Download Chapt14-16ReviewNC

Principles
Learn The Method
Principles
Basics should be automatic
Memorize and Practice!
The angular velocity of the
disk (r = 1m) in rad/s is
(A) 6 k
J

5.129
i
(B)
(C)
(D)
(E)
-6 k
6i
- 3k
3k
wDisk = vo / r = -3/1 = -3 rad/s
The angular accel. of the
disk (r = 2 m) in rad/s2 is
(A) 5 k
J

i
(B) - 5 k
(C) - 10 k
(D) 10k
(E) 2.5 k
aDisk = ao / r = 5/2 = 2.5 rad/s2
5.129
Arm AB with Length L
moves the slotted
rod BD in x-direction
only.
B
J
L
AB
i
A
 (t)
D
vD(t)
X() =L*cos
How do we determine
the velocity vD of
point D as a function
of angle (t)?
  w 
angular _ vel
Arm AB with Length L
moves the slotted
rod BD in x-direction
only.
B
J
L
AB
i
A
 (t)
D
vD(t)
X() =L*cos
How do we determine
the velocity vD of
point D as a function
of angle (t)?
  w 
angular _ vel
X-dot() =-L*sin*_dot
Chapter 14
Energy Methods
 
dW  F  dr
Scalar _ Pr oduct
Only Force components in direction of
motion do WORK
The work-energy relation: The relation
between the work done on a particle by the
forces which are applied on it and how its
kinetic energy changes follows from Newton’s
second law.
2. If the pendulum is released from the
horizontal position, the velocity of its
bob in the vertical position is _____
A) 3.8 m/s.
B) 6.9 m/s.
C) 14.7 m/s.
D) 21 m/s.
The work done is mgh
½*m*v2 = mgh or
V = sqrt(2gh) = sqrt(14.7)
Power
E dE Units of power:
P

t
dt J/sec = N-
m/sec = Watts
1 hp = 746 W
The potential energy V is defined
as:
V  - W  -  F * dr
Conservative Forces
T1 + V1 = T2 + V2
Potential Energy
Potential energy is energy which
results from position or configuration.
An object may have the capacity for
doing work as a result of its position
in a gravitational field. It may have
elastic potential energy as a result of
a stretched spring or other elastic
deformation.
Potential Energy
elastic potential
energy as a result of
a stretched spring or
other elastic
deformation.
Potential Energy
Potential Energy
y
Ex. Problem 14.26
Procedure
1. Frame, start and end
points
2. Constraint equations?
3. Apply the Energy Principle
(only ext. Forces!)
2 𝑥𝑐 + 𝑥𝐴 = 𝐿
T1 = 0, start from rest
0 + F*xc = 0.5*m*(vA)2
Chapter 16
Rigid Body Kinematics
16.1
16.3 Rot. about Fixed Axis Memorize!
Vector Product is NOT
commutative!
Cross Product
i
a  b  ax
j
ay
k
az
bx
by
bz

a  b  a y bz  a z by
a z bx  a x bz
a x by  a y bx

Derivative of a Rotating Vector
• vector r is rotating around the origin, maintaining
a fixed distance
• At any instant, it has an angular velocity of ω
dr
 ωr
dt
r
ω r
ω
Page 317:
 dr
v
 ωr
dt
an = w x ( w x r)
at = a x r
General Motion = Translation + Rotation
Vector sum vA = vB
+
vA/B
16.4 Motion Analysis
Approach
1. Geometry: Definitions
Constants
Variables
Make a sketch
2a. Analysis (16.4)
Derivatives (velocity
2b. Rel. Motion (16.5)
and acceleration)
3. Equations of Motion
4. Solve the Set of Equations. Use
Computer Tools.
Example
Bar BC rotates at
constant wBC.
Find the angular
Veloc. of arm
BC.
Step 1: Define the
Geometry
Example
B
J
vA(t)
A
AC
O
A
Bar BC rotates at
constant wBC.
Find the ang.
Veloc. of arm
BC.
Step 1: Define the
Geometry
 (t)
O
i
(t)
 (t)
C
Geometry: Compute all lengths and
angles as f((t))
All angles and
distance AC(t)
are time-variant
B
J
vA(t)
A
Velocities: w
= -dot is
given.
O
A
AC
 (t)
O
Vector Analysis: wOA  rA
i
(t)
 (t)
C
vCOLL  wBC  rAC
Rigid Body Acceleration
Chapter 16.7
Stresses and Flow Patterns in a Steam Turbine
FEA Visualization (U of Stuttgart)
General Procedure
A
a (t)
B
J
AB
BD
i
 (t) = 45
deg
D
vD(t)= const
1. Compute all velocities
and angular velocities.
2. Start with centripetal
acceleration: It is
ALWAYS oriented inward
towards the center.
General procedure
A
a (t)
B
J
AB
BD
i
 (t) = 45
deg
D
vD(t)= const
1. Compute all velocities
and angular velocities.
2. Start with centripetal
acceleration: It is
ALWAYS oriented inward
towards the center.
3. The angular accel is NORMAL to the
Centripetal acceleration.
General Procedure
A
a (t)
B
J
AB
BD
i
 (t) = 45
deg
D
vD(t)= const
1. Compute all velocities
and angular velocities.
2. Start with centripetal
acceleration: It is
ALWAYS oriented inward
towards the center.
3. The angular accel is NORMAL to the
Centripetal acceleration. The direction of
the angular acceleration is found from the
mathematical analysis.
wExample
1.
Find
all
v
and
w
(Ch.
16.5)
AB = -11.55k
i
i
HIBBELER 16-125
wBC = -5k
2. View from A: aB =
aABXrB – wAB2*rB
3. View from D: aB = aC
+ aBCXrB/C – wBC2*rB/C
Centripetal Terms: We know magnitudes and
HIB 16-125
directions
We now can solve two simultaneous vector
equations for wAB and wBC
– wBC2*rB/C
– wAB2*rB
aD
aABXrB – wAB2*rB = aC
+ aBCXrB/C – wBC2*rB/C
HIB 16-125
aABXrB – wAB2*rB = aC
+ aBCXrB/C – wBC2*rB/C
16.8 Relative Motion
aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE
fig_05_11
16.8 Relative Motion
aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE
16.8 Relative Motion
From Ch. 12.8
radial
aP = 𝒓*ur + ωx(ωxr) +
(α x r) + 2 (ω x 𝒓)
tangential
End of Review
Chapters 14 and 16