208322 Mechanical Vibrations
Lesson 4-1
6 Kinetics of Rigid Body: Force and Acceleration......................................11
6.1 Mass Moment of Inertia.....................................................................11
6.2 Planar Kinetic Equations of Motion ..................................................11
6.3 Friction Problems ..............................................................................12
7 Kinetics of Rigid Body: Work and Energy ..............................................12
7.1 Work of Various Forces ....................................................................12
7.2 Principle of Work and Energy ...........................................................12
7.3 Power .................................................................................................12
8 Kinetics of Rigid Body: Impulse and Momentum....................................12
8.1 Momentum.........................................................................................12
8.2 Principle of Impulse and Momentum ................................................13
8.3 Conservation of Momentum ..............................................................13
8.4 Eccentric Impact ................................................................................13
9 Miscellany.................................................................................................13
9.1 Gravitational Constant .......................................................................13
9.2 Systems of Units ................................................................................13
9.3 Vector Analysis .................................................................................13
References....................................................................................................14
Dynamics Review [1]
1 Kinematics of Particle.................................................................................2
1.1 Rectilinear Motion: Continuous Motion..............................................2
1.2 Rectilinear Motion: Erratic Motion .....................................................2
1.3 Curvilinear Motion: Rectangular Components ....................................2
1.4 Curvilinear Motion: Normal and Tangential Components ..................3
1.5 Curvilinear Motion: Polar Components...............................................3
1.6 Absolute-Motion Analysis of Dependent Motion................................3
1.7 Relative-Motion Analysis Using Translating Axes .............................4
2 Kinetics of Particle: Force and Acceleration ..............................................4
2.1 Newton’s Laws of Motion ...................................................................4
2.2 Equation of Motion for a System of Particles......................................4
2.3 Equation of Motion: Rectangular Coordinates ....................................5
2.4 Equation of Motion: Normal and Tangential Coordinates ..................5
2.5 Equation of Motion: Cylindrical Coordinates......................................5
3 Kinetics of Particle: Work and Energy .......................................................5
3.1 The Work of a Force ............................................................................5
3.2 Principle of Work and Energy .............................................................6
3.3 Principle of Work and Energy for a System of Particles .....................6
3.4 Power and Efficiency...........................................................................6
4 Kinetics of Particle: Impulse and Momentum ............................................6
4.1 Principle of Linear Impulse and Momentum .......................................7
4.2 Principle of Linear Impulse and Momentum
for a System of Particles ............................................................................7
4.3 Conservation of Linear Momentum for a System of Particles ............7
4.4 Impact ..................................................................................................7
4.5 Angular Momentum.............................................................................7
4.6 Angular Impulse and Momentum Principles .......................................8
5 Kinematics of Rigid Body ..........................................................................8
5.1 Absolute-Motion Analysis ...................................................................8
5.2 Relative-Motion Analysis for General Plane Motion:
Translating Axes ........................................................................................9
5.3 Relative-Motion Analysis for General Plane Motion:
Rotating Axes...........................................................................................10
1
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
For Constant Acceleration
1 Kinematics of Particle
We treat an object as a particle when we only consider the motion
at its mass center and when any rotation of the body is neglected.
v2 = v1 + at ,
1
s2 = s1 + v1t + at 2 ,
2
2
2
v2 = v1 + 2a ( s2 − s1 ) .
1.1 Rectilinear Motion: Continuous Motion
This section discusses the case when position, velocity, and
acceleration are described using mathematical functions.
Only two of the above three equations are independent of one
another.
Position vector, s , is used to specify the location of a particle at any given
time.
Displacement vector, s2 − s1 , is defined as the change in position vector.
1.2 Rectilinear Motion: Erratic Motion
Average velocity vector, ( s2 − s1 ) / t , is the displacement vector divided by
When position, velocity, and acceleration are described using
graphs.
time.
Instantaneous velocity vector is v = ds / dt.
Slope of s - t graph equals velocity ( ds / dt = v. )
Slope of v - t graph equals acceleration ( dv / dt = a. )
Distance, a scalar quantity, equals the total traveling distance of a particle.
Average speed, a scalar quantity, equals distance divided by time.
Instantaneous speed, a scalar quantity, is magnitude of the instantaneous
velocity, v = ds / dt.
∫
Area under v - t graph equals change in displacement ( v dt = s2 − s1. )
∫
Area under a - t graph equals change in velocity ( a dt = v2 − v1. )
1 2 2
( v2 − v1 ) . )
2
Signs
Area under a - s graph ( a ds =
First, set an origin and set positive direction of the position vector.
Velocity and acceleration’s positive directions follow that of the position
vector.
Velocity multiplies slope of v - s graph equals acceleration ( a = v
∫
For Non-constant Acceleration
dv
.)
ds
1.3 Curvilinear Motion: Rectangular Components
r = xi + yj + zk ,
+ yj
+ zk
,
v = xi
a = xi + yj + zj .
ds
,
dt
dv
a= ,
dt
v dv = a ds.
v=
2
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
1.5 Curvilinear Motion: Polar Components
Motion of a Projectile
Set coordinates for correct signs, x - axis for horizontal and y - axis
for vertical.
uθ
ur
ax = 0,
a y = − g.
P
r
Then apply equations for constant acceleration in Section 1.1.
1.4 Curvilinear Motion: Normal and Tangential
Components
θ
un
v = vr ur + vθ uθ = r ur + rθ uθ ,
a = ar ur + aθ uθ = r − rθ 2 ur + rθ + 2rθ uθ .
(
When r = f
P
(θ )
)
(
)
is given, use the chain rule to find velocity and
acceleration components.
When in 3 dimensions (cylindrical coordinates), add z component as
in Section 1.3.
Sometimes, 2 or more types of coordinates can be used to help
solve a problem.
ut
v = v ut = s ut ,
2
v a = at ut + anun = v ut + un ,
1.6 Absolute-Motion Analysis of Dependent Motion
ρ
When the motion of one particle depends on the corresponding
motion of another particle, set position coordinates then relates each
coordinates geometrically, that is,
at ds = v dv,
1 + ( dy / dx ) 2 

ρ=
2
2
( d y / dx )
3/ 2
.
sa + sb = c,
∆sa + ∆sb = 0,
When in 3 dimensions, include ub = ut × un as the third-component
va + vb = 0,
unit vector.
aa + ab = 0.
3
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
1.7 Relative-Motion Analysis Using Translating Axes
F=
When an observer A is on a translating axes.
Gm1m2
,
r2
G = 66.73 × 10−12
y
B
again, obtained from experiments. Weight is found from the law above,
therefore, weight is not absolute.
Inertial frame of reference is a coordinate system that does not
rotate and is either fixed or translates with a constant velocity. Whenever
the Newton’s law is applied, the acceleration must be made from the
inertial frame of reference. Even if the earth is rotating, motions on or near
the surface of the earth can approximately use the earth as the inertial
frame of reference.
Y
x
A
rB
vB
aB
m3
,
kg ⋅ s 2
X
= rA + rB / A ,
= v A + vB / A ,
= a A + aB / A .
2.2 Equation of Motion for a System of Particles
Center of mass of a system of particles is defined by
mrG = ∑ mi ri ,
2 Kinetics of Particle: Force and Acceleration
therefore,
This method is suitable for solving kinetic problems that involve
force and acceleration.
maG = ∑ mi ai .
2.1 Newton’s Laws of Motion
For one particle in the system, we have
From experiments, we have Newton’s second law of motion
Fi + fi = mi ai ,
∑ F = ma.
where Fi is an external force and f i is an internal force of the system.
Mass is found from the law above, therefore, mass is an absolute quantity,
which means it is constant disregarding location.
External attraction force between two bodies is governed by
Newton’s law of gravitational attraction,
For a system of particles, we have
F
∑ i = maG .
Only external forces of the system are taken into account since all the
internal forces are cancelled out.
4
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
r = f (θ )
2.3 Equation of Motion: Rectangular Coordinates
Free-Body Diagram (FBD)
Draw all external forces (
Tangent
∑F )
in the left-hand-side diagram.
i
ϕ
Draw all inertia forces ( maG ) in the right-hand-side diagram.
∑F
∑F
∑F
x
= max ,
y
= ma y ,
z
= maz .
r
θ
Signs of quantities in the FBD
tan ϕ =
Any positive values are in the same direction as in the FBD. Any
negative values are in the opposite direction as in the FBD (including a. )
Friction force always acts on the FBD such that it opposes the
motion of the particle relative to the surface it contacts.
3 Kinetics of Particle: Work and Energy
This method is suitable for solving kinetic problems that involve
velocity, force, and displacement.
2.4 Equation of Motion: Normal and Tangential
Coordinates
3.1 The Work of a Force
∑ F = ma ,
∑ F = ma ,
∑ F = 0.
t
t
n
n
F
ds
θ
dr
r1
r2
b
2.5 Equation of Motion: Cylindrical Coordinates
dU = F ⋅ dr ,
dU = F ds cos θ .
∑ Fr = mar ,
∑ Fθ = maθ ,
∑ F = ma .
z
r
.
dr / dθ
Signs of Work
z
depends on angle between F and dr ,
5
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
if 0 < θ < 90, work U is positive,
if 90 < θ < 180, work U is negative,
if θ = 90, work U is zero.
3.2 Principle of Work and Energy
dU = dT ,
∑U
Work of a Variable Force
s2
U1−2 = ∫ F cosθ ds.
1− 2
= T2 − T1 =
1 2 1 2
mv2 − mv1 .
2
2
The principle of work and energy cannot be used to determine
normal force. Instead, it can be used to find v and then apply
s1
∑F
n
Work of a Constant Force Moving along a Straight Line
U1−2 = FC cos θ ( s2 − s1 ) .
= man = mv 2 / ρ .
3.3 Principle of Work and Energy for a System of
Particles
Work of a Weight
∑U
U1−2 = ± mg ∆y.
1− 2
= ∑ T2 − ∑ T1.
Signs depend on whether moving up (negative work) or moving
down (positive work.)
Work and kinetic energy are of all internal and external forces
acting on the particles of the system because each particle does not travel
the same path, therefore internal forces are not cancelled out.
Work of a Spring Force
3.4 Power and Efficiency
U1−2
Power is given by
1
1

= ±  ks22 − ks12  ,
2
2

P=
when s1 and s2 are distance from its unstretched position.
dU dT =
= F ⋅ v.
dt
dt
Efficiency is given by
For signs, if the direction of spring force and the direction of the
displacement vector are the same, work is positive. If opposite, work is
negative.
ε=
output power output energy
=
.
input power
input energy
4 Kinetics of Particle: Impulse and Momentum
This method is suitable for solving kinetic problems that involve
velocity, force, and time.
6
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
For the y direction (plane of contact),
4.1 Principle of Linear Impulse and Momentum
∑∫
t2
t1
mAv Ay 2 − mAv Ay1 = 0,
F dt = mv2 − mv1.
mB vBy 2 − mB vBy1 = 0.
4.2 Principle of Linear Impulse and Momentum for a
System of Particles
∑∫
t2
t1
Energy loss during collision equals
∑U
F dt = ∑ mi vi 2 − ∑ mi vi1 = mvG 2 − mvG1.
i i2
i i1
= ∑ T2 − ∑ T1.
y
v A2
4.3 Conservation of Linear Momentum for a System
of Particles
∑m v − ∑m v
1− 2
vB 2
Line of impact
A
x
B
= 0.
v A1
Internal impulses are always cancelled out. If the time period over
which the motion is studied is very short, we can neglect some external
forces (nonimpulsive forces.)
v B1
Plane of contact
4.5 Angular Momentum
4.4 Impact
z
Central Impact
From conservation of linear momentum,
mAv A1 + mB vB1 = mAv A 2 + mB vB 2 .
y
Coefficient of restitution is
∫R
e=
∫R
1
2
r
v −v
= B 2 A 2 , ( 0 ≤ e ≤ 1) .
dt v A1 − vB1
dt
x
d
mv
Oblique Impact
For the x direction (line of impact), use 2 formulas above.
7
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
( H o ) z = ( d )( mv ) ,
5.1 Absolute-Motion Analysis
H o = r × mv ,
∑ M o = H o ,
F
∑ = L.
Translation
All points in the rigid body subjected to either curvilinear or
rectilinear translations move with the same velocity and acceleration.
v A = vB ,
a A = aB .
For system of particles, only external forces and moments are
considered since internal forces and moments are cancelled out.
Rotation about a Fixed Axis
4.6 Angular Impulse and Momentum Principles
All points in the rigid body subjected to rotation about a fixed axis
move with the same angular velocity and angular acceleration.
For non-constant acceleration,
t2 M
dt
=
H
−
H
(
)
(
∑∫ o
o 2
o )1 .
t1
dθ
,
dt
dω
α=
,
dt
α dθ = ω d ω.
ω=
Conservation of Angular Momentum
For a particle,
(H ) = (H )
o 1
o 2
.
For system of particles,
For constant acceleration,
∑ ( H o )1 = ∑ ( H o )2 .
ω2 = ω1 + α t ,
1
2
In some cases, particle’s angular momentum is conserved but linear
momentum is not, for example, particle subjected to central forces.
θ 2 = θ1 + ω1t + α t 2 ,
ω22 = ω12 + 2α (θ 2 − θ1 ) .
5 Kinematics of Rigid Body
There are 3 types of rigid-body motions: translation (rectilinear or
curvilinear), rotation about a fixed axis, and general plane motion.
Signs are similar to those in Section 1.1 but with clockwise and counterclockwise directions.
Motion of a point P is given by
v = ω × r,
a = at + an = α × r + ω × (ω × r ) .
8
Copyright  2007 by Withit Chatlatanagulchai
•
•
•
General Plane Motion
The general plane motion is a combination of the translation and
rotation motions and can be described by
208322 Mechanical Vibrations
Lesson 4-1
mechanism linkages with pivoted joints
fixed slides
rolling without slip
s = f (θ ) .
s is measured from a fixed origin and is directed along the
straight-line path. θ is the angular position of a line lying in the body.
Differentiating by chain rule gives velocity and acceleration.
5.2 Relative-Motion Analysis for General Plane
Motion: Translating Axes
When an observer A is on translating axes and B is another point
on the rigid body. The rigid body rotates with an angular velocity ω. The
general plane motion can be broken down to translation of point A and
rotation of the rigid body.
y
Points, which are coincident at a pin, move with the same
acceleration.
Points, which are in contact without slipping and move along
different paths, have the same velocity and tangential component of
acceleration but have different normal components of acceleration.
ω
B
Y
ω1
x
A
ω2
an1
X
rB = rA + rB / A ,
vB = v A + vB / A = v A + ω × rB / A ,
aB = a A + ( aB / A )t + ( aB / A )n = a A + α × rB / A + ω × (ω × rB / A ) .
an 2
at v
This analysis is suitable for
9
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
Rotation without slip can be broken down to pure translation and
pure rotation (Chasle’s theorem.)
5.3 Relative-Motion Analysis for General Plane
Motion: Rotating Axes
When an observer A is on axes that are both rotating and
translating.
α
ω
=
+
y
B
Y
Ω, Ω
For rotation without slip, velocity at center is vG = rω i and acceleration at
center is aG = α ri .
x
A
Instantaneous Center of Zero Velocity
A
A
IC
vA
B
X
vA
rB = rA + rB / A ,
vB = v A + Ω × rB / A + ( vB / A ) xyz ,
aB = a A + Ω
× rB / A + Ω × ( Ω × rB / A ) + 2Ω × ( vB / A ) xyz + ( aB / A ) xyz .
IC
vB
B
IC
vB
A
B
vA
vB
10
Copyright  2007 by Withit Chatlatanagulchai
•
•
•
F1
This analysis is suitable for
rigid body with free sliding at its connection
motions of 2 points that are on different rigid bodies
kinematics of particle when the particle is moving along a rotating
path
F2
M1
y
208322 Mechanical Vibrations
Lesson 4-1
maG
y
G
I Gα
G
mg
6 Kinetics of Rigid Body: Force and Acceleration
This method is suitable for solving kinetic problems that involve
force and acceleration.
x
P
6.1 Mass Moment of Inertia
Free-body diagram
I = ∫ r 2 dm = ∫ r 2 ρ dV .
Kinetic diagram
F
∑ = maG ,
M
=
M
(
∑ P ∑ k )P .
V
Parallel-axis theorem states that
I = I G + md 2 .
To prevent sign errors, when writing the FBD and the kinetic
diagram, acceleration vectors α and aG must be in the same sense of
Radius of gyration is defined as k , when
direction and friction forces must oppose the relative movement of the
contact surfaces.
I Gα and couple moment M 1 are free vectors and can therefore
I = mk .
2
For composite bodies, finding the mass moment of inertia is done
by adding algebraically the moments of inertia of all composite shapes
about an axis.
Mass center is given by
y=
x
F3
Mass moment of inertia about an axis is given by
m
P
act at any point.
For rotation about a fixed axis at point O,
M
∑ o = I oα = IGα + maG d = ( I G + md 2 )α .
)
∑ ( ym
.
∑m
6.2 Planar Kinetic Equations of Motion
For a general plane motion, we can write an FBD and a kinetic
diagram of a rigid body as follows.
11
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
6.3 Friction Problems
The work is positive when M and the angular displacement have the
same direction and negative if not.
Work of a constant force, weight, and spring are the same as those
in Section 3.
Forces that do no work are as follows
• Forces that act at the instantaneous center, IC, for example, F and
N.
• Forces that act at fixed points, for example, at pins.
• Forces with direction perpendicular to their displacement.
F
N
7.2 Principle of Work and Energy
Rolling without Slip
When F ≤ µ S N , then aG can be related to α by the kinematics
equation aG = α r.
∑U
1− 2
= T2 − T1 ,
T=
1 2 1
mvG + I Gω 2 .
2
2
Rolling with Slip
When F > µ S N , then aG cannot be related to α and the
relationship F = µ K N can be used instead.
Unlike the case of system of particles, the work of the body’s
internal forces does not have to be considered since the body is rigid and
they cancelled out.
7 Kinetics of Rigid Body: Work and Energy
7.3 Power
This method is suitable for solving kinetic problems that involve
velocity, force, and displacement.
P=
7.1 Work of Various Forces
dU dT =
= F ⋅ v.
dt
dt
8 Kinetics of Rigid Body: Impulse and Momentum
Work of a variable force is given by
This method is suitable for solving kinetic problems that involve
velocity, force, and time.
U F = ∫ F cos θ ds.
S
8.1 Momentum
Work of a couple moment is given by
Linear momentum is given by
LG = mvG .
θ2
U M = ∫ M dθ .
θ1
Angular momentum is given by
12
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
H G = I Gω .
when v A and vB are velocities in the direction of the line of impact.
9 Miscellany
Angular momentum at any point P is given by
H P = I Gω + ( d ) ( mvG ) .
9.1 Gravitational Constant
In SI unit, g = 9.807 m / s .
2
8.2 Principle of Impulse and Momentum
∑∫
t2
t1
∑∫
t2
t1
F dt = ( LG )2 − ( LG )1 or
M P dt = ( H P )2 − ( H P )1 or
In US customary unit, g = 32.17 ft / s .
2
∑ F = L.
9.2 Systems of Units
System
length
M
=
H
.
∑
time
mass
8.3 Conservation of Momentum
SI
Metric
m
s
kg
m
s
metric
slug
 kg ⋅ s 


 m 
kg
For system of rigid bodies,
British
ft
s
slug
poundal
 lb ⋅ s 2 


 ft 
= 32.2 lb
lb = 16 oz
2
0 = ∑ ( LG )2 − ∑ ( LG )1 ,
0 = ∑ ( H P ) 2 − ∑ ( H P )1 .
force
N
 kg ⋅ m 
 2 
 s 
US
ft
s
lb
 lb ⋅ ft 
 2 
 s 
Internal impulses are always cancelled out. If the time period over which
the motion is studied is very short, we can neglect some external forces
(non-impulsive forces.)
9.3 Vector Analysis
8.4 Eccentric Impact
Vector is constant if and only if both its magnitude and direction are
constant.
The eccentric impact occurs when the line connecting the mass
centers of the two bodies does not coincide with the line of impact.
Use the conservation of momentum above together with the
coefficient of restitution
e=
dA
=
dt
d ( A× B)
=
dt
d ( A⋅ B)
=
dt
( vB )2 − ( vA )2
, ( 0 ≤ e ≤ 1) ,
( vA )1 − ( vB )1
13
(
d A uA
) = d A u
dt
dt
dA dB
× B + A×
,
dt
dt
dA dB
⋅ B + A⋅ .
dt
dt
duA
,
A + A
dt
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-1
References
[1]
Engineering Mechanics: Dynamics, by Russell C. Hibbeler, Prentice
Hall, 2006
14
Copyright  2007 by Withit Chatlatanagulchai