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

Date: __________________

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RELATIVE MOTION ANALYSIS: ACCELERATION

Today’s objectives: Students

will be able to

1 Resolve the acceleration of

a point on a body into

components of translation

and rotation.

2 Determine the acceleration

of a point on a body by

using a relative

acceleration analysis.

In-class activities:

• Reading Quiz

• Applications

• Translation and Rotation

Components of

Acceleration

• Relative Acceleration

Analysis

• 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 planemotion,(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 plane

motion. Since point A moves along a

curved path, it has two components of

acceleration while point B, sliding in a

straight track, has only one.

• The components of acceleration of

these points can be inferred since their

motions are known.

• How can we determine the

accelerations of the links in the

mechanism?

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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 relative

acceleration 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 with

ωAB = 3 rad/s, αAB = 2 rad/s2 at this

instant.

• Find: The velocity and acceleration of

the slider block C.

• Plan: Follow the solution procedure!

• Note that Point B is rotating. So what

components of acceleration will it be

experiencing?

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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/s

aB = (−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 between

points B and C.

aC = aB +αBC × rC/B − ω2

BCrC/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 component

equations

−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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