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CHAPTER 12
K I N E M AT I C S
OF A
PA R T I C L E
12.5 Curvilinear Motion: Rectangular
12
Components
Occasionally the motion of a particle can best be described along a path
that can be expressed in terms of its x, y, z coordinates.
z
Position. If the particle is at point (x, y, z) on the curved path s shown
in Fig. 12–17a, then its location is defined by the position vector
s
k
i
r = xi + yj + zk
z
r ! xi " yj " zk
j
y
x
y
x
Position
(12–10)
When the particle moves, the x, y, z components of r will be functions of
time; i.e., x = x1t2, y = y1t2, z = z1t2, so that r = r1t2.
At any instant the magnitude of r is defined from Eq. C–3 in
Appendix C as
(a)
r = 4x2 + y2 + z2
And the direction of r is specified by the unit vector ur = r>r.
Velocity. The first time derivative of r yields the velocity of the
z
particle. Hence,
v =
s
v ! vxi " vyj " vzk
y
x
Velocity
When taking this derivative, it is necessary to account for changes in both
the magnitude and direction of each of the vector’s components. For
example, the derivative of the i component of r is
d
dx
di
1xi2 =
i + x
dt
dt
dt
(b)
Fig. 12–17
dr
d
d
d
=
1xi2 +
1yj2 +
1zk2
dt
dt
dt
dt
The second term on the right side is zero, provided the x, y, z reference
frame is fixed, and therefore the direction (and the magnitude) of i does
not change with time. Differentiation of the j and k components may be
carried out in a similar manner, which yields the final result,
v =
dr
= vxi + vy j + vzk
dt
(12–11)
where
#
#
#
vx = x vy = y vz = z
(12–12)
12.5
CURVILINEAR MOTION: RECTANGULAR COMPONENTS
35
# # #
The “dot” notation x, y, z represents the first time derivatives of
x = x1t2, y = y1t2, z = z1t2, respectively.
The velocity has a magnitude that is found from
12
v = 4v2x + v2y + v2z
and a direction that is specified by the unit vector uv = v>v. As discussed
in Sec. 12–4, this direction is always tangent to the path, as shown in
Fig. 12–17b.
z
Acceleration. The acceleration of the particle is obtained by taking
the first time derivative of Eq. 12–11 (or the second time derivative of
Eq. 12–10). We have
s
a ! axi " ayj " azk
dv
= axi + ay j + azk
a =
dt
(12–13)
x
Acceleration
(c)
where
#
$
ax = vx = x
#
$
ay = vy = y
#
$
az = vz = z
(12–14)
Here ax , ay , a z represent, respectively, the first time derivatives of
vx = vx1t2, vy = vy1t2, vz = vz1t2, or the second time derivatives of the
functions x = x1t2, y = y1t2, z = z1t2.
The acceleration has a magnitude
a = 4a2x + a2y + a2z
and a direction specified by the unit vector ua = a>a. Since a represents
the time rate of change in both the magnitude and direction of the
velocity, in general a will not be tangent to the path, Fig. 12–17c.
y
36
12
CHAPTER 12
K I N E M AT I C S
OF A
PA R T I C L E
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.
• In general, the 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 axes do 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.
Procedure for Analysis
Coordinate System.
• A rectangular coordinate system can be used to solve problems
for which the motion can conveniently be expressed in terms of
its x, y, z components.
Kinematic Quantities.
• Since rectilinear motion occurs along each coordinate axis, the
motion along each axis is found using v = ds>dt and a = dv>dt;
or in cases where the motion is not expressed as a function of
time, the equation a ds = v dv can be used.
• In two dimensions, the equation of the path y = f(x) can be used
to relate the x and y components of velocity and acceleration by
applying the chain rule of calculus. A review of this concept is
given in Appendix C.
• Once the x, y, z components of v and a have been determined, the
magnitudes of these vectors are found from the Pythagorean
theorem, Eq. B–3, and their coordinate direction angles from the
components of their unit vectors, Eqs. B–4 and B–5.
12.5
37
CURVILINEAR MOTION: RECTANGULAR COMPONENTS
EXAMPLE 12.9
12
At any instant the horizontal position of the weather balloon in
Fig. 12–18a is defined by x = 18t2 ft, where t is in seconds. If the
equation of the path is y = x2>10, determine the magnitude and
direction of the velocity and the acceleration when t = 2 s.
y
B
y!
SOLUTION
Velocity. The velocity component in the x direction is
d
#
vx = x =
18t2 = 8 ft>s :
dt
x2
10
x
A
16 ft
To find the relationship between the velocity components we will use
the chain rule of calculus. (See Appendix A for a full explanation.)
(a)
d 2
#
#
vy = y =
1x >102 = 2xx>10 = 21162182>10 = 25.6 ft>s c
dt
When t = 2 s, the magnitude of velocity is therefore
v = 4(8 ft>s22 + (25.6 ft>s22 = 26.8 ft>s
Ans.
v ! 26.8 ft/s
The direction is tangent to the path, Fig. 12–18b, where
uv = tan-1
vy
vx
= tan-1
25.6
= 72.6°
8
Ans.
Acceleration. The relationship between the acceleration components
is determined using the chain rule. (See Appendix C.) We have
uv ! 72.6"
B
(b)
d
#
ax = vx =
182 = 0
dt
d
#
#
# #
$
a y = vy =
12xx>102 = 21x2x>10 + 2x1x2>10
dt
Thus,
= 21822>10 + 21162102>10 = 12.8 ft>s2 c
a = 4(02 2 + (12.822 = 12.8 ft> s2
a ! 12.8 ft/s2
B
The direction of a, as shown in Fig. 12–18c, is
-1 12.8
ua = tan
0
= 90°
ua ! 90"
Ans.
(c)
Ans.
NOTE: It is also possible to obtain vy and a y by first expressing
y = f1t2 = 18t22>10 = 6.4t2 and then taking successive time derivatives.
Fig. 12–18
38
12
CHAPTER 12
K I N E M AT I C S
OF A
PA R T I C L E
EXAMPLE 12.10
For a short time, the path of the plane in Fig. 12–19a is described by
y = (0.001x2) m. If the plane is rising with a constant velocity of 10 m>s,
determine the magnitudes of the velocity and acceleration of the plane
when it is at y = 100 m.
y
x
SOLUTION
When y = 100 m, then 100 = 0.001x2 or x = 316.2 m. Also, since
vy = 10 m>s, then
y = vyt;
100 m = (10 m>s) t
t = 10 s
Velocity. Using the chain rule (see Appendix C) to find the
relationship between the velocity components, we have
d
#
#
vy = y =
(0.001x2) = (0.002x)x = 0.002xvx
dt
y
(1)
Thus
y ! 0.001x2
100 m
x
10 m>s = 0.002(316.2 m)(vx)
vx = 15.81 m>s
The magnitude of the velocity is therefore
(a)
v = 4vx2 + vy2 = 4(15.81 m>s)2 + (10 m>s)2 = 18.7 m>s
Ans.
Acceleration. Using the chain rule, the time derivative of Eq. (1)
gives the relation between the acceleration components.
#
#
#
ay = vy = 0.002xvx + 0.002xvx = 0.002(v2x + xax)
#
When x = 316.2 m, vx = 15.81 m>s, vy = ay = 0,
y
a
100 m
0 = 0.002((15.81 m>s)2 + 316.2 m(a x))
ax = - 0.791 m>s2
v
vy
vx
x
(b)
Fig. 12–19
The magnitude of the plane’s acceleration is therefore
a = 4a2x + a2y = 4(- 0.791 m>s222 + (0 m>s222
= 0.791 m>s2
These results are shown in Fig. 12–19b.
Ans.