

• With P’ of the moving reference frame coinciding

with P, the velocity of the point P is found from

( )

i R j Rk r v

k L j Ri L jr v

vvv

AP DP

P

PPP

22

11

ω ω ω

ω ω Ω

−=×=×=

−=+×=×=

+=

′

′

F F

F

k Li RvP

12 ω ω −−=




• The acceleration of P is found from

cPPP aaaa

++= ′ F

( ) ( ) i Lk L jr aP

2111 ω ω ω Ω Ω −=−×=××=′

( ) j Ri Rk

r a AP D DP

2222 ω ω ω −=−×=

××= F F F

( ) k Ri R j

va Pc

2121 22 ω ω ω ω =−×=

×= F

k R j Ri LaP

21

22

21 2 ω ω ω ω +−−=

• Angular velocity and acceleration of the disk,

F DΩ += k j 21 ω ω ω +=

( )

( )k j j

211 ω ω ω

ω Ω ω α

+×=

×+=F

i

21α =

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