Announcements
• Week-of-prayer schedule (10:45-11:30)
Introduction and Rectilinear Kinematics:
Continuous Motion - Sections 12.1-2
Today’s Objectives:
Students will be able to find the
kinematic quantities (position,
displacement, velocity, and
acceleration) of a particle traveling
along a straight path.
Engr222 Spring 2004 Chapter 12
In-Class Activities:
• Reading quiz
• Applications
• Relations between s(t), v(t), and
a(t) for general rectilinear
motion
• Relations between s(t), v(t),
and a(t) when acceleration is
constant
• Concept quiz
• Group problem solving
• Attention quiz
1
Reading Quiz
1. In dynamics, a particle is assumed to have _________.
A) both translation and rotational motions
B) only a mass
C) a mass but the size and shape cannot be neglected
D) no mass or size or shape, it is just a point
2. The average speed is defined as __________.
A) ∆r/∆t
B) ∆s/∆t
C) sT/∆t
D) None of the above
Applications
The motion of large objects,
such as rockets, airplanes, or
cars, can often be analyzed
as if they were particles.
Why?
If we measure the altitude
of this rocket as a function
of time, how can we
determine its velocity and
acceleration?
Engr222 Spring 2004 Chapter 12
2
Applications - continued
A train travels along a straight length of track.
Can we treat the train as a particle?
If the train accelerates at a constant rate, how can we
determine its position and velocity at some instant?
An Overview of Mechanics
Mechanics: the study of how bodies
react to forces acting on them
Statics: the study
of bodies in
equilibrium
Engr222 Spring 2004 Chapter 12
Dynamics:
1. Kinematics – concerned
with the geometric aspects of
motion
2. Kinetics - concerned with
the forces causing the motion
3
Position and Displacement
A particle travels along a straight-line path
defined by the coordinate axis s.
The position of the particle at any instant,
relative to the origin, O, is defined by the
position vector r, or the scalar s. Scalar s
can be positive or negative. Typical units
for r and s are meters (m) or feet (ft).
The displacement of the particle is
defined as its change in position.
Vector form: ∆ r = r’ - r
Scalar form: ∆ s = s’ - s
The total distance traveled by the particle, sT, is a positive scalar
that represents the total length of the path over which the particle
travels.
Velocity
Velocity is a measure of the rate of change in the position of a particle.
It is a vector quantity (it has both magnitude and direction). The
magnitude of the velocity is called speed, with units of m/s or ft/s.
The average velocity of a particle during a
time interval ∆t is
vavg = ∆r/∆t
The instantaneous velocity is the time-derivative of position:
v = dr/dt
Speed is the magnitude of velocity: v = ds/dt
Average speed is the total distance traveled divided by elapsed time:
(vsp)avg = sT/ ∆t
Engr222 Spring 2004 Chapter 12
4
Acceleration
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. Typical units are m/s2 or ft/s2.
The instantaneous acceleration is the time
derivative of velocity.
Vector form: a = dv/dt
Scalar form: a = dv/dt = d2s/dt2
Acceleration can be positive (speed
increasing) or negative (speed decreasing).
As the book indicates, the derivative equations for velocity and
acceleration can be manipulated to get
a ds = v dv
Summary of Kinematic Relationships:
Rectilinear Motion
• Differentiate position to get velocity and acceleration.
v = ds/dt ;
a = dv/dt or a = v dv/ds
• Integrate acceleration for velocity and position.
Position:
Velocity:
v
vo
t
v
s
s
o
vo
so
so
dv = a dt or v dv = a ds
t
ds = v dt
o
• Note that so and vo represent the initial position and
velocity of the particle at t = 0.
Engr222 Spring 2004 Chapter 12
5
Constant Acceleration
The three kinematic equations can be integrated for the special case
when acceleration is constant (a = ac) to obtain very useful equations.
A common example of constant acceleration is gravity; i.e., a body
freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2
ft/s2 downward. These equations are:
v
t
dv = ac dt
vo
o
s
t
ds = v dt
so
v
yields
v = vo + act
yields
s = s o + v ot + (1/2)a ct 2
yields
v2 = (vo )2 + 2ac(s - so)
o
s
v dv = ac ds
vo
so
Example
Given: A motorcyclist travels along a straight road at a speed
of 27 m/s. When the brakes are applied, the
motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it stops.
Plan: Establish the positive coordinate s in the direction the
motorcycle is traveling. Since the acceleration is given
as a function of time, integrate it once to calculate the
velocity and again to calculate the position.
Engr222 Spring 2004 Chapter 12
6
Solution:
Example - continued
1) Integrate acceleration to determine the vvelocity.
t
=
dv
(−6t )dt
a = dv / dt => dv = a dt =>
vo
o
=> v – vo = -3t2 => v = -3t2 + vo
2) We can now determine the amount of time required for
the motorcycle to stop (v = 0). Use vo = 27 m/s.
0 = -3t2 + 27 => t = 3 s
3) Now calculate the distance traveled in 3s by integrating the
velocity using so = 0:
s
t
=
ds
( −3t 2 + vo) dt
v = ds / dt => ds = v dt =>
so
o
=> s – so = -t3 + vot
=> s – 0 = -(3)3 + (27)(3) => s = 54 m
3 m/s
Concept Quiz
5 m/s
t=2s
t=7s
1. A particle moves along a horizontal path with its velocity
varying with time as shown. The average acceleration of the
particle is _________.
A) 0.4 m/s2
B) 0.4 m/s2
C) 1.6 m/s2
D) 1.6 m/s2
2. A particle has an initial velocity of 30 ft/s to the left. If it
then passes through the same location 5 seconds later with a
velocity of 50 ft/s to the right, the average velocity of the
particle during the 5 s time interval is _______.
A) 10 ft/s
B) 40 ft/s
C) 16 m/s
D) 0 ft/s
Engr222 Spring 2004 Chapter 12
7
Group Problem Solving
Given:Ball A is released from rest
at a height of 40 ft at the
same time that ball B is
thrown upward, 5 ft from the
ground. The balls pass one
another at a height of 20 ft.
Find: The speed at which ball B
was thrown upward.
Plan: Both balls experience a constant downward acceleration
of 32.2 ft/s2. Apply the formulas for constant
acceleration, with ac = -32.2 ft/s2.
Group Problem Solving - continued
Solution:
1) First consider ball A. With the origin defined at the ground,
ball A is released from rest (vA)o = 0 at a height of 40 ft
(sA )o = 40 ft. Calculate the time required for ball A to drop to
20 ft (sA = 20 ft) using a position equation.
sA = (sA )o + (vA)ot + (1/2)act2
20 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t2) => t = 1.115 s
2) Now consider ball B. It is throw upward from a height of 5 ft
(sB)o = 5 ft. It must reach a height of 20 ft (sB = 20 ft) at the
same time ball A reaches this height (t = 1.115 s). Apply the
position equation again to ball B using t = 1.115s.
sB = (sB)o + (vB)ot + (1/2) ac t2
20 ft = 5 + (vB)o(1.115) + (1/2)(-32.2)(1.115)2
=> (vB)o = 31.4 ft/s
Engr222 Spring 2004 Chapter 12
8
Attention Quiz
1. A particle has an initial velocity of 3 ft/s to the left at
s0 = 0 ft. Determine its position when t = 3 s if the
acceleration is 2 ft/s2 to the right.
A) 0.0 ft
C) 18.0 ft
B) 6.0 ft
D) 9.0 ft
2. A particle is moving with an initial velocity of v = 12 ft/s
and constant acceleration of 3.78 ft/s2 in the same direction
as the velocity. Determine the distance the particle has
traveled when the velocity reaches 30 ft/s.
A) 50 ft
C) 150 ft
B) 100 ft
D) 200 ft
Textbook Problem 12-
Engr222 Spring 2004 Chapter 12
9
Announcements
• Correction - example 1 (page 13) from last lecture
• Mid-term #1 will be Tuesday, April 20 – more details
Friday
Rectilinear Kinematics: Erratic Motion
Today’s Objectives:
Section 12.3
Students will be able to
determine position, velocity,
and acceleration of a particle
using graphs.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• s-t, v-t, a-t, v-s, and a-s
diagrams
• Concept quiz
• Group problem solving
• Attention quiz
Engr222 Spring 2004 Chapter 12
10
Reading Quiz
1. The slope of a v-t graph at any instant represents instantaneous
A) velocity.
B) acceleration.
C) position.
D) jerk.
2. Displacement of a particle in a given time interval equals the
area under the ___ graph during that time.
A) a-t
B) a-s
C) v-t
C) s-t
Applications
In many experiments, a
velocity versus position (v-s)
profile is obtained.
If we have a v-s graph for
the rocket sled, can we
determine its acceleration at
position s = 300 meters ?
How?
Engr222 Spring 2004 Chapter 12
11
Graphing
Graphing provides a good way to handle complex
motions that would be difficult to describe with
formulas. Graphs also provide a visual description of
motion and reinforce the calculus concepts of
differentiation and integration as used in dynamics.
The approach builds on the facts that slope and
differentiation are linked and that integration can be
thought of as finding the area under a curve.
S-T Graph
Plots of position vs. time can be
used to find velocity vs. time
curves. Finding the slope of the
line tangent to the motion curve at
any point is the velocity at that
point (or v = ds/dt).
Therefore, the v-t graph can be
constructed by finding the slope at
various points along the s-t graph.
Engr222 Spring 2004 Chapter 12
12
V-T Graph
Plots of velocity vs. time can be used to
find acceleration vs. time curves.
Finding the slope of the line tangent to
the velocity curve at any point is the
acceleration at that point (or a = dv/dt).
Therefore, the a-t graph can be
constructed by finding the slope at
various points along the v-t graph.
Also, the distance moved
(displacement) of the particle is the
area under the v-t graph during time ∆t.
A-T Graph
Given the a-t curve, the change
in velocity (∆v) during a time
period is the area under the a-t
curve.
So we can construct a v-t graph
from an a-t graph if we know the
initial velocity of the particle.
Engr222 Spring 2004 Chapter 12
13
A-S Graph
A more complex case is presented by
the a-s graph. The area under the
acceleration versus position curve
represents the change in velocity
(recall a ds = v dv ).
½ (v1² – vo²) =
s2
s1
a ds = area under the
a-s graph
This equation can be solved for v1, allowing you to solve for
the velocity at a point. By doing this repeatedly, you can
create a plot of velocity versus distance.
V-S Graph
Another complex case is presented
by the v-s graph. By reading the
velocity v at a point on the curve
and multiplying it by the slope of
the curve (dv/ds) at this same point,
we can obtain the acceleration at
that point.
a = v (dv/ds)
Thus, we can obtain a plot of a vs. s
from the v-s curve.
Engr222 Spring 2004 Chapter 12
14
Example
Given: v-t graph for a train moving between two stations
Find: a-t graph and s-t graph over this time interval
Think about your plan of attack for the problem!
Example - continued
Solution: For the first 30 seconds the slope is constant
and is equal to:
a0-30 = dv/dt = 40/30 = 4/3 ft/s2
Similarly,
a30-90 = 0
and a90-120 = -4/3 ft/s2
a(ft/s2)
4
3
t(s)
-4
3
Engr222 Spring 2004 Chapter 12
15
Example - continued
s(ft)
The area under the v-t graph
represents displacement.
3600
3000
∆s0-30 = ½ (40)(30) = 600 ft
∆s30-90 = (60)(40) = 2400 ft
∆s90-120 = ½ (40)(30) = 600 ft
600
30
90
t(s)
120
Concept Quiz
1. If a particle starts from rest and
accelerates according to the graph
shown, the particle’s velocity at
t = 20 s is
A) 200 m/s
B) 100 m/s
C) 0
D) 20 m/s
2. The particle in Problem 1 stops moving at t = _______.
A) 10 s
B) 20 s
C) 30 s
D) 40 s
Engr222 Spring 2004 Chapter 12
16
Group Problem Solving
Given: The v-t graph shown
Find: The a-t graph, average
speed, and distance
traveled for the 30 s
interval
Plan: Find slopes of the curves and draw the v-t graph. Find
the area under the curve--that is the distance traveled.
Finally, calculate average speed (using basic
definitions!).
Group Problem Solving - continued
Solution:
For 0
t
For 10
10
t
30
a = dv/dt = 0.8 t ft/s²
a = dv/dt = 1 ft/s²
a(ft/s²)
8
1
10
Engr222 Spring 2004 Chapter 12
30
t(s)
17
Group Problem Solving - continued
∆s0-10 = v dt = (1/3) (.4)(10)3 = 400/3 ft
∆s10-30 = v dt = (0.5)(30)2 + 30(30) – 0.5(10)2 – 30(10)
= 1000 ft
s0-30 = 1000 + 400/3 = 1133.3 ft
vavg(0-30) = total distance / time
= 1133.3/30
= 37.78 ft/s
Attention Quiz
1. If a car has the velocity curve shown, determine the time t
necessary for the car to travel 100 meters. v
A) 8 s
B) 4 s
C) 10 s
D) 6 s
75
6s
t
2. Select the correct a-t graph for the velocity curve shown.
a
A)
a
t
B)
a
C)
Engr222 Spring 2004 Chapter 12
v
t
a
t
D)
t
t
18
Textbook Problem 12-
Announcements
• Correction - example 1 (page 13) from last lecture
• Mid-term #1 will be Tuesday, April 20 – more details
Friday
• Section 12-3 will not be covered
• HW #7 can be turned in on Friday
Engr222 Spring 2004 Chapter 12
19
Curvilinear Motion: Rectangular Components
Sections 12.4-5
Today’s Objectives:
Students will be able to:
a) Describe the motion of a
particle traveling along
a curved path.
b) Relate kinematic
quantities in terms of
the rectangular
components of the
vectors.
In-Class Activities:
• Reading quiz
• Applications
• General curvilinear motion
• Rectangular components of
kinematic vectors
• Concept quiz
• Group problem solving
• Attention quiz
Reading Quiz
1. In curvilinear motion, the direction of the instantaneous
velocity is always
A)
B)
C)
D)
tangent to the hodograph.
perpendicular to the hodograph.
tangent to the path.
perpendicular to the path.
2. In curvilinear motion, the direction of the instantaneous
acceleration is always
A)
B)
C)
D)
tangent to the hodograph.
perpendicular to the hodograph.
tangent to the path.
perpendicular to the path.
Engr222 Spring 2004 Chapter 12
20
Applications
The path of motion of each plane in
this formation can be tracked with
radar and their x, y, and z coordinates
(relative to a point on earth) recorded
as a function of time.
How can we determine the velocity
and acceleration of each plane at any
instant?
Should they be the same for each
aircraft?
Applications - continued
A roller coaster car travels down
a fixed, helical path at a constant
speed.
How can we determine its
position and acceleration at
any instant?
If you are designing the track, why is it important to be
able to predict the acceleration of the car?
Engr222 Spring 2004 Chapter 12
21
Position and Displacement
A particle moving along a curved path undergoes curvilinear motion.
Since the motion is often three-dimensional, vectors are used to
describe the motion.
A particle moves along a curve
defined by the path function, s.
The position of the particle at any instant is designated by the vector
r = r(t). Both the magnitude and direction of r may vary with time.
If the particle moves a distance ∆s along the
curve during time interval ∆t, the
displacement is determined by vector
subtraction: ∆ r = r’ - r
Velocity
Velocity represents the rate of change in the position of a
particle.
The average velocity of the particle
during the time increment ∆t is
vavg = ∆r/∆t .
The instantaneous velocity is the
time derivative of position
v = dr/dt .
The velocity vector, v, is always
tangent to the path of motion.
The magnitude of v is called the speed. Since the arc length ∆s
approaches the magnitude of ∆r as t 0, the speed can be
obtained by differentiating the path function (v = ds/dt). Note
that this is not a vector!
Engr222 Spring 2004 Chapter 12
22
Acceleration
Acceleration represents the rate of change in the
velocity of a particle.
If a particle’s velocity changes from v to v’ over a
time increment ∆t, the average acceleration during
that increment is:
aavg = ∆v/∆t = (v - v’)/∆t
The instantaneous acceleration is the timederivative of velocity:
a = dv/dt = d2r/dt2
A plot of the locus of points defined by the arrowhead
of the velocity vector is called a hodograph. The
acceleration vector is tangent to the hodograph, but
not, in general, tangent to the path function.
Rectangular Components - Position
It is often convenient to describe the motion of a particle in
terms of its x, y, and z or rectangular components, relative to a
fixed frame of reference.
The position of the particle can be
defined at any instant by the
position vector
r=xi+yj+zk
The x, y, and z components may all
be functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
The magnitude of the position vector is: r = (x2 + y2 + z2)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
Engr222 Spring 2004 Chapter 12
23
Rectangular Components - Velocity
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt
Since the unit vectors i, j, and k are constant in magnitude and
direction, this equation reduces to v = vx i + vy j + vz k
•
•
•
where vx = x = dx/dt, vy = y = dy/dt, vz = z = dz/dt
The magnitude of the velocity
vector is
v = [(vx)2 + (vy)2 + (vz)2]0.5
The direction of v is tangent
to the path of motion.
Rectangular Components - Acceleration
The acceleration vector is the time derivative of the
velocity vector (second derivative of the position vector):
a = dv/dt = d2r/dt2 = ax i + ay j + az k
where
•
•
•
••
••
ax = vx = x
= dvx /dt, ay = vy = y
= dvy /dt,
••
az = vz = z = dvz /dt
The magnitude of the acceleration vector is
a = [(ax)2 + (ay)2 + (az)2 ]0.5
The direction of a is usually
not tangent to the path of the
particle.
Engr222 Spring 2004 Chapter 12
24
Example
Given: The motion of two particles (A and B) is described
by the position vectors:
rA = [3t i + 9t(2 – t) j] m
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
Find: The point at which the particles collide and their
speeds just before the collision.
Plan: 1) The particles will collide when their position
vectors are equal, or rA = rB .
2) Their speeds can be determined by differentiating
the position vectors.
Example - continued
rA = [3t i + 9t(2 – t) j] m
Solution:
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
1) The point of collision requires that rA = rB,
so xA = xB and yA = yB .
x-components:
Simplifying:
Solving:
=> t = 2 or 1 s
3t = 3(t2 – 2t + 2)
t2 – 3t + 2 = 0
t = {3 ± [32 – 4(1)(2)]0.5}/2(1)
y-components: 9t(2 – t) = 3(t – 2)
Simplifying: 3t2 – 5t – 2 = 0
Solving:
t = {5 ± [52 – 4(3)(–2)]0.5}/2(3)
=> t = 2 or – 1/3 s
So, the particles collide when t = 2 s. Substituting this value into rA
or rB yields
xA = xB = 6 m
Engr222 Spring 2004 Chapter 12
and yA = yB = 0
25
Example - continued
rA = [3t i + 9t(2 – t) j] m
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
2) Differentiate rA and rB to get the velocity vectors.
vA = drA/dt =
.
.
x. A i + yA j
= [3i + (18 – 18t) j] m/s
At t = 2 s: vA = [3i – 18j] m/s
vB = drB/dt = x•Bi + y•Bj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
Speed is the magnitude of the velocity vector:
vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5 = 6.71 m/s
Concept Quiz
1. If the position of a particle is defined by
r = [(1.5t2 + 1) i + (4t – 1) j] (m), its speed at t = 1 s is
A) 2 m/s
B) 3 m/s
C) 5 m/s
D) 7 m/s
2. The path of a particle is defined by y = 0.5x2. If the
component of its velocity along the x-axis at x = 2 m is
vx = 1 m/s, its velocity component along the y-axis at this
position is
A) 0.25 m/s
B) 0.5 m/s
C) 1 m/s
D) 2 m/s
Engr222 Spring 2004 Chapter 12
26
Group Problem Solving
Given: A particle travels along a path described by the
parabola y = 0.5x2. The x-component of velocity is
given by vx = (5t) ft/s. When t = 0, x = y = 0.
Find: The particle’s distance from the origin and the
magnitude of its acceleration when t = 1 s.
Plan: Note that vx is given as a function of time.
1) Determine the x-component of position and
acceleration by integrating and differentiating vx,
respectively.
2) Determine the y-component of position from the
parabolic equation and differentiate to get ay.
3) Determine the magnitudes of the position and
acceleration vectors.
Group Problem Solving - continued
Solution:
1) x-components:
•
Velocity:
vx = x = dx/dt = (5t) ft/s
x
dx =
Position:
0
t
5 t dt
=> x = (5/2)t2 = (2.5t2) ft
0
••
•
Acceleration: ax = x = vx = d/dt (5t) = 5 ft/s2
2) y-components:
Position:
y = 0.5x2 = 0.5(2.5t2)2 = (3.125t4) ft
Velocity:
vy = dy/dt = d (3.125t4) /dt = (12.5t3) ft/s
Acceleration: ay = vy = d (12.5t3) /dt = (37.5t2) ft/s2
Engr222 Spring 2004 Chapter 12
27
Group Problem Solving - continued
3) The distance from the origin is the magnitude of the
position vector:
r = x i + y j = [2.5t2 i + 3.125t4 j] ft
At t = 1 s, r = (2.5 i + 3.125 j) ft
Distance: d = r = (2.52 + 3.1252) 0.5 = 4.0 ft
The magnitude of the acceleration vector is calculated as:
Acceleration vector: a = [5 i + 37.5t2 j ] ft/s2
Magnitude: a = (52 + 37.52)0.5 = 37.8 ft/s2
Attention Quiz
1. If a particle has moved from A to B along the circular path
in 4s, what is the average velocity of the particle ?
y
A) 2.5 i m/s
B) 2.5 i +1.25j m/s
C) 1.25 π i m/s
R=5m
A
x
B
D) 1.25 π j m/s
2. The position of a particle is given as r = (4t2 i – 2t j) m.
Determine the particle’s acceleration.
A) (4 i +8 j ) m/s2
B) (8 i -16 j ) m/s2
C) (8 i) m/s2
D) (8 j ) m/s2
Engr222 Spring 2004 Chapter 12
28
Textbook Problem 12-
Announcements
• Mid-term #1 will be Wednesday, April 21
• Closed book. Cheat sheet (review sheet) and calculator
allowed.
• Review sheet on class web page
Engr222 Spring 2004 Chapter 12
29
Motion of a Projectile - Section 12.6
Today’s Objectives:
Students will be able to
analyze the free-flight motion of
a projectile.
In-Class Activities:
• Reading quiz
• Applications
• Kinematic equations
for projectile motion
• Concept quiz
• Group problem solving
• Attention quiz
Reading Quiz
1. The downward acceleration of an object in free-flight motion is
A) zero
B) increasing with time
C) 9.81 m/s2
D) 9.81 ft/s2
2. The horizontal component of velocity remains _________
during a free-flight motion.
A) zero
B) constant
C) at 9.81 m/s2
D) at 32.2 ft/s2
Engr222 Spring 2004 Chapter 12
30
Applications
A kicker should know at what angle, θ, and initial velocity,
vo, he must kick the ball to make a field goal.
For a given kick “strength”, at what angle should the ball
be kicked to get the maximum distance?
Applications - continued
A fireman wishes to know the maximum height on the wall
he can project water from the hose. At what angle, θ, should
he hold the hose?
Engr222 Spring 2004 Chapter 12
31
Concept of Projectile Motion
Projectile motion can be treated as two rectilinear motions, one in
the horizontal direction experiencing zero acceleration and the other
in the vertical direction experiencing constant acceleration (i.e.,
gravity).
For illustration, consider the two balls on the
left. The red ball falls from rest, whereas the
yellow ball is given a horizontal velocity. Each
picture in this sequence is taken after the same
time interval. Notice both balls are subjected to
the same downward acceleration since they
remain at the same elevation at any instant.
Also, note that the horizontal distance between
successive photos of the yellow ball is constant
since the velocity in the horizontal direction is
constant.
Kinematic Equations – Horizontal Motion
Since ax = 0, the velocity in the horizontal direction remains
constant (vx = vox) and the position in the x direction can be
determined by:
x = xo + (vox)(t)
Why is ax equal to zero (assuming movement through the air)?
Engr222 Spring 2004 Chapter 12
32
Kinematic Equations – Vertical Motion
Since the positive y-axis is directed upward, ay = -g. Application
of the constant acceleration equations yields:
vy = voy – g(t)
y = yo + (voy)(t) – ½g(t)2
vy2 = voy2 – 2g(y – yo)
For any given problem, only two of these three
equations can be used. Why?
Example
Given: vo and
Find: The equation that defines
y as a function of x.
Plan: Eliminate time from the
kinematic equations.
Solution: Using
vx = vo cos
We can write: x = (vo cos )t
or
and
t =
vy = vo sin
x
vo cos
y = (vo sin )t – ½ g(t)2
By substituting for t:
y = (vo sin )
Engr222 Spring 2004 Chapter 12
(
x
vo cos
) ( )(
–
g
2
x
vo cos
2
)
33
Example - continued
Simplifying the last equation, we get:
y = (x tanθ) –
(
g x2
2vo2
)
(1 + tan2θ)
The above equation is called the “path equation” which describes the
path of a particle in projectile motion. The equation shows that the
path is parabolic.
Example
Given: Snowmobile is going 15
m/s at point A.
Find: The horizontal distance it
travels (R) and the time in
the air.
Solution:
First, place the coordinate system at point A. Then write the
equation for horizontal motion.
xB = xA + vAxtAB and vAx = 15 cos 40° m/s
Now write a vertical motion equation. Use the distance equation.
yB = yA + vAytAB – 0.5gctAB2 vAy = 15 sin 40° m/s
Note that xB = R, xA = 0, yB = -(3/4)R, and yA = 0.
Solving the two equations together (two unknowns) yields
R = 42.8 m tAB = 3.72 s
Engr222 Spring 2004 Chapter 12
34
Concept Quiz
1. In a projectile motion problem, what is the maximum
number of unknowns that can be solved?
A) 1
B) 2
C) 3
D) 4
2. The time of flight of a projectile, fired over level ground
with initial velocity Vo at angle , is equal to
A) (vo sin θ)/g
B) (2vo sin θ)/g
C) (vo cos θ)/g
D) (2vo cos θ)/g
Group Problem Solving
Given: Skier leaves the ramp
at θA = 25o and hits the
slope at B.
Find: The skier’s initial speed vA.
Plan: Establish a fixed x,y coordinate system (in the
solution here, the origin of the coordinate system is
placed at A). Apply the kinematic relations in x- and
y- directions.
Engr222 Spring 2004 Chapter 12
35
Group Problem Solving - continued
Solution:
Motion in x-direction:
Using
xB = xA + vox(tAB)
tAB=
(4/5)100
vA (cos 25)
=
88.27
vA
Motion in y-direction:
yB = yA + voy(tAB) – ½ g(tAB)2
Using
-64 = 0 + vA(sin 45)
80
vA (cos 25)
– ½ (9.81)
88.27
vA
2
vA = 19.42 m/s
Attention Quiz
1. A projectile is given an initial velocity
vo at an angle φ above the horizontal.
The velocity of the projectile when it
hits the slope is ____________ the
initial velocity vo.
A) less than
C) greater than
B) equal to
D) None of the above.
2. A particle has an initial velocity vo at angle θ with respect to the
horizontal. The maximum height it can reach is when
A) θ = 30°
B) θ = 45°
C) θ = 60°
D) θ = 90°
Engr222 Spring 2004 Chapter 12
36
Textbook Problem 12-95
Determine the horizontal velocity va of a tennis ball at A
so that it just clears the net at B. Also, find the distance s
where the ball strikes the ground.
Announcements
Engr222 Spring 2004 Chapter 12
37
Curvilinear Motion: Normal and Tangential
Components - Section 12.7
Today’s Objectives:
Students will be able to
determine the normal and
tangential components of
velocity and acceleration of a
particle traveling along a
curved path.
In-Class Activities:
• Reading quiz
• Applications
• Normal and tangential
components of velocity
and acceleration
• Special cases of motion
• Concept quiz
• Group problem solving
• Attention quiz
Reading Quiz
1. If a particle moves along a curve with a constant speed, then
its tangential component of acceleration is
A) positive.
B) negative.
C) zero.
D) constant.
2. The normal component of acceleration represents
A) the time rate of change in the magnitude of the velocity.
B) the time rate of change in the direction of the velocity.
C) magnitude of the velocity.
D) direction of the total acceleration.
Engr222 Spring 2004 Chapter 12
38
Applications
Cars traveling along a clover-leaf
interchange experience an
acceleration due to a change in
speed as well as due to a change in
direction of the velocity.
If the car’s speed is increasing at a
known rate as it travels along a
curve, how can we determine the
magnitude and direction of its total
acceleration?
Why would you care about the total acceleration of the car?
Applications - continued
A motorcycle travels up a
hill for which the path can
be approximated by a
function y = f(x).
If the motorcycle starts from rest and increases its speed at a
constant rate, how can we determine its velocity and
acceleration at the top of the hill?
How would you analyze the motorcycle'
s “flight” at the top of
the hill?
Engr222 Spring 2004 Chapter 12
39
Normal and Tangential Components
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, ρ, 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.
Engr222 Spring 2004 Chapter 12
40
Velocity in the N-T Coordinate System
The velocity vector is always
tangent to the path of motion
(t-direction).
The magnitude is determined by taking the time derivative of
the path function, s(t).
.
v = vut where v = s = ds/dt
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.
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 = vut + (v2/ρ)un = atut + anun.
Engr222 Spring 2004 Chapter 12
41
Acceleration in the N-T Coordinate System
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/ρ
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5
Special Cases of Motion
There are some special cases of motion to consider.
1) The particle moves along a straight line.
.
ρ
∞ => an = v2/ρ = 0 => a = at = v
The tangential component represents the time rate of
change in the magnitude of the velocity.
2) The particle moves along a curve at constant speed.
.
at = v = 0 => a = an = v2/ρ
The normal component represents the time rate of change
in the direction of the velocity.
Engr222 Spring 2004 Chapter 12
42
Special Cases of Motion - continued
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vot + (1/2)(at)ct2
v = vo + (at)ct
v2 = (vo)2 + 2(at)c(s – so)
As before, so and vo are the initial position and velocity of the
particle at t = 0. How are these equations related to projectile
motion equations? Why?
4) The particle moves along a path expressed as y = f(x).
The radius of curvature, ρ, at any point on the path can be
calculated from
[ 1 + (dy/dx)2 ]3/2
ρ = ________________
d2y/dx 2
Three-Dimensional Motion
If a particle moves along a space
curve, the n and t axes are defined as
before. At any point, the t-axis is
tangent to the path and the n-axis
points toward the center of curvature.
The plane containing the n and t axes
is called the osculating plane.
A third axis can be defined, called the binomial axis, b. The
binomial unit vector, ub, is directed perpendicular to the osculating
plane, and its sense is defined by the cross product ub = ut x un.
There is no motion, thus no velocity or acceleration, in the
binomial direction.
Engr222 Spring 2004 Chapter 12
43
Example
Given: Starting from rest, a motorboat
travels around a circular path of
ρ = 50 m at a speed that
increases with time,
v = (0.2 t2) m/s.
Find: The magnitudes of the boat’s
velocity and acceleration at
the instant t = 3 s.
Plan: The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 3s using v(t).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
Example - continued
Solution:
1) The velocity vector is v = v ut , where
the magnitude is given by v = (0.2t2)
m/s. At t = 3s:
v = 0.2t2 = 0.2(3)2 = 1.8 m/s
.
2) The acceleration vector is a = atut + anun = vut + (v2/ρ)un.
.
Tangential component: at = v = d(.2t2)/dt = 0.4t m/s2
At t = 3s: at = 0.4t = 0.4(3) = 1.2 m/s2
Normal component: an = v2/ρ = (0.2t2)2/(ρ) m/s2
At t = 3s: an = [(0.2)(32)]2/(50) = 0.0648 m/s2
The magnitude of the acceleration is
a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2
Engr222 Spring 2004 Chapter 12
44
Concept Quiz
1. A particle traveling in a circular path of radius 300 m has an
instantaneous velocity of 30 m/s and its velocity is
increasing at a constant rate of 4 m/s2. What is the
magnitude of its total acceleration at this instant?
A) 3 m/s2
B) 4 m/s2
C) 5 m/s2
D) -5 m/s2
2. If a particle moving in a circular path of radius 5 m has a
velocity function v = 4t2 m/s, what is the magnitude of its
total acceleration at t = 1 s?
A) 8 m/s
B) 8.6 m/s
C) 3.2 m/s
D) 11.2 m/s
Group Problem Solving
Given: A jet plane travels along a vertical
parabolic path defined by the
equation y = 0.4x2. At point A, the
jet has a speed of 200 m/s, which is
increasing at the rate of 0.8 m/s2.
Find: The magnitude of the plane’s
acceleration when it is at point A.
Plan: 1) The change in the speed of the plane (0.8 m/s2) is the
tangential component of the total acceleration.
2) Calculate the radius of curvature of the path at A.
3) Calculate the normal component of acceleration.
4) Determine the magnitude of the acceleration vector.
Engr222 Spring 2004 Chapter 12
45
Group Problem Solving - continued
Solution:
1) The tangential component of acceleration is the rate of
.
increase of the plane’s speed, so at = v = 0.8 m/s2.
2) Determine the radius of curvature at point A (x = 5 km):
dy/dx = d(0.4x2)/dx = 0.8x, d2y/dx2 = d (0.8x)/dx = 0.8
At x =5 km, dy/dx = 0.8(5) = 4, d2y/dx2 = 0.8
[ 1 + (dy/dx) ] = [1 + (4)2]3/2/(0.8) = 87.62 km
=> ρ = ________________
2
3/2
d2y/dx 2
3) The normal component of acceleration is
an = v2/ρ = (200)2/(87.62 x 103) = 0.457 m/s2
4) The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5 = [(0.8)2 + (0.457)2]0.5 = 0.921 m/s2
Attention Quiz
1. The magnitude of the normal acceleration is
A) proportional to radius of curvature.
B) inversely proportional to radius of curvature.
C) sometimes negative.
D) zero when velocity is constant.
2. The directions of the tangential acceleration and velocity are
always
A) perpendicular to each other. B) collinear.
C) in the same direction.
Engr222 Spring 2004 Chapter 12
D) in opposite directions.
46
Textbook Problem 12-
Announcements
Engr222 Spring 2004 Chapter 12
47
Curvilinear Motion: Cylindrical Coordinates
Section 12.8
Today’s Objectives:
Students will be able to
determine velocity and
acceleration components
using cylindrical coordinates.
In-Class Activities:
• Reading quiz
• Applications
• Velocity Components
• Acceleration Components
• Concept quiz
• Group problem solving
• Attention quiz
Reading Quiz
1. In a polar coordinate system, the velocity
vector can
.
. be
.
written as v = vrur + v u = rur + rθuθ. The term θ is
called
A) transverse velocity.
B) radial velocity.
C) angular velocity.
D) angular acceleration.
2. The speed of a particle in a cylindrical coordinate system is
.
B) rθ
.
A) r
C)
Engr222 Spring 2004 Chapter 12
.
.
(rθ)2 + (r)2
D)
.
.
.
(rθ)2 + (r)2 + (z)2
48
Applications
The cylindrical coordinate
system is used in cases
where the particle moves
along a 3-D curve.
In the figure shown, the boy
slides down the slide at a
constant speed of 2 m/s.
How fast is his elevation
from the ground
. changing
(i.e., what is z )?
Applications - continued
A polar coordinate system is a 2-D representation of the
cylindrical coordinate system.
When the particle moves in a plane (2-D), and the radial
distance, r, is not constant, the polar coordinate system can
be used to express the path of motion of the particle.
Engr222 Spring 2004 Chapter 12
49
Position – Polar Coordinates
We can express the location of P in polar coordinates as r = rur.
Note that the radial direction, r, extends outward from the fixed
origin, O, and the transverse coordinate, θ, is measured counterclockwise (CCW) from the horizontal.
Velocity – Polar Coordinates
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
du
.
v = rur + r dt r
Using the chain rule:
dur/dt = (dur/dθ)(dθ/dt)
.
We can prove that dur/d
θ
=
u
so
du
/dt
=
θ
u
r
.
.
Therefore: v = rur + rθu
.
Thus, the velocity vector has two components:
r,
.
called the radial component, and rθ, called the
transverse component. The speed of the particle at
any given instant is the sum of the squares of both
components or
v=
Engr222 Spring 2004 Chapter 12
.
.
(r θ )2 + ( r )2
50
Acceleration – Polar Coordinates
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rur + rθu )
After manipulation, the acceleration can be
expressed as
..
..
.. .
a = (r – rθ2)ur + (rθ + 2rθ)u
.. .
The term (r – rθ2) is the radial acceleration
or ar.
..
..
The term (rθ + 2rθ) is the transverse
acceleration or aθ
..
..
.. .
The magnitude of acceleration is a = (r – rθ2)2 + (rθ + 2rθ)2
Cylindrical Coordinates
If the particle P moves along a space
curve, its position can be written as
rP = rur + zuz
Taking time derivatives and using
the chain rule:
Velocity:
.
.
.
vP = rur + rθu + zuz
..
..
.. .
..
Acceleration: aP = (r – rθ2)ur + (rθ + 2rθ)u + zuz
Engr222 Spring 2004 Chapter 12
51
Example
Given: r. = 5 cos(2θ) (m)
θ = 3t2 (rad/s)
θo = 0
Find: Velocity and acceleration at θ = 30°.
.
..
Plan: Apply chain rule to determine r and r
and evaluate at θ = 30°.
t
Solution:
θ=
At θ = 30°,
θ=
.
θ dt =
t o= 0
t
3t2 dt = t3
0
π
= t3. Therefore: t = 0.806 s.
6
.
θ = 3t2 = 3(0.806)2 = 1.95 rad/s
Example - continued
..
θ = 6t = 6(0.806) = 4.836 rad/s2
r = 5 cos(2θ) = 5 cos(60) = 2.5m
r = -10 sin(2θ)θ = -10 sin(60)(1.95) = -16.88 m/s
.
.
r = -20 cos(2θ)θ2 – 10 sin(2θ)θ
.
..
2 – 10 sin(60)(4.836)
.. = -20 cos(60)(1.95)
= -80 m/s2
Substitute in the equation
. for velocity
.
v = rur + rθu
v = -16.88ur + 2.5(1.95)u
v = (16.88)2 + (4.87)2 = 17.57 m/s
Engr222 Spring 2004 Chapter 12
52
Example - continued
Substitute in the equation for acceleration:
..
..
.. .
a = (r – rθ2)ur + (rθ + 2rθ)u
a = [-80 – 2.5(1.95)2]ur + [2.5(4.836) + 2(-16.88)(1.95)]u
a = -89.5ur – 53.7u m/s2
a = (89.5)2 + (53.7)2 = 104.4 m/s2
Concept Quiz
1. If r. is zero for a particle, the particle is
A) not moving.
B) moving in a circular path.
C) moving on a straight line.
D) moving with constant velocity.
2. If a particle moves in a circular path with constant velocity, its
radial acceleration is
A) zero.
B) ..
r.
C) -rθ. 2.
..
D) 2rθ.
Engr222 Spring 2004 Chapter 12
53
Group Problem Solving
Given: The car’s speed is constant at
1.5 m/s.
Find: The car’s acceleration (as a
vector).
Hint:
The tangent to the ramp at any
point is at an angle
12
φ = tan-1(
) = 10.81°
2π(10)
Also, what is the relationship between φ and θ?
Plan: Use cylindrical coordinates. Since r is constant, all
derivatives of r will be zero.
Solution: Since r .is constant the velocity only has 2 components:
.
vθ = rθ = v cosφ and vz = z = v sinφ
Group Problem Solving - continued
.
v cosφ
Therefore: θ = (
) = 0.147 rad/s
r
..
θ = 0
.
vz = z = v sinφ = 0.281 m/s
..
z = 0
.
..
r = r = 0
..
..
.. .
..
a = (r – rθ2)ur + (rθ + 2rθ)u + zuz
.
a = (-rθ2)ur = -10(0.147)2ur = -0.217ur m/s2
Engr222 Spring 2004 Chapter 12
54
Attention Quiz
1. The radial component of velocity of a particle moving in a
circular path is always
A) zero.
B) constant.
C) greater than its transverse component.
D) less than its transverse component.
2. The radial component of acceleration of a particle moving in
a circular path is always
A) negative.
B) directed toward the center of the path.
C) perpendicular to the transverse component of acceleration.
D) All of the above.
Textbook Problem 12-
Engr222 Spring 2004 Chapter 12
55
Announcements
Absolute Dependent Motion Analysis of Two
Particles - Section 12.9
Today’s Objectives:
Students will be able to relate
the positions, velocities, and
accelerations of particles
undergoing dependent motion.
In-Class Activities:
• Reading quiz
• Applications
• Define dependent motion
• Develop position, velocity,
and acceleration
relationships
• Concept quiz
• Group problem solving
• Attention quiz
Engr222 Spring 2004 Chapter 12
56
Reading Quiz
1. When particles are interconnected by a cable, the motions
of the particles are ______.
A) always independent
B) always dependent
C) dependent, but not always
D) None of the above
2. If the motion of one particle is dependent on that of
another particle, each coordinate axis of the particles
_______.
A) should be directed along the path of motion
B) can be directed anywhere
C) should have the same origin
D) None of the above
Applications
The cable and pulley system shown
here can be used to modify the speed
of block B relative to the speed of the
motor. It is important to relate the
various motions in order to determine
the power requirements for the motor
and the tension in the cable.
If the speed of the cable coming onto
the motor pulley is known, how can
we determine the speed of block B?
Engr222 Spring 2004 Chapter 12
57
Applications - continued
Rope and pulley arrangements
are often used to assist in lifting
heavy objects. The total lifting
force required from the truck
depends on the acceleration of
the cabinet.
How can we determine the
acceleration and velocity of
the cabinet if the acceleration
of the truck is known?
Dependent Motion
In many kinematics problems, the motion of one object will
depend on the motion of another object.
The blocks in this figure are
connected by an inextensible cord
wrapped around a pulley. If block
A moves downward along the
inclined plane, block B will move
up the other incline.
The motion of each block can be related mathematically by
defining position coordinates, sA and sB. Each coordinate axis is
defined from a fixed point or datum line, measured positive along
each plane in the direction of motion of each block.
Engr222 Spring 2004 Chapter 12
58
Dependent Motion - continued
In this example, position
coordinates sA and sB can be
defined from fixed datum lines
extending from the center of
the pulley along each incline
to blocks A and B.
If the cord has a fixed length, the position coordinates sA
and sB are related mathematically by the equation
sA + lCD + sB = lT
Here lT is the total cord length and lCD is the length of cord
passing over arc CD on the pulley.
Dependent Motion - continued
The velocities of blocks A and B
can be related by differentiating
the position equation. Note that
lCD and lT remain constant, so
dlCD/dt = dlT/dt = 0
dsA/dt + dsB/dt = 0
=>
vB = -vA
The negative sign indicates that as A moves down the incline
(positive sA direction), B moves up the incline (negative sB
direction).
Accelerations can be found by differentiating the velocity
expression. Prove to yourself that aB = -aA .
Engr222 Spring 2004 Chapter 12
59
Example
Consider a more complicated
example. Position coordinates (sA
and sB) are defined from fixed
datum lines, measured along the
direction of motion of each block.
Note that sB is only defined to the
center of the pulley above block
B, since this block moves with the
pulley. Also, h is a constant.
The red colored segments of the cord remain constant in length
during motion of the blocks.
Example - continued
The position coordinates are related
by the equation
2sB + h + sA = l
Where l is the total cord length minus
the lengths of the red segments.
Since l and h remain constant during
the motion, the velocities and
accelerations can be related by two
successive time derivatives:
2vB = -vA and 2aB = -aA
When block B moves downward (+sB), block A moves to the
left (-sA). Remember to be consistent with the sign convention!
Engr222 Spring 2004 Chapter 12
60
Dependent Motion - Procedures
These procedures can be used to relate the dependent motion of
particles moving along rectilinear paths (only the magnitudes of
velocity and acceleration change, not their line of direction).
1. Define position coordinates from fixed datum lines,
along the path of each particle. Different datum lines can
be used for each particle.
2. Relate the position coordinates to the cord length.
Segments of cord that do not change in length during the
motion may be left out.
3. If a system contains more than one cord, relate the
position of a point on one cord to a point on another
cord. Separate equations are written for each cord.
4. Differentiate the position coordinate equation(s) to relate
velocities and accelerations. Keep track of signs!
Example
Given:In the figure on the
left, the cord at A is
pulled down with a
speed of 8 ft/s.
Find: The speed of block B.
Plan: There are two cords involved in the motion in this
example. The position of a point on one cord must be
related to the position of a point on the other cord. There
will be two position equations (one for each cord).
Engr222 Spring 2004 Chapter 12
61
Example - continued
Solution:
1) Define the position coordinates from a fixed datum line. Three
coordinates must be defined: one for point A (sA), one for block B
(sB), and one relating positions on the two cords. Note that pulley
C relates the motion of the two cords.
DATUM
sA
sC
sB
• Define the datum line through the top
pulley (which has a fixed position).
• sA can be defined to the center of the
pulley above point A.
• sB can be defined to the center of the
pulley above B.
• sC is defined to the center of pulley C.
• All coordinates are defined as
positive down and along the direction
of motion of each point/object.
Example - continued
DATUM
sA
sC
sB
2) Write position/length equations
for each cord. Define l1 as the
length of the first cord, minus
any segments of constant length.
Define l2 in a similar manner for
the second cord:
Cord 1: 2sA + 2sC = l1
Cord 2: sB + (sB – sC) = l2
3) Eliminating sC between the two
equations, we get
2sA + 4sB = l1 + 2l2
4) Relate velocities by differentiating this expression. Note that l1 and l2
are constant lengths.
2vA + 4vB = 0 => vB = - 0.5vA = - 0.5(8) = - 4 ft/s
The velocity of block B is 4 ft/s up (negative sB direction).
Engr222 Spring 2004 Chapter 12
62
Concept Quiz
1. Determine the speed of block B.
A) 1 m/s
B) 2 m/s
C) 4 m/s
D) None of the above.
2. Two blocks are interconnected by a
cable. Which of the following is
correct ?
A) vA= - vB
B) (vx)A= - (vx)B
y
C) (vy)A= - (vy)B D) All of the above.
x
Group Problem Solving
Given:In this pulley system,
block A is moving
downward with a
speed of 4 ft/s while
block C is moving up
at 2 ft/s.
Find: The speed of block B.
Plan: All blocks are connected to a single cable, so only one
position/length equation will be required. Define position
coordinates for each block, write out the position relation,
and then differentiate it to relate the velocities.
Engr222 Spring 2004 Chapter 12
63
Group Problem Solving - continued
Solution:
1) A datum line can be drawn through the upper, fixed,
pulleys and position coordinates defined from this line to
each block (or the pulley above the block).
2) Defining sA, sB, and sC
as shown, the position
relation can be written:
DATUM
sA + 2sB + 2sC = l
sA
sC
sB
3) Differentiate to relate
velocities:
vA + 2vB + 2vC = 0
4 + 2vB + 2(-2) =0
vB = 0
Attention Quiz
1. Determine the speed of block B when
block A is moving down at 6 ft/s while
block C is moving down at 18 ft/s .
A) 24 ft/s
B) 3 ft/s
C) 12 ft/s
D) 9 ft/s
vC=18 ft/s
vA=6 ft/s
2. Determine the velocity vector of
block A when block B is moving
downward with a speed of 10 m/s.
A) (8i + 6j) m/s
B) (4i + 3j) m/s
C) (-8i - 6j) m/s D) (3i + 4j) m/s
Engr222 Spring 2004 Chapter 12
j
vB=10 m/s
i
64
Textbook Problem 12-
Announcements
Engr222 Spring 2004 Chapter 12
65
Relative Motion Analysis - Section 12.10
Today’s Objectives:
Students will be able to:
a) Understand translating
frames of reference.
b) Use translating frames of
reference to analyze
relative motion.
In-Class Activities:
• Reading quiz
• Applications
• Relative position, velocity
and acceleration
• Vector & graphical methods
• Concept quiz
• Group problem solving
• Attention quiz
Reading Quiz
1. The velocity of B relative to A is defined as
A) vB – vA .
B) vA – vB .
C) vB + vA .
D) vA + vB .
2. Since vector addition forms a triangle, there can be at
most _________ unknowns (either magnitudes and/or
directions of the vectors).
A) one
B) two
C) three
D) four
Engr222 Spring 2004 Chapter 12
66
Applications
When you try to hit a moving
object, the position, velocity,
and acceleration of the object
must be known. Here, the boy
on the ground is at d = 10 ft
when the girl in the window
throws the ball to him.
If the boy on the ground is
running at a constant speed of 4
ft/s, how fast should the ball be
thrown?
Applications - continued
When fighter jets take off
or land on an aircraft
carrier, the velocity of the
carrier becomes an issue.
If the aircraft carrier travels at a forward velocity of 50 km/hr
and plane A takes off at a horizontal air speed of 200 km/hr
(measured by someone on the water), how do we find the
velocity of the plane relative to the carrier?
How would you find the same thing for airplane B?
How does the wind impact this sort of situation?
Engr222 Spring 2004 Chapter 12
67
Relative Position
The absolute position of two
particles A and B with respect to
the fixed x, y, z reference frame
are given by rA and rB. The
position of B relative to A is
represented by
rB/A = rB – rA
Therefore, if rB = (10 i + 2 j ) m
and
rA = (4 i + 5 j ) m,
then
rB/A = (6 i – 3 j ) m.
Relative Velocity
To determine the relative velocity of B
with respect to A, the time derivative
of the relative position equation is
taken.
vB/A = vB – vA
or
vB = vA + vB/A
In these equations, vB and vA are called absolute velocities
and vB/A is the relative velocity of B with respect to A.
Note that vB/A = - vA/B .
Engr222 Spring 2004 Chapter 12
68
Relative Acceleration
The time derivative of the relative
velocity equation yields a similar
vector relationship between the
absolute and relative accelerations of
particles A and B.
aB/A = aB – aA
or
aB = aA + aB/A
Solving Problems
Since the relative motion equations are vector equations,
problems involving them may be solved in one of two ways.
For instance, the velocity vectors in vB = vA + vB/A could be
written as Cartesian vectors and the resulting scalar
equations solved for up to two unknowns.
Alternatively, vector problems can be solved “graphically” by
use of trigonometry. This approach usually makes use of the
law of sines or the law of cosines.
Could a CAD system be used to solve these types of
problems?
Engr222 Spring 2004 Chapter 12
69
Laws of Sines and Cosines
C
a
B
Since vector addition or subtraction forms
a triangle, sine and cosine laws can be
applied to solve for relative or absolute
velocities and accelerations. As review,
their formulations are provided below.
b
A
c
Law of Sines:
a
sin A
Law of Cosines:
b
sin B
=
=
c
sin C
a 2 = b 2 + c 2 − 2 bc cos A
2
2
b = a + c
2
c = a
2
2
− 2 ac cos B
2
+ b − 2 ab cos C
Example
Given:
Find:
Plan:
vA = 600 km/hr
vB = 700 km/hr
vB/A
a) Vector Method: Write vectors vA and vB in
Cartesian form, then determine vB – vA
b) Graphical Method: Draw vectors vA and vB from a
common point. Apply the laws of sines and cosines
to determine vB/A.
Engr222 Spring 2004 Chapter 12
70
Example - continued
Solution:
a) Vector Method:
vA = 600 cos 35 i – 600 sin 35 j
= (491.5 i – 344.1 j ) km/hr
vB = -700 i km/hr
vB/A = vB – vA = (- 1191.5 i + 344.1 j ) km/hr
(1191. 5 )2 + ( 344.1 )2 = 1240. 2 km
hr
− 1 344 .1
θ
) = 16.1°
where θ = tan (
1191. 5
vB /A =
Example - continued
0
r
/h
km
/A
60
vB = 700 km/hr
θ
145 °
vB
=
vA
b) Graphical Method:
Note that the vector that measures
the tip of B relative to A is vB/A.
Law of Cosines:
2
vB/2A = ( 700 ) 2 + ( 600 ) − 2 ( 700)(600 )cos 145 °
vB/A = 1240 . 2 km
hr
Law of Sines:
vB/A
vA
=
sin(145° )
sin θ
Engr222 Spring 2004 Chapter 12
or
θ = 16 . 1 °
71
Concept Quiz
1. Two particles, A and B, are moving in
the directions shown. What should be
the angle θ so that vB/A is minimum?
A) 0°
B) 180°
C) 90°
D) 270°
B
A
ft
vB = 4 s
θ
vA = 3 ft s
2. Determine the velocity of plane A with respect to plane B.
A) (400 i + 520 j ) km/hr
B) (1220 i - 300 j ) km/hr
C) (-181 i - 300 j ) km/hr
30°
D) (-1220 i + 300 j ) km/hr
Group Problem Solving
y
x
Given: vA = 10 m/s
vB = 18.5 m/s
at)A = 5 m/s2
aB = 2 m/s2
Find: vA/B
aA/B
Plan: Write the velocity and acceleration vectors for A and B
and determine vA/B and aA/B by using vector equations.
Solution:
The velocity of A is:
vA = 10 cos(45)i – 10 sin(45)j = (7.07i – 7.07j) m/s
Engr222 Spring 2004 Chapter 12
72
Group Problem Solving - continued
The velocity of B is:
vB = 18.5i (m/s)
y
x
The relative velocity of A with respect to B is (vA/B):
vA/B = vA – vB = (7.07i – 7.07j) – (18.5i) = -11.43i – 7.07j
or
vB/A =
(11.43)2 + (7.07)2 = 13.4 m/s
θ = tan-1(
7.07
) = 31.73°
11.43
θ
Group Problem Solving - continued
The acceleration of A is:
aA = (at)A + (an)A = [5 cos(45)i – 5 sin(45)j]
+ [-(
102
102
) sin(45)i – (
) cos(45)j]
100
100
aA = 2.83i – 4.24j (m/s2)
The acceleration of B is:
aB = 2i (m/s2)
The relative acceleration of A with respect to B is:
aA/B = aA – aB = (2.83i – 4.24j) – (2i) = 0.83i – 4.24j
aA/B = (0.83)2 + (4.24)2 = 4.32 m/s2
β
β = tan-1( 4.24 ) = 78.9°
0.83
Engr222 Spring 2004 Chapter 12
73
Attention Quiz
1. Determine the relative velocity of particle B with respect to
particle A.
y
A) (48i + 30j) km/h
B
vB=100 km/h
B) (- 48i + 30j ) km/h
C) (48i - 30j ) km/h
30°
x
D) (- 48i - 30j ) km/h
A
vA=60 km/h
2. If theta equals 90° and A and B start moving from the same
point, what is the magnitude of rB/A at t = 5 s?
A)
B)
C)
D)
20 ft
15 ft
18 ft
25 ft
B
θ
A
ft
vB = 4 s
vA = 3 ft s
Textbook Problem 12-
Engr222 Spring 2004 Chapter 12
74