Ch 12 (4) Curvilinear Motion X-Y Coordinate.pptx

KINEMATICS OF PARTICLES
Kinematics of particles
Road Map
Rectilinear motion Curvilinear motion
x-y coord. n-t coord. r- coord.
Engineering Dynamics

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
or acceleration of each plane at any
instant?

APPLICATIONS (continued)
A roller coaster car travels down
a fixed, helical path at a constant
speed.
How can we determine its
position or acceleration at any
instant?

General Curvilinear Motion
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 Ds along the
curve during time interval Dt, the
displacement is determined by vector
subtraction: D 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 Dt is
vavg = Dr/Dt .
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. 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 velocity changes from v to v’ over a
time increment Dt, the average acceleration during
that increment is:
aavg = Dv/Dt = (v’ - v)/Dt
The instantaneous acceleration is the time-
derivative of velocity:
a = dv/dt = d2r/dt2

Given: A particle travels along the curve from A to B in 2s. It takes 4s for it
to go from B to C and then 3s to go from C to D.
Find: Determine its average speed when it goes from A to D.
r1=10m, d=15m, r2=5m.
Solution:
56
.
38
)
5
)(
2
(
4
1
15
)
10
)(
2
(
4
1



 

t
s
s
m
t
s
v
AvgSpeed
t
t
sp 28
.
4
3
4
2
56
.
38
)
( 





Given: A car travelling along the straight portion of the road has the
velocities indicated in the figure when it arrives at a points A,B and
C. If it takes 3s to go from A to B and 5s from B to C.
Find: Determine the average acceleration between points A-B and A-C.
tAB=3s. tBC=5s. vA=20m/s, vB=30m/s vC=40m/s, θ =45 deg.
i
vA 20

j
i
vB 21
.
21
21
.
21 

i
vC 40

3
20
21
.
21
21
.
21 i
j
i
t
v
aAB



D
D

2
}
07
.
7
404
.
0
{ s
m
j
i
aAB 

i
i
i
t
v
aAC 50
.
2
8
20
40



D
D

Solution:
If a particle’s velocity changes
from v to v’ over a time increment
Dt, the average acceleration during
that increment is:
aavg = Dv/Dt = (v’ - v)/Dt

Kinematics of particles
Road Map
Rectilinear motion Curvilinear motion
x-y coord. n-t coord. r- coord.
Used to solve problems for
which the motion can be
conveniently expressed in
terms of its x,y & z compts

Road Map

When position vector
of particle P is given
by its rectangular
components

Velocity has a magnitude
First time derivative of r gives velocity

Direction of a is not tangent to the path of the particle
ua = a/a

Curvilinear Motion: Important Points
Curvilinear motion can cause changes in both the magnitude and
direction of the position, velocity and acceleration vectors
The velocity vector is always directed tangent to the path
acceleration vector is not tangent to the path but rather, it is tangent
to the hodograph
If the motion is described using rectangular coordinates, then the
components along each of the axis does not change direction, only
their magnitude and sense (algebraic sign) will change
By considering the component motions, the change in magnitude
and direction of the particle’s position and velocity are automatically
taken into account

V = { 16t2i + 4t3j + (5t + 2)k }

Given: The flight path of the helicopter as it takes off from A is defined by
the equations x=2t2 and y=0.04t3, where t is the time in seconds
after takeoff.
Find: Determine the distance the helicopter is from point A and the
magnitude of its velocity and acceleration when t= 10s.
2
2t
x  3
04
.
0 t
y 
At t =10s. x = 200m y =40m
.
204
)
40
(
)
200
( 2
2
m
d 


Solution:
t
dt
dx
vx 4


4


dt
dv
a x
x
2
12
.
0 t
dt
dy
vy 

t
dt
dv
a
y
y 24
.
0


At t =10s.
s
m
v 8
.
41
)
12
(
)
40
( 2
2



2
2
2
66
.
4
)
4
.
2
(
)
4
( s
m
a 



Projectile motion is a form of motion in which an object or
particle (called a projectile) is thrown obliquely near the
earth's surface, and it moves along a curved path under the
action of gravity.
Projectile motion only occurs when there is one force applied
at the beginning of the trajectory, after which there is no force
in operation apart from gravity
The path followed by a projectile is called its trajectory which
is directly influenced by gravity
Free flight motions are studied in terms of rectangular
component since projectile’s acceleration always acts vertically
Motion of a Projectile
(Rectangular Components of Velocity & Acceleration)

A projectile is any object propelled through space by the exertion of a
force which ceases after launch
APPLICATIONS

APPLICATIONS
A kicker should know at what angle, , 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, , should he hold the
hose?

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)
s = so + vot + (1/2) act2 v = vo + act v2 = (vo)2 + 2ac (s - so)
Consider a projectile launched at (xo, yo)
Only force acting on the projectile is its weight, resulting in
constant downward acceleration
Air resistance is neglected

KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = -g. Application of
the constant acceleration equations yields:
For any given problem, only two
of these three equations can be
used.
vy = voy – g(t)
y = yo + (voy)(t) – ½g(t)2
vy
2 = voy
2 – 2g(y – yo)
v = vo + act s = so + vot + (1/2) act2 v2 = (vo)2 + 2ac (s - so)

Horizontal motion: Velocity in x direction is const vx =vo)x
Verticle Motion,

Example 2
Solving the two equations together (two unknowns) yields
R = 42.78 m tAB = 3.72 s
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 – 0.5gctAB
2 vAy = 15 sin 40° m/s
Note that xB = R, xA = 0, yB = -(3/4)R, and yA = 0.
Given: Snowmobile is going 15
m/s at point A.
Find: The horizontal distance it
travels (R) and the time in
the air.

GROUP PROBLEM SOLVING
Given: Skier leaves the ramp at
A = 25o and hits the
slope at B.
Find: The skier’s initial speed vA.

GROUP PROBLEM SOLVING (continued)
Motion in x-direction:
Using xB = xA + vox(tAB)
=
tAB= (4/5)100
vA (cos 25)
88.27
vA
vA = 41.03 m/s
-64 = 0 + vA(sin 25)
88.27
vA
– ½ (9.81)
88.27
vA
2
Motion in y-direction:
Using yB = yA + voy(tAB) – ½ g(tAB)2
Solution:

Example
Given: vo and θ
Find: The equation that defines
y as a function of x.
Solution: Using vx = vo cos θ
y = (vo sin θ) x g x
vo cos θ 2 vo cos θ
–
2
( ) ( )( )
By substituting for t:
vy = vo sin θ
and
We can write: x = (vo cos θ)t
and y = (vo sin θ)t – ½ g(t)2

Example 1 (continued):
The above equation is called the “path equation” which describes the
path of a particle in projectile motion.
Simplifying the last equation, we get:
y = (x tan) –
g x2
2vo
2
(1 + tan2)
( )

Given: The fireman standing on the ladder directs the flow of water from his
hose to the fire at B.
Find: Determine the velocity of water at A if it is observed that the hose is
held at θ =20o.
Solution:
t
v
X
X ox
o 

t
v o
A )
20
cos
(
0
20 

A
v
t
28
.
21

2
2
1
t
a
t
v
Y
Y c
oy
o 


)
( 

)
( 

2
28
.
21
)
81
.
9
(
2
1
28
.
21
20
sin
0
10 



















A
A
o
A
v
v
v
1
.
2221
)
20
(sin
28
.
21
10
2
2



 A
A v
v 
s
m
vA 57
.
28

10
20

Engineering Statics
ASSIGNMENT NO: 1 FIRST QUESTION

Ch 12 (4) Curvilinear Motion X-Y Coordinate.pptx

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Ch 12 (4) Curvilinear Motion X-Y Coordinate.pptx