CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
Today’s Objectives:
Students will be able to:
1. Determine velocity and
acceleration components
using cylindrical
coordinates.
In-Class Activities:
• Check Homework
• Reading Quiz
• 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 = vrur + vθuθ = rur + rquq. 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
.
.
A) r
C)
B) rq
.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 box
slides down the helical ramp.
How would you find the
box’s velocity components to
know if the package will fly
off the ramp?
CYLINDRICAL COMPONENTS
(Section 12.8)
We can express the location of P in polar coordinates as r = r ur.
Note that the radial direction, r, extends outward from the fixed
origin, O, and the transverse coordinate, q, is measured counterclockwise (CCW) from the horizontal.
VELOCITY in POLAR COORDINATES)
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
dur
.
v = rur + r
dt
Using the chain rule:
dur/dt = (dur/dq)(dq/dt)
.
We can prove that dur/d
. q = uθ so dur/dt = quθ
.
Therefore: v = rur + rquθ
.
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
.
.
v = (r q )2 + ( r )2
ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rur + rquθ)
After manipulation, the acceleration can be
expressed as
..
.
..
..
2
a = (r – rq )ur + (rq + 2rq)uθ
.
..
The term (r – rq 2) is the radial acceleration
or ar .
..
The term (rq + 2rq) is the transverse
acceleration or aq .
.. . .
.2 2
..
The magnitude of acceleration is a = (r – rq ) + (rq + 2rq) 2
..
CYLINDRICAL COORDINATES
If the particle P moves along a space
curve, its position can be written as
rP = rur + zuz
Taking time derivatives and using
the chain rule:
Velocity:
.
.
.
vP = rur + rquθ + zuz
..
.. . 2
..
..
Acceleration: aP = (r – rq )ur + (rq + 2rq)uθ + zuz
EXAMPLE
Given: A car travels along acircular path.
.
r = 300 ft, q = 0.4 (rad/s),
..
q = 0.2 (rad/s2)
Find: Velocity and acceleration
Plan: Use the polar coordinate system.
Solution:
..
.
. ..
r = 300 ft, r = r = 0, and q = 0.4 (rad/s), q = 0.2 (rad/s2)
Substitute in the equation
for velocity
.
.
v = r ur + rq uθ = 0 ur + 300 (0.4) uθ
v = (0)2 + (120)2 = 120 ft/s
EXAMPLE
(continued)
Substitute in the equation for acceleration:
.. . .
.
..
2
a = (r – rq )ur + (rq + 2rq)uθ
a = [0 – 300(0.4)2] ur + [300(0.2) + 2(0)(0.4)] uθ
a = – 48 ur + 60 uθ ft/s2
a =
(– 48)2 + (60)2 = 76.8 ft/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) − rq 2.
..
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).
Plan:
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.
GROUP PROBLEM SOLVING (continued)
Solution: Since r is
has 2 components:
. constant, the velocity only
.
vq = rq = v cos f and vz = z = v sin f
.
Therefore: q = (
..
v cosf
) = 0.147 rad/s
r
q = 0
.
vz = z = v sinf = 0.281 m/s
..
z = 0
.
..
r = r = 0
.
.. . .
..
..
2
a = (r – rq )ur + (rq + 2rq)uθ + zuz
.
a = (-rq 2)ur = -10(0.147)2ur = -0.217ur m/s2
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.