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
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
ME 242 Chapter 12
Ya pili
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
ME 242 Chapter 12
Ya kwanza
Solution:160 mi/h = 235 ft/sWe use v*dv = a*dxIntegrate: 1/2v2 = a*d, where d is the length of the runway, and the start velocity = 0
22
22
/92300
235*5.0
*300*5.0
sfta
aftv
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
A (x0,y0)
B (d,h)v0
g
horiz.
distance = dx
yh
Chapter 12-5 Curvilinear Motion X-Y Coordinates
Here is the solution in Mathcad
Example: Hit target at Position (360’, -80’)
0 100 200 300100
50
0
50
92.87
100
h1 t( )
h2 t( )
3600 d1 t( ) d2 t( )
Two solutions exist (Tall Trajectory and flat Trajectory).The Given - Find routine finds only one solution, depending on the guessvalues chosen. Therefore we must solve twice, using multiple guessvalues. We can also solve explicitly, by inserting one equation into thesecond:
Example: Hit target at Position (360, -80)
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 position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.
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.
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.
..
. a = v ut + (v2/r) un = at ut + an un.
After mathematical manipulation, the acceleration vector can be expressed as:
ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
So, there are two components to the acceleration vector:
a = at ut + an un
• The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r
• 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 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 CoordinatesVelocity Page 53 tusv *
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
LB
A
i
JvA = const
vA is given as shown.Find vB
Approach: Use rel. Velocity:vB = vA +vB/A
(transl. + rot.)
Vectors and Geometry
j
ix
y
(t)
r(t)
Make a sketch: A V_rel
v_Truck
BThe rel. velocity is:
V_Car/Truck = v_Car -vTruck
12.10 Relative (Constrained) Motion
V_truck = 60V_car = 65
Example Vector equation: Sailboat tacking at 50 deg. against Northern Wind
(blue vector)
BoatWindBoatWind VVV /
We solve Graphically (Vector Addition)
Example Vector equation: Sailboat tacking at 50 deg. against Northern Wind
BoatWindBoatWind VVV /
An observer on land (fixed Cartesian Reference) sees Vwind and vBoat .
Land
ABAB VVV /
Plane Vector Addition is two-dimensional.
12.10 Relative (Constrained) Motion
vB
vA
vB/A
Example cont’d: Sailboat tacking against Northern Wind
BoatWindBoatWind VVV /
2. Vector equation (1 scalar eqn. each in i- and j-direction). Solve using the given data (Vector Lengths and orientations) and Trigonometry
500
150
i
Chapter 12.10 Relative Motion
BABA rrr /
Vector Addition
BABA VVV /
Differentiating gives:
ABAB VVV /
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