Lecture curvilinear

CURVILINEAR MOTION: RECTANGULAR COMPONENTS
(Sections 12.4-12.5)
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:
•Applications
• General curvilinear motion
• Rectangular components of
kinematic vectors
• Group problem solving

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?
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 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?

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!

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 time-
derivative of velocity:
a = dv/dt = d2
r/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, 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 = 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) .
The magnitude of the position vector is: r = (x2
+ y2
+ z2
)0.5
The direction of r is defined by the unit vector: λr = (1/r)r
Note: I’m used to λ for a unit vector book uses u

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.
Since the unit vectors i, j, k are constant in magnitude and
direction, this equation reduces to v = vxi + vyj + vzk
where vx = = dx/dt, vy = = dy/dt, vz = = dz/dtx y z
•••
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt

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 = d2
r/dt2
= axi + ayj + azk
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
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:
x-components: 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
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 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
vA = drA/dt = = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s
jyAixA. +
vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
. .
• •

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:
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
Velocity: vx = x = dx/dt = (5t) ft/s
Position: = => x = (5/2)t2
= (2.5t2
) ft
Acceleration: ax = x = vx = d/dt (5t) = 5 ft/s2
••
∫
x
dx
0
∫
t
0
dtt5
•
•

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
Acceleration vector: a = [5 i + 37.5t2
j ] ft/s2
Magnitude: a = (52
+ 37.52
)0.5
= 37.8 ft/s2
The magnitude of the acceleration vector is calculated as:

Lecture curvilinear

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