Emm3104 chapter 1

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EMM 3104 : Dynamics 1
Topic review:
• A particle moving along a straight line is subjected
to a deceleration a=(-2v3) m/s2, where v is in m/s. If
it has a velocity v=8 m/s and a position s=10 m
when t = 0, determine its velocity and position
when t = 4 s.

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 or
acceleration of the plane at any
instant?
CURVILINEAR MOTION:
GENERAL & RECTANGULAR COMPONENTS

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?
If you are designing the track, why is it important to be able
to predict the acceleration of the car?

GENERAL CURVILINEAR MOTION
A particle moving along a curved path undergoes curvilinear motion. Since
the motion is often three-dimensional, vectors are used to describe the
motion.
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.
A particle moves along a curve defined
by the path function, s.
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. Since the arc length Ds
approaches the magnitude of Dr 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
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.

CURVILINEAR MOTION: RECTANGULAR COMPONENTS
It is often convenient to describe the motion of a particle in terms of
its x, y, z or rectangular components, relative to a fixed frame of
reference.
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
The position of the particle can be
defined at any instant by the position
vector
r = x i + y j + z k .
The x, y, z components may all be
functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .

RECTANGULAR COMPONENTS: VELOCITY
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.
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, k are constant in magnitude and direction,
this equation reduces to v = vx i + vy j + vz k
where vx = = dx/dt, vy = = dy/dt, vz = = dz/dtx y z
•••

RECTANGULAR COMPONENTS: ACCELERATION
The direction of a is usually not
tangent to the path of the particle.
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 = = = dvx /dt, ay = = = dvy /dt,
az = = = dvz /dt
vx x vy y
vz z
• •• ••
•••
•
The magnitude of the acceleration vector is
a = [(ax)2 + (ay)2 + (az)2 ]0.5

EXAMPLE
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(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)
1) The point of collision requires that rA = rB,
so xA = xB and yA = yB .
Solution:
Set the 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 equal: 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 (only common time).
Substituting this value into rA or rB yields
xA = xB = 6 m and yA = yB = 0

EXAMPLE (continued)
2) Differentiate rA and rB to get the velocity vectors.
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
vB = drB/dt = xB i + yB j = [ (6t – 6) i + 3 j ] m/s
At t = 2 s: vB = [ 6 i + 3 j ] m/s
• •
vA = drA/dt = = [ 3 i + (18 – 18t) j ] m/s
At t = 2 s: vA = [ 3i – 18 j ] m/s
jyAixA. +
• •

EXAMPLE
Given: The velocity of the particle 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.
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 the magnitude of the acceleration vector using
t = 2 s.

Solution:
1) x-components:
2) y-components:
Velocity known as: vx = x = dx/dt = (16 t2 ) m/s
Position: = (16 t2) dt  x = (16/3)t3 = 42.7 m at t = 2 s
Acceleration: ax = x = vx = d/dt (16 t2) = 32 t = 64 m/s2••

x
dx
0

t
0
•
•
Velocity known as: vy = y = dy/dt = (4 t3 ) m/s
Position: = (4 t3) dt  y = t4 = (16) m at t = 2 s
Acceleration: ay = y = vy = d/dt (4 t3) = 12 t2 = 48 m/s2••

y
dy
0

t
0
•
•
EXAMPLE (continued)

3) z-components:
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 m/s2
4) The position vector and magnitude of the acceleration vector
are written using the component information found above.
Velocity is known as: vz = z = dz/dt = (5 t + 2) m/s
Position: = (5 t + 2) dt  z = (5/2) t2 + 2t = 14 m at t=2s
Acceleration: az = z = vz = d/dt (5 t + 2) = 5 m/s2••

z
dz
0

t
0
•
•
EXAMPLE (continued)

Example

APPLICATIONS
A good kicker instinctively knows at what angle, q, 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?
MOTION OF A PROJECTILE

A basketball is shot at a certain angle. What parameters should the
shooter consider in order for the basketball to pass through the
basket?
Distance, speed, the basket location, … anything else ?

A firefighter needs to know the maximum height on the wall he
can project water from the hose. What parameters would you
program into a wrist computer to find the angle, q, that he should
use to hold the hose?

MOTION OF A PROJECTILE
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).
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)?

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 – ½ g t2
vy
2 = voy
2 – 2 g (y – yo)
For any given problem, only two of these three equations can be
used because the last equation is the elimination of t from the
first two equations.

Given: vo and θ
Find: The equation that defines y
as a function of x.
Plan: Eliminate time from the
kinematic equations.
Solution: Using vx = vo cos θ and vy = vo sin θ
We can write: x = (vo cos θ)t or
y = (vo sin θ) t – ½ g (t)2
t =
x
vo cos θ
y = (vo sin θ) { } { }2x g x
vo cos θ 2 vo cos θ
–
By substituting for t:
EXAMPLE I

EXAMPLE I
(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 tanq) –
g x2
2vo
2
(1 + tan2q)

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 (in this solution, the
origin of the coordinate system is placed at A). Apply the
kinematic relations in x- and y-directions.

EXAMPLE II (continued)
Solving for tAB first, tAB = 19.89 s.
Then, R = 120 tAB = 120 (19.89) = 2387 m
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 + vAy tAB – 0.5 g tAB
2
where yB = – 150, yA = 0, and vAy = 150(3/5) m/s
We get the following equation: –150 = 90 tAB + 0.5 (– 9.81) tAB
2

EXAMPLE III
Given: A skier leaves the ski
jump ramp at qA = 25o
and hits the slope at B.
Find: The skier’s initial speed vA.
Plan:
Establish a fixed x,y coordinate system (in this solution, the
origin of the coordinate system is placed at A). Apply the
kinematic relations in x and y-directions.
x
y

Motion in x-direction:
Using xB = xA + vox(tAB) => (4/5)100 = 0 + vA (cos 25) tAB
=tAB= 80
vA (cos 25)
88.27
vA
vA = 19.42 m/s
– 64 = 0 + vA(sin 25) { }
88.27
vA
– ½ (9.81) { }288.27
vA
Motion in y-direction:
Using yB = yA + voy(tAB) – ½ g(tAB)2
Solution: EXAMPLE III (continued)

Example

Exercise 1.2
• Problems no (page 46):
– 12.71, 12.74, 12.77, 12.83, 12.87, 12.89, 12.91, 12.93,
12.95.

Topic review

Emm3104 chapter 1

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