Transcript ME 242 Chapter 13

Chapter 12 Kinematics
ME 242 Chapter 12
Question 1 We obtain the acceleration fastest
(A)By taking the derivative of x(t)
(B)By Integrating x(t) twice
(C)By integrating the accel as function of displacement
(D)By computing the time to liftoff, then choosing the
accel such that the velocity is 160 mph
ME 242 Chapter 12
160 mi/h = 235 ft/s
Question 2 The acceleration is approximately
(A)92 ft/s2
(B)66 ft/s2
(C)85.3 ft/s2
(D)182 ft/s2
(E)18.75 ft/s2
Ya pili
ME 242 Chapter 12
Solution:
160 mi/h = 235 ft/s
We use v*dv = a*dx
1/ v2 = a*d,
Integrate:
2
Question 2 The acceleration is approximately
where d is the length
of the runway, and the
2
(A)92 ft/s
start velocity = 0
(B)66 ft/s2
(C)85.3 ft/s2
(D)182 ft/s2
(E) 18.75 ft/s2
0.5 * v22  300 ft * a
0.5 * 2352
a
 92 ft / s 2
300
Ya kwanza
Question 3 Road map: We obtain the velocity fastest
(A)By Taking the derivative of a(t)
(B)By Integrating a(t)
(C)By integrating the accel as function of displacement
(D)By computing the time to bottom, then computing the
velocity.
ME242 Tutoring
• Graduate Assistant Ms. Yang Liu will be
available to assist with homework preparation
and answer questions.
• Coordination through the Academic Success
Center in TBE-A 207 Tuesday and Friday
mornings.
• Contact hours: MW after class
ME242 Reading Assignments
• Look up the next Homework assignment on
Mastering
• Example: your second assignment covers
sections 12.5 and 12.6
• Study the text and practice the examples in the
book
• An I-Clicker reading test on each chapter
section will be given at the start of each lecture
• More time for discussion and examples
Supplemental Instruction
ME 242
• Questions
– Yang Liu – PhD student in ME
– [email protected]
– Lab: SEB 4261
Chapter 12-5 Curvilinear Motion
X-Y Coordinates
v
g
B (d,h)
0
y

A (x0,y0)
x
horiz.
distance = d
h
Here is the solution in
Mathcad
Example: Hit target at Position (360’, -80’)
Example: Hit target at Position (360, -80)
Two s olutions exist (Tall Trajectory and flat Trajectory).
The Given - Find routine finds only one s olution, depending on the gues s
values chos en. Therefore we m ust s olve twice, using m ultiple gues s
values. We can als o s olveexplicitly, by inserting one equation into the
s econd:
92.87
50
h1 ( t)
h2 ( t)
0
 50
 100
 100
0
0
100
200
d1 ( t) d2 ( t)
300
360
NORMAL AND TANGENTIAL COMPONENTS
(Section 12.7)
When a particle moves along a curved path, it is sometimes convenient
to describe its motion using coordinates other than Cartesian. When the
path of motion is known, normal (n) and tangential (t) coordinates are
often used.
In the n-t coordinate system, the
origin is located on the particle
(the origin moves with the
particle).
The t-axis is tangent to the path (curve) at the instant considered,
positive in the direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction
toward the center of curvature of the curve.
NORMAL AND TANGENTIAL COMPONENTS
(continued)
The positive n and t directions are
defined by the unit vectors un and ut,
respectively.
The center of curvature, O’, always
lies on the concave side of the curve.
The radius of curvature, r, is defined
as the perpendicular distance from
the curve to the center of curvature at
that point.
The position of the particle at any
instant is defined by the distance, s, along the curve from a
fixed reference point.
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change
of velocity:
.
.
a = dv/dt = d(vut)/dt = vut + vut
.
Here v represents the change in
.
the magnitude of velocity and ut
represents the rate of change in
the direction of ut.
After mathematical manipulation,
the acceleration vector can be
expressed as:
.
a = v ut + (v2/r) un = at ut + an un.
ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
So, there are two components to the
acceleration vector:
a = at ut + an un
• The tangential component is tangent to the curve and in the
direction of increasing or decreasing velocity.
.
at = v or at ds = v dv
• The normal or centripetal component is always directed
toward the center of curvature of the curve. an = v2/r
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5
NORMAL AND TANGENTIAL COMPONENTS
(Section 12.7)
When a particle moves along a curved path, it is sometimes convenient
to describe its motion using coordinates other than Cartesian. When the
path of motion is known, normal (n) and tangential (t) coordinates are
often used.
In the n-t coordinate system, the
origin is located on the particle
(the origin moves with the
particle).
The t-axis is tangent to the path (curve) at the instant considered,
positive in the direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction
toward the center of curvature of the curve.
Normal and Tangential
Coordinates
Velocity
Page 53

v  s * ut
Normal and Tangential Coordinates
‘e’ denotes unit vector
(‘u’ in Hibbeler)
‘e’ denotes unit vector
(‘u’ in Hibbeler)
Learning Techniques
•Complete Every Homework
•Team with fellow students
•Study the Examples
•Ask: Ms. Yang, peers, me
•Mathcad provides structure and
numerically correct results
Course Concepts
•Math
•Think Conceptually
•Map your approach
BEFORE starting work
12.8 Polar coordinates
Polar coordinates
‘e’ denotes unit vector
(‘u’ in Hibbeler)
Polar coordinates
‘e’ denotes unit vector
(‘u’ in Hibbeler)
.
.
.
.
.
.
.
.
.
12.10 Relative (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.)
Vectors and Geometry
r(t)
j
y

(t)
i
x
12.10 Relative (Constrained) Motion
V_truck = 60
V_car = 65
Make a sketch:
A
V_rel
v_Truck
B
The rel. velocity is:
V_Car/Truck = v_Car -vTruck
Example Vector equation: Sailboat
tacking at 50 deg. against Northern Wind
(blue vector)



VWind  VBoat  VWind / Boat
We solve Graphically (Vector Addition)
Example Vector equation: Sailboat
tacking at 50 deg. against Northern Wind



VWind  VBoat  VWind / Boat
An observer on land (fixed Cartesian
Reference) sees Vwind and vBoat .
Land
12.10 Relative (Constrained) Motion

 
VB  VA  VB / A
Plane Vector Addition is two-dimensional.
vA
vB
vB/A
Example cont’d: Sailboat tacking against
Northern Wind



VWind  VBoat  VWind / Boat
2. Vector equation (1 scalar
eqn. each in i- and jdirection). Solve using the
given data (Vector Lengths
and orientations) and
Trigonometry
500
150
i
Chapter 12.10 Relative Motion
Vector Addition
  
rA  rB  rA / B

 
VA  VB  VA/ B
Differentiating gives:

 
VB  VA  VB / A
Exam 1
• We will focus on Conceptual Solutions.
Numbers are secondary.
• Train the General Method
• Topics: All covered sections of Chapter
12
• Practice: Train yourself to solve all
Problems in Chapter 12
Exam 1
Preparation:
Start now! Cramming won’t work.
Questions: Discuss with your peers. Ask
me.
The exam will MEASURE your
knowledge and give you objective
feedback.
Exam 1
Preparation:
Practice:
Step 1: Describe Problem Mathematically
Step2:
Calculus and Algebraic Equation Solving
And here a few visual
observations about
contemporary forms of
socializing, sent to me by a
colleague at the Air Force
Academy.
Enjoy!
Having coffee with friends.
A day at the beach.
Cheering on your team.
Out on an intimate date.
Having a conversation with your BFF
A visit to the museum
Enjoying the sights