applications relative motion analysis ......cee 271: applied mechanics ii, dynamics – lecture 23:...
TRANSCRIPT
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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