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

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Wednesday, October 19, 2011

1 / 21

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

2 / 21

READING QUIZ

1 When a relative-motion analysis involving two sets ofcoordinate axes is used, the x′ − y′ coordinate system 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)

3 / 21

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?

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

4 / 21

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?

5 / 21

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, drAis displacement due to translation, anddrB/A is displacement due to rotation.

6 / 21

RELATIVE MOTION ANALYSIS: VELOCITY

• The velocity at B is given as:

(drB/dt) = (drA/dt) + (drB/A/dt)

vB = vA + vB/A

• Since the body is taken as rotating about A,

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 / 21

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.

8 / 21

RELATIVE MOTION ANALYSIS: VELOCITY (cont’d)

vB = vA + ω × rB/A

• When a wheel rollswithout slipping, point Ais oftenselected to be at thepoint of contact with the ground.

• Since there is no slipping, point Ahas zero velocity.

• Furthermore, point B at thecenter of the wheel moves alonga horizontal path. Thus, vB has aknown direction, e.g., parallel tothe surface.

9 / 21

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.

10 / 21

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.

11 / 21

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.

12 / 21

EXAMPLE I(Solution)

Kinematic diagram:

y

Express the velocity vectors in CVN

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

13 / 21

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

14 / 21

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 + ωAB × rB/A

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

vBj = −3.6i+ 0.5196ωABj+ 0.3ωABi

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

15 / 21

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 / 21

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.

17 / 21

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 rotatingat ωR = 3 rad/s, the velocity atpoint A will be ;vA = −3(160)i = −480i mm/s

• Also note that vB = 0 since thegear S is held fixed ωS = 0.

• Applying the relative velocityequation to points A and B;

vB = vA + ωP × rB/A

0 = −480i+ (ωPk)× (−80j)⇒ 0 = −480i+ 80ωP i

ωP = 6 rad/s18 / 21

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) = −240imm/s

• Note that the shaft A has a circularmotion with the radius of 120mm. Theangular velocity of the shaft isωA = vC/r = −240/120 = −2 rad/s

• The shaft A is rotating incounter-clockwise direction!

19 / 21

GROUP PROBLEM SOLVING (Solution)

• Apply the relative velocity equation at point B and C toGear P in order to find the velocity at B.vC = vB + ωP × rC/B = 0 + (6k)× (40j) = −240imm/s

P

y

x

3 rad/s

160mm

A

B

vB = 0vA

R

S

80mm

C

• Note that the shaft A has a circularmotion with the radius of 120mm. Theangular velocity of the shaft isωA = vC/r = −240/120 = −2 rad/s

• The shaft A is rotating incounter-clockwise direction!

20 / 21

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

center of the gear, C, if the velocity vA is known?

(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)21 / 21