ME 220 Engineering MechanicsDynamics
12.11 INTRODUCTION &
12
12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION
Objectives:
Obj
ti
Students will be able to find the kinematic quantities
(position displacement
(position,
displacement, velocity,
velocity and acceleration) of a
particle traveling along a straight path.
• 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.
constant
APPLICATIONS
The motion of large objects,
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?
APPLICATIONS
(continued)
A sports car travels along a straight road.
Can we treat the car as a particle?
p
If the car accelerates at a constant rate, how can we
determine its position and velocity at some instant?
An Overview of Mechanics
Mechanics: The study
y of how bodies
react to forces acting on them.
Statics: The study of
bodies in equilibrium.
D
Dynamics:
i
1. Kinematics – concerned with
the
h geometric
i aspects off motion
i
2. Kinetics - concerned with
th forces
the
f
causing
i the
th motion
ti
RECTILINEAR KINEMATICS: CONTINIOUS MOTION
(Section 12.2)
A particle travels along a straight-line
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
iti vector
t r, or the
th scalar
l s. Scalar
S l s
can be positive or negative. Typical units
for r and s are meters (m) or feet (ft).
(ft)
The displacement of the particle is
defined as its change in position.
Vector form: Δ r = r’ - r
Scalar form: Δ s = s’ - s
The total distance traveled by the particle, sT, is a positive scalar
p
the total length
g of the ppath over which the particle
p
that represents
travels.
VELOCITY
Velocity is a measure of the rate of change in the position of a particle.
quantityy (it
( has both magnitude
g
and direction).
) The
It is a vector q
magnitude of the velocity is called speed, with units of m/s or ft/s.
The average velocity
Th
l it off a particle
ti l during
d i a
time interval Δt is
vavg = Δr / Δt
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 / Δt
ACCELERATION
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity.
i Typical
T i l units
i are m/s
/ 2 or ft/s
f / 2.
The instantaneous acceleration is the time
derivative of velocity.
Vector form: a = dv / dt
S l fform: a = dv
Scalar
d / 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 RELATIONS:
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
t
v
s
d = ∫ a dt
d or ∫ v dv
d = ∫ a ds
d
∫ dv
s
t
d = ∫ v dt
d
∫ ds
vo
o
vo
so
so
o
• Note that
h so and
d vo represent the
h initial
i i i l position
i i and
d
velocity of the particle at t = 0.
CONSTANT ACCELERATION
The three kinematic equations can be integrated for the special case
when
h acceleration
l i is
i constant (a
( = ac) to obtain
b i very useful
f l equations.
i
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 = ∫ a
dt
yields
v = vo + act
∫ ds = ∫ v dt
yields
s = s o + v ot + (1/2) a c t 2
vo
o
s
t
so
v
c
o
s
∫ v dv = ∫ ac ds
vo
so
yields
v
2=
(vo ) + 2ac(s - so)
2
EXAMPLE 12.1
12 1
Given: A car moves in a straight line to the right
with a velocity of v = (0.9 t2 +0.6 t) m/s where t is
in seconds
seconds. Also
Also, s = 0 when t = 0.
0
Find: The position and acceleration of the particle
when t = 3 s.
Plan: Establish the positive coordinate, s, in the direction the
car is traveling
traveling. Since the velocity is given as a
function of time, take a derivative of it to calculate the
acceleration. Conversely, integrate the velocity
function to calculate the position.
p
Solution:
EXAMPLE
(continued)
1) Take a derivative of the velocity to determine the acceleration.
acceleration
a = dv / dt = d(0.9 t2 +0.6 t) / dt =1.8 t + 0.6
=> a = 6 m/s
/ 2 (or
( in
i the
h Î direction)
di i ) when
h t=3s
2) Calculate the distance traveled in 4s by integrating the
g so = 0:
velocityy using
s
t
v = ds / dt => ds = v dt => ∫ ds = ∫ (0.9 t2 +0.6 t) dt
so
o
=> s – so = 0.3 t3 + 0.3 t2
=> s – 0 = 0.3 (3)2 + 0.3 (3)3 => s = 10.8 m ( or Î)
EXAMPLE
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
ground.
another at a height of 20 ft.
Find:The speed at which ball B was
thrown upward
upward.
Plan: Both balls experience
p
a constant downward acceleration
of 32.2 ft/s2 due to gravity. Apply the formulas for
constant acceleration
acceleration, with ac = -32.2
32 2 ft/s2.
Solution:
EXAMPLE
(continued)
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 fft (s
( A = 20 ft)
f ) using
i a position
i i equation.
i
sA = (sA )o + (vA)o t + (1/2) ac t2
So,
20 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t2) => t = 1.115 s
Solution:
EXAMPLE
(continued)
2) Now consider ball B. It is throw upward from a height of 5 ft
(( B)o = 5 ft).
((s
ft) It mustt reachh a height
h i ht off 20 ft (s
( B = 20 ft) att th
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
So,
So
20 ft = 5 + (vB)o(1.115) + (1/2)(-32.2)(1.115)2
=> (vB)o = 31.4 ft/s
EXAMPLE 12
12..2
During a test a rocket travels upwards at 75 m/s, and when it is 40 m from
the ground its engine fails. Determine the maximum height sB reached by
the
h rocket
k and
d iits speedd just
j before
b f
it
i hits
hi the
h ground.
d While
hil in
i motion
i the
h
rocket is subjected to a constant downward acceleration of 9.81 m/s2 due to
gravity. Neglect the effect of air resistance.
EXAMPLE 12
12..2 (continued)
12. 3 RECTILINEAR KINEMATICS: ERRATIC MOTION
Objectives:
Students will be able to determine position, velocity, and
acceleration of a particle using graphs.
s-t, v-t, a-t, v-s, and a-s diagrams
APPLICATION
In many experiments
experiments, a
velocity versus position (v-s)
profile is obtained.
graph
p for the
If we have a v-s g
tank truck, how can we
determine its acceleration at
position s = 1500 feet?
ERRATIC MOTION
(Section 12.3)
Graphing
G
hi provides
id a goodd way to
handle complex motions that
would
ld bbe diffi
difficult
lt to
t describe
d
ib
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
S
T GRAPH
Plots of position vs.
vs time can be
used to find velocity vs. time
curves. Finding the slope of the
line tangent to the motion curve at
any
y point
p
is the velocity
y at that
point (or v = ds/dt).
Therefore, the v-t graph can be
Therefore
constructed by finding the slope at
various points along the ss-tt graph.
graph
V-T GRAPH
Plots of velocity vs. time can be used to
fi d acceleration
find
l ti vs. time
ti curves.
Finding the slope of the line tangent to
th velocity
the
l it curve att any point
i t is
i the
th
acceleration at that point (or a = dv/dt).
Therefore, the acceleration vs. time (or
a t) graph can be constructed by
a-t)
finding the slope at various points
along the v-t
v t graph.
graph
Also,, the distance moved
(displacement) of the particle is the
area under the v-t graph during time Δt.
A-T
A
T GRAPH
Given the acceleration vs. time
or a-t curve, the change in
velocity (Δv) 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.
A-S GRAPH
A more complex
p case is ppresented byy
the acceleration versus position or a-s
graph. The area under the a-s curve
represents the change in velocity
((recall ∫ a ds = ∫ v dv )).
s2
½ (v1² – vo²) = ∫ a ds = area under the
s1
a-s graph
This equation
Thi
i can be
b solved
l d for
f v1,
allowing you to solve for the velocity
att a point.
i t By
B doing
d i this
thi repeatedly,
t dl
you can create a plot of velocity
versus distance
distance.
V-S
V
S GRAPH
Another complex case is presented
by the velocity vs.
vs distance or vv-ss
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,
point
we can obtain the acceleration at
that point. Recall the formula
a = v (dv/ds).
(
)
Thus, we can obtain an a-s plot
p
from the v-s curve.
EXAMPLE from F12-12
Given: The s-t graph for a sports car moving along a straight road.
Find: The v-t graph and a-t graph over the time interval
0 ≤ t ≤ 10s
EXAMPLE (continued)
(
)
Solution: The v-t graph can be constructed by finding the slope
off th
the s-tt graphh att key
k points.
i t What
Wh t are those?
th ?
when
h 0 < t < 5 s;
v0-5 = ds/dt
d /dt = d(3t2)/dt = 6 t m/s
/
when 5 < t < 10 s; v5-10 = ds/dt = d(30t−75)/dt
d(30t 75)/dt = 30 m/s
v(m/s)
v-t graph
30
t(s)
5
10
EXAMPLE (continued)
(
)
Similarly, the a-t graph can be constructed by finding the slope
att various
i
points
i t along
l
the
th v-tt graph.
h
when
h 0 < t < 5 s;
a0-5 = dv/dt
d /dt = d(6t)/dt = 6 m/s
/2
when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2
a(m/s2)
a-t graph
6
t(s)
5
10
EXAMPLE from P12-51
P12 51
Given: The v-t graph shown.
Find: The a-t graph, average
speed,
p , and distance
traveled for the 0 - 90 s
interval.
Plan:
Find
Fi
d slopes
l
off the
h v-t curve and
d ddraw the
h a-t graph.
h
Find the area under the curve. It is the distance traveled.
Fi ll calculate
Finally,
l l t average speedd (using
( i basic
b i definitions!).
d fi iti !)
Solution:
Find the a–t graph:
For 0 ≤ t ≤ 30
EXAMPLE
(continued)
a = dv/dt = 1.0 m/s
m/s²
For 30 ≤ t ≤ 90 a = dv/dt = -0.5 m/s²
a(m/s²)
a-t g
graph
p
1
30
-0.5
05
90 t(s)
EXAMPLE (continued)
Now find the distance traveled:
Δs0-30 = ∫ v dt = (1/2) (30)2 = 450 m
Δs30-90 = ∫ v dt
= (1/2) (-0.5)(90)2 + 45(90) – (1/2) (-0.5)(30)2 – 45(30)
= 900 m
s0-90 = 450 + 900 = 1350 m
vavg(0-90) = total distance / time
= 1350 / 90
= 15 m/s
/
EXAMPLE 12
12..8
The v-s graph describing the motion of a motorcycle is shown in Fig.1215a. Construct the a-s graph of the motion and determine the time needed
f the
for
h motorcycle
l to reachh the
h position
i i s = 120 m.
EXAMPLE 12
12..8 (continued)
12-4 CURVILINEAR MOTION:
GENERAL & RECTANGULAR COMPONENTS
Objectives:
Students will be able to:
1. Describe the motion of a
particle traveling along a
curved path.
2 Relate
2.
R l t kinematic
ki
ti quantities
titi
in terms of the rectangular
components of the vectors.
vectors
• General Curvilinear Motion
g
Components
p
of
• Rectangular
Kinematic Vectors
APPLICATIONS
The path of motion of a plane can
be tracked with radar and its x, y,
and z coordinates (relative to a
point on earth) recorded as a
function of time.
How can we determine the velocity
y
or acceleration of the plane at any
instant?
APPLICATIONS
(continued)
A roller coaster car travels down
a fixed,
fi d helical
h li l path
th att a constant
t t
speed.
How can we determine its
position or acceleration at any
instant?
If you are designing the track, why is it important to be
able
bl tto predict
di t th
the acceleration
l ti off the
th car??
GENERAL CURVILINEAR MOTION
(Section 12.4)
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,
function s.
s
The position
Th
iti off th
the particle
ti l att any iinstant
t t iis ddesignated
i t d by
b the
th vector
t
r = r(t). Both the magnitude and direction of r may vary with time.
If the particle moves a distance Δs along the
curve during time interval Δt, the
displacement is determined by vector
subtraction: Δ r = r’ - r
VELOCITY
Velocity represents the rate of change in the position of a
particle.
particle
The average velocity of the particle
during the time increment Δt is
vavg = Δr/Δt .
Th instantaneous
The
i t t
velocity
l it is
i the
th
time-derivative of position
v = dr/dt .
The velocity vector, v, is always
tangent to the
h pathh off motion.
i
The magnitude
g
of v is called the speed.
p
Since the arc length
g Δs
approaches the magnitude of Δr as t→0, the speed can be
obtained by differentiating the path function (v = ds/dt). Note
that this is not a vector!
ACCELERATION
Acceleration represents the rate of change in the
velocity of a particle.
If a particle’s
ti l ’ velocity
l it changes
h
from
f
v to
t v’’ over a
time increment Δt, the average acceleration during
that increment is:
aavg = Δv/Δt = (v - v’)/Δt
The instantaneous acceleration is the timetime
derivative 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,
not,
general tangent to the path function.
function
CURVILINEAR MOTION: RECTANGULAR COMPONENTS
(Section 12.5)
It is often convenient to describe the motion of a p
particle in
terms of its x, y, 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, z components may all be
functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
05
Th magnitude
The
i d off the
h position
i i vector is:
i r = (x
( 2 + y2 + z2)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
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,
i j,
j k are constant in magnitude and
direction, this equation reduces to v = vx i + vy j + vz k
•
•
•
where vx = x = dx/dt,
dx/dt vy = y = dy/dt,
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 = v• z = 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.
EXAMPLE I
Given:The motion of two particles (A and B) is described by
the position vectors
rA = [3t i + 9t(2 – t) j] m and
rB = [[3(t
( 2 –2t +2)) i + 3(t
( – 2)) j] m.
Find: The point at which the particles collide and their
speeds
d just
j t before
b f
the
th collision.
lli i
y
t=0 s
A
-6
6
B
x
EXAMPLE I (continued)
(
)
Solution:
1)) The ppoint of collision requires
q
that rA = rB,
so xA = xB and yA = yB .
Set the x-components
x components equal: 3t = 3(t2 – 2t + 2)
Simplifying: t2 – 3t + 2 = 0
Solving:
t = {3 ± [32 – 4(1)(2)]0.5}/2(1)
=> t = 2 or 1 s
Set the y-components
y components equal: 9t(2 – t) = 3(t – 2)
Simplifying:
3t2 – 5t – 2 = 0
Solving: t = {5 ± [52 – 4(3)(–2)]
4(3)( 2)]0.5}/2(3)
=> t = 2 or – 1/3 s
So the particles collide when t = 2 s (only common
So,
time). Substituting this value into rA or rB yields
xA = xB = 6 m
and
d yA = yB = 0
y
t=2 s
A
6
B
-6
6
x
EXAMPLE I (continued)
2) Differentiate rA and rB to get the velocity vectors.
•
•
vA = drA/dt = .xA i + yA j = [ 3 i + ((18 – 18t)) j ] m/s
At t = 2 s: vA = [ 3i – 18 j ] m/s
vB = drB/dt = xB i + yB j = [ (6t – 6) i + 3 j ] m/s
•
•
A t = 2 s: vB = [ 6 i + 3 j ] m/s
At
/
Speed is the magnitude of the velocity vector
vector.
vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5 = 6.71 m/s
EXAMPLE II (P12
(P12-74)
74)
Given: The velocity
y of the pparticle is
v = [ 16 t2 i + 4 t3 j + (5 t + 2) k] m/s.
When t = 0, x = y = z = 0.
Find: The particle’s coordinate position and the magnitude of
its acceleration when t = 2 s.
s
Plan:
Note that velocity vector is given as a function of time.
1) Determine the position and acceleration by
integrating and differentiating v, respectively, using
the initial conditions.
2)) Determine
i the
h magnitude
i d off the
h acceleration
l i vector
using t = 2 s.
EXAMPLE II (P12-74)
(P12 74) (continued)
Solution:
1) x
x-components:
components:
•
Velocity known as: vx = x = dx/dt = (16 t2 ) m/s
x
Position:
∫
0
t
∫
dx = (16 t2) dt ⇒ x = (16/3)t3 = 42.7 m at t = 2 s
0
••
•
Acceleration: ax = x = vx = d/dt (16 t2) = 32 t = 64 m/s2
2) y-components:
•
Velocityy known as: vy = y = dy/dt
y = ((4 t3 ) m/s
y
Position:
∫
0
t
∫
dy = (4 t3) dt ⇒ y = t4 = (16) m at t = 2 s
0
••
•
Acceleration: ay = y = vy = d/dt (4 t3) = 12 t2 = 48 m/s2
EXAMPLE II (P12-74)
(P12 74) (continued)
3) z-components:
•
Velocity is known as: vz = z = dz/dt = (5 t + 2) m/s
z
Position:
∫
0
t
∫
dz = (5 t + 2) dt ⇒ z = (5/2) t2 + 2t = 14 m at t=2s
0
••
•
Acceleration: az = z = vz = d/dt (5 t + 2) = 5 m/s2
4) The position vector and magnitude of the acceleration vector
are written using the component information found above.
above
Position vector: r = [ 42.7 i + 16 j + 14 k] m.
Acceleration vector: a = [ 64 i + 48 j + 5 k] m/s2
Magnitude: a = (642 + 482 +52)0.5 = 80.2
80 2 m/s2
EXAMPLE 12.9
12 9 At any instant the horizontal position of
the weather balloon in Fig 12-18a is defined by x = (9t) m,
where t is in seconds.
seconds If the equation of the path is
y=x2/30, determine the magnitude and direction of the
velocity and the acceleration when t = 2s.
2s
EXAMPLE 12.9
12 9 (continued)
12-6 MOTION OF A PROJECTILE
Objectives:
Students will be able to analyze the free-flight motion of a
projectile.
projectile
• Kinematic Equations for
Projectile Motion
APPLICATIONS
A good kicker instinctively knows at what angle, θ, and initial velocity, vA,
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 bbasketball
k b ll is
i shot
h at a certain
i angle.
l What
Wh parameters should
h ld the
h shooter
h
consider in order for the basketball to pass through the basket?
Distance,, speed,
p , the basket location,, … anything
y
g else ?
APPLICATIONS (continued)
(
)
A fi
firefighter
fi h needs
d to know
k
the
h maximum
i
height
h i h on the
h wall
ll she
h can
project water from the hose. What parameters would you program into a
wrist computer to find the angle, θ, that she should use to hold the hose?
MOTION OF A PROJECTILE
((Section 12.6))
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.,
from gravity)
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) andd th
the position
iti in
i the
th x direction
di ti can be
b determined
d t
i d by:
b
x = xo + (vox) t
Why is ax equal to zero (assuming movement through the air)?
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 + (voyy) t – ½ g t2
vy2 = voy2 – 2 g (y – yo)
For any given problem, only two of these three equations can be used.
Why?
y
EXAMPLE I
Given: vo and θ
Find: The equation
q
that defines y as a
function of x.
Plan: Eliminate time from the
ki
kinematic
ti equations.
ti
Solution: Using vx = vo cos θ
W can write:
We
it
and
x = (v
( o cos θ)t
vy = vo sin θ
or
t =
y = ((vo sin θ)) t – ½ g ((t))2
x
vo cos θ
By substituting for t:
y = (vo sin θ) {
x }
vo cos θ 2
–{ g
vo cos θ
}2x
EXAMPLE I
(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.
Th equation
The
ti shows
h
that
th t the
th path
th is
i parabolic.
b li
EXAMPLE II
Given: Projectile is fired with vA=150 m/s at point
A.
Find:
The horizontal distance it travels (R)
and the time in the air.
Plan:
Establish a fixed x,, y coordinate system
y
(in
( this solution,, the origin
g
of the coordinate system is placed at A). Apply the kinematic
relations in x- and y-directions.
EXAMPLE II (continued)
Solution:
1) Place the coordinate system at point A.
Then, write the equation for horizontal motion.
+ xB = xA + vAx tAB
where xB = R, xA = 0, vAx = 150 (4/5) m/s
Range, R will be R = 120 tAB
2) Now write a vertical motion equation. Use the distance equation.
+ yB = yA + vAyy tAB – 0.5 g tAB2
where yB = – 150, yA = 0, and vAy = 150(3/5) m/s
We get the following equation: –150
150 = 90 tAB + 0.5 ((– 9.81) tAB2
Solving for tAB first, tAB = 19.89 s.
Then, R = 120 tAB = 120 (19.89) = 2387 m
12-7 CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS
Objectives:
Students will be able to determine the normal and tangential
components of velocity and acceleration of a particle
traveling along a curved path.
• Normal
N
l andd Tangential
T
ti l
Components of Velocity and
Acceleration
• Special Cases of Motion
APPLICATIONS
Cars traveling along a clover-leaf
interchange experience an
acceleration due to a change in
velocity as well as due to a change
in direction of the velocity.
If the car’s speed
p
is increasing
g 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 roller
ll coaster travels
l down
d
a
hill for which the path can be
approximated
i t d by
b a function
f ti
y = f(x).
The roller coaster starts from rest
and increases its speed at a
constant rate.
How can we determine its velocity
and acceleration at the bottom?
Why would we want to know
these values?
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
g is located on the p
particle
origin
(the origin moves with the
particle).
The tt-axis
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.
respectively
The center of curvature
curvature, O
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.
VELOCITY IN THE n-t
n t COORDINATE SYSTEM
The velocity
y vector is always
y
tangent to the path of motion
(t-direction).
The magnitude is determined by taking the time derivative of
the path function
function, s(t).
s(t)
v = v ut where v = .s = ds/dt
Here v defines the magnitude of the velocity (speed) and
y vector.
ut defines the direction of the velocity
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change
of
velocity:
.
.
a = dv/dt
d /dt = d(vu
d( t)/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,
p
,
the acceleration vector can be
expressed as:
.
a = v ut + (v2/ρ) un = at ut + an un.
ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
So, there are two components to the
So
acceleration vector:
a = at ut + an un
• The
Th tangential
i l component is
i tangent to the
h curve andd in
i the
h
direction of increasing or decreasing velocity.
.
at = v or at ds
d = v dv
d
• The normal or centripetal component is always directed
t
toward
d the
th center
t off curvature
t
off the
th 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.
.
2
ρ
∞ => an = v /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
p
represents
p
the time rate of change
g in the
direction of the velocity.
SPECIAL CASES OF MOTION (continued)
(
)
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vo t + (1/2) (at)c t2
v = vo + (at)c t
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.
0 How are these equations related to projectile
motion equations? Why?
4) Th
The particle
i l moves along
l
a pathh expressedd as y = f(x).
f( )
The radius of curvature, ρ, at any point on the path can be
calculated
l l t d from
f
[ 1 + (dy/dx)2 ]3/2
ρ = ________________
d2y/dx
/d 2
THREE-DIMENSIONAL MOTION
If a particle moves along a space
curve the n and t axes are defined as
curve,
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 p
perpendicular
p
to the osculating
g
plane, and its sense is defined by the cross product ub = ut × un.
There is no motion, thus no velocity or acceleration, in the
binomial direction.
EXAMPLE I
Given: A boat travels around a
circular
i l path,
th ρ = 40 m, att a
speed that increases with
ti
time,
v = (0.0625
(0 0625 t2) m/s.
/
Find:
The magnitudes
g
of the boat’s
velocity and acceleration at
the instant t = 10 s.
Plan:
The boat
Th
b t starts
t t from
f
restt (v
( = 0 when
h t = 0).
0)
1) Calculate the velocity at t = 10 s using v(t).
2) Calculate
C l l t th
the ttangential
ti l andd normall components
t off
acceleration and then the magnitude of the
acceleration vector.
vector
EXAMPLE (continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is
given by v = (0.0625t2) m/s. At t = 10s:
v = 0.0625 t2 = 0.0625 (10)2 = 6.25 m/s
.
2) The acceleration vector is a = atut + anun = vut + (v2/ρ)un.
.
Tangential component: at = v = d(.0625 t2 )/dt = 0.125 t m/s2
At t = 10s: at = 0.125t = 0.125(10) = 1.25 m/s2
Normall component: an = v2/ρ
N
/ m/s
/2
At t = 10s: an = (6.25)2 / (40) = 0.9766 m/s2
The magnitude of the acceleration is
0 5 = [(1.25)
0 5 = 1.59
a = [(a
[( t)2 + (a
( n)2]0.5
[(1 25)2 + (0.9766)
(0 9766)2]0.5
1 59 m/s
/2
EXAMPLE II
Given: A roller coaster travels along a
vertical parabolic path defined by
the equation y = 0.01x2. At point B,
it has a speed of 25 m/s, which is
increasing at the rate of 3 m/s2.
1
1.
2.
3.
4.
Find: The magnitude of the roller coaster’s
acceleration when it is at point B.
Plan:
The change in the speed of the car (3 m/s2) is the
tangential component of the total acceleration.
Calculate the radius of curvature of the path at B.
Calculate the normal component of acceleration.
Determine the magnitude
g
of the acceleration vector.
EXAMPLE II
(continued)
Solution:
1) The tangential component of acceleration is the
rate
of
.
increase of the roller coaster’s speed, so at = v = 3 m/s2.
2) Determine the radius of curvature at point B (x = 30 m):
dy/dx = d(0.01x2)/dx = 0.02x, d2y/dx2 = d (0.02x)/dx = 0.02
At x =30 m, dy/dx = 0.02(30) = 0.6, d2y/dx2 = 0.02
=> ρ =
[[1+(dy/dx)
( y )2]3/2
d2y/dx2
= [1 + (0.6)2]3/2/(0.02) = 79.3 m
3) The normal component of acceleration is
an = v2/ρ
ρ = ((25))2/(79.3)
(
) = 7.881 m/s2
4) The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5 = [(3)2 + (7.881)
(7 881)2]0.5 = 8.43
8 43 m/s2
12-8
12
8 CURVILINEAR
C
A MOTION:
O O
CYLINDRICAL COMPONENTS
Objectives:
Students will be able to:
1. Determine velocity and acceleration components using
cylindrical coordinates.
• Velocity Components
• Acceleration Components
A
APPLICATIONS
CA O S
The cylindrical coordinate
system is used in cases
where the particle moves
along a 3-D curve.
In the figure shown, the box
slides
lid down
d
the
h helical
h li l ramp.
How would you find the
b ’ velocity
box’s
l it components
t to
t
know if the package will fly
off
ff th
the ramp??
CYLINDRICAL COMPONENTS
(Section 12.8)
transverse coordinate, θ
radial coordinate, r
We can express the location of P in polar coordinates as r = r ur.
Note that the radial direction
direction, r,
r extends outward from the fixed
origin, O, and the transverse coordinate, θ, is measured counterclockwise (CCW) from the horizontal.
horizontal
VELOCITY in POLAR COORDINATES)
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
d ur
.
v = rur + r
dt
Using the chain rule:
dur/dt = (dur/dθ)(dθ/dt)
.
We can prove that dur/d
θ = uθ so dur/dt = θuθ
.
.
Therefore: v = rur + rθuθ
.
Thus, the velocity vector has two components:
r,
.
called the radial component,
component and rθ called the
transverse component. The speed of the particle at
any
y ggiven instant is the sum of the squares
q
of both
components or
.
v = (r θ
)2
. 2
+(r)
ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rur + rθuθ)
After manipulation,
Af
i l i the
h acceleration
l i can be
b
expressed as
..
.
..
..
2
a = (r – rθ )ur + (rθ + 2rθ)uθ
.
..
The term (r – rθ 2) is the radial acceleration
or ar .
..
The term (rθ + 2rθ) is the transverse
acceleration
l i or aθ .
.. . .
.2 2
..
The magnitude of acceleration is a = (r – rθ ) + (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
h chain
h i rule:
l
Velocity:
.
.
.
vP = rur + rθuθ + zuz
..
.. . 2
..
..
Acceleration: aP = (r – rθ )ur + (rθ + 2rθ)uθ + zuz
EXAMPLE I
Given: A car travels along acircular path.
path
.
r = 300 ft, θ = 0.4 (rad/s),
..
θ = 0.2
0 2 (rad/s
( d/ 2)
Find: Velocity and acceleration
Plan: Use the polar coordinate system.
Solution:
..
.
. ..
r = 300 ft,
ft r = r = 0,
0 and θ = 0.4
0 4 (rad/s),
(rad/s) θ = 0.2
0 2 (rad/s2)
Substitute in the equation
for velocity
.
.
v = r ur + rθ uθ = 0 ur + 300 (0.4) uθ
v = (0)2 + (120)2 = 120 ft/s
ft/
EXAMPLE I
(continued)
Substitute in the equation for acceleration:
.. . .
.
..
2
a = (r – rθ )ur + (rθ + 2rθ)uθ
a = [0 – 300(0.4)2] ur + [300(0.2) + 2(0)(0.4)] uθ
a = – 48 ur + 60 uθ ft/s2
a = (– 48)2 + (60)2 = 76.8 ft/s2
12 -9
9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF
TWO PARTICLES
Objectives:
Students will be able to relate the positions, velocities, and
accelerations of particles undergoing dependent motion.
• Define
D fi Dependent
D
d t Motion
M ti
• Develop Position, Velocity, and
Acceleration Relationships
APPLICATIONS
The cable and pulley system shown
can be used to modify the speed of
the mine car, A, relative to the speed
of the motor
motor, M.
M
It is important to establish the
relationships between the various
motions in order to determine the
power requirements for the motor
and the tension in the cable.
For instance, if the speed of the cable (P) is known because we
k
know
th
the motor
t characteristics,
h
t i ti how
h can we determine
d t
i the
th
speed of the mine car? Will the slope of the track have any
i
impact
t on the
th answer??
APPLICATIONS
(continued)
Rope andd pulley
R
ll arrangements
are often used to assist in lifting
h
heavy
objects.
bj t The
Th total
t t l lifting
lifti
force required from the truck
depends on both the weight and
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 (Section 12.9)
In many kinematics problems, the motion of one object will
d
depend
d on the
th motion
ti off another
th object.
bj t
The blocks in this figure are
connected by an inextensible cord
wrapped around a pulley.
pulley
If block A moves downward along
the inclined plane,
plane block B will
move up the other incline.
The motion of each block can be related mathematically by
d fi i position
defining
iti coordinates,
di t sA and
d sB. Each
E h coordinate
di t axis
i is
i
defined from a fixed point or datum line, measured positive along
each
h plane
l
in
i the
th direction
di ti off motion
ti off each
h block.
bl k
DEPENDENT MOTION
(continued)
IIn this
hi example,
l position
ii
coordinates sA and sB can be
d fi d from
defined
f
fixed
fi d datum
d t
lines
li
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 the 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
/d = dlT/dt
/d = 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 = -aaA .
expression.
DEPENDENT MOTION 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.
block
Note that sB is only
y 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.
blocks
DEPENDENT MOTION EXAMPLE (continued)
The position coordinates are related by
th equation
the
ti
2sB + h + sA = lT
Wh lT is
Where
i the
th total
t t l cordd length
l th minus
i
the lengths of the red segments.
Since lT 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 your sign convention!
DEPENDENT MOTION EXAMPLE (continued)
(
)
This example can also be worked
by defining the position coordinate
for B (sB) from the bottom pulley
instead of the top pulley.
pulley
The position,
position velocity
velocity, and
acceleration relations then become
2(h – sB) + h + sA = lT
and
2vB = vA
2aB = aA
Prove to yourself that the results are the same
same, even if the sign
conventions are different than the previous formulation.
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 D
1.
Define
fi position
i i coordinates
di
f
from
fi d datum
fixed
d
lines,
li
along the path of each particle. Different datum lines can
b usedd ffor each
be
h particle.
ti l
2. Relate the position coordinates to the cord length.
S
Segments
t off cordd that
th t do
d nott change
h
in
i length
l th during
d i the
th
motion may be left out.
3 If a system contains
3.
i more than
h one cord,
d relate
l the
h
position of a point on one cord to a point on another
cord.
d Separate
S
t equations
ti
are written
itt for
f eachh cord.
d
4. Differentiate the position coordinate equation(s) to relate
velocities
l iti andd accelerations.
l ti
K
Keep
ttrack
k off signs!
i !
EXAMPLE
Given: In the figure on the left, the
cord at A is pulled down
with a speed of 2 m/s.
Find: The speed of block B.
Plan:
There are two cords involved
in the motion in this example.
There will be two position
equations (one for each cord).
Write these two equations,
q
,
combine them, and then
differentiate them.
EXAMPLE (continued)
(
)
Solution:
1) Define the position coordinates from a fixed datum line
line. Three
coordinates must be defined: one for point A (sA), one for block B
(sB), andd one
o e for
o block
b oc C (sC).
• Define the datum line through the top
pulley (which has a fixed position).
• sA can be defined to the 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)
2) Write position/length equations for
each cord. Define l1 as the length of
the first cord, minus any segments of
constant length.
length Define l2 in a similar
manner for the second cord:
Cord
C
d 1:
1 sA + 2s
2 C = l1
Cord 2: sB + (sB – sC) = l2
3) Eliminating sC between the two
equations, we get
sA + 4s
4 B = l1 + 2l2
4) Relate velocities by differentiating this expression. Note that l1 and l2
are constant lengths.
vA + 4vB = 0 => vB = – 0.25vA = – 0.25(2) = – 0.5 m/s
The
h velocity
l i off block
bl k B is
i 0.5 m/s
/ up (negative
(
i sB direction).
di i )
12-10
12
10 RELATIVE
RELATIVE-MOTION
MOTION ANALYSIS OF TWO
PARTICLES USING TRANSLATING AXES
Objectives:
Students will be able to:
1 Understand translating frames of reference.
1.
reference
2. Use translating frames of reference to analyze relative
motion.
• Relative Position, Velocity and
Acceleration
• Vector & Graphical Methods
APPLICATIONS
When you try to hit a
moving
i object,
bj
the
h position,
ii
velocity, and acceleration of
th object
the
bj t all
ll have
h
to
t be
b
accounted for by your mind.
You are smarter than you
thought!
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 is underway with 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?
RELATIVE POSITION (Section 12.10)
12 10)
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
Therefore,
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.
A
N t that
Note
th t vB/A = - vA/B .
RELATIVE ACCELERATION
The time derivative of the relative
velocity equation yields a similar
vector relationship
p between the
absolute and relative accelerations
of particles A and B.
Th
These
dderivatives
i ti
yield:
i ld aB/A = aB – aA
or
aB = aA + aB/A
SOLVING PROBLEMS
Since the relative motion equations are vector equations
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 two dimensional (2-D) Cartesian vectors and the
resulting 2-D scalar component equations solved for up to
two unknowns.
Alternatively, vector problems can be solved “graphically” by
use of trigonometry
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?
LAWS OF SINES AND COSINES
Since vector addition or subtraction forms
a triangle, sine and cosine laws can be
applied
pp
to solve for relative or absolute
velocities and accelerations. As a review,
their formulations are provided below.
C
b
a
B
A
c
Law of Sines:
a
sin A
Law of Cosines:
2
b
=
sin B
2
a = b + c
2
2
2
2
c
=
sin C
− 2 bc cos A
b = a + c − 2 ac cos B
2
2
2
c = a + b − 2 ab cos C
EXAMPLE
Given:
vA = 650 km/h
vB = 800 km/h
Find:
vB/A
Plan:
a) Vector Method: Write vectors vA and vB in Cartesian
form then determine vB – vA
form,
b) Graphical Method: Draw vectors vA and vB from a
common point. Apply the laws of sines and cosines to
determine vB/A.
EXAMPLE (continued)
Solution:
a) Vector Method:
vA = (650 i ) km/h
vB = –800 cos 60 i – 800 sin 60 j
= ( –400 i – 692.8 j) km/h
vB/A = vB – vA = (–1050 i – 692.8 j) km/h
vB /A = (−1050)2 +(−692.8)2 = 1258 km/h
θ=
tan-1(
692.8
692
8
) = 33.4°
1050
θ
EXAMPLE (continued)
(
)
b) Graphical Method:
N that
Note
h the
h vector that
h measures the
h tip
i off B relative
l i to A is
i vB/A.
vA
120°
vB
vB/A
Law of Cosines:
(vB/A)2 = (800) 2 + (650) 2 − (800) (650) cos 120
120°
vB/A = 1258 km/h
L off Sines:
Law
Si
vB/A
sin(120 °)
=
vA
sin θ
or θ = 33.4
33 4°
EXAMPLE II
Given: vA = 30 mi/h
vB = 20 mi/h
aB = 1200 mi/h2
aA = 0 mi/h2
Find:
vB/A
aB/A
Plan: Write the velocity and acceleration vectors for A and B
and determine vB/A and aB/A by using vector equations
equations.
Solution:
The velocity of B is:
( ) i + 20 cos(30)
( ) j = ((–10 i + 17.32 j) mi/h
vB = –20 sin(30)
EXAMPLE II (continued)
The velocity of A is:
vA = –30
30 i (mi/h)
Th relative
The
l ti velocity
l it off B with
ith respectt to
t A is
i (v
( B/A):
)
vB/A = vB – vA = (–10i
( 10i + 17.32j)
17 32j) – (–30i)
( 30i) = (20 i + 17.32
17 32 j) mi/h
i/h
or
vB/A =
(20)2 + (17.32)
(17 32)2 = 26.5
26 5 mi/h
θ = tan-1(
17.32
) = 40.9
40 9°
20
θ
EXAMPLE II (continued)
The acceleration of B is:
aB = (at)B + (an)B = [–
[ 1200 sin(30) i +1200 cos(30) j]
2
2
20
20
+[(
) cos(30) i +(
) sin(30) j]
0.3
0.3
aB = 554.7
554 7 i +1706 j (mi/h2)
The acceleration of A is zero : aA = 0
The relative
Th
l ti acceleration
l ti off B with
ith respectt to
t A is:
i
aB/A = aB – aA = 554.7 i +1706 j (mi/h2)
aA/B = (554.7)2 + (1706)2 = 1790 mi/h2
β = ttan-11(1706 / 554.7)
554 7) = 72°
β