unit 5 rigid body dynamics
TRANSCRIPT
ENGINEERING MECHANICSUnit – V
Rigid Body Dynamics
by
S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India
Email : [email protected]
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Kinematics of Rigid BodiesA rigid body has size that is not negligible and does not deform (distance between two points on body is constant). (Idealisation)
Rigid body motion involves translation and/or rotationTypes of Rigid Body Plane Motion
Translation: - No rotation of any line in the body- All points in body have same velocity and acceleration- No relative motion between any two particles
Rectilinear translation
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Translation
Every line segment on the body remains parallel to its original direction during the motion
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Fixed-axis rotation:- All points move in circular paths about axis of rotation
Curvilinear translation
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Rotation about fixed axis
All particles of the body move along circular paths except those which lie on the axis of rotation
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General plane motion
Combination of translation and rotation
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General plane motion- Both translation and rotation occur- Distances between particles are fixed
Note: We will consider plane motion only.
- Relative motion of one particle to another will always be a circular motion
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General Plane Motion is the summation of a Translation and a Rotation
Consider the motion of the rigid bar AB:
General Motion
B1
B2
A1 A2
Rotation about AA2
B’1
B2
We could break this motion down another way:
General Motion
B1
B2
A1 A2
Translation with B
B1
B2
A1
A’1
Rotation about BA’1
A2
B2
A1 A2
B1 B’1
Translation with A
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Rigid Bodies:
Why are Rigid Bodies so different from Particles?
- Size negligible compared to motion
Particles:
mg
N
F
- All forces act through center of gravity
- Neglect rotation about center of gravity
R2R1F
mg
- Points of application, and lines of action of forces are important
- Rotation and Moments about center of gravity are important
Rigid Bodies Vs Particles
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Types of rigid body planar motion
Translation – only linear direction
Rotational about fixed axis – rotational motion
General plane motion – consists of both linear and rotational motion
Rigid-Body Motion
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Example
Rectilinear translation Rotation about a fixed axis
Curvilinear translationGeneral plane motion
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Translation
ABAB /rrr Position
AB
AB
ABAB
constdtd
vv.r
/rvv
/
/
Velocity
Acceleration
AB aa
All points move with same velocity and acceleration14/12/2014
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Summary• Time dependent acceleration
dtdsv
)(ts
2
2
dtsd
dtdva
dvvdsa
• Constant acceleration
tavv c 0
200 2
1 tatvss c
)(2 020
2 ssavv c
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Rotation About a Fixed axis
Angular Position ( q )
Defined by the angle q measured between a fixed reference line and rMeasured in rad
Angular DisplacementMeasured as dqVector quantityMeasured in radians or revolutions1 rev = 2 p rad
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qq dtd
Angular velocity ( )“the time rate of change in the angular position”
q dtd
Angular acceleration“the time rate of change of the angular velocity”
qq 2
2
dtd
= f(q)
q dd 14/12/2014
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Constant Angular Acceleration
)(2 02
02 qq c
tc 0
200 2
1 tt cqq
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Comparison
)(221
02
02
200
0
c
c
c
tt
t
dtdsv )(ts
2
2
dtsd
dtdva
dvvdsa
tavv c 0
200 2
1 tatvss c
)(2 020
2 ssavv c
dtdq 2
2
dtd
dtd q
q dd
)(tq
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Motion of Point P
Prxv
Position :
qrs The arc-length is
Is defined by the position vector r
tvdtds
( )qrdtd
rdtdq
rVelocity
“tangent to the path”
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Acceleration
tadtd
( )rdtd
dtdr
r
ran
2
rr 2)(
r2
Direction of an is always toward O
“rate of change in the velocity’s magnitude”
“rate of change in the velocity’s direction”
a 22rt aa ( ) ( )222 rr 42 r
Motion of Point P
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Restat = 4t m/s2
=?q=?
ra tP )(
2/20
)2.0()4(
sradt
t
tdtd 20
sradt
dttdt
/10
20
2
0 0
210 tdtd
q
radt
dttdt
3
0 0
2
33.3
10
q
Problem 1
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Problem 2
Cable C has a constant acceleration of 225 mm/s2 and an initial velocity of 300 mm/s, both directed to the right.
Determine (a) the number of revolutions of the pulley in 2 s, (b) the velocity and change in position of the load B after 2 s, and (c) the acceleration of the point D on the rim of the inner pulley at t = 0.
SOLUTION:• Due to the action of the cable, the
tangential velocity and acceleration of D are equal to the velocity and acceleration of C. Calculate the initial angular velocity and acceleration.
• Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the pulley after 2 s.
• Evaluate the initial tangential and normal acceleration components of D.
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Problem 2SOLUTION:• The tangential velocity and acceleration of D are equal to the
velocity and acceleration of C. ( ) ( )( )
( )srad4
75300
smm300
00
00
00
rv
rvvv
D
D
CD
( )( )
( ) 2srad33
225
2/225
ra
rasmmaa
tD
tD
CtD
• Apply the relations for uniformly accelerated rotation to determine velocity and angular position of pulley after 2 s.
( )( ) srad10s 2srad3srad4 20 t
( )( ) ( )( )rad 14
s 2srad3s 2srad4 22212
21
0
tt q
( ) revs ofnumber rad 2
rev 1rad 14
p
N rev23.2N
( )( )( )( )rad 14mm 125
srad10mm 125
q
ry
rv
B
B
m 75.1sm25.1
B
B
yv
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Problem 2• Evaluate the initial tangential and normal acceleration
components of D.
( ) 2smm225CtD aa
( ) ( )( ) 2220 smm1200srad4mm 57 DnD ra
( ) ( ) 22 smm1200smm225 nDtD aa
Magnitude and direction of the total acceleration,
( ) ( )22
22
1200225
nDtDD aaa2smm1220Da
( )( )
2251200
tan
tD
nD
aa
4.7914/12/2014
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Problem 3
The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s.
Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.
SOLUTION:• The displacement of the gear center in
one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities.
• The velocity for any point P on the gear may be written as
Evaluate the velocities of points B and D.
APAAPAP rkvvvv
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Problem 3SOLUTION:• The displacement of the gear center in one revolution is
equal to the outer circumference.
For xA > 0 (moves to right), < 0 (rotates clockwise).
qpq
p 122rx
rx
AA
Differentiate to relate the translational and angular velocities.
m0.150sm2.1
1
1
rvrv
A
A
( )kk srad8
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Problem 3• For any point P on the gear, APAAPAP rkvvvv
Velocity of the upper rack is equal to velocity of point B:
( ) ( ) ( )( ) ( )ii
jki
rkvvv ABABR
sm8.0sm2.1m 10.0srad8sm2.1
( )ivR sm2
Velocity of the point D:
( ) ( ) ( )iki
rkvv ADAD
m 150.0srad8sm2.1
( ) ( )sm697.1
sm2.1sm2.1
D
D
vjiv
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Slider Crank Mechanism
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Slider Crank Mechanism consists of 1. Crank shaft – Pure Rotation2. Connecting rod – Both Translation and
Rotation3. Piston – Pure Rotation
The motion of Connecting rod depends on motion of crank shaftSimilarly the motion of piston depends on motion of connecting rod.
Slider Crank MechanismSlider Crank Mechanism
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Slider Crank MechanismMotion of Crank AB
VB = VA + VB/A here VA = 0 because A is fixed
therefore
VB = VB/A = rAB . ωAB
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Slider Crank MechanismMotion of Connecting Rod:
When crank rotates in clockwise direction, connecting rod rotates in anticlockwise direction.Also VC/B is perpendicular to the axis of the connecting rod
Apply sine and cosine rule to find the magnitude and direction the velocity of each component
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Problem 4
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In the reciprocating engine shown in the figure, the crank AB has a constant angular velocity of 2000 rpm. For the crank position indicated determinei). Angular velocity of Crank ABii). Angular Velocity of the Connecting Rod BCiii). Velocity of Piston
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Problem 4
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References1. Ferdinand P Beer & E.Russell Johnston “VECTOR MECHANICS FOR ENGINEERS STATICS & Dynamics”, (Ninth Edition) Tata McGraw Hill Education Private Limited, New Delhi.2. Engineering Mechanics – Statics & Dynamics by S.Nagan, M.S.Palanichamy, Tata McGraw-Hill (2001).
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Thank you
Any Queries contact
S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India
Email : [email protected]
02/01/2017