Dynamics FE Review
For rectilinear motion - motion in a straight line - where the position is defined by s:
Mechanics
Response of mass (body) to mechanical disturbance
Statics
Dynamics
Analysis of body at rest
Kinematics
Video 2011
v
Analysis of body in motion
Kinetics
Geometry of motion— no concern for forces that caused motion
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Find: v(t) and a(t) Solution:
dv dt
Where v is the instantaneous velocity, a is the instantaneous acceleration, and t is time. 2011
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Given: Position of a car is described by s 3t 3 t 2 m.
For a particle whose position is defined by the vector r:
a
a ds v dv
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• Fundamental equations of motion
dr dt
dv dt
Example Problem:
Kinematics of Particles
v
a
Note: Instead of s, the position could be defined by x, y, etc.
Relation between force, mass, and motion
Figures and problems taken from the textbook Dynamics, 5th edition, Meriam and Kraige, Wiley.
ds dt
2
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v
ds : dt
v 9t 2 2t m/s
a
dv : dt
a 18t 2
m/s 2 4
1
Example Problem:
So, we start with
Given: Acceleration of a car is given by:
a(t )
dv dt
Note we wrote the acceleration as a(t) to emphasize the fact that a is a function of t.
Then: dv a( t ) dt
a( t ) 3t 2 5t 1 m/ m/s2
v
t
v0
0
dv a(t ) dt
At t = 0, v0 = 4 m/s Find: Velocity (v) when t=3 s.
t
v v0 a( t ) dt 0
This gives us v as a function of time or v(t) 2011
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t
v( t ) v0 a(t ) dt 0
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t
First, look for a fundamental equation that contains both a, t and v.
v( t ) v0 a(t ) dt 0
So for a(t) = 3t2+5t+1 5t 1, v0 = 4 m/s, and t = 3 s
v
ds dt
a
dv dt
t
v 4 (3t 2 5t 1) dt
a ds v dv
0
3
3t 3 5t 2 t v 4 2 3 0 v 56.5 m/s 2011
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For uniformly accelerated rectilinear (UARM) motion (a=constant) the following equations apply:
– If the acceleration is constant, we can apply the UARM equations in the x and y directions. For the x direction
v v0 a (t t0 )
a (t t0 ) s s0 v0 (t t0 ) 2
vy vy ayt
vx vx 0 axt
2
at x x0 vx 0 t x 2
v v 2a s s0 2
For the y direction 0
2
y y0 v y t
ayt 2
0
2
vx 2 vx 0 2ax x x0 v y 2 v y 0 2a y y y0
2 0
2
2
THESE ONLY APPLY IF THE ACCELERATION IS CONSTANT!!! 2011
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Note: This is for t0 = 0. If t0 = 0 then replace t with t-t0
– Projectile Motion using Rectangular Coordinates
• Curvilinear Motion Using Rectangular Coordinates (x-y) – Useful when the position (r) is given in rectangular coordinates
ax = 0 ay = -g
Fig 2/7 Meriam and Kraige
v
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dr xi y j dt
Fig 2/8 Meriam and Kraige
For the x direction
vx vx 0
dv a xi yj dt
x x0 vx 0 t
For the
y direction
v y v y gt 0
y y0 v y t 0
gt 2 2
v y 2 v y 2 g y y0 2
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0
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• Curvilinear Motion Using NormalTangential (Path) Coordinates (n-t)
Example: Given: Projectile fired off a cliff as shown
– Useful when the path is given, especially the curvature of the path
y
et
180 m/s
o
ymax
30º
x
en
150 m
x at impact Fig g 2/9 Meriam and Kraige g
Find: x at impact and
v vet
ymax 2011
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a
v2
en vet
an
v2
at v
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– Special Case: Circular motion using n-t coordinates = constant = r Angular position given i b by
v r v2 r 2 v r at v r an
note: and Fig 2/12 Meriam and Kraige 2011
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Example: Car on the circular part of track
For Curvilinear Motion Using Polar Coordinates:
Given: FAS, r = 200 m, v = 50 m/s, at = 2 m/s2
v rer reθ
et
er
e
en
For circular motion:
r 0 r 0
r
note:
Find: a Fig P2/144 Meriam and Kraige
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a r r 2 er r 2r eθ
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and
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Kinetics of Particles • Kinetics: Relations between forces and motion. • Newton’s Second Law: The acceleration of a particle is proportional to the resulting force acting on it and is in the direction of this force. force (assumes m is constant) F ma
• FBD: You must be able to draw good free body diagrams! 2011
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• Rectangular Coordinates (Cartesian)
• Polar Coordinates (Radial/Transverse)
For particle P: y
F2
For particle A:
F ma F ma i ma j
F3
x
F3
F2
y
F ma F ma e
r r
ma e
P F1
Scalar components: x
F F
x y
ma y
r
Fig 2-13 Meriam and Kraige
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• Normal/Tangential Coordinates (Path) F2
F1
n n
r
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– Units
For particle C:
F ma F ma e
F ma m r r F ma m r 2r 2
ma x
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F3
Scalar components:
F1
Force A Acceleration l ti Mass g
mat et
Scalar components:
SI
US
N m/s / 2 kg 9.81 m/s2
lb 2 ft/ ft/sec slug (32.2 lbm) 32.2 ft/sec2
Fig 2-9 Meriam and Kraige
F
n
ma n m
F ma t
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t
v2
a = 1 m/s2
F=1N m=1 kg
mv 22
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a = 1 ft/sec2 F
F = 1 lb m=1 slug
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Example Problem: Textbook 3/12
Example Problem: Textbook 3/1
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3/1
Given:
Find:
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3/12
FAS v0 = 7 m/s at x0 = 0 k = 0.4, m = 50 kg
Given:
FAS, W = 100 lb ft/sec c2 up incline nc n a = 5 ft/s k = 0.25
Find:
P
t and x when v = 0
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Example Problem: Textbook 3/54
Example Problem:Textbook 3/50
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3/50
Given:
FAS, m = 2 kg vB = 3.5 m/s = 2.4 24m
Find:
NB and vA such that NA = 0
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3/54
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Given:
FAS, W = 0.2 lb = 30 = 3 rad/sec ccw r = -4 ft/sec
Find:
N
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Example Problem: Textbook 3/56
Work/Energy • We have been using the direct application of Newton’s Second Law to solve kinetics problems.
F ma
Forces
Acceleration
Motion
This method of solution can be very difficult sometimes! 2011
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3/56
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• Work/Energy methods: – These methods will make it MUCH EASIER to solve some kinetics problems! – Definition of work:
Given:
FAS,, W = 3000 lb r = 100 ft, v = 35 mi/hr
Find:
aN, FN
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U
• Component of force acting in the direction of motion times the displacement. • Units: SI N•m = J US
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ft•lb
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– Conservative force: Work done by a conservative force is independent of path.
Consider a particle moving along the path from A to A r2
• The work only depends on the starting and ending positions! • When a p particle moves under the influence f of fa conservative force:
SCALAR!
U F dr r1
If we let |dr| = ds
U 1 2 V1 V2
s2
U Ft ds Sign convention: Positive if active force (Ft) is in the direction of motion and negative if it is in the opposite direction
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Then:
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V g mgh from the datum. + if above datum - if below datum
Ve
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T1 V g 1 Ve1 U 1 2 T2 V g 2 Ve 2
h is measured
Spring:
T1 V g 1 Ve1 U 1 2 T2 V g 2 Ve 2
This is the fundamental equation for applying the work/energy method.
• Energy available due to position
datum
U12
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– Potential Energy: (V) Gravity:
T1 V1 T2 V2
•Work done by nonconservative forces:
s1
Fig 3-2 Meriam and Kraige
and
The FE reference handbook gives the equation in this form:
1 2 kx 2
T1 U 1 W1 2 T2 U 2
Fig 33-6 6 Meriam and Kraige
- Kinetic Energy: (T) 2011
T
Whenever you have conservative forces doing work -- gravity and springs -- consider using the work/energy method.
1 mv 2 2 38
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40
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Example Problem: Textbook Sample 3/17
Example Problem: Textbook 3/104
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Sample 3/17
3/104
Given: FAS, vA = 5 m/s, hA = 0, hB = 0.8 m Find: vB
Given: FAS, mg = 6 lb, k = 2 lb/in, unstretched length = 24 in, vA = 0 Find:
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vB
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Example Problem: Textbook 3/144
Impulse/Momentum
• In some situations, the FORCES are described as acting over an interval of TIME. Impulse/Momentum methods work well in these cases.
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3/144
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•Linear Impulse - Linear Momentum Define:
t2
Fdt Linear Impulse
VECTOR!
t1
m v G Linear Momentum Then:
Given: FAS, m = 4 kg, vA = 0 unstretched length = 24 in Find:
vB, x at max deformation
t2
Fdt G
2
VECTOR!
G 1 G
t1
Linear Impulse = Change of Linear Momentum 2011
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Rearranging:
3/179
t2
G 1 Fdt G 2 t1
Scalar Components:
t2
G x1 Fx dt G x2
Given:
FAS, For projectile: M=75 g, v1=600 m/s For block: M=50 kg v1 = 0
Find:
E during impact
t1
t2
G y1 Fy dt G y2 t1
Conservation of Linear Momentum:
If
If F 0
then G 1 G 2
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Example Problem: Textbook 3/179
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Example Problem: Textbook 3/188
50
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3/188
y
Given: FAS, For tanker: M=10.43x106 slugs, v1=0, Cable tension = 50,000 lb Find: Time required to bring speed of tanker to 1 knot
O x
Fig 3-11 Meriam and Kraige
For plane motion in the x-y plane:
H o r mv mvr sin k 2011
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• Angular Impulse - Angular Momentum
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Define Angular Impulse:
– Angular Momentum: H0 • The moment of the linear momentum about a point
It can be shown:
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t2
M
o
dt Angular Impulse
o
dt H O2 H O1 H O
t1
t2
M t1
Rearranging:
O
t2
H O1 M o dt H O2 t1
Conservation of Angular Momentum:
If
M
O
0 then H O 1 H O 2
Fig 3-11 Meriam and Kraige 2011
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Kinetics of Particles Impact
Example Problem: Textbook 3/227
• Impact: Collision between two bodies. bodies Direct Central Impact: Centers of mass located on Line L ne of Impact (LOI). Velocities in direction of LOI. 2011
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3/277
LOI
LOI
Fig 3-14 Meriam and Kraige
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If we have no external impulsive forces, TOTAL linear momentum of the system is conserved.
G Before G After Along the LOI:
m 1 v1 m 2 v 2 m 1 v1' m 2 v 2' Coefficient of Restitution (e)
Given: FAS, N1 = 0 N2 = 150 rpm T = 20 N Find:
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e
t 58
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Relative velocity after Relative velocity before
v2' v1' v1 v2 60
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Plane Kinematics of Rigid Bodies - Plane Motion
Example Problem: Textbook 3/247
• Rigid Body: System of particles for which the distances between the particles remain unchanged. • Plane Motion: All parts of the body move in parallel planes. planes – Plane of Motion: Plane that contains the center of mass. 2011
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3/247
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• Types of Plane Motion: Translation: All points on the rigid body have the same velocity and acceleration. l Kinematic concepts from Chapter 2 apply
Gi Given: FAS e = 0 FAS, 0.6 6 Find:
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v1’ and v2’
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Fig 5-1 Meriam and Kraige
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– Rotation Concepts:
Example Problem: Textbook 5/2
d dt d dt d d
Fig 5-2 Meriam and Kraige
Angular velocity
Angular acceleration
For = constant:
O (t t 0 ) O2 2 O 1 2
O (t t 0 ) (t t 0 ) 2 2011
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5/2
– Fixed Axis Rotation: For any point on the rigid body,
v r
a n r 2 v
2
a t r
r
v
In vector form,
Given: FAS, =10 rad/s
v r ω r Fig 5-3 Meriam and Kraige
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Find: F
a n ω ω r 2 r
vA an and aA
at α r 66
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Relative Motion -Translating Axes (Velocity)
Plane Kinematics of Rigid Bodies Relative Motion Method •G Generall Plane Pl Motion M ti can be b considered id d as Translation + Rotation
(Translating Frame)
Absolute velocity of A wrt fixed frame (X,Y)
y
vA vB vA B
A/B
Y
x
(Fixed Frame)
X
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Relative Motion:
rA rB rA rA rB rA rA rB rA
If A and B are on the same rigid body, vA/B = rA/B
y ((Translating ranslat ng Frame)
A/B
B
B
Absolute acceleration of A wrt fixed f frame (X,Y)
x
B
or v A v B v A or a A a B a A
B
aA aB aA B
Y (Fixed Frame)
B
X
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Relative R l velocity l of f A wrt a translating frame (x,y) attached to B. 71
Relative Motion -Translating Axes (Acceleration)
XY Fixed xy Translating with B
Fig. g 2-17 Meriam and Kraige g
Fig 5/6 Meriam and Kraige
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Absolute velocity of origin of the translating frame at B wrt fixed frame (X,Y)
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Fig 5/9 Meriam and Kraige
Absolute acceleration of origin of the t translating l ti frame at B wrt fixed frame (X,Y)
Relative acceleration of A wrt a translating frame (x,y) attached to B. Due to rotation about B 72
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Relative Motion Translating Axes. To summarize:
5/120
vA vB vA B or v A v B ω rA B
y (Translating Frame)
or A/B
x
v A v B ω rrel aA aB aA B or
Y (Fixed Frame)
X
a A a B ω ω rA B α rA B
Given:
FAS, = 2 rad/s, = 0, a0=3 m/s2
Find:
aA when = 0º, 90º, and 180º
or a A a B ω ω rrel α rrel
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Example Problem: Textbook 5/120
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Example Problem: Textbook 5/141
100
60
180
Y
80
80
X
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5/141
• Relative Motion - Rotating Axes – This method works best when sliding occurs relative to two rigid bodies. – Consider the following:
100 180
60 80Y
80
A
Given:
FAS,
P
OA = 10 rad/s CCW
Find:
AB
Fig 5/11 Meriam and Kraige 2011
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5/141
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Relative Motion -Rotating Axes (Velocity)
100 180
60 80Y
v A v B v P B v A/ P
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Or:
v A v B ω rrel v rel Given:
FAS,
OA = 10 rad/s CCW
Find:
AB
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5/174
Relative Motion -Rotating Axes (Acceleration)
a A aB aP B a A/ P Given:
Or:
a A a B α rrel ω ω rrel 2ω v rel a rel 2011
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Find:
FAS,
OA = 10 rad/s
CW = 30 BC , arel
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Example Problem: Textbook 5/174
Given:
Find:
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FAS,
OA = 10 rad/s
CW = 30 BC , arel
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IC Method Steps: 1. Identify directions of velocity vectors of two points. 2. At these two points draw lines perpendicular to the velocity vectors. 3. These lines intersect at the IC point C. Given:
Find:
Fig 5/7 Meriam and Kraige
FAS, OA = 10 rad/s CW = 30 BC , arel
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4. If we know the magnitude of vA or vb you can solve for . 85
v A rA v B rB
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Example Problem: Textbook 5/97
Plane Kinematics of Rigid Bodies - Instantaneous Center of Zero Velocity Method For plane motion, at any instant, the motion may be considered as pure rotation about a point called the instantaneous center of zero velocity (IC) Works well if we know the directions of the velocity vectors of two points on the rigid body 2011
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5/97
For general Plane Motion:
Video
F ma M Iα : M I G
G
Given: Find:
Or:
FAS, FAS OB = 0.8 rad/sec cw vA and vC
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F ma M I
Where C refers to the center of mass, shown as G in the figures 89
C
C C
α
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Plane Kinetics of Rigid Bodies Newton’s Second Law
If we take moments about an arbitrary point, P, then the moment equation becomes:
• All of the concepts of particle kinetics apply to kinetics of rigid bodies.
M M
• However, we must account for the rotational effects of the rigid body.
P
Iα ρ m a
P
I mad
Or:
• We will use the overbar to indicate a quantity reverenced to the center of mass, G.
M
P
I C α ρ PC m a C
Where C refers to the center of mass 2011
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6/33
For Fixed Axis Rotation about point O:
F ma M Iα M I M I α M I
Given:
G
Find:
G
O
O
O
O
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FAS, M = 20 kg, released from rest Reaction forces at pin O
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Example Problem: Textbook 6/33
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Example Problem: Textbook 6/77
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6/77
Kinetic Energy: Translation:
T
1 mv 2 2
T
1 I O 2 2
T
1 1 mv 2 I 2 2 2
T
1 I IC 2 2
Fixed Axis Rotation:
Given:
General Plane Motion:
FAS, released from rest, =40 s=0.3, =40, =0 3 k=0.2 =0 2 aG, friction force
Find:
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Plane Kinetics of Rigid Bodies Work/Energy Methods
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Potential Energy: Same as for Particles
Gravity:
h is measured
• All of the work/energy concepts of particle kinetics apply to kinetics of rigid bodies.
from the datum. + if above datum - if below datum
• However, we must account for additional rotational effects . • Recall: R ll
T1 V g 1 Ve1 U 1 2 T2 V g 2 Ve 2
Spring:
Or: 2011
T1 U 1 W1 2 T2 U 2
V g mgh
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Ve
1 2 kx 2 100
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Example Problem: Textbook 6/122
Plane Kinetics of Rigid Bodies Impulse/Momentum Method • All of the Impulse/Momentum concepts of particle kinetics apply to kinetics of rigid bodies. • However, once again, we must account for additional rotational effects . • Linear Impulse/Momentum t2
G 1 r: Fdt G 2
G mv
t1
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6/122
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Angular Momentum: Angular Impulse:
HG I t2
M
G
dt
t1
Then: Given:
FAS, W=12 lb kspring = 3 lb/in
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H G1 M G dt H G2 t1
kO= 10 in, 1=90, 2=0, 1 = 0 2 = 4 rad/s M
Find:
t2
For Fixed Axis Rotation About O: 102
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t2
H O I O H O1 M O dt H O2 t1
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Example Problem: Textbook 6/173
Vibration and Time Response • Mechanical and structural systems are often oft n su subjected j ct to vibratory ratory mot motion. on. – – – –
Automobiles on a rough road Power lines and bridges on a windy day Aircraft wings experiencing flutter Buildings during an earthquake
• Here we have a brief introduction to undamped, free vibration of particles. 2011
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6/173
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Undamped Free Vibration: Consider what happens when the spring mounted cart is disturbed from its equilibrium position a distance x.
107
From the FBD and Newton’s Second Law:
F
x
ma x
O Or:
kx mx
mx kx 0 Given: Find:
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FAS
at t=4s for each case
Fig 8/1 Meriam and Kraige
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mx kx 0 Let’s define the following:
n k m
Fig 8/2 Meriam and Kraige
Then:
Displacement
x x 0
x x0 cos n t
2 n
Fig 8/1 Meriam and Kraige
This equation describes simple harmonic motion. The acceleration is proportional to the displacement, but of opposite sign.
Let
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A x0
x0
n
and
Natural Frequency
sin n t
B
x 0
n
n k m
Period
n 2
n
Then the amplitude is
C
A2 B 2
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Example Problem: Textbook 8/4 - p 601
x n2 x 0 This is a linear, homogeneous, second order, differential equation The solution is as equation. follows: Initial velocity
Fig 8/1 Meriam and Kraige
x x0 cos n t
x0
n
sin n t
Position Initial displacement 2011
Natural frequency
Time 110
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8/4
8/17
Given: FAS, x0 = -2 in v0 = 7 in/sec
Given: Weight = 120 lb Deflection = 0.9 in Find: n
Find: Amplitude
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Torsional Vibration
Example Problem: Textbook 8/17
0 cos n t
0 sin n t n
n kt I kt GJ
L
I = Mass moment of inertia G = Shear modulus J = Polar moment of inertia 2011
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