CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
(Section 12.8)
Today’s Objectives:
In-Class Activities:
Students will be able to
• Check homework, if any
determine velocity and
• Reading quiz
acceleration components
using cylindrical coordinates.
• Applications
• Velocity Components
• Acceleration Components
• Concept quiz
• Group problem solving
• Attention quiz
READING QUIZ
1. In a polar coordinate system, the velocity
vector can
be
.
.
.
written as v = vrer + vθeθ = rer + rqeq. The term q is called
A) transverse velocity.
B) radial velocity.
C) angular velocity.
D) angular acceleration.
2. The speed of a particle in a cylindrical coordinate system is
.
B) rq
.
A) r
C)
. 2 .2
(rq) + (r)
D)
. 2 . 2 .2
(rq) + (r) + (z)
APPLICATIONS
The cylindrical coordinate
system is used in cases
where the particle moves
along a 3-D curve.
In the figure shown, the boy
slides down the slide at a
constant speed of 2 m/s.
How fast is his elevation
from the ground
. changing
(i.e., what is z )?
APPLICATIONS (continued)
A polar coordinate system is a 2-D representation of the
cylindrical coordinate system.
When the particle moves in a plane (2-D), and the radial
distance, r, is not constant, the polar coordinate system can
be used to express the path of motion of the particle.
POSITION (POLAR COORDINATES)
We can express the location of P in polar coordinates as r = rer. Note
that the radial direction, r, extends outward from the fixed origin, O,
and the transverse coordinate, q, is measured counter-clockwise
(CCW) from the horizontal.
VELOCITY (POLAR COORDINATES)
The instantaneous velocity is defined as:
v = dr/dt = d(rer)/dt
der
.
v = rer + r dt
Using the chain rule:
der/dt = (der/dq)(dq/dt)
.
We can prove that der/d. q = eθ so der/dt = qeθ
.
Therefore: v = rer + rqeθ
.
Thus, the velocity vector has two components:
r,
.
called the radial component, and rq, called the
transverse component. The speed of the particle at
any given instant is the sum of the squares of both
components or
. 2
.
v = (r q ) + ( r )2
ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rer + rqeθ)
After manipulation, the acceleration can be
expressed as
..
..
.. . 2
a = (r – rq )er + (rq + 2rq)eθ
.. . 2
The term (r – rq ) is the radial acceleration
or ar.
..
..
The term (rq + 2rq) is the transverse
acceleration or aq
..
.. 2
.. . 2 2
The magnitude of acceleration is a = (r – rq ) + (rq + 2rq)
CYLINDRICAL COORDINATES
If the particle P moves along a space
curve, its position can be written as
rP = rer + zez
Taking time derivatives and using
the chain rule:
Velocity:
.
.
.
vP = rer + rqeθ + zez
..
..
.. . 2
..
Acceleration: aP = (r – rq )er + (rq + 2rq)eθ + zez
EXAMPLE
Given: r. = 5 cos(2q) (m)
q = 3t2 (rad/s)
qo = 0
Find: Velocity and acceleration at q = 30°.
.
..
Plan: Apply chain rule to determine r and r
and evaluate at q = 30°.
t
Solution:
t
.
q =  q dt =  3t2 dt = t3
t o= 0
At q = 30°,
q=
0
p
= t3. Therefore: t = 0.806 s.
6
.
q = 3t2 = 3(0.806)2 = 1.95 rad/s
EXAMPLE (continued)
..
q = 6t = 6(0.806) = 4.836 rad/s2
r = 5 cos(2q) = 5 cos(60) = 2.5m
.
.
r = -10 sin(2q)q = -10 sin(60)(1.95) = -16.88 m/s
.
..
..
r = -20 cos(2q)q2 – 10 sin(2q)q
= -20 cos(60)(1.95)2 – 10 sin(60)(4.836) = -80 m/s2
Substitute in the equation
for velocity
.
.
v = rer + rqeθ
v = -16.88er + 2.5(1.95)eθ
v = (16.88)2 + (4.87)2 = 17.57 m/s
EXAMPLE (continued)
Substitute in the equation for acceleration:
..
..
.. . 2
a = (r – rq )er + (rq + 2rq)eθ
a = [-80 – 2.5(1.95)2]er + [2.5(4.836) + 2(-16.88)(1.95)]eθ
a = -89.5er – 53.7eθ m/s2
a = (89.5)2 + (53.7)2 = 104.4 m/s2
CONCEPT QUIZ
.
1. If r is zero for a particle, the particle is
A) not moving.
B) moving in a circular path.
C) moving on a straight line.
D) moving with constant velocity.
2. If a particle moves in a circular path with constant velocity, its
radial acceleration is
A) zero.
.
C) -rq2.
..
B) r.
..
D) 2rq.
GROUP PROBLEM SOLVING
Given: The car’s speed is constant at
1.5 m/s.
Find: The car’s acceleration (as a
vector).
Hint:
The tangent to the ramp at any
point is at an angle
12
-1
f = tan (
) = 10.81°
2p(10)
Also, what is the relationship between f and q?
Plan: Use cylindrical coordinates. Since r is constant, all
derivatives of r will be zero.
Solution: Since r .is constant the velocity only has 2 components:
.
vq = rq = v cosf and vz = z = v sinf
GROUP PROBLEM SOLVING (continued)
.
v cosf
Therefore: q = (
) = 0.147 rad/s
r
..
q = 0
.
vz = z = v sinf = 0.281 m/s
..
z = 0
.
..
r = r = 0
..
..
.. . 2
..
a = (r – rq )er + (rq + 2rq)eθ + zez
.2
a = (-rq )er = -10(0.147)2er = -0.217er m/s2
Example
SOLUTION:
• Evaluate time t for q = 30o.
• Evaluate radial and angular positions,
and first and second derivatives at
time t.
Rotation of the arm about O is defined
by q = 0.15t2 where q is in radians and t
in seconds. Collar B slides along the
arm such that r = 0.9 - 0.12t2 where r is
in meters.
After the arm has rotated through 30o,
determine (a) the total velocity of the
collar, (b) the total acceleration of the
collar, and (c) the relative acceleration
of the collar with respect to the arm.
• Calculate velocity and acceleration in
cylindrical coordinates.
• Evaluate acceleration with respect to
arm.
SOLUTION:
• Evaluate time t for q = 30o.
q = 0.15 t
2
= 30  = 0 . 524 rad
t = 1 . 869 s
• Evaluate radial and angular positions, and first
and second derivatives at time t.
r = 0 . 9  0 . 12 t
2
= 0 . 481 m
r =  0 . 24 t =  0 . 449 m s
r =  0 . 24 m s
2
2
q = 0 . 15 t = 0 . 524 rad
q = 0 . 30 t = 0 . 561 rad s
q = 0 . 30 rad s
2
• Calculate velocity and acceleration.
v r = r =  0 . 449 m s
vq = r q = ( 0 . 481 m )( 0 . 561 rad s ) = 0 . 270 m s
v=
2
2
 = tan
v r + vq
 1 vq
vr
v = 0 . 524 m s
 = 31 . 0 
2
a r = r  r q
=  0 . 240 m s  ( 0 . 481 m )( 0 . 561 rad s )
2
=  0 . 391 m s
aq = r q + 2 rq
2
(
= ( 0 . 481 m ) 0 . 3 rad s
=  0 . 359 m s
a =
2
2
a r + aq
2
2
) + 2 (  0 .449 m s )(0 .561 rad
s)
2
 = tan
 1 aq
ar
a = 0 . 531 m s
 = 42 . 6 
• Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is rectilinear
and defined by coordinate r.
aB
OA
= r =  0 . 240 m s
2
ATTENTION QUIZ
1. The radial component of velocity of a particle moving in a
circular path is always
A) zero.
B) constant.
C) greater than its transverse component.
D) less than its transverse component.
2. The radial component of acceleration of a particle moving in
a circular path is always
A) negative.
B) directed toward the center of the path.
C) perpendicular to the transverse component of acceleration.
D) All of the above.