ME 012 Lecture 5

ME 012 Engineering Dynamics: Lecture 5

J. M. Meyers, Ph.D. ([email protected])

ME 012 Engineering Dynamics

Lecture 5

Curvilinear motion: Normal, Tangential and Cylindrical Components

(Chapter 12, Sections 7 and 8)

Tuesday,

Jan. 29, 2013


J. M. Meyers, Ph.D. ([email protected]) 2

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



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.




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?



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 perpendicular to the t-axis with the

positive direction toward the center of

curvature of the curve.


The �-axis is tangent to the path (curve) at the

instant considered, positive in the direction of

the particle’s motion.



•The positive �and � directions are defined by the

unit vectors ��and ��, respectively.

•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, �.


•The center of curvature, �’, always lies on the

concave side of the curve.

POSITION

RADIUS OF CURVATURE

•The position of the particle at any instant is

defined by the distance, �, along the curve from a

fixed reference point.




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, �(�).

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!

� = �� = �

�(where � = ��)




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:

� = �� + �2

�� = �� + ��

The acceleration vector can now be expressed as:

� = �

�= (��)

�= �� = �� + ��

�� = �� =��

�� =

�

��




ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)

There are two components to the acceleration

vector:

• The normal or centripetal (center seeking) component is always directed

toward the center of curvature of the curve:

• The tangential component is tangent to the curve

and in the direction of increasing or decreasing

velocity:

• The magnitude of the acceleration vector is:

� = �� + ��

�� = �� = � �or

�� =��

�

� = �� + ��




SPECIAL CASES OF MOTION

There are some special cases of motion to consider.

2) The particle moves along a curve at constant speed:

The normal component represents the time rate of change in the direction of

the velocity.

1) The particle moves along a straight line:

The tangential component represents the time rate of change in the

magnitude of the velocity.

� → ∞ ⇒ �� =��

�= 0 ⇒ � = �� = ��

�� = �� = 0 ⇒ � = �� =��

�



SPECIAL CASES OF MOTION (continued)

3) The tangential component of acceleration is constant, �� = �� "Integrating:

4) The particle moves along a path expressed as � = �(�).The radius of curvature, �, at any point on the path can be calculated from

� =

1 + � �

�$�%

��

� = �& + �&� +1

2�� "�

�

� = �& +1

2�� "�

�� = �&� + 2 �� " � − �&

As before, �& and �& are the initial position and velocity of the particle at � = 0.


� = �/ �

�� = �/ �

�� = � �



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!)




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 0.8m/s2. Determine magnitude of the plane’s acceleration

when it is at point +.



EXAMPLE 1: Solution





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.



EXAMPLE 2: Solution




APPLICATIONS

12.8 Curvilinear motion: Cylindrical Components

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, �, is not constant,

the polar coordinate system can be used

to express the path of motion of the

particle.

The cylindrical coordinate system is used

in cases where the particle moves along

a 3-D curve.




Note that the radial direction, �, extends

outward from the fixed origin, �, and the

transverse coordinate, � , is measured

counter-clockwise (CCW) from the

horizontal.

POLAR COORDINATES - POSITION

. = ��.

We can express the location of / in polar

coordinates as:




POLAR COORDINATES - VELOCITY

The instantaneous velocity is defined as:

We can prove that:

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:

� = .

�= .� =

��0 �

= �� 0 + � �0 �

� = ��+ ��

�0 �= �0 �

�

�Using the chain rule:

Therefore:

�0 �= �1

�0 �= ��1

� = �0�0 + �1�1�1 = ��0 = ��

Where:




POLAR COORDINATES - ACCELERATION

After manipulation (work out for yourself with assistance

from book), the acceleration can be expressed as:

The magnitude of acceleration is:

The instantaneous acceleration is defined as:

� = �� = �

�=

��0 + ��1

� = ��0 + ��1

� = �2 − ��+ ��2 − 2��

�

�� = �2 − ��

�1 = ��2 − 2��Where:




If the particle P moves along a space curve, its

position can be written as

Velocity:

CYLINDRICAL COORDINATES (� − � − 3)

.4 = ��0 + 3�5Taking time derivatives and using the chain

rule:

�4 = ��0 + ��1 + 3��5

�4 = �2 − �� 0 + ��2 − 2�� 1 + 32�5

Acceleration:




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�2rad/s,

determine the radial and transverse components of the

particle’s velocity and acceleration at the instant

� = 30deg. Note, when � = 0�, � = 0°.




EXAMPLE 3: Solution




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.




EXAMPLE 4: Solution

ME 012 Lecture 5

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