Unit 5 rigid body dynamics

ENGINEERING MECHANICS
Unit – V
Rigid Body Dynamics
by
S.Thanga Kasi Rajan
Assistant Professor
Department of Mechanical Engineering
Kamaraj College of Engineering & Technology,
Virudhunagar – 626001.
Tamil Nadu, India
Email : stkrajan@gmail.com

Kinematics of Rigid Bodies
A 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 rotation
Types 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

Fixed-axis rotation:
- All points move in circular paths about axis of rotation
Curvilinear translation

Rotation about fixed axis
All particles of the body move along circular paths
except those which lie on the axis of rotation

General plane motion
Combination of translation and rotation

- 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

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 A
A2
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 B
A’1
A2
B2
A1 A2
B1 B’1
Translation with A

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
R2R1
F
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

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

Example
Rectilinear translation Rotation about a fixed axis
Curvilinear translation

Translation
ABAB /rrr 
Position
AB
AB
ABAB
const
dtd
vv
.r
/rvv
/
/



Velocity
Acceleration
AB aa 
All points move with same velocity and acceleration
14/12/2014S. ThangaKasiRajan, stkrajan@gmail.com12

Summary
• Time dependent acceleration
dt
ds
v 
)(ts
2
2
dt
sd
dt
dv
a 
dvvdsa 
• Constant acceleration
tavv c 0
2
00
2
1
tatvss c
)(2 0
2
0
2
ssavv c 

Rotation About a Fixed axis
Angular Position ( q )
Defined by the angle q measured between a fixed
reference line and r
Measured in rad
Angular Displacement
Measured as dq
Vector quantity
Measured in radians or revolutions
1 rev = 2 p rad

q
q
 
dt
d
Angular velocity ( )
“the time rate of change in the angular position”
q

 
dt
d
Angular acceleration
“the time rate of change of the angular velocity”
q
q
  2
2
dt
d
 = f(q)
q dd 

Constant Angular Acceleration
)(2 0
2
0
2
qq  c
tc  0
2
00
2
1
tt cqq 

Comparison
)(2
2
1
0
2
0
2
2
00
0
qq
qq




c
c
c
tt
t
dt
ds
v )(ts 2
2
dt
sd
dt
dv
a 
dvvdsa 
tavv c 0
2
00
2
1
tatvss c
)(2 0
2
0
2
ssavv c 
dt
dq
  2
2
dt
d
dt
d q
 
q dd 
)(tq

Motion of Point P
Prxv 
Position :
qrs The arc-length is
Is defined by the position vector r
tv
dt
ds
 ( )qr
dt
d
 r
dt
dq
 r
Velocity
“tangent to the path”

Acceleration
ta
dt
d
 ( )r
dt
d

dt
d
r

 r
r
an
2


r
r 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 22
rt aa  ( ) ( )222
 rr  42
  r
Motion of Point P

2211 rrS qq 
2211 rr  
2211 rra  
r1 r2
s , v, a
r1
r2

Rest
at = 4t m/s2
=?
q=?
ra tP )(
2
/20
)2.0()4(
sradt
t




t
dt
d
20


sradt
dttd
t
/10
20
2
0 0

 



2
10 t
dt
d

q

radt
dttd
t
3
0 0
2
33.3
10

 
q
q
q
Problem 1

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.
14/12/2014 S. ThangaKasiRajan, stkrajan@gmail.com

Problem 2
SOLUTION:
• The tangential velocity and acceleration of D are equal to the
velocity and acceleration of C.
( ) ( )
( )
( )
srad4
75
300
smm300
0
0
00
00



r
v
rv
vv
D
D
CD



( )
( )
( ) 2
srad3
3
225
2/225



r
a
ra
smmaa
tD
tD
CtD



• Apply the relations for uniformly accelerated rotation to
determine velocity and angular position of pulley after 2 s.
( )( ) srad10s2srad3srad4 2
0  t
( )( ) ( )( )
rad14
s2srad3s2srad4 22
2
12
2
1
0

 tt q
( ) revsofnumber
rad2
rev1
rad14 






p
N rev23.2N
( )( )
( )( )rad14mm125
srad10mm125


q

ry
rv
B
B
m75.1
sm25.1


B
B
y
v


Problem 2
• Evaluate the initial tangential and normal acceleration
components of D.
( ) 
2
smm225CtD aa

( ) ( )( ) 222
0 smm1200srad4mm57  DnD ra
( ) ( )  22
smm1200smm225 nDtD aa

Magnitude and direction of the total acceleration,
( ) ( )
22
22
1200225 
 nDtDD aaa
2
smm1220Da
( )
( )
225
1200
tan


tD
nD
a
a

 4.79
14/12/2014 S. ThangaKasiRajan, stkrajan@gmail.com

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

 

Problem 3
SOLUTION:
• The displacement of the gear center in one revolution is
equal to the outer circumference.
For xA > 0 (moves to right),  < 0 (rotates clockwise).
q
p
q
p 1
22
rx
r
x
A
A 
Differentiate to relate the translational and angular
velocities.
m0.150
sm2.1
1
1


r
v
rv
A
A


( )kk

srad8 

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



sm8.0sm2.1
m10.0srad8sm2.1


 
( )ivR

sm2
Velocity of the point D:
( ) ( ) ( )iki
rkvv ADAD


m150.0srad8sm2.1 
 
( ) ( )
sm697.1
sm2.1sm2.1


D
D
v
jiv


Slider Crank Mechanism
Slider Crank Mechanism consists of
1. Crank shaft – Pure Rotation
2. Connecting rod – Both Translation and
Rotation
3. Piston – Pure Rotation
The motion of Connecting rod depends on
motion of crank shaft
Similarly the motion of piston depends on
motion of connecting rod.
Slider Crank MechanismSlider Crank Mechanism

Motion of Crank AB
VB = VA + VB/A
here VA = 0 because A is fixed
therefore
VB = VB/A
= rAB . ωAB

Motion 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

Problem 4
In the reciprocating engine shown in the figure, the crank AB has a constant angular
velocity of 2000 rpm. For the crank position indicated determine
i). Angular velocity of Crank AB
ii). Angular Velocity of the Connecting Rod BC
iii). Velocity of Piston

Problem 4

References
1. 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).
02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 35

Thank you
Any Queries contact
S.Thanga Kasi Rajan
Assistant Professor
Department of Mechanical Engineering
Kamaraj College of Engineering & Technology,
Virudhunagar – 626001.
Tamil Nadu, India
Email : stkrajan@gmail.com
02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 36

Unit 5 rigid body dynamics

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