ME 012 Engineering Dynamics
Lecture 5
Curvilinear motion: Normal, Tangential and Cylindrical Components
(Chapter 12, Sections 7 and 8)
Tuesday,
Jan. 29, 2013
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
CURVILINEAR MOTION: NORMAL, TANGENTIAL and CYLIND. COMPONENTS
Today’s Objectives:
1. [12.7] Determine the normal and tangential
components of velocity and acceleration of
a particle traveling along a curved path.
2. [12.8] Determine velocity and acceleration
components using cylindrical coordinates.
In-Class Activities:
• Applications
• Normal and Tangential Components of
Velocity and Acceleration
• Special Cases of Motion
• Example Problems
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
2
12.7 Curvilinear motion: Normal and Tangential Components
APPLICATIONS
Cars traveling along a clover-leaf interchange
experience an acceleration due to a change in
speed as well as due to a change in direction
of the velocity.
If the car’s speed is increasing at a known rate
as it travels along a curve, we can then
determine the magnitude and direction of its
total acceleration.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
APPLICATIONS (continued)
A motorcycle travels up a hill for
which the path can be approximated
by a function = ( ).
If the motorcycle starts from rest and increases its speed at a constant rate, how can
we determine its velocity and acceleration at the top of the hill?
How would you analyze the motorcycle's “flight” at the top of the hill?
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
When a particle moves along a curved path, it is sometimes convenient to describe its
motion using coordinates other than Cartesian. When the path of motion is known,
normal ( ) and tangential ( ) coordinates are often used.
In the − coordinate system, the origin is
located on the particle (origin moves with the
particle).
The -axis is tangent to the path (curve) at the
instant considered, positive in the direction of
the particle’s motion.
The -axis is perpendicular to the t-axis with the
positive direction toward the center of
curvature of the curve.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
POSITION
•The position of the particle at any instant is
defined by the distance, , along the curve from a
fixed reference point.
•The positive and directions are defined by the
unit vectors and , respectively.
•The center of curvature,
concave side of the curve.
’, always lies on the
RADIUS OF CURVATURE
•The radius of curvature, , is defined as the
perpendicular distance from the curve to the center
of curvature at that point.
•Curve can be constructed into differential segments
of path length
which defines an arc segment of
constant radius of curvature, .
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
VELOCITY
• The velocity vector is always tangent to the path
of motion ( -direction).
• The magnitude is determined by taking the time
derivative of the path function, ( ).
=
=
(where
= )
Here defines the magnitude of the velocity (speed) and defines the direction of
the velocity vector. Again, the velocity acts only tangential to the path!
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change of velocity
=
=
(
)
=
=
+
•How do we find the normal contribution, ?
• Note that particle moves
over interval .
• Using infinitesimal relations of differentials and the
relation
=
we can find that:
=
=
=
The acceleration vector can now be expressed as:
2
=
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
+
=
+
8
12.7 Curvilinear motion: Normal and Tangential Components
ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
There are two components to the acceleration
vector:
=
+ • The tangential component is tangent to the curve
and in the direction of increasing or decreasing
velocity:
=
or
=
• The normal or centripetal (center seeking) component is always directed
toward the center of curvature of the curve:
=
• The magnitude of the acceleration vector is:
=
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1)
The particle moves along a straight line:
→∞⇒
=
=0⇒
=
=
The tangential component represents the time rate of change in the
magnitude of the velocity.
2)
The particle moves along a curve at constant speed:
=
=0⇒
=
=
The normal component represents the time rate of change in the direction of
the velocity.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
SPECIAL CASES OF MOTION (continued)
3)
The tangential component of acceleration is constant, =
Integrating:
1
= &+ & +
= /
2
=
/
4)
&
and
"
1
= &+
2
=
=
As before,
"
&
&
+2
"
"
−
&
are the initial position and velocity of the particle at = 0.
The particle moves along a path expressed as = ( ).
The radius of curvature, , at any point on the path can be calculated from
$%
1+
=
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
THREE-DIMENSIONAL MOTION
If a particle moves along a space curve, the n and t
axes are defined as before. At any point, the -axis
is tangent to the path and the -axis points toward
the center of curvature. The plane containing the n
and t axes is called the osculating plane.
A third axis can be defined, called the binomial axis, (. The binomial unit vector, (, is
directed perpendicular to the osculating plane, and its sense is defined by the cross
product ( = × . (RIGHT HAND RULE FOR DIRECTION!)
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
12
12.7 Curvilinear motion: Normal and Tangential Components
EXAMPLE 1
A jet plane travels along a vertical parabolic path defined
by the equation = 0.4 . At point +, the jet has a
speed of 200 m/s, which is increasing at the rate of 0.8
m/s2. Determine magnitude of the plane’s acceleration
when it is at point +.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
EXAMPLE 1: Solution
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
14
12.7 Curvilinear motion: Normal and Tangential Components
EXAMPLE 2
At a given instant the train engine at - has speed
(20 m/s) and acceleration (14 m/s2) acting in
the direction shown. Determine the rate of
increase in the train's speed and the radius of
curvature of the path for = 75 deg.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.7 Curvilinear motion: Normal and Tangential Components
EXAMPLE 2: Solution
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
16
12.8 Curvilinear motion: Cylindrical Components
APPLICATIONS
A polar coordinate system is a 2-D
representation
of
the
cylindrical
coordinate system.
The cylindrical coordinate system is used
in cases where the particle moves along
a 3-D curve.
When the particle moves in a plane (2-D),
and the radial distance, , is not constant,
the polar coordinate system can be used
to express the path of motion of the
particle.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.8 Curvilinear motion: Cylindrical Components
POLAR COORDINATES - POSITION
We can express the location of / in polar
coordinates as:
. = .
Note that the radial direction, , extends
outward from the fixed origin, , and the
transverse coordinate,
, is measured
counter-clockwise
(CCW)
from
the
horizontal.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
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12.8 Curvilinear motion: Cylindrical Components
POLAR COORDINATES - VELOCITY
The instantaneous velocity is defined as:
.
0
=
=.=
=
0
Using the chain rule:
0
We can prove that:
Therefore:
=
0 0
+
+
0
0
=
=
0
0
1
1 1
=
Where:
1
0
=
1
=
Thus, the velocity vector has two components: , called the
radial component, and
, called the transverse component.
The speed of the particle at any given instant is the sum of the
squares of both components or:
=
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
+
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12.8 Curvilinear motion: Cylindrical Components
POLAR COORDINATES - ACCELERATION
The instantaneous acceleration is defined as:
=
=
=
0
+
1
After manipulation (work out for yourself with assistance
from book), the acceleration can be expressed as:
=
0
+
1
= 2−
Where:
1
=
2 −2
The magnitude of acceleration is:
=
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
2−
+
2 −2
20
12.8 Curvilinear motion: Cylindrical Components
CYLINDRICAL COORDINATES ( −
− 3)
If the particle P moves along a space curve, its
position can be written as
.4 =
0
+3
5
Taking time derivatives and using the chain
rule:
Velocity:
4
=
0
+
1
+3
5
Acceleration:
4
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
=
2−
0
+
2 −2
1
+ 32
5
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12.8 Curvilinear motion: Cylindrical Components
EXAMPLE 3
A particle travels along a portion of the “four-leaf rose”
defined by the equation = 5 cos 2 m. If the angular
velocity of the radial coordinate line is ; = 3 2 rad/s,
determine the radial and transverse components of the
particle’s velocity and acceleration at the instant
= 30deg. Note, when = 0 , = 0°.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
22
12.8 Curvilinear motion: Cylindrical Components
EXAMPLE 3: Solution
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
23
12.8 Curvilinear motion: Cylindrical Components
EXAMPLE 4
A boy slides down a slide at a constant speed = 2 m/s.
The slide is in the form of a helix, defined by the
equations:
= 1.5[m] and 3 = −(2 )/(2D) [m],
Determine the boy’s angular velocity about the 3-axis, ′
and the magnitude of his acceleration.
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
24
12.8 Curvilinear motion: Cylindrical Components
EXAMPLE 4: Solution
ME 012 Engineering Dynamics: Lecture 5
J. M. Meyers, Ph.D. (jmmeyers@uvm.edu)
25