2 Kinematics of a Rigid Body
description
Transcript of 2 Kinematics of a Rigid Body
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Motion Relative to Point-Attached Reference Frame
Relative Motion Velocity
Relative Motion - Acceleration
Instantaneous Centre of Zero
Velocity
Wheel Rolling Without Slip
Sliding Contact Between Two
Bodies
2 KinematicsofaRigidBody
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x i", "y j", "
Bodyattachedframe
Pointattachedframe
x i' ,
y j' ,
O x i,
y j,
Basicfixedframe
A( x yA A, )
Referencepoint
d.o.f = 3
Independent coordinates: xA yA
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Planar motion of a rigid slab
(1) Translation
(IF = constant)(2) Rotation about A
(IF xA, yA are constant)
Independent coordinates: xA yA
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A-xytranslates
with A
Slab rotatesabout A-xy
Absolute vel=Re + En
A
O
BrBA
x'
y '
x
y
v A
Relative vel.
Entrained vel.
AB vv '
v A
Entraining point B on A-xy:
B
vBvB A/
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AO
BrBA
x'
y'
x
y
v A
AABB vvv /
v A
vB A/vB
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RelativeMotionAnalysis:Velocity
Velocity vB/A hasamagnitudeof vB/A =rB/A anda
directionwhichisperpendiculartorB/AABAB /vvv
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RelativeMotionAnalysis:VelocityVelocity vB isdeterminedbyconsideringtheentire
bodytotranslatewithavelocityofvA,androtateaboutA withanangularvelocity
Vectoradditionofthesetwoeffects,appliedtoB,yieldsvB
vB/A representstheeffectofcircularmotion,aboutA.Itcanbeexpressedbythecrossproduct
ABAB /rvv
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RelativeMotionAnalysis:VelocityVelocity PointAonlinkABmustmovealonga
horizontalpath,whereaspointBmovesonacircularpath
Thewheelrolls withoutslippingwherepointAcanbeselectedattheground
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AO
y
B
aB A/aB
x'
y'
a A
a A
x
AABB aaa /
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RelativeMotionAnalysis:Acceleration
Thetermscanberepresentedgraphically
= +
ABABAB /2
/ rraa
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v rB A BA/ a r rB A BA BA/ ( ) Relative motion: (Circular)
Entrained motion
Av
Aa
,
AABB vvv / AABB aaa /
Absolute = Relative + Entrained
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A wheel rolling without slip
Example
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Example
G0.45 m
0.3 m
i
j
P v aGiven
Determine
Rolling without slipping
C
A
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C
G
C
Av
0.45 m
0.3 m
i
j
Rolling without slipping vC = vC = 0
ACr CCAA vvv /
0 v
A relative to C= 0.75=1.33v
(AC)
(ans)(CW)
For the instant
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ia janA iaG (AG) jAG 2GGAA aaa
/i=
C
A
(AG)t GAa /AGan GA 2/
a 0.75=1.33a
(ans)(CW)
G iaG 45.0
? ??
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?Ca
C
A G iaG 45.0
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Instantaneous center of zero velocity
For a slab in general planar motion,
there exists a point C on the slab
at any instant
with vC = 0.
C : Instantaneous Center of Zero Velocity
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GC
A
i
j
vC = 0
Av
ACr CCAA vvv /
0
Av (AC)
Instant Center of zero vel.
Pure rotationAbout I.C.
Proportional to AC
For the instant
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GC
A
i
j
vC = 0
ACr CCAA vvv /
0
Av
Av (AC)Pure rotationAbout I.C.
Instant Center of vel.
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GC
A
i
j
vC = 0
ACr CCAA vvv /
0
Av
Av (AC)Pure rotationAbout I.C.
Instant Center of vel.
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InstantaneousCenterofZeroVelocity VelocityofanypointB locatedonarigidbodycan
beobtainedifbasepointA haszerovelocity SincevA =0,thereforevB = xrB/A. PointA iscalledtheinstantaneouscenterofzero
velocity (IC)anditliesontheinstantaneousaxisofzerovelocity
Magnitude ofvB isrB/IC Duetocircularmotion,
direction ofvB mustbeperpendicular torB/IC
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InstantaneousCenterofZeroVelocity
Forwheelrollingwithoutslipping,thepointofcontactwiththegroundhaszerovelocity
HencethispointrepresentstheIC forthewheelLocationoftheIC Velocity ofapointonthebodyisalways
perpendicular totherelativepositionvectorextendingfromtheIC tothepoint
ICislocatedalongthelinedrawnperpendiculartovA,distancefromA totheICisrA/IC =vA/
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InstantaneousCenterofZeroVelocityLocationoftheIC ConstructAand Blinesegmentsthatare
perpendiculartovA andvB. Extendingtheseperpendiculartotheirpointof
intersection asshownlocatesthe IC ICisdeterminedbyproportionaltriangles
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f fixed P vP
P
Sliding contact of a point along the boundary of a rigid slab
P fixed
f moves P
vP
f moves f moves
Velocity of P tangential to the curve
Velocity of P tangential to the curve,tooNo penetration
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BC
200 mm
600 mm
G 30
= 0.5 rad/s
700 mm
D on the wall
D on AB
Example
AB?A
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BC
600 mm
G 30
= 0.5 rad/sD on AB
D on the wall
vDA
= ?
Example
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P
f
P fixed
P vP
f
P movesvP
vP
vP/P
Condition of sliding contact
vP/P nvP/P tangential to the curve
nNo penetration
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fP vP/P
P
If f and P are both moving
f f
vP/P tangential to the curve always
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BC
600 mm
G 30
= 0.5 rad/sD on AB
Example 2-5
D on the wall
vDA
I.C.
Sliding contact
Example
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AB
C
200 mm
400 mm
G
D
30
0.5 rad/s AB
vA
vD
E
45
15 D
27.4 27.4
700 mm
I.C. of rod AB
AEvA
AB
)(BEv ABB
Example
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Motion Relative to a Body-Attached
Reference Frame
Relative Motion Velocity
Relative Motion -
Acceleration
2 KinematicsofaRigidBody
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P belongs to the Body fA
O
x
y
x
y
Av
Pv
P Coincideswith P
at the instant
PfP vv /P
PAAP rvv '
f
PAr
P Entraining pointP'Pv
?
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'/ PPv PfP vv / 'Pv
Relative to f Relative to P=
'/ PfPP vvv
Abs. = Rel. + Ent.
Principle of Velocity Combination
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A point / particle relative to a body
Acceleration of
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AO x
y
A body f
f
f
P A particle P
GivenPa
Pa
f
f
Aa
Aa
f
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AO
P
x
y
f
Pa
fAa
APAP aaa /
P relative to A
point A attachedTrans. Frame A-xy
x
y
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AO
f
x
y
Pa
What is accel. of Prelative to f ?
fPa /
x
y
Body attachedFrame A-xy
P
f
Aaf
AP aa
If f is in translation+ rotation
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AO x
y
Pa
x
y
Body attachedFrame A-xy
Pf
Aa
Introduce
P Entraining pointP
ff
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fPP vv / 'Pv
Abs = Rel + Ent
Do we also have a aP P f / 'Pa ?Abs = Rel + Ent
Since
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xy
P
O
erererera PrPPrPP 2 2
fPa /
xUsing Body attached Ref.
(O- x,y)
yPr
Relative accel.
rPer
-
xy
P
O
erP
rP er 2
erererera PrPPrPP 2 2 aP'
tPa '
nPa 'Entrained Accel.
P entraining point
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xy
P
O
erererera PrPPrPP 2 2
CPa
erP 2
Coriolis Accel. fPf v /2
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Acceleration of P
a aP P f / 'PaAbs = Relative + Entrained
fPf v /2
+ Coriolis
Principle of Acceleration Combination