# CH16NC

```Angular Momentum


H o  r  mv


G  mv
Linear Momentum
Always work from Principle!
Ex: Work of gravity
mg
y
Principle: dW = F * ds
Here: dW = - mg * dy
Always work from Principle!
Ex: Slider arm Kinematics
A
aJA
Principle: x-Position of A:
xA = b*cos(t)
vA
b

O
i
C
D
The velocity x-dot is
the derivative: x-dot =
- b* *sin
Accel x-ddot is the 2nd
derivative: x-ddot = b*2 *cos
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
bx

j
ay
by
k
az
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 =  x (  x r)
at = a x r
Rotation Kinematics
Similar to translation:
   a * dt
    * dt
and
 * d  a * d
Important!
Memorize and Practice!
General Motion = Translation + Rotation
Vector sum vA = vB
+
vA/B
Any rigid body motion can be
viewed as a pure rotation about
an “Instantaneous Center”
(Chapter 16.6)
fig_05_011
fig_05_012
fig_05_013
16.4 Motion Analysis
16.4 Motion Analysis
http://courses.engr.illino
is.edu/tam212/aml.xhtml
http://www.mekanizmalar.com/fourbar01.html
http://iel.ucdavis.edu/ch
html/toolkit/mechanism/
1. A body subjected to general plane motion undergoes a/an
A) translation.
B) rotation.
C) simultaneous translation and rotation.
D) out-of-plane movement.
2. In general plane motion, if the rigid body is represented by a
slab, the slab rotates
A) about an axis perpendicular to the plane.
B) about an axis parallel to the plane.
C) about an axis lying in the plane.
D) None of the above.
1. The position, s, is given as a function of angular position, ,
as s = 10 sin 2. The velocity, v, is
A) 20 cos 2
B) 20 sin 2
C) 20  cos 2
D) 20  sin 2
2. If s = 10 sin 2, the acceleration, a, is
A) 20 a sin 2
B) 20 a cos 2 − 40 2 sin 2
C) 20 a cos 2
D) -40 a sin2 
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 BC. Find
the angular Veloc.
of arm OA.
Step 1: Define the
Geometry
Bar BC rotates at
constant BC. Find
Example
the angular Veloc.
of arm OA.
B
J
vA(t)
A
O
A
AC
 (t)
O
Step 1: Define the
Geometry
i
(t)
 (t)
C
16.5 Relative Motion
Analysis
General Motion = Translation + Rotation
Vector sum vB = vA
+ vB/A
Geometry: Compute all lengths and
angles as f((t))
All angles and
distance AC(t)
are time-variant
B
J
vA(t)
A
Velocities: 
= -dot is
given.
O
A
AC
 (t)
O
Vector Analysis: OA  rA
i
(t)
 (t)
C
vCOLL  BC  rAC
Analysis: Solve the rel. Veloc.
Vector equation
Seen from O:
vA = OA x OA
J
B v (t)
A
A
i
O
Vector Analysis: OA  rA
(t)
O
A
AC
 (t)
 (t)
C
vCOLL  BC  rAC
Analysis: Solve the rel. Veloc.
Vector equation
B vA,rel = JvColl
Seen from C:
vCollar + BC x AC(t)
A
BC x AC(t)
i
O
Vector Analysis: OA  rA
(t)
O
A
AC
 (t)
 (t)
C
vCOLL  BC  rAC
Analysis: Solve the rel. Veloc.
Vector equation numerically
B
J
vA(t)
A
i
(t)
O
A
AC
Vector Analysis: OA  rA vCOLL  BC  rAC
Mathcad does not evaluate cross products symboli
 (t)
RIGHT sides of the above
equation (t)are listed below
O
terms yields two equations for the unknownsC OA
Enter vectors:
 0 
 l cos ( ) 
 0 
 l cos ( ) 
OA   0  rA   l sin ( )  BC   0  rAC   l sin ( ) 








0
 wOA 
 0 
 wBC 


Analysis: Solve the rel. Veloc.
Vector equation numerically
B
A
i
O
A
 (t)
(t)
Vector Analysis: OA  rA
vA(t)
AC
Here: BC is given as
Find OA
O
J
 (t)
C
vCOLL  BC  rAC
LEFT_i  l wOA  sin ( )
RIGHT_i  l wBC sin ( )  vcoll cos ( )
LEFT_j  l wOA  cos ( )
RIGHT_j  l wBC sin ( )  vcoll sin ( )
J
B Veloc.
Analysis: Solve the rel.
v (t)
A
Vector equation numerically
A
i
O
(t)
O
A
AC
 (t)
 (t)
Solve the two vector (i and j) equations :
Given
l wOA  sin ( )
l wOA  cos ( )
l wBC sin ( )  vcoll cos ( )
l wBC sin ( )  vcoll sin ( )
vec  Find ( wOA vcoll)
 4.732 
vec  

 0.568 
C
Recap: The analysis
is becoming more
complex.
B
vA(t)
A
i
 (t)
O
(t)
O
A
AC
•To succeed: Try
Clear Organization
from the start
J
 (t)
C
•Vector Equation = 2 simultaneous
equations, solve simultaneously!
Relative Velocity
vA = vB + vA/B
vRot =  x r
Relative Velocity
vA = vB + vA/B
= VB (transl)+ vRot
fig_05_011
Seen from A:
vB = vA + AB x rB/A
Seen from O:
vB = o x r
Seen from A:
vB = vA + AB x rB/A
Seen from O:
vB = o x r
Visualization
http://courses.engr.illino
is.edu/tam212/aml.xhtml
http://www.mekanizmalar.com/fourbar01.html
http://iel.ucdavis.edu/ch
html/toolkit/mechanism/
Example
Given are: BC wOA  6 (counterclockwise), Geometry: l triangle with
OA  4 inches. OC  12 inches.
Angle   30

180
Collar slides rel. to bar BC.
Guess
Values:
(outward
motion of
collar is
positive)
Crank and
Slider Pin
part 1
Geometry
wBC  1
vcoll  1
gamma  1
AC  1
Vector Analysis: OA  rA vCOLL  BC  rAC
Mathcad does not evaluate cross products symbolically, so the LEFT and
RIGHT sides of the above equation are listed below. Equaling the i- and jterms yields two equations for the unknowns BC and vCOLL
Step 1: Geometry: Find length AC and angle gamma at C.
Law of cosines and law of sines:
Given
2
AC
2
2
OA  OC  2 OA  OC cos ( )
AC sin ( gamma)
OA  sin ( )
vec  Find ( ACgamma)
 8.767 

 0.23 
vec  
gamma_deg  vec 
180
1 
gamma_deg  13.187
AC  vec
0
AC  8.767
gamma  vec
1
LEFT_i  OA  wOA  sin ( ) RIGHT_i  AC wBC  sin ( gamma)  vcoll cos ( gamma)
LEFT_j  OA  wOA  cos ( ) RIGHT_j  AC wBC  sin ( gamma)  vcoll sin ( gamma)
Examples
Crank
and Pin
part 2:
Solving
the vector
equations
Solve the two vector (i and j) equations :
Given
OA  wOA  sin ( )
AC wBC sin ( gamma)  vcoll cos ( gamma)
OA  wOA  cos ( )
AC wBC  cos ( gamma)  vcoll sin ( gamma)
vec  Find ( wBC vcoll)
 1.996 

 16.425 
vec  
(cw).
The pin moves radially outward at vcoll = 16.425 in/s
Vector Analysis Concepts: Always start from default assumptions, i.e. assume positive
rotations and velocities. While magnitudes and signs are not initially known, all vector angles
are known from the given geometry.
tor Analysis:
OA  rA
vCOLL  BC  rAC
hcad does not evaluate cross products symbolically, so the LEFT a
THT
sides of the above equation
 rOA are listed below. Equaling the i- a
msExamples
yields two equations for the unknowns
OA and
OA X
Crank
and
slider
Pin
part 3
Graphical
Solution

O
C
A
rAC X BC
RIGHT ARM
BC: VA =BC X rAC
Pin slides rel. to Arm BC at
velocity vColl. The angle of
vector vColl is = 13o
B



Left ARM OA:
VA = OA X rOA
A
J
L
vA = const

i
B
Given Velocity
V_A = const as shown at
left
The velocity of Point B is
(A) constant, same as V_A
(B) constant, but different
from V_A
(C)VB(t) is variable
(D) None of the above
Given: Geometry and
VA
Find: vB and AB
J
i
A
vA = const
L
ABCounter
clockw.

B
vB
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
In order to connect points A
and B, we define vector rB/A.
J
i
A
r defines the position of
r
vA = const
AB
(we use the symbol 'r' as short notation)

B
point B relative to point A.
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
vA +
AB
x
vA = const
J
Graphical Solution
Veloc. of B
iA
vA = const
r
vB = ?
ABCounterclo
ckw.

B
vB
vA is
given
r
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
vA +
AB
x
r
vA = const
J
Graphical Solution
Veloc. of B
iA
vA = const
r
ABCounterclo
ckw.

B
denotes a
AB x rrotation
AB x r
vB
vB = ?
vA is
given
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
vA +
AB
x
vA = const
J
Graphical Solution
Veloc. of B
vA = const
r
ABCounterclo
ckw.

B
AB x r
vB
vB = ?
vA is
given

iA
AB x r
r
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
vA +
AB
x
vA = const
J
iA
r
ABCounterclo
ckw.

B
AB x r
vB
vA is
given

vA = const
Solution:
vB = vA + AB X r
AB x r
r
16.6 INSTANTANEOUS CENTER OF ZERO VELOCITY
Today’s Objectives:
Students will be able to:
1. Locate the instantaneous center of
zero velocity.
2. Use the instantaneous center to
determine the velocity of any point
on a rigid body in general plane
motion.
Rigid Body Acceleration
Chapter 16.7
Stresses and Flow Patterns in a Steam Turbine
FEA Visualization (U of Stuttgart)
Rigid Body Acceleration
Conceptual Solution
Using Vector Graphics
Propulsion Mechanism of a Baldwin Steam Locomotive
Baldwin Locomotive 60,000
Q: Is this a Freight or
Passenger Locomotive ?
A: We can tell from the wheel
diameter.
The internal forces (accelerations) in the piston
mechanism limit the maximum speed (10 m/s
max. Piston velocity).
Rigid
Body
Acceleration
Given: Geometry and
VA,aA, vB, AB
Find: aB and aAB
J

i
First:
Find all velocities.
A
vA = const
L
ABCounterclo
ckw.
B
vB
Given: Geometry and
VA,aA, vB, AB
Law: aB = aA +
Transl +
Find: aB and aAB
J
aB/A
Centripetal +
angular accel
The relative motion equation aB = aA + aB/A
connects the unknown acceleration at B to the known
(given) acceleration at A.
i
A
vA = const
L
AB
Counterclockw.

B
vB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
In order to connect points A
and B, we define vector rB/A.
J
i
A
r defines the position of
r
vA = const
AB
(we use the symbol 'r' as short notation)
Centrip. r* AB 2

B
point B relative to point A.
aB = aA + aB/A,centr+ aB/A,angular
Given: Geometry and
VA,aA, vB, AB
Find: aB and aAB
r* AB2 + r* a
J
Look at the Accel. of B relative to A:
iA
vA = const
r
AB
Counterclockw
.

B
vB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
J
Look at the Accel. of B relative to A:
iA
We know:
vA = const
r

B
Centrip.
r* AB 2
1. Centripetal: magnitude r2 and
direction (inward). If in doubt, compute
the vector product x(*r)
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
J
Look at the Accel. of B relative to A:
iA
We know:
vA = const
r
Centrip. r*
AB 2

B
r* a
1. Centripetal: magnitude r2 and
direction (inward). If in doubt, compute
the vector product x(*r)
2. The DIRECTION of the angular accel
(normal to bar AB)
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
J
Look at the Accel. of B relative to A:
iA
We know:
vA = const
r
Centrip. r*
AB 2

B
Angular r* a
aB
1. Centripetal: magnitude r2 and
direction (inward). If in doubt, compute
the vector product x(*r)
2. The DIRECTION of the angular accel
(normal to bar AB)
3. The DIRECTION of the accel of point B
(horizontal along the constraint)
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
Find: aB and aAB
r is the vector from reference We know: aA =0
point A to point B
J
i
A
r
vA = const
AB
Angular r* a
Centrip. r* AB 2

B
aB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
Find: aB and aAB
r is the vector from reference
point A to point B
J
i
A
r
vA = const
AB
Angular r* a
Centrip. r* AB 2

B
aB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
Find: aB and aAB
r is the vector from reference
point A to point B
J
i
A
r
vA = const
AB
Angular r* a
Centrip. r* AB 2

B
aB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
Find: aB and aAB
r is the vector from reference
point A to point B
J
i
A
r
vA = const
AB
Angular r* a
Centrip. r* AB 2

B
aB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
Find: aB and aAB
r is the vector from reference
point A to point B
J
i
A
r
vA = const
AB
The vectors form a triangle
with aB as the hypotenuse.
We can therefore determine
the magnitudes and directions
of both aB and r* a
Angular r* a
Centrip. r* AB 2

B
aB
Given: Geometry and
VA,aA, vB, AB
Find: aB and aAB
r is the vector from
reference point A to point B
J
i
aB = aA + aB/A,centr+ aB/A,angular
Now
Complete the
Triangle:
r* a
r* AB2
aB
A
vA = const
r
AB
Centrip. r* AB2

B
Result:
a is &lt; 0 (clockwise)
aB is negative (to the
left)
General Procedure
A
a (t)
B
J
AB
BD
i
 (t) = 45
deg
D
vD(t)= const
1. Compute all velocities
and angular velocities.
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.
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.
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.
Example
1.
Find
all
v
and

(Ch.
16.5)
AB = -11.55k
i
i
HIBBELER 16-125
2*r
2.
aB
=
a
Xr
–

AB
B
AB
B
 = -5k
BC
3. aB = aC
+ aBCXrB/C – BC2*rB/C
Centripetal Terms: We know magnitudes and
HIB 16-125
directions
We now can solve two simultaneous vector
equations for aAB and aBC
– BC2*rB/C
– AB2*rB
aC
aABXrB – AB2*rB = aC
+ aBCXrB/C – BC2*rB/C
HIB 16-125
aABXrB – AB2*rB = aC
+ aBCXrB/C – BC2*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
Seen from B:
vA = vB,rel +
AB x rA/B
Velocities
Seen from P:
vA = o x r
Seen from P:
Seen from B:
aA = aA,rel + 2xvrel aA = ao x r+centr.
aAB x rA/B+centrip.
Accelerations
Midterm #2 Preparation
Posted:
• Collection of Problems
• Practice exam #2
• Powerpoint Slides
• Four questions will be on
Chapter 16, 2Q. on Ch. 14
Stresses and Flow Patterns in a Steam Turbine
FEA Visualization (U of Stuttgart)
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