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These notes are a supplement to FE Reference Handbook, 9.5 Version, for Computer- ... See motion equations in the 9.5 Handbook on pp. 78‐79.
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by
Stephen F. Felszeghy
CSULA Prof. Emeritus of ME
These notes are a supplement to FE Reference Handbook , 9.5 Version, for Computer-
Based Testing, NCEES, June 2018, pp. 77-84.
These notes were prepared for the FE/EIT Exam Review Course class meeting held on
Feb. 8, 2020, 9:00 a.m. to 4:00 p.m.
Dynamics
Kinematics – deals with motion alone apart from considerations of
force and mass.
Kinetics – relates unbalanced forces with changes in motion.
Kinematics of Particles
Rectilinear Motion of a Particle
Suppose
v = v x
; apply “Chain Rule”:
dv
dt
= a =
dv
dx
dx
dt
→ a =
dv
dx
v
Determination of Motion of a Particle
Integrate differential relations:
Motion
Kinematics
Kinetics
Unbalanced Forces
Position coordinate
(Rectilinear displacement):
Velocity:
v =
dx
dt
x
Acceleration:
a =
dv
dt
d
2
x
dt
2
= ˙ x ˙
dx = v dt
dv = a dt
v dv = a dx
x
Rectangular Components
Application: See projectile motion in the 9.5 Handbook on p. 79.
Motion Relative to Translating Reference Axes
v (tangent to path)
patpath)
a
r
y
x
Path
Position vector:
€
r = r ( t )
(Vector function)
Velocity:
v =
d r
dt
r
Acceleration:
a =
d v
dt
d
2
r
dt
2
= ˙ r ˙
y
z
x
j
k
i
Position vector:
€
r = x i + y j + z k
Velocity:
€
v = x ˙ i + y ˙ j + ˙ z k
Acceleration:
€
a = ˙ x ˙ i + ˙ y ˙ j + ˙ z ˙ k
We write:
v
x
x , etc.
a
x
x , etc.
y
A
A / B
x
y’
x’
B
“Translating” means x’ – y’ axes
move but remain parallel to
x – y axes.
r
A
= r
B
A / B
r
A
r
B
r
A / B
v
A
= v
B
A / B
a
A
= a
B
A / B
Tangential and Normal Components
v =
d r
dt
ds
dt
e
t
= v e
t
a =
d v
dt
d
2
r
dt
2
d
2
s
dt
2
e
t
v
2
ρ
e
n
dv
dt
e
t
v
2
ρ
e
n
t
t
n
n
t
n
Radial and Transverse Components
a =
v =
r =
r − r
θ
2
( )
e
r
θ + 2
r
θ
( )
e
θ
r
v
θ
= r
θ
a
r
= ˙ r ˙ − r
θ
2
a
θ
= r
θ + 2 r ˙
θ
n
r
y
x
Path
s
t
Center of
curvature
CP = ρ = radius of curvature
e
t
= unit vector tangent to path
e
n
= unit vector normal to path
pointing to C
s = directed distance along path
Polar coordinates of P :
r , θ
r
e
θ
= unit vector perpendicular to r
in direction of increasing θ
r
v =
r =
r e
r
θ e
θ
y
x
θ
r
θ
Path
r
In either system, W = mg , where
W = weight
g = acceleration due to gravity
At surface of earth: (SI) g = 9.807 m/s
2
(USCS) g = 32.174 ft/s
2
AVOID: lbf, lbm
Equations of Motion: Rectangular Components
x
∑
= m a
x
y
∑
= m a
y
Equations of Motion: Tangential and Normal Components
y
y
x
x
y
∑
x
∑
m a
y
x
t
∑
= m a
t
n
∑
= m a
n
y y
x
x
n
∑
t
∑
n
t
Path
Equations of Motion: Radial and Transverse Components
Kinetics of Particles: Energy Methods
Above result is the principle of work and energy. Units: (SI) N⋅m = J; (USCS) ft⋅lb
y
x x
r
r
θ
θ
Path
y
θ
∑
r
∑
θ
r
r
∑
= m a
r
θ
∑
= m a
θ
y
x
Position 2
Position 1
2
r
1
s
Path
The work done by F on the particle
during a finite movement of the
particle along a curved path from
position 1 to position 2 is
1 → 2
1 → 2
= F ⋅ d r
r
1
r
2
∫
(Line integral)
It can be shown:
1 → 2
t
s
1
s
2
∫
ds
mv
2
2
mv
1
2
Let T =
mv
2
= kinetic energy of particle
Then,
1 → 2
2
1
or T
2
1
1 → 2
Position 2
Position 1
2
1
t
n
y
x
s
Summary
The work‐energy equation can now be written as:
1 → 2
g
e
where
1 → 2
is the work done on the particle by forces other than
gravitational and spring forces.
If
1 → 2
above is zero, then:
2
g
( )
2
e
( )
2
1
g
( )
1
e
( )
1
This is the law of conservation of total mechanical energy.
Power and Efficiency
Power is the time rate of doing work by a force on a particle.
Power = F ⋅ v Units:
N ⋅ m s = J s = W; USCS
hp = 550
ft ⋅ lb
s
η =
power output
power input
= mechanical efficiency
Kinetics of Particles: Momentum Methods
Define angular momentum
O
of particle about O :
O
y
x
v
1
2
Path
r
Recall Newton’s 2
nd
law:
F = m a =
d
dt
m v
where
F = resultant force
Then,
O
= r × m a = r × F = M
O
or
O
O
where M
O
= sum of the moments about O of
all forces acting on particle
Equations of Impulse and Momentum
What is the cumulative effect of integrating
F and
O
with respect to time
over an interval from
1
to
2
F dt
t
1
t 2
∫
= d m v
= m v
2
m v
1
m v 2
∫
− m v
1
or
m v
1
t
1
t
2
∫
= m v
2
Graphical interpretation:
or
mv
x
1
x
t 1
t
2
∫
dt = mv
x
2
mv
y
( )
1
y
t 1
t
2
∫
dt = mv
y
( )
2
Units:
kg ⋅
m
s
= N ⋅ s; USCS
lb ⋅ s
Recall
O
d H
O
dt
O
t
1
t 2
∫
dt = d H
O
O
2 H
O
( )
1
H O
( )
2
∫
O
1
y
x
1
F dt
t 1
t
2
∫
2
Initial
linear
momentum
Linear
impulse
Final
linear
momentum
Note: Angular impulses from internal forces of action and reaction cancel.
If no external forces act from time
1
to
2
, then
O
( )
1
O
( )
2
∑ ∑
and the total angular momentum of the particles is conserved.
Graphical interpretation:
Direct Central Impact
Before
Impact
After
Total linear momentum is conserved during impact:
m
1
v
1
2
v
2
= m
1
v ′
1
2
v ′
2
x x
x
y
y y
m
1
v
1
1
m
2
v
2
1
m
3
v
3
1
m
1
v
1
2
m
2
v
2
2
m
3
v
3
2
F dt
t 1
t
2
∫
∑
O
dt
t
1
t 2
∑ ∫
m
1
m
1
m
1
m
2
m
2
m
2
v
1
v
2
v ′
1
v
2
v
1
v
2
v
1
v
2
v
1
v
2
u
Coefficient of restitution : e =
velocity of separation
velocity of approach
v
2
v
1
v
1
− v
2
If total kinetic energy is conserved, impact is said to be perfectly elastic and
e = 1 .
If particles stick together after impact,
v
1
v
2
, impact is said to be perfectly
plastic , and
e = 0.
For all other impact cases,
0 ≤ e ≤ 1.
A special case occurs when
m
1
= m
2
, collision is elastic ,
v
1
0 , and
v
2
Then,
v
1
= 0 and
v
2
= v
1
Kinematics of Rigid Bodies
Types of plane motion:
Rectilinear translation
Curvilinear translation
Fixed‐axis rotation
1
1
2
2
1
1
2
2
1
2
θ
Note: In r θ coordinates,
v
r
v
θ
= ω r
a
r
= − ω
2
r
a
θ
= α r
In t n axes,
v = ω r
a
n
= ω
2
r
a
t
= α r
General Plane Motion – Absolute and Relative Velocity and Acceleration
Graphical interpretation:
Plane Motion Translation with
v
B
Rotation about B
with ω
Plane Motion Translation with
a
B
Rotation about B
with ω and α
Axes x – y translate with their
origin attached to particle B.
r A
= r B
v
A
= v
B
rel
a A
= a B
α × r rel
ω × (ω × r rel
= a B
− ω
2
r rel
r
A
y
x
€
r
B
€
r
rel
α
ω
ω × r
rel
ω × r
rel
v
A
v
A
v
B
v
B
v
B
v
B
a
A
a
B
a
B
a
B
a
A
a
B
α × r
rel
α × r rel
− ω
2
r
rel
Instantaneous Center of Rotation in Plane Motion
Suppose
v
B
in the previous analysis. Then,
v A
= ω × r rel
This result implies the body is rotating for an instant about point B. Such a
point is called an instantaneous center of rotation (I.C.R.). Such a point can be
determined, as follows, if the velocities of two different particles in a body
are known.
Note: The location of the I.C.R. changes with time in general. Hence,
a
ICR
in general!
Plane Motion of a Particle Relative to a Rotating Frame
x
y
r
A
r
r
rel
ω
α
Axes x – y are body‐fixed
axes, which have angular
velocity ω and angular
acceleration α.
Particle A moves relative to
the body‐fixed axes x – y.
The relative position vector
of A referenced to the x – y
axes is
r
rel
= x i + y j
v
A
v
B
v
A
v
B
v
A
v
B
where: m = total mass
c = center of mass
c
= mass moment of inertia about axis
through c parallel to z axis
p = any moment center in x – y plane
In component form:
x
∑
= ma
cx
y
∑
= ma
cy
c
∑
c
Noncentroidal Rotation
q
∑
c
α + ρ
qc
× m a
ct
q
α
where
q
= mass moment of inertia about axis through q
parallel to z axis.
Laws of Friction
g
Block is initially at rest when force P is
applied and its magnitude is progressively
increased from zero. As long as
P = F < μ
s
the block will not slide.
μ
s
= coefficient of static friction.
c c
1
2
c
α
ρ qc
(bearing
force)
Center
of
rotation
y
x
y
x
q q
m a
ct
m a
cn
When
P = F = μ
s
N , the block starts to slide, and F becomes:
F = μ
k
where
μ
k
= coefficient of kinetic friction ,
μ
k
< μ
s
Kinetics of Rigid Bodies: Energy Methods
For a body in plane motion, the work done on the body by all external forces
i
is
1 → 2
i
r
i
( )
1
r
i
( )
2
∑ ∫
⋅ d r
i
when the body is displaced from position 1 to position 2.
For a body in plane motion, the kinetic energy is
mv
c
2
c
ω
2
For a body in plane motion, the work done on the body by a couple M is
1 → 2
= M d θ
θ
1
θ
2
∫
when the body is displaced from position 1 to position 2.
In general,
1 → 2
m v
c
2
2
c
ω
2
2
m v
c
1
2
c
ω
1
2
2
1
or
2
1
1 → 2
If a gravitational force W acts on the body, and/or a linearly‐elastic spring
force, then the work‐energy equation can be written as:
1 → 2
g
e
where
1 → 2
now excludes gravitational and spring forces. If
1 → 2
above is
zero, total mechanical energy is conserved.
