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ME 012 Engineering Dynamics: Lecture 26
J. M. Meyers, Ph.D. ([email protected])
ME 012 Engineering Dynamics
Lecture 26
FINAL EXAM REVIEW
(Chapter 12, 13, 14, 15, 16, and 17)
Tuesday,
Apr. 30, 2013
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OVERVIEW
Chapter 12: Kinematics of a Particle
• 12.1 Introduction
• 12.2 Rectilinear Kinematics: Continuous Motion
• 12.3 Rectilinear Kinematics: Erratic Motion
• 12.4 General Curvilinear Motion
• 12.5 Curvilinear Motion: Rectangular Components
• 12.6 Motion of a Projectile
• 12.7 Curvilinear Motion: Normal and Tangential
Components
• 12.8 Curvilinear Motion: Cylindrical Components
• 12.9 Absolute Dependent Motion Analysis of Two Particles
• 12.10 Relative-Motion of Two Particles Using Translating
Axes
Chapter 13: Kinematics of a Particle: Force and Acceleration
• 13.1 Newton’s Second Law of Motion
• 13.2 The Equation of Motion
• 13.3 Equation of Motion for a System of Particles
• 13.4 Equations of Motion: Rectangular Coordinates
• 13.5 Equations of Motion: Normal and Tangential Coordinates
• 13.6 Did not cover
• 13.7 Did not cover
• Chapter 12 and 13 material is, in a sense,
introductory material
• The material represents the building blocks
for all subsequent chapters.
• Each and every problem from chapter 14 to
chapter 17 requires the foundations
developed in these chapters
• Go over Exam 1 review
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EXAM FORMAT
Chapter 14: Kinetics of a Particle: Work and Energy
• 14.1 The Work of a Force
• 14.2 Principle of Work and Energy
• 14.3 Principle of Work and Energy for a System of
Particles
• 14.4 Power and Efficiency
• 14.5 Conservative Forces and Potential Energy
• 14.6 Conservation of Energy
Chapter 15: Kinetics of a Particle: Impulse and Momentum
• 15.1 Principle of Linear Impulse and Momentum
• 15.2 Principle of Linear Impulse and Momentum for a System of Particles
• 15.3 Conservation of Linear Momentum for a System of Particles
• 15.4 Impact
• 15.5 Angular Momentum
• 15.6 Relation Between Moment of a Force and Angular Momentum
• 15.7 Angular Impulse and Moment Principles
• 15.8 Did not cover
• 15.9 Did not cover
• This material will not be reviewed as this has
already been done
• Go over Exam 2 review
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EXAM FORMAT
Chapter 16: Planar Kinematics of a Rigid Body
• 16.1 Rigid-Body Motion
• 16.2 Translation
• 16.3 Rotation About a Fixed Axis
• 16.4 Absolute Motion Analysis
• 16.5 Relative Motion Analysis: Velocity
• 16.6 Instantaneous Center of Zero Velocity
• 16.7 Relative Motion Analysis: Acceleration
• 16.8 Did not cover
Chapter 17: Planar Kinematics of a Rigid Body: Force and Acceleration
• 17.1 Moment of Inertia
• 17.2 Planar Kinetic Equations of Motion
• 17.3 Equations of Motion: Translation
• 17.4 Equations of Motion: Rotation About a Fixed Axis
• 17.5 Equations of Motion: General Plane Motion
• Today’s Review will mainly cover material
not already covered by a review session
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There are three types of planar rigid body motion:
1) Translation (Rectilinear and Curvilinear)
2) Rotation about a fixed axis 3) General plane motion
16.1 Rigid Body Motion
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16.2 Rigid-Body Motion: Translation
The positions of two points � and � on a translating body
can be related by:
Note, all points in a rigid body subjected to translation move with the same velocity
and acceleration.
The velocity at B is:
POSITION
VELOCITY
�� = ��+ ��/�where �� & �� are the absolute position vectors defined
from the fixed x-y coordinate system, and ��/� is the
relative-position vector between � and �.
�� = �� + ��/��
Now drB/A/dt = 0 since rB/A is constant, ergo, vB = vA
By following similar logic, aB = aA.
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16.3 Rigid-Body Motion: Rotation About a Fixed Axis
• Angular velocity ( ): obtained by taking the time derivative of angular
displacement:
ANGULAR MOTION
• Angular acceleration (�): if positive, accelerating, if negative, decelerating:
• Angular position (�): angular position of radial line �with units of either
radians or revolutions related by: 1 revolution = 2π radians
= ��
� = � = ��
��• Useful relation between the three quantities: �� =
When a body rotates about a fixed axis, any point P in the body travels along a
circular path.
If the angular acceleration of the body is constant, � = ��:
= � + ��� � = �� + �� + 12���� � = � � + 2�� � − ��
Note these equations are very similar to the constant acceleration relations
developed for the rectilinear motion of a particle.
CONSTANT ANGULAR ACCELERATION
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The magnitude of the velocity of P is equal to r (the text provides the
derivation). The velocity’s direction is tangent to the circular path of P.
In the vector formulation, the magnitude and direction of v can be
determined from the cross product of � and rrrr�. Here rrrr� is a vector from
any point on the axis of rotation to P.
vvvv = �× rrrr� = �× rrrr
MOTION OF POINT P: VELOCITY
Scalar form:
Vector form:
v = � MOTION OF POINT P: ACCELERATION
Using the vector formulation, the acceleration of P can also be defined by
differentiating the velocity.
� = �� = �
� × �� +�× ���
� = � × �� − ��� = aaaa! + aaaa"a! = �r a" = �rScalar form:
Vector form:
a = a! � + a" �
O a!
a"
��
P
MOTION OF POINT P: POSITION
The position of P is defined by the position vector r which extends from Oto P.
16.3 Rigid-Body Motion: Rotation About a Fixed Axis
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16.4 Absolute Motion Analysis
Absolute motion analysis (also called the parametric method) can be used to study planar motion.
• Once a relationship in the form of sP = f(θ) is established, the velocity and acceleration of point Pare obtained in terms of the angular velocity, ω, and angular acceleration, α, of the rigid body by
taking the first and second time derivatives of the position function.
#� = $(�, �) &� = [$ � ] � = $ �, �, )� = �[$(�) ]
�� = $ �, �, , �
Book Example
16.3
* = 2� cos �
Using trigonometry, a relation between the rotational motion
of OA and rectilinear translation of rod R (measured from
fixed point O):
Using the chain rule:
*� = −2� (sin �) �� & = −2� (sin �)
&� = −2�
� (sin �) − 2� (cos �) �� ) = −2� � (sin �) + �(cos �)
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16.5 Relative Motion Analysis: Velocity
0� = 0� + 0�/�
= +
0� = 0� +�× 1�/�
When a wheel rolls without slipping, point
A is often selected to be at the point of
contact with the ground. Since there is no
slipping, point A has zero velocity and
&� = �
• This equation is formulated such that point � moves
along a circular path about point �.
• The relative velocity, 0�/�, acts perpendicular to the
relative displacement vector between � and �… thus
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16.6 Instantaneous Center of Zero Velocity
If the location of this point can be determined, the velocity analysis can be
simplified because the body appears to rotate about this point at that
instant.
For any body undergoing planar motion, there always exists a point in the
plane of motion at which the velocity is instantaneously zero (if it were
rigidly connected to the body) known as the instantaneous center of zero
velocity (IC).
A third case is when the magnitude and direction of two parallel velocities
at � and� are known.
Here the location of the IC is determined by proportional triangles. As a
special case, note that if the body is translating only (vA = vB), then the 23would be located at infinity. Then equals zero, as expected.
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16.6 Instantaneous Center of Zero Velocity
Being able to solve for using 23 analysis can help in solving more complicated relative-
motion analysis problems involving acceleration.
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16.7 Relative Motion Analysis: Acceleration
The equation relating the accelerations of two points on the body is determined by
differentiating the velocity equation with respect to time:
Measured from a set of fixed
x and y axes.
Will develop tangential and
normal components because of
circular path motion!!!!
0�� = 0�
� + 0�/��
Absolute acceleration
of point �Absolute acceleration
of point �= + Acceleration of point �with respect to point �
4� = 4� + 4�/� = 4� + 4�/� ! + 4�/� "
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B
A
B
A
B
A
4� = 4� + � × 1�/� − �1�/�
Shape doesn’t matter so long as the
body is rigid!
Analysis is still thought of as a rigid
bar connecting points A and B
B
A
16.7 Relative Motion Analysis: Acceleration
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4� = 4� + � × 1�/� − �1�/�
B
A
B
A
)�/� "
)�/� !
y’
x’
16.7 Relative Motion Analysis: Acceleration
�4�
4�
1�/�
)�5)�6)�7=
)�5)�6)�7+
�5�6�7×
��/�5��/�6��/�7− �
��/�5��/�6��/�7
iiii jjjj kkkk�5 �6 �7��/�5 ��/�6 ��/�7
You can think of this cross-product
as the determinant of this relation
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17.1 Mass Moment of Inertia
• The figures on the right show the mass moment of inertia formulations for two
flat plate shapes commonly used when working with three dimensional
bodies.
• The shapes are often used as the differential element being integrated over
the entire body.
• Moment of Inertia is a measure of the resistance of a body to angular
acceleration (: = 2�) in the same way that mass is a measure of the body’s
resistance to acceleration (; = <))
• In other words, the mass moment of inertia is a measure of an object’s
resistance to rotation… mathematically:
2 = =�� < = =�� >?< ?
• Center of gravity and mass moment of inertia of these and other homogenous solid geometries are
given inside the back cover of your text (will be supplied for exam)
• Most problems only rely on the analysis on a collective of simple geometries
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PARALLEL-AXIS THEOREM
If the mass moment of inertia of a body about an axis passing
through the body’s mass center is known, then the moment of
inertia about any other parallel axis may be determined by using
the parallel axis theorem:
2� 7 = 2@ 7A +<�
17.1 Mass Moment of Inertia
If a body is constructed of a number of simple shapes, such as disks, spheres, or
rods, the mass moment of inertia of the body about any axis can be determined
by algebraically adding together all the mass moments of inertia, found about the
same axis, of the different shapes.
The mass moment of inertia of a body about a specific axis can be defined using the radius of
gyration (k). The radius of gyration has units of length and is a measure of the distribution of the
body’s mass about the axis at which the moment of inertia is defined
COMPOSITE BODIES
RADIUS OF GYRATION
2 = <B� ⟹ B = 2/<
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17.2 Planar Kinetic Equations of Motion
• Translation (Section 17.3)
• Rectilinear (x-y coordinates)
• Curvilinear (n-t coordinates)
• Pure Rotation (Section 17.4)
• General Plane Motion ((Section 17.5)
Our study of planar kinetics covered three (actually four!) types of motion:
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17.3 Equations of Motion: Translation
When a rigid body undergoes only translation, all
the particles of the body have the same
acceleration so )D = )and � = 0. The equations
of motion become:
RECTILINEAR TRANSLATION
Σ;5 = < )@ 5Σ;6 = < )@ 6Σ:@ = 0
1
2
3
When a rigid body is subjected to curvilinear
translation, it is best to use an n-t coordinate
system. Then apply the equations of motion,
as written below, for n-t coordinates.
CURVIILINEAR TRANSLATION
Σ;" = < )@ "Σ;! = < )@ !Σ:@ = 0Σ:� = Σ ℳG �Σ:� = H ∙ < )@ !−ℎ ∙ < )@ "
1
2
3
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17.4 Equations of Motion: Rotation About a Fixed Axis
• When a rigid body rotates about a fixed axis perpendicular to the plane
of the body at point K, the body’s center of gravity D moves in a
circular path of radius rD.
• Thus, the acceleration of point D can be represented by a tangential
component and a normal component:
aaaa@ ! = rrrr@� aaaa@ " = rrrr@ �acceleration magnitudes
acts in direction
consistent with �acts in direction
from point D to K• Angular acceleration and velocity (� and ) are caused by the external
forces (FFFF) and couple moment system on the body
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17.4 Equations of Motion: Rotation About a Fixed Axis
=
Free Body Diagram Kinetic Diagram
ΣF" = < )@ " = < ��@ΣF! = < )@ ! = <��@
ΣM@ = ID�
ΣF" = < )@ " = < ��@ΣF! = < )@ ! = <��@
ΣM� = Σ(ℳG)�= 2@� + �@< )@ != 2@ +< �@ � �=IK�
• It may be convenient to sum these moments
about the pin point K• Here the moment about pin point K will be
equal to the sum of kinematic moments about
pin point K (Σ(ℳG)�)
• The body experiences an angular acceleration so its
inertia creates a moment of ID� that is equal to the
moment of the external forces about point D.
• The body’s moment of inertia (ID) is calculated about
an axis which is perpendicular to the page and passes
through D
parallel-axis
theorem
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17.5 Equations of Motion: General Plane Motion
17.4) Rotation About
a Fixed Axis
17.5) General Plane
Motion
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17.5 Equations of Motion: General Plane Motion
P
ΣF6 = < )@ 6
ΣF5 = < )@ 5
ΣM� = Σ ℳG �
ΣF6 = < )@ 6
ΣF5 = < )@ 5
ΣM@ = ID�
Sometimes, it may be convenient to write the
moment equation about some point P other
than D. In this case, Σ ℳG �represents the
sum of the moments of ID� and <�@ about
point P.
Using an x-y inertial coordinate system, the
equations of motions about the center of
mass, D, may be written as:
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17.5 Equations of Motion: General Plane Motion
FRICTIONAL ROLLING PROBLEMS
When analyzing the rolling motion of wheels, cylinders, or disks, it may not be known if the body rolls
without slipping or if it slides as it rolls.
• For example, consider a disk with mass < and radius �, subjected to a
known force PPPP.
• The equations of motion will be:
There are 4 unknowns in
these three equations:
F, N, �, and )DΣF6 = < )@ 6ΣF5 = < )@ 5 ⇒ T− U = <)@
⇒ V−<W = 0ΣM@ = ID� ⇒ U� = ID�
Case 1:
• Assume no slipping and use )@ = �� as the 4th equation
• DO NOT use static friction: FX = YZN• After solving, you will need to verify that the assumption was
correct by checking if FX ≤ YZNCase 2:
• Assume slipping and use FX = YGNas the 4th equation.
• In this case, )@ ≠ ��
Hence, we have to make an assumption to provide another equation.
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EXAM FORMAT
• 20 Short Answer Word Problems (your quiz problems and previous short
answer problems and variations thereof may make another appearance)
• 6 HW Style Problems: if the following are not applied, you will lose points:
• Define what is given before you proceed
• Define what is to be found before you proceed
• Show your FBD and KD when applicable
• Show your relative motion diagrams when applicable (pulley problems)
• Show full version of equation used before you put in numbers
• Be clear about your approach… this helps the grading process but also
helps your thought process through the problem
• You will be supplied with:
• Equation Sheets (regular equations and formulations for mass moments of
simple geometries)
• Paper (you will not be using your own)
• You can bring one “cheat sheet” filled out on both sides