CEE 271: Applied Mechanics II, Dynamics
– Lecture 6: Ch.12, Sec.8 –
Prof. Albert S. Kim
Civil and Environmental Engineering, University of Hawaii at Manoa
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CURVILINEAR MOTION: CYLINDRICAL
COMPONENTS
Today’s objectives: Students
will be able to
1
Determine velocity and
acceleration components
using cylindrical
coordinates.
In-class activities:
• Reading Quiz
• Applications
• Velocity Components
• Acceleration Components
• Concept Quiz
• Group Problem Solving
• Attention Quiz
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READING QUIZ
1
In a polar coordinate system, the velocity vector can be
written as v = vr ûr + vθ ûθ = ṙûr + rθ̇ûθ . The term θ̇ is
called
(a)
(b)
(c)
(d)
2
transverse velocity.
radial velocity.
angular velocity.
angular acceleration.
ANS: (c)
The speed of a particle in a cylindrical coordinate system is
(a) ṙ
(b) r
rθ̇
2
ṙ2 + rθ̇
r
2
(d)
ṙ2 + rθ̇ + ż 2
(c)
ANS: (d)
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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?
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CYLINDRICAL COMPONENTS (Section 12.8)
• We can express the location of P in polar coordinates as
r = rûr . Note that the radial direction, r, extends outward
from the fixed origin, O, and the transverse coordinate, θ,
is measured counter-clockwise (CCW) from the horizontal.
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VELOCITY in POLAR COORDINATES
• The instantaneous velocity is defined as:
v=
d(rûr )
dûr
dr
=
= ṙûr + r
dt
dt
dt
• Using the chain rule:
dûr
dûr dθ
=(
)( )
dt
dθ dt
• We can prove that
dûr
= ûθ
dθ
so
dûr
= θ̇ûθ
dt
• Therefore:
v = ṙûr + rθ̇ûθ
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VELOCITY in POLAR COORDINATES (cont’d)
• Thus, the velocity vector has two components: ṙ, called the
radial component, and rθ̇ called the transverse component.
• The speed of the particle at any given instant is the sum of
the squares of both components
q
v = (rθ̇)2 + (ṙ)2
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ACCELERATION (POLAR COORDINATES)
• The instantaneous acceleration
is defined as:
a = dv/dt = (d/dt)(ṙûr + rθ̇ûθ )
• After manipulation, the
acceleration can be expressed as
a = (r̈ − rθ̇2 )ûr + (rθ̈ + 2ṙθ̇)ûθ
• The term (r̈ − r θ̇ 2 ) is the radial
acceleration or ar .
• The term (r θ̈ + 2ṙ θ̇) is the
transverse acceleration or aθ .
• The q
magnitude of acceleration is
a=
(r̈ − rθ̇2 )2 + (rθ̈ + 2ṙθ̇)2
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CYLINDRICAL COORDINATES
• If the particle P moves along a space curve, its position
can be written as rP = rûr + z ûz
• Taking time derivatives and using the chain rule.
vP
= ṙûr + rθ̇ûθ + ż ûz
aP
= (r̈ − rθ̇2 )ûr + (rθ̈ + 2ṙθ̇)ûθ + z̈ ûz
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EXAMPLE
• Given: A car travels along a
circular path. r = 300 ft,
θ̇ = 0.4 (rad/s), θ̈ = 0.2 (rad/s2 )
• Find: Velocity and acceleration
• Plan: Use the polar coordinate
system.
• Solution: r = 300 ft, ṙ = r̈ = 0, and θ̇ = 0.4 rad/s,
θ̈ = 0.2 rad/s2
• Substitute in the equation for velocity
v = ṙûr + rθ̇ûθ = 0ûr + 300(0.4)ûθ
p
v =
02 + 1202 = 120 ft/s
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EXAMPLE (continued)
• Substitute in the equation for acceleration:
a = (r̈ − rθ̇2 )ûr + (rθ̈ + 2ṙθ̇)ûθ
= [0 − 300(0.4)2 ]ûr + [300(0.2) + 2(0)(0.4)]ûθ
= −48ûr + 60ûθ ft/s2
a =
p
(−48)2 + (60)2
= 76.8 ft/s2
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CONCEPT QUIZ
1
If ṙ is zero for a particle, the particle is
(a)
(b)
(c)
(d)
2
not moving.
moving in a circular path.
moving on a straight line.
moving with constant velocity.
ANS: (b)
If a particle moves in a circular path with constant velocity,
its radial acceleration is
(a)
(b)
(c)
(d)
zero.
r̈.
−rθ̇2 .
2ṙθ̇.
ANS: (c)
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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
−1
φ = tan
12
2π(10)
= 10.81◦
• Also, what is the relationship between φ and θ?
• Plan: Use cylindrical coordinates. Since r is constant, all
derivatives of r will be zero.
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GROUP PROBLEM SOLVING (continued)
• Solution:Since r is constant, the velocity only has 2
components: vθ = rθ̇ = v cos φ and vz = ż = v sin φ
Therefore:
φ
• θ̇ = ( v cos
r ) = 0.147 rad/s
• θ̈ = 0
• vz = ż = v sin φ = 0.281 m/s
• z̈ = 0 and ṙ = r̈ = 0
• a = (r̈ − r θ̇ 2 )ûr + (r̈θ + 2ṙ θ̇)ûθ + z̈ ûz
= −(rθ̇2 )ûr = −10(0.1472 )ûr = −0.217ûr m/s2
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ATTENTION QUIZ
1
The radial component of velocity of a particle moving in a
circular path is always
(a)
(b)
(c)
(d)
2
zero.
constant.
greater than its transverse component.
less than its transverse component.
ANS: (a)
The radial component of acceleration of a particle moving
in a circular path is always
(a)
(b)
(c)
(d)
negative.
directed toward the center of the path.
perpendicular to the transverse component of acceleration.
All of the above.
ANS: (d)
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Note
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