CEE 271: Applied Mechanics II, Dynamics– Lecture 23: Ch.16, Sec.5-6 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Thursday, Nov. 1, 2012

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RELATIVE MOTION ANALYSIS: VELOCITY

Today’s objectives: Studentswill be able to

1 Describe the velocity of arigid body in terms oftranslation and rotationcomponents.

2 Perform a relative-motionvelocity analysis of a pointon the body.

In-class activities:• Reading Quiz• Applications• Translation and Rotation

Components of Velocity• Relative Velocity Analysis• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 When a relative-motion analysis involving two sets ofcoordinate axes is used, the x′ − y′ (moving) coordinatesystem will(a) be attached to the selected point for analysis.(b) rotate with the body.(c) not be allowed to translate with respect to the fixed frame.(d) None of the above.

ANS: (a)

2 In the relative velocity equation, vB/A is(a) the relative velocity of B with respect to A.(b) due to the rotational motion.(c) ω × rB/A .(d) All of the above.

ANS: (d)

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APPLICATIONS

• As the slider block A moves horizontally to the left with vA,it causes the link CB to rotate counterclockwise. Thus vBis directed tangent to its circular path.

• Which link is undergoing general plane motion? Link ABor link BC? AB=general, BC=angular only

• How can the angular velocity ω of link AB be found?

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APPLICATIONS(continued)

• Planetary gear systems are usedin many automobile automatictransmissions. By locking orreleasing different gears, thissystem can operate the car atdifferent speeds.

• How can we relate the angularvelocities of the various gears inthe system?

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RELATIVE MOTION ANALYSIS (Section 16.5)

When a body is subjected to general plane motion, itundergoes a combination of translation and rotation.

Point A is called the base point in this analysis. It generally hasa known motion. The x′ − y′ frame translates with the body, butdoes not rotate. The displacement of point B can be written:

drB = drA + drB/A

where drB is displacement due to translation and rotation,drA is displacement due to translation only, anddrB/A is displacement due to rotation only.

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RELATIVE MOTION ANALYSIS: VELOCITY

• The velocity at B is given as vB = vA + vB/A or

drBdt

=drAdt

+drB/A

dt(1)

• Since the body is taken as rotating about A, i.e.,|rA − rB| = const,

vB/A =drB/A

dt= ω × rB/A

• Here ω will only have a k component since the axis ofrotation is perpendicular to the plane of translation. 7 / 40

RELATIVE MOTION ANALYSIS: VELOCITY (cont’d)

vB = vA + ω × rB/A

• When using the relative velocityequation, points A and B shouldgenerally be points on the bodywith a known motion. Often thesepoints are pin connections inlinkages.

• For example, point A on link AB mustmove along a horizontal path, whereaspoint B moves on a circular path.

• The directions of vA and vB are knownsince they are always tangent to theirpaths of motion.

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RELATIVE MOTION ANALYSIS: VELOCITY (cont’d)

vB = vA + ω × rB/A

• When a wheel rolls without slipping,point A is often selected to be at thepoint of contact with the ground.

• Since there is no slipping, point A haszero velocity.

• Furthermore, point B at the center ofthe wheel moves along a horizontalpath. Thus, vB has a known direction,e.g., parallel to the surface.

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PROCEDURE FOR ANALYSIS

The relative velocity equation can be applied using either aCartesian vector analysis or by writing scalar x− and y−component equations directly.

• Scalar Analysis:1 Establish the fixed x− y coordinate directions and draw a

kinematic diagram for the body. Then establish themagnitude and direction of the relative velocity vector vB/A.

2 Write the equation vB = vA + vB/A. In the kinematicdiagram, represent the vectors graphically by showing theirmagnitudes and directions underneath each term.

3 Write the scalar equations from the x and y components ofthese graphical representations of the vectors. Solve for theunknowns.

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PROCEDURE FOR ANALYSIS(contd.)

• Vector Analysis:1 Establish the fixed x− y coordinate directions and draw the

kinematic diagram of the body, showing the vectors vA, vB ,rB/A and ω. If the magnitudes are unknown, the sense ofdirection may be assumed.

2 Express the vectors in Cartesian vector form and substitutethem into vB = vA + ω × rB/A. Evaluate the cross productand equate respective i and j components to obtain twoscalar equations.

3 If the solution yields a negative answer, the sense ofdirection of the vector is opposite to that assumed.

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EXAMPLE I

• Given: Roller A is moving to theright at 3m/s.

• Find: The velocity of B at theinstant θ = 30o.

• Plan:1 Establish the fixed x− y directions and draw a kinematic

diagram of the bar and rollers.2 Express each of the velocity vectors for A and B in terms of

their i, j,k components and solve vB = vA + ω × rB/A.

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EXAMPLE I(Solution)

Kinematic diagram:

y

Express the velocity vectors.

vB = vA + ω × rB/A

−vBj = 3i+ ωk × rB/A

rB/A = (−1.5 cos 30◦i+ 1.5 sin 30◦j)

−vBj = 3i− 1.299ωj − 0.75ωi

Equating the i and j componentsgives:

0 = 3− 0.75ω

−vB = −1.299ωSolving: ω = 4 rad/s or ω = 4 rad/sk

vB = 5.2m/s or vB = −5.2m/s j

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EXAMPLE II

• Given: Crank rotates OA with anangular velocity of 12 rad/s.

• Find: The velocity of piston B andthe angular velocity of rod AB.

• Plan : Notice that point A moves on a circular path. Thedirections of vA is tangent to its path of motion. Draw akinematic diagram of rod AB and use

vB = vA + ωAB × rB/A

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EXAMPLE II (Solution)

Figure: Kinematic diagram of AB

• Since crank OA rotates withan angular velocity of 12 rad/s,the velocity at A will be:

vA = −0.3(12)i = −3.6im/s• Rod AB. Write the relative-

velocity equation:vB = vA + ωA × rB/A

vBj = −3.6i+ ωAk× (0.6 cos 30◦i− 0.6 sin 30◦j)

vBj = −3.6i+ 0.5196ωAj+ 0.3ωAi

• By comparing the i, j components:i : 0 = −3.6 + 0.3ωA ⇒ ωA = 12rad/sj : vB = 0.5196ωA ⇒ vB = 6.24m/s

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CHECK YOUR UNDERSTANDING QUIZ

1 If the disk is moving with a velocity at point O of 15 ft/s andω = 2 rad/s, determine the velocity at A.

A

2 ft

v = 15 ft/s

ω

O

(a) 0 ft/s

(b) 4 ft/s

(c) 15 ft/s

(d) 11 ft/s

ANS: (d)2 If the velocity at A is zero, then determine the angular

velocity, ω.(a) 30 rad/s(b) 0.0 rad/s(c) 7.5 rad/s(d) 15 rad/s

ANS: (c)16 / 40

GROUP PROBLEM SOLVING

• Given: The ring gear R is rotatingat ωR = 3 rad/s, and the sun gearS is held fixed, ωS = 0.

• Find: The angular velocity of theeach of the planet gears P and ofshaft A.

• Plan: Draw the kinematic diagramof gears. Then, apply the relativevelocity equations to the gearsand solve for unknowns.

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GROUP PROBLEM SOLVING (Solution)

P

y

x

3 rad/s

160mm

A

B

vB = 0vA

R

S

80mm

C

Figure: Kinematic diagramof gears

• Since the ring gear R is rotating atωR = 3 rad/s, the velocity at point Awill be:vA = −3(160)i = −480i mm/s

• Also note that vB = 0 since the gearS is held fixed: ωS = 0.

• Applying the relative velocityequation to points A and B:

vA = vB + ωP × rA/B

−480i = 0 + (ωPk)× (80j)

⇒ −480i = −80ωP i

ωP = 6 rad/s18 / 40

GROUP PROBLEM SOLVING (Solution)

P

y

x

3 rad/s

160mm

A

B

vB = 0vA

R

S

80mm

C

• Apply the relative velocity equation atpoint B and C to Gear P in order to findthe velocity at B.

vC = vB + ωP × rC/B

= 0 + (6k)× (40j) = −240 imm/s

• Note that the shaft A has a circularmotion with the radius of 120mm. Theangular velocity of the shaft is

ωshaft =vCr

= −240/120 = −2 rad/s

• The shaft A is rotating incounter-clockwise direction!

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ATTENTION QUIZ1 Which equation could be used to find the velocity of the

center of the gear, C, if the velocity vA (e.g., 2 ft/s) isknown?

(a) vB = vA+ωgear×rB/A

(b) vA = vC+ωgear×rA/C

(c) vB = vC+ωgear×rC/B

(d) vA = vC+ωgear×rC/A

ANS: (b)

2 If the bar’s velocity at A is 3m/s, what basepoint (first term on the RHS of the velocityequation) would be best used to simplify findingthe bar’s angular velocity when θ = 60◦?

(a) A(b) B

(c) C(d) No difference.

ANS: (a)

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INSTANTANEOUS CENTER OF ZERO VELOCITY

Today’s objectives: Studentswill be able to

1 Locate theinstantaneous center ofzero velocity.

2 Use the instantaneouscenter to determine thevelocity of any point on arigid body in general planemotion.

In-class activities:• Reading Quiz• Applications• Location of the

Instantaneous Center• Velocity Analysis• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 If applicable, the method of instantaneous center can beused to determine the of any point on a rigidbody.(a) velocity(b) acceleration(c) velocity and acceleration(d) force

ANS: (a)

2 The velocity of any point on a rigid body is tothe relative position vector extending from the IC to thepoint.(a) always parallel(b) always perpendicular(c) in the opposite direction(d) in the same direction

ANS: (b)

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APPLICATIONS

• The instantaneous center (IC) ofzero velocity for this bicyclewheel is at the point in contactwith ground. The velocitydirection at any point on the rimis perpendicular to the lineconnecting the point to the IC.

• Which point on the wheel has the maximum velocity?• Does a larger wheel mean the bike will go faster for the

same rider effort in pedaling than a smaller wheel?

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APPLICATIONS(continued)

• As the board slides down the wall (tothe left), it is subjected to general planemotion (both translation and rotation).

• Since the directions of the velocities ofends A and B are known, the IC islocated as shown.

• How can this result help you analyzeother situations?

• What is the direction of the velocity ofthe center of gravity of the board?

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INSTANTANEOUS CENTER OF ZERO VELOCITY(Section 16-6)

• For any body undergoing planar motion, there alwaysexists a point in the plane of motion at which the velocity isinstantaneously zero (if it is rigidly connected to the body).

• This point is called the instantaneous center (IC) of zerovelocity. It may or may not lie on the body!

• If the location of this point can be determined, the velocityanalysis can be simplified because the body appears torotate about this point at that instant.

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LOCATION OF THE INSTANTANEOUS CENTER

• To locate the IC, we can use the fact that the velocity of apoint on a body is always perpendicular to therelative position vector from the IC to the point. Severalpossibilities exist.

• First, consider the case when velocity vA ofa point A on the body and the angularvelocity ω of the body are known.

• In this case, the IC is located along the linedrawn perpendicular to vA at A, a distancerA/IC = vA

ω from A: vIC = 0

• Note that the IC lies up and to the right of Asince vA must cause a clockwise angularvelocity ω about the IC.

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LOCATION OF THE INSTANTANEOUS CENTER(continued)

• A second case is when thelines of action of two non-parallelvelocities, vA and vB, are known.

• First, construct line segments from Aand B perpendicular to vA and vB. Thepoint of intersection of these two linesegments locates the IC of the body.

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LOCATION OF THE INSTANTANEOUS CENTER(continued)

• A third case is when the magnitude and direction oftwo parallel velocities at A and B are known. Here thelocation of the IC is determined by proportional triangles.

• As a special case, note that if the body is translating only(vA = vB), then the IC would be located at infinity. Then ωequals zero, as expected.

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VELOCITY ANALYSIS

• The velocity of any point on a body undergoing generalplane motion can be determined easily once theinstantaneous center of zero velocity of the body is located.

• Since the body seems to rotate about the ICat any instant, as shown in this kinematicdiagram, the magnitude of velocity of anyarbitrary point is v = ω × r, where r is theradial distance from the IC to the point.

• The velocities line of action is perpendicularto its associated radial line.

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EXAMPLE I

• Given: A linkage undergoingmotion as shown. The velocity ofthe block, vD, is 3m/s.

• Find: The angular velocities oflinks AB and BD.

• Plan : Locate the instantaneous center of zero velocity oflink BD and then solve for the angular velocities.

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EXAMPLE I

• Solution: Since D moves to the right, it causes link AB torotate clockwise about point A. The instantaneous centerof velocity for BD is located at the intersection of the linesegments drawn perpendicular to vB and vD. Note that vBis perpendicular to link AB. Therefore we can see that theIC is located along the extension of link AB.

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EXAMPLE I (continued)

• Using these facts,

rB/IC = 0.4 tan 45◦ = 0.4m

rD/IC = 0.4/ cos 45◦ = 0.566m

• Since the magnitude of vD is known,the angular velocity of link BD can befound from vD = (ωBD)

(rD/IC

).

ωBD =vD

rD/IC=

3

0.566= 5.3rad/s

• Link AB is subjected to rotation about A.

ωAB =vBrB/A

= (rB/IC)ωBD/rB/A

= 0.4(5.3)/0.4 = 5.3rad/s32 / 40

EXAMPLE II

• Given: The wheel rolls on its hubwithout slipping on the horizontalsurface with vC = 2 ft/s →

• Find: The velocities of points Aand B at the instant shown.

• Plan : Locate the IC of thewheel. Then calculate thevelocities at A and B.

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EXAMPLE II(continued)

• Solution:

• Note that the wheel rolls withoutslipping. Thus the IC is at the contactpoint with the surface.

• The angular velocity of the wheel canbe found fromω = (vC)/

(rC/IC

)= 2/3 = 0.667rad/s

or ω = −0.667k(rad/s)• The velocity at A and B will bevA = ω × rA/IC = (−0.667)k× (3i+ 3j) = (2i− 2j)in/svB = ω × rB/IC = (−0.667)k× (11j) = 7.34i in/s

vA =√

(22 + 22) = 2.83in/s, vB = 7.34in/s

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CONCEPT QUIZ

1 When the velocities of two points on a body are equal inmagnitude and parallel but in opposite directions, the IC islocated at(a) infinity.(b) one of the two points.(c) the midpoint of the line connecting the two points.(d) None of the above.

ANS: (c)

2 When the direction of velocities of two points on a body areperpendicular to each other, the IC is located at(a) infinity.(b) one of the two points.(c) the midpoint of the line connecting the two points.(d) None of the above.

ANS: (d)

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GROUP PROBLEM SOLVING

• Given: The four bar linkage ismoving with ωCD equal to 6 rad/sCCW.

• Find: The velocity of point E onlink BC and angular velocity oflink AB.

• Plan: This is an example of the second case in the lecturenotes. Since the direction of Point B’s velocity must beperpendicular to AB, and Point C ’s velocity must beperpendicular to CD, the location of the instantaneouscenter, I, for link BC can be found.

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GROUP PROBLEM SOLVING (continued)

0.6m

ωCD = 6 rad/s

yC = 0.6(6) = 3.6m/s

Link CD:

Link BC:

B CE

vB

vE

I

D

C

60◦

vC = 3.6m/s

ωBD

30◦

A

B

vBLink AB:

1.2m

30◦

⇐ From triangle CBI

IC = BC/ tan 60◦ = 0.346m

IB = 0.6 sin 60◦ = 0.693m

vC = (IC)ωBC

ωBC = vC/IC = 3.6/0.346

= 10.39 rad/s !37 / 40

GROUP PROBLEM SOLVING (continued)

• vB = (IB)ωBC = 0.693(10.39) = 7.2m/s

• From link AB, vB is also equal to 1.2ωAB.Therefore 7.2 = 1.2ωAB ⇒ ωAB = 6 rad/s"

• vE = (IE)ωBC where distanceIE =

√0.32 + 0.3462 = 0.458meter

• vE = 0.458(10.39) = 4.76m/swhere θ = tan−1(0.3/0.346) = 40.9◦

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ATTENTION QUIZ

1 The wheel shown has a radius of 15 in and rotatesclockwise at a rate of ω = 3 rad/s. What is vB?

(a) 5 in/s(b) 15 in/s

(c) 0 in/s(d) 45 in/s

ANS: (d)

2 Point A on the rod has a velocity of 8m/s to the right.Where is the IC for the rod?

(a) Point A.(b) Point B.

(c) Point C.(d) Point D.

ANS: (c)

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Note

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