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.
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.
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
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
‘e’ denotes unit vector
(‘u’ in Hibbeler)
‘e’ denotes unit vector
(‘u’ in Hibbeler)
12.8 Polar coordinates
Polar coordinates
‘e’ denotes unit vector
(‘u’ in Hibbeler)
Polar coordinates
‘e’ denotes unit vector
(‘u’ in Hibbeler)
12.8 Polar coordinates
In a polar coordinate system, the velocity
vector .can be written as
v
=
v
u
+
v
u
=
r
r
θ
θ
.
.
rur +ru. The term  is called
A) transverse velocity.
B) radial velocity.
C) angular velocity.
D) angular acceleration
.
.
.
.
.
.
.
.
.
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
Result
A
Given: vectors A and B
as shown. The RESULT
B vector is:
•(A) RESULT = A - B
•(B) RESULT = A + B
•(C) None of the above
Result
A
B
Given: vectors A and B
as shown. The RESULT
vector is:
•(A) RESULT = A - B
•(B) RESULT = A + B
•(C) None of the above
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
12.10 Relative (Constrained) Motion
Make a sketch:
A
V_river
v_boat
B
The velocity is:
(A)V_total = v+boat – v_river
(B)V_total = v+boat + v_river
12.10 Relative (Constrained) Motion
Make a sketch:
A
V_river
v_boat
B
The velocity is:
(A)V_total = v+boat – v_river
(B)V_total = v+boat + v_river
Rel. Velocity example:
Solution
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