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FE Review
Dynamics
G. Mauer
UNLV
Mechanical Engineering
Point Mass Dynamics
X-Y Coordinates
v
g
B (d,h)
0
y

A (x0,y0)
x
horiz.
distance = d
h
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
A ball is thrown horizontally
from A and passes
through B(d=20,h = -20)
meters.
The travel time t to Point B is
Use g = 10 m/s2 (A) t = 4 s
(B) t = 1 s
(C) t = 0.5 s
(D) t = 2 s
Use g = 10 m/s2
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
y (t ) Use
 v0 * sin(
t  0m/s
.5 * g *2t 2
g =) *10
 20m  0.5 *10m / s 2 * t 2
t  2s
A ball is thrown horizontally
from A and passes
through B(d=20,h = -20)
meters.
The travel time t to Point
B is
(A) t = 4 s
(B) t = 1 s
(C) t = 0.5 s
(D) t = 2 s
Use g = 10 m/s2
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
Use g = 10 m/s2
A ball is thrown horizontally
from A and passes through
B(d=20,h = -20) meters at
time t = 2s.
The start velocity v0 is
(A) v0
= 40 m/s
(B) v0 = 20 m/s
(C) v0 = 10 m/s
(D) v0 = 5 m/s
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
x(t )  v0 * cos( ) * t
20m  v0 *1* 2s
v0  10m / s
Use g = 10 m/s2
A ball is thrown horizontally
from A and passes through
B(d=20,h = -20) meters at
time t = 2s.
The start velocity v0 is
(A) v0
= 40 m/s
(B) v0 = 20 m/s
(C) v0 = 10 m/s
(D) v0 = 5 m/s
12.7 Normal and Tangential Coordinates
ut : unit tangent to the path
un : unit normal to the path
Normal and Tangential
Coordinates
Velocity
Page 53

v  s * ut
Normal and Tangential Coordinates
Fundamental Problem 12.27
a
v
2

* un  at * ut
The boat is traveling along the
circular path with  = 40m and a
speed of v = 0.5*t2 , where t is
in seconds. At t = 4s, the normal
acceleration is:
(A) constant
•(B) 1 m/s2
•(C) 2 m/s2
•(D) not enough information
•(E) 4 m/s2
Fundamental Problem 12.27
a
v
2

* un  at * ut
The boat is traveling along the
circular path with  = 40m and a
speed of v = 0.5*t2 , where t is
in seconds. At t = 4s, the normal
acceleration is:
at  dv / dt  2 * 0.5 * t
At __ t  2s : _ at  2 *1m / s 2
(A) constant
•(B) 1 m/s2
•(C) 2 m/s2
•(D) not enough information
•(E) 4 m/s2
Polar coordinates
Polar coordinates
Polar coordinates
Polar Coordinates
Point P moves on a
counterclockwise circular
path, with r =1m, dot(t)
= 2 rad/s. The radial and
tangential accelerations
are:
•(A) ar = 4m/s2 a = 2 m/s2
•(B) ar = -4m/s2 a = -2 m/s2
•(C) ar = -4m/s2 a = 0 m/s2
•(D) ar = 0 m/s2 a = 0 m/s2
Polar Coordinates
Point P moves on a
counterclockwise circular
path, with r =1m, dot(t)
= 2 rad/s. The radial and
tangential accelerations
are:
•(A) ar = 4m/s2 a = 2 m/s2
•(B) ar = -4m/s2 a = -2 m/s2
•(C) ar = -4m/s2 a = 0 m/s2
•(D) ar = 0 m/s2 a = 0 m/s2
disk = 10 rad/s
B
C
r
e
er
Unit vectors
Point B moves radially
outward from center C,
with r-dot =1m/s, dot(t)
= 10 rad/s. At r=1m, the
radial acceleration is:
•(A) ar = 20 m/s2
•(B) ar = -20 m/s2
•(C) ar = 100 m/s2
•(D) ar = -100 m/s2
disk = 10 rad/s
B
C
r
e
er
Unit vectors
Point B moves radially
outward from center C,
with r-dot =1m/s, dot(t)
= 10 rad/s. At r=1m, the
radial acceleration is:
•(A) ar = 20 m/s2
•(B) ar = -20 m/s2
•(C) ar = 100 m/s2
•(D) ar = -100 m/s2
Example cont’d: Problem 2.198 Sailboat
tacking against Northern Wind



VWind  VBoat  VWind/ Boat
2. Vector equation (1 scalar eqn. each in
i- and j-direction)
500
150
i
Given:
r(t) = 2+2*sin((t)), _dot= constant
The radial velocity is
(A) 2+2*cos((t ))*-dot,
(B) -2*cos((t))*-dot
(C) 2*cos((t))*-dot
(D) 2*cos((t))
(E) 2* +2*cos((t ))*-dot
Given:
r(t) = 2+2*sin((t)), _dot= constant
The radial velocity is
(A) 2+2*cos((t ))*-dot,
(B) -2*cos((t))*-dot
(C) 2*cos((t))*-dot
(D) 2*cos((t))
(E) 2* +2*cos((t ))*-dot
2.9 Constrained Motion
A
J
L
vA = const
vA is given
as shown.
Find vB

i
B
Approach:
Use rel.
Velocity:
vB = vA +vB/A
(transl. + rot.)
V
r = 150 mm
The conveyor belt is moving to the left at v =
6 m/s. The angular velocity of the drum
(Radius = 150 mm) is
6 m/s
(B) 40 rad/s
(C) -40 rad/s
(D) 4 rad/s
(E) none of the above
(A)
V
r = 150 mm
The conveyor belt is moving to the left at v =
6 m/s. The angular velocity of the drum
(Radius = 150 mm) is
6 m/s
(B) 40 rad/s
(C) -40 rad/s
(D) 4 rad/s
(E) none of the above
(A)
Omit all constants!
XB
xA
A
B
c
yE
Xc
The rope length between points
A and B is:
•(A) xA – xB + xc
•(B) xB – xA + 4xc
•(C) xA – xB + 4xc
•(D) xA + xB + 4xc
Omit all constants!
XB
xA
A
B
c
yE
Xc
The rope length between points
A and B is:
•(A) xA – xB + xc
•(B) xB – xA + 4xc
•(C) xA – xB + 4xc
•(D) xA + xB + 4xc
NEWTON'S LAW OF INERTIA
A body, not acted on by any force, remains in
uniform motion.
NEWTON'S LAW OF MOTION
Moving an object with twice the mass will require
twice the force.
Force is proportional to the mass of an object and
to the acceleration (the change in velocity).
F=ma.
Dynamics
M1: up as positive:
Fnet = T - m1*g = m1 a1
M2: down as positive.
F
=
F
=
m
*g
T
=
m
a2
net
2
2
3. Constraint equation:
a1 = a2 = a
Equations
From previous:
T - m1*g = m1 a
 T = m1 g + m1 a
Previous for Mass 2:
m2*g - T = m2 a
Insert above expr. for T
m2 g - ( m1 g + m1 a ) = m2 a
( m2 - m1 ) g = ( m1 + m2 ) a
( m1 + m2 ) a = ( m2 - m1 ) g
a = ( m 2 - m 1 ) g / ( m1 + m 2 )
Rules
1. Free-Body Analysis, one for
each mass
2. Constraint equation(s):
Define connections.
You should have as many
equations as Unknowns.
COUNT!
3. Algebra:
Solve system of equations
for all unknowns
J
g
m
M*g*sin*i
i
0 = 30
M*g
0
-M*g*cos*j
Mass m rests on the 30
deg. Incline as shown.
Step 1: Free-Body
Analysis. Best
approach: use
coordinates tangential
and normal to the path
of motion as shown.
J
g
m
M*g*sin*i
i
0 = 30
M*g
Mass m rests on the 30
deg. Incline as shown.
Step 1: Free-Body
Analysis.
Step 2: Apply Newton’s
Law in each Direction:
0
N
-M*g*cos*j
 (Forces _ x)  m * g * sin  * i  m * x
 (Forces _ y)  N - m * g * cos * j  0(static _ only)
Friction F = mk*N:
Another horizontal
reaction is added in
negative x-direction.
J
g
m
M*g*sin*i
i
0 = 30
M*g
0
mk*N
N
-M*g*cos*j
 (Forces _ x)  (m * g * sin   mk * N ) * i  m * x
 (Forces _ y)  N - m * g * cos * j  0(static _ only)
Problem 3.27 in Book:
Find accel of Mass A
Start with:
(A)Newton’s Law for A.
(B)Newton’s Law for A and
B
(C) Free-Body analysis of
A and B
(D) Free-Body analysis of
A
Problem 3.27 in Book:
Find accel of Mass A
Start with:
(A)Newton’s Law for A.
(B)Newton’s Law for A and
B
(C) Free-Body analysis of
A and B
(D) Free-Body analysis of
A
Problem 3.27 in Book
cont’d
Newton applied to mass B
gives:
(A) SFu = 2T = mB*aB
(B) SFu = -2T + mB*g = 0
(C) SFu = mB*g-2T = mB*aB
(D) SFu = 2T- mB*g-2T = 0
Problem 3.27 in Book
cont’d
Newton applied to mass B
gives:
(A) SFu = 2T = mB*aB
(B) SFu = -2T + mB*g = 0
(C) SFu = mB*g-2T = mB*aB
(D) SFu = 2T- mB*g-2T = 0
Problem 3.27 in Book
cont’d
Newton applied to mass A
gives:
(A) SFx = T +F= mA*ax ; SFy = N - mA*g*cos(30o) = 0
(B) SFx = T-F= mA*ax SFy = N- mA*g*cos(30o) = mA*ay
(C) SFx = T = mA*ax ; SFy = N - mA*g*cos(30o) =0
(D) SFx = T-F = mA*ax ; SFy = N-mA*g*cos(30o) =0
Problem 3.27 in Book
cont’d
Newton applied to mass A
gives:
(A) SFx = T +F= mA*ax ; SFy = N - mA*g*cos(30o) = 0
(B) SFx = T-F= mA*ax SFy = N- mA*g*cos(30o) = mA*ay
(C) SFx = T = mA*ax ; SFy = N - mA*g*cos(30o) =0
(D) SFx = T-F = mA*ax ; SFy = N-mA*g*cos(30o) =0
Energy Methods
 
dW  F  dr
Scalar _ Pr oduct
Only Force components in direction of
motion do WORK
Work
of
Gravity
Work
of a
Spring
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.
A car is traveling at 20 m/s on a level
road, when the brakes are suddenly
applied and all four wheels lock. mk =
0.5. The total distance traveled to a
full stop is (use Energy Method, g =
10 m/s2)
(A) 40 m
(B) 20 m
(C) 80 m
(D) 10 m
(E) none of the above
Collar A is compressing the
spring after dropping
vertically from A. Using the
y-reference as shown, the
work done by gravity (Wg)
and the work done by the
compression spring (Wspr)
are
y
(A) Wg <0, Wspr <0
(B) Wg >0, Wspr <0
(C) Wg <0, Wspr >0
(D) Wg >0, Wspr >0
Conservative Forces
A conservative force is one for which
the work done is independent of the
path taken
Another way to state it:
The work depends only on the initial
and final positions,
not on the route taken.
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
A child of mass 30 kg is sliding downhill
while the opposing friction force is 50 N
along the 5m long incline (3m vertical
drop). The work done by friction is
(A) -150 Nm
(B) 150 Nm
(C) 250 Nm
(D) -250 Nm
(E) 500 Nm
h
d

(Use Energy Conservation) A 1
kg block slides d=4 m down a
frictionless plane inclined at
=30 degrees to the horizontal.
The speed of the block at the
bottom of the inclined plane is
(A) 1.6 m/s
(B) 2.2 m/s
(C) 4.4 m/s
(D) 6.3 m/s
(E) none of the
above
Angular Momentum


H o  r  mv


G  mv
Linear Momentum
Rot. about Fixed Axis Memorize!
Page 336:
 dr
v
 ωr
dt
an =  x (  x r)
at = a x r
Meriam Problem 5.71
Given are: BC wBC  2 (clockwise), Geometry: equilateral triangle
with l  0.12 meters. Angle   60

180
Collar slides rel. to bar AB.
Mathcad EXAMPLE
Guess
Values:
(outward
motion of
collar is
positive)
wOA  1
vcoll  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 OA and vCOLL
Mathcad Example
part 2:
Solving the vector
equations
16.4 Motion Analysis
http://gtrebaol.free.fr/d
oc/flash/four_bar/doc/
Approach
1. Geometry: Definitions
Constants
Variables
Make a sketch
2. Analysis: Derivatives (velocity 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
BC.
Step 1: Define the
Geometry
Example
B
J
vA(t)
A
AC
O
A
Bar BC rotates at
constant BC.
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: 
= -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 conceptually
B
Seen from O: vA =  x OA
J
vA(t)
A
i
O
Vector Analysis: OA  rA
(t)
O
A
AC
 (t)
 (t)
C
vCOLL  BC  rAC
Analysis: Solve the rel. Veloc.
Vector equation
Seen from O: vA =  x OA
B
J
vA(t)
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 J
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 symbol
 (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
-2 rad/s (clockwise).
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
•Mathcad
 (t)
O
(t)
O
A
AC
•To succeed: Try
Clear Organization
from the start
J
 (t)
C
•Vector Equation = 2 simultaneous
equations, solve simultaneously!
16.6 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
Rigid Body Acceleration
Stresses and Flow Patterns in a Steam Turbine
FEA Visualization (U of Stuttgart)
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)
We can add graphically:
Start with Centipetal
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
We can add graphically:
Start with Centipetal
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 < 0 (clockwise)
aB is negative (to the
left)
E
F
A
a (t)
B
H
AB
The instantaneous center
of Arm BD is located at
Point:
(A) F
(B) G
(C) B
(D) D
(E) H
BD
J
 (t)
O
i
G
D
vD(t)
Plane Motion
3 equations:
S Forces_x
S Forces_y
S Moments about G
fig_06_002
Plane Motion
3 equations:
S Forces_x
S Forces_y
S Moments about G
Translatio n :  Fx  m * x
..................... Fy  m * y
Rotation : ..... M G  I G *a
fig_06_002
Parallel Axes Theorem
Pure rotation about fixed point P
I P  IG  m * d
fig_06_005
2
Constrained Motion:
The system no longer has all three
Degrees of freedom
Describe the constraint(s) with an
Equation