CEE 271: Applied Mechanics II, Dynamics– Lecture 23: Ch.16, Sec.7 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Tuesday, Nov. 8, 2012

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RELATIVE MOTION ANALYSIS: ACCELERATION

Today’s objectives: Studentswill be able to

1 Resolve the acceleration ofa point on a body intocomponents of translationand rotation.

2 Determine the accelerationof a point on a body byusing a relativeacceleration analysis.

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

Components ofAcceleration

• Relative AccelerationAnalysis

• Roll-Without-Slip Motion• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ1 If two bodies contact one another without slipping, and the

points in contact move along different paths, the tangentialcomponents of acceleration will be and thenormal components of acceleration will be .(a) the same, the same(b) the same, different(c) different, the same(d) different, different

ANS: (b)2 When considering a point on a rigid body in general plane

motion,(a) It’s total acceleration consists of both absolute acceleration

and relative acceleration components.(b) It’s total acceleration consists of only absolute acceleration

components.(c) It’s relative acceleration component is always normal to the

path.(d) None of the above.

ANS: (a) 3 / 26

APPLICATIONS

• In the mechanism for a window, link ACrotates about a fixed axis through C,and AB undergoes general planemotion. Since point A moves along acurved path, it has two components ofacceleration while point B, sliding in astraight track, has only one.

• The components of acceleration ofthese points can be inferred since theirmotions are known.

• How can we determine theaccelerations of the links in themechanism?

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APPLICATIONS(continued)

• In an automotive engine, the forcesdelivered to the crankshaft and theangular acceleration of the crankshaft,depend on the speed and accelerationof the piston.

• How can we relate the accelerations ofthe piston, connection rod, andcrankshaft to each other?

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RELATIVE MOTION ANALYSIS: ACCELERATION(Section 16-7)

• The equation relating the accelerations of two points on thebody is determined by differentiating the velocity equationwith respect to time.

dvBdt

=dvAdt

+dvB/A

dt

• dvBdt and dvA

dt are absolute accelerations of points A and B,which are measured from a set of fixed x− y axes.

• dvB/A

dt is the acceleration of B with respect to A and includesboth tangential and normal components.

• The result is

aB = aA + (aB/A)t + (aB/A)n

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RELATIVE MOTION ANALYSIS: ACCELERATION(continued)

• Graphically: aB = aA + (aB/A)t + (aB/A)n

• The relative tangential acceleration component (aB/A)t is(α× rB/A) and perpendicular to rB/A.

• The relative normal acceleration component (aB/A)n is(−ω2rB/A) and the direction is always from B towards A.

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RELATIVE MOTION ANALYSIS: ACCELERATION(continued)

• Since the relative acceleration components can be as(aB/A)t = (α× rB/A) and (aB/A)n = (−ω2rB/A),the relative acceleration equation becomes

aB = aA +α× rB/A − ω2rB/A

• Note that the last term in the relative acceleration equationis not a cross product. It is the product of a scalar (square ofthe magnitude of angular velocity, ω2) and the relativeposition vector, rB/A.

• Another expression can be

aB/A = α× rB/A − ω2rB/A

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APPLICATION OF THE RELATIVE ACCELERATIONEQUATION

• In applying the relative acceleration equation, the twopoints used in the analysis (A and B) should generally beselected as points which have a known motion, such aspin connections with other bodies.

• In this mechanism, point B is known to travel along acircular path, so aB can be expressed in terms of itsnormal and tangential components. Note that point B onlink BC will have the same acceleration as point B on AB.

• Point C, connecting link BC and the piston, moves along astraight-line path. Hence, aC is directed horizontally.

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PROCEDURE FOR ANALYSIS

1 Establish a fixed coordinate system.2 Draw the kinematic diagram of the body.3 Indicate on its aA, aB, ω, α, and rB/A. If the points A andB move along curved paths, then their accelerationsshould be indicated in terms of their tangential and normalcomponents, i.e.,aA = (aA)t + (aA)n and aB = (aB)t + (aB)n.

4 Apply the relative acceleration equation:aB = aA +α× rB/A − ω2rB/A

5 If the solution yields a negative answer for an unknownmagnitude, this indicates that the sense of direction of thevector is opposite to that shown on the diagram.

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EXAMPLE I

• Given: Point A on rod AB has anacceleration of 5m/s2 and a velocity of6m/s at the instant shown.

• Find: The angular acceleration of therod and the acceleration at B at thisinstant.

• Plan Follow the problem solvingprocedure!

• Solution: First, we need to find the angular velocity of therod at this instant. Locating the instant center (IC) for rodAB, we can determine ω:

ω =vA

rA/IC=vA3

= 2 rad/s

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EXAMPLE I(continued)

• Since points A and B both move alongstraight-line paths,aA = −5j m/s2aB = aBi m/s

2

• Applying the relative acceleration equation

aB = aA +α× rB/A − ω2rB/A

aBi = −5j+ αk× (3i− 4j)− 22(3i− 4j)

= −5j+ 4αi+ 3αj− (12i− 16j)12 / 26

EXAMPLE I(continued)

• So with aBi = −5j+ 4αi+ 3αj− (12i− 16j) , we can solve.

By comparing the i, j components;

aB = 4α− 12

0 = 11 + 3α

• Solving:

aB = −26.7m/s2(←)

α = −3.67 rad/s2(!)

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BODIES IN CONTACT

1 Consider two bodies in contact with one anotherwithout slipping, where the points in contact move alongdifferent paths.

2 In this case, the tangential components of acceleration willbe the same, i. e.,

(aA)t = (aA′)t (which implies αBrB = αCrC ) .3 The normal components of acceleration will not be the

same.(aA)n 6= (aA′)n so aA 6= aA′

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ROLLING MOTION

1 Another common type of problem encountered in dynamicsinvolves rolling motion without slip; e.g., a ball, cylinder, ordisk rolling without slipping. This situation can be analyzedusing relative velocity and acceleration equations.

2 As the cylinder rolls, point G (center) moves along astraight line. If ω and α are known, the relative velocity andacceleration equations can be applied to A, at the instantA is in contact with the ground. The point A is theinstantaneous center of zero velocity, however it is not apoint of zero acceleration.

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ROLLING MOTION(continued)1 Velocity: Since no slip occurs, vA = 0 when A is in contact

with ground. From the kinematic diagram:

vG = vA + ω × rG/A (1)vGi = 0 + (−ωk)× (rj) (2)vG = ωri (3)

2 Acceleration: Since G moves along a straight-line path, aGis horizontal. Just before A touches ground, its velocity isdirected downward, and just after contact, its velocity isdirected upward. Thus, point A accelerates upward as itleaves the ground.

aG = aA +α× rG/A − ω2rG/A

aGi = aAj+ (−αk)× (rj)− ω2(rj)

3 Evaluating and equating i and j components:aG = αri and aA = ω2rj

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EXAMPLE II

• Given: The gear rolls on the fixed rack.• Find: The accelerations of point A at

this instant.• Plan : Follow the solution procedure!

• Solution: Since the gear rolls on the fixed rack without slip,aO is directed to the right with a magnitude of:

aO = αr = (6 rad/s2)(0.3m) = 1.8m/s2

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EXAMPLE II (continued)

• So now with aO = 1.8m/s2, we can apply the relativeacceleration equation between points O and A.

aA = aO +α× rA/O − ω2rA/O

aA = 1.8i+ (−6k)× (0.3j)− 122(0.3j)

= (3.6i− 43.2j)m/s2

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CONCEPT QUIZ1 If a ball rolls without slipping, select the tangential and

normal components of the relative acceleration of point Awith respect to G.

(a) +αri+ ω2rj

(b) −αri+ ω2rj

(c) +ω2ri− αrj

(d) Zero.

ANS: (b)

2 What are the tangential and normal components of therelative acceleration of point B with respect to G.(a) −ω2ri− αrj(b) −αri+ ω2rj(c) +ω2ri− αrj(d) Zero.

ANS: (a)19 / 26

GROUP PROBLEM SOLVING

• Given: The member AB is rotating withωAB = 3 rad/s, αAB = 2 rad/s2 at thisinstant.

• Find: The velocity and acceleration ofthe slider block C.

• Plan: Follow the solution procedure!• Note that Point B is rotating. So what

components of acceleration will it beexperiencing?

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GROUP PROBLEM SOLVING (solution)

Since Point B is rotating, its velocity and acceleration will be:

vB = (ωAB)rB/A = (3)7 = 21 in/s

aBn = (ωAB)2rB/A = (3)27 = 63 in/s2

aBt = (αAB)rB/A = (2)7 = 14 in/s2

vB = (−21i) in/saB = (−14i− 63j) in/s2

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GROUP PROBLEM SOLVING (continued)

• Now apply the relative velocity equation between points Band C to find the angular velocity of link BC.

vC = vB + ωBC × rC/B

(−0.8vC i− 0.6vCj) = (−21i) + ωBCk× (−5i− 12j)

= (−21 + 12ωBC)i− 5ωBCj

• By comparing the i, j components;

−0.8vC = −21 + 12ωBC

−0.6vC = −5ωBC

• Solving gives

ωBC = 1.125 rad/s

vC = 9.375 in/s

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GROUP PROBLEM SOLVING (continued)

• Now, apply the relative acceleration equation betweenpoints B and C.

aC = aB +αBC × rC/B − ω2BCrC/B

(−0.8aC i− 0.6aCj) = (−14i− 63j) + αBCk× (−5i− 12j)

−(1.125)2(−5i− 12j)

−0.8aC i− 0.6aCj = (−14 + 12αBC + 6.328)i

+(−63− 5αBC + 15.19)j

• By comparing the i, j components;

−0.8aC = −7.672 + 12αBC

−0.6aC = −47.81− 5αBC

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GROUP PROBLEM SOLVING (continued)

• Solving these two i, j componentequations

−0.8aC = −7.672 + 12αBC

−0.6aC = −47.81− 5αBC

yields

αBC = −3.0 rad/s2

aC = 54.7 in/s2

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ATTENTION QUIZ

1. Two bodies contact one another without slipping. If thetangential component of the acceleration of point A ongear B is 100 ft/s2, determine the tangential component ofthe acceleration of point A′ on gear C.

(a) 50 ft/s2

(b) 100 ft/s2

(c) 200 ft/s2

(d) None of above.

ANS: (b)

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ATTENTION QUIZ

2. If the tangential component of the acceleration of point Aon gear B is 100 ft/sec2, determine the angularacceleration of gear B.(a) 50 rad/sec

2

(b) 100 rad/sec2

(c) 200 rad/sec2

(d) None of above.ANS: (a)

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