Transcript Chapter 1 - Portal UniMAP

ENT 252 DYNAMICS
Program: B. Eng. (Mechanical)
School of Mechatronic Engineering
Universiti Malaysia Perlis
UniMAP
Contents
• Course Description
• Syllabus
• Textbooks
• Contact Information
Course Description
Unit Name: ENT 252 Dynamics.
Lecturer: Mr. Mohd Noor Arib Md Rejab.
Contact Hours: 5 hours per week.
Tuition Pattern: 3 hours lecture, 2 hours tutorial/laboratory.
Credits: 4 credits
Pre-Requisites : None
Objective: To introduce the students to the basic theories and
application of engineering mechanics.
Syllabus
Chapter 1. Kinematics of a Particle
1.1 Introduction.
1.2 Rectilinear Kinematics: Continuous Motion.
1.3 Rectilinear Kinematics: Erratic Motion
1.4 General Curvilinear Motion
1.5 Curvilinear Motion: Rectangular Component
1.6 Motion of a Projectile
1.7 Curvilinear Motion: Normal and Tangential Components
Chapter 2. Kinetics of a Particle: Force and Acceleration
2.1 Newton’s Law of Motion
2.2 The Equation of Motion
2.3 Equations of Motion for a System of Particles
2.4 Equations of Motion: Rectangular Coordinates
2.5 Equation of Motion: Normal and Tangential Coordinates
Chapter 3. Kinetics of a Particle: Work and Energy
3.1 The Work of a Force
3.2 Principle of Work and Energy
3.3 Principle of Work and Energy for a System of Particles
3.4 Power and Efficiency
3.5 Conservative Forces and Potential Energy
3.6 Conservation of Energy.
Chapter 4. Kinetics of a Particle: Impulse and Momentum
4.1 Principle of Linear Impulse and Momentum
4.2 Principle of Linear and Momentum for a System of Particles
4.3 Conservation of Linear Momentum for a System of Particles
4.4 Impact
Chapter 5. Planar Kinematics of a Rigid Body
5.1 Rigid-Body Motion
5.2 Translation
5.3 Rotation About a Fixed Axis
5.4 Relative-Motion Analysis: Velocity
5.5 Relative-Motion Analysis: Acceleration
Chapter 6. Planer Kinetics of a Rigid Body: Force and Acceleration
6.1 Moment of Inertia
6.2 Planar Kinetic Equations of Motion
6.3 Equation of Motion: Translation
Chapter 7. Planar Kinetics of a Rigid Body: Work
and Energy
7.1 Kinetic Energy
7.2 The Work of a Force
7.3 The Work of Couple
7.4 Principle of Work and Energy
7.5 Conservation of Energy
Chapter 8. Planar Kinetics of a Rigid Body:
Impulse and Momentum
8.1 Linear and Angular Momentum
8.2 Principle of Impulse and Momentum
8.3 Conservation of Momentum
Required Text :
• R. C. Hibbler, Engineering Mechanics: Dynamics, 3 rd Ed.,
Prentice Hall, 2004.
Recommended Texts:
• J. L. Meriam and L. Glenn Kraige, Engineering
Mechanics: Dynamics, 2001, John Willey & Sons, Inc.
• F. P. Beer, E. R. Johnston and W. E. Clausen, Vector
Mechanics for Engineers: Dynamics, 2004, Mc Graw
Hill.
Evaluation
• Final examination
= 50%
• course work
= 50%
a) Prectical
= 25%
B) Theoretical test
= 15%
c) Assignment/quiz: = 10%
Total ……………………………………..= 100%
Contact Information
Mr. Mohd Noor Arib Bin Md Rejab
Email: [email protected]
HP : 017-4126695
CHAPTER 1
KINEMATICS OF A PARTICLE
1.1 Introduction & 1.2 Rectilinear Kinetics: Continuous
Motion
Today’s Objectives:
Students will be able to find
the kinematic quantities
(position, displacement,
velocity, and acceleration) of
a particle traveling along a
straight path.
In-Class Activities:
• Applications
• Relations between s(t), v(t),
and a(t) for general
rectilinear motion
• Relations between s(t), v(t),
and a(t) when acceleration is
constant
APPLICATIONS
The motion of large objects,
such as rockets, airplanes, or
cars, can often be analyzed
as if they were particles.
Why?
If we measure the altitude
of this rocket as a function
of time, how can we
determine its velocity and
acceleration?
APPLICATIONS (continued)
A train travels along a straight length of track.
Can we treat the train as a particle?
If the train accelerates at a constant rate, how can we
determine its position and velocity at some instant?
An Overview of Mechanics
Mechanics: the study of how bodies
react to forces acting on them
Statics: the study
of bodies in
equilibrium
Dynamics:
1. Kinematics – concerned
with the geometric aspects of
motion
2. Kinetics - concerned with
the forces causing the motion
POSITION AND DISPLACEMENT
A particle travels along a straight-line path
defined by the coordinate axis s.
The position of the particle at any instant,
relative to the origin, O, is defined by the
position vector r, or the scalar s. Scalar s
can be positive or negative. Typical units
for r and s are meters (m) or feet (ft).
The displacement of the particle is
defined as its change in position.
Vector form:  r = r’ - r
Scalar form:  s = s’ - s
The total distance traveled by the particle, sT, is a positive scalar
that represents the total length of the path over which the particle
travels.
VELOCITY
Velocity is a measure of the rate of change in the position of a particle.
It is a vector quantity (it has both magnitude and direction). The
magnitude of the velocity is called speed, with units of m/s or ft/s.
The average velocity of a particle during a
time interval t is
vavg = r/t
The instantaneous velocity is the time-derivative of position.
v = dr/dt
Speed is the magnitude of velocity: v = ds/dt
Average speed is the total distance traveled divided by elapsed time:
(vsp)avg = sT/  t
ACCELERATION
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. Typical units are m/s2 or ft/s2.
The instantaneous acceleration is the time
derivative of velocity.
Vector form: a = dv/dt
Scalar form: a = dv/dt = d2s/dt2
Acceleration can be positive (speed
increasing) or negative (speed decreasing).
As the book indicates, the derivative equations for velocity and
acceleration can be manipulated to get
a ds = v dv
SUMMARY OF KINEMATIC RELATIONS:
RECTILINEAR MOTION
• Differentiate position to get velocity and acceleration.
v = ds/dt ;
a = dv/dt or a = v dv/ds
• Integrate acceleration for velocity and position.
Position:
Velocity:
v
t
v
s
 dv =  a dt or  v dv =  a ds
s
t
 ds =  v dt
vo
o
vo
so
so
o
• Note that so and vo represent the initial position and
velocity of the particle at t = 0.
CONSTANT ACCELERATION
The three kinematic equations can be integrated for the special case
when acceleration is constant (a = ac) to obtain very useful equations.
A common example of constant acceleration is gravity; i.e., a body
freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2
ft/s2 downward. These equations are:
v
t
 dv =  a dt
c
vo
o
s
t
 ds =  v dt
so
v
v = vo + act
yields
s = s o + v ot + (1/2)a ct 2
yields
v2 = (vo )2 + 2ac(s - so)
o
s
 v dv =  ac ds
vo
yields
so
EXAMPLE
Given: A motorcyclist travels along a straight road at a speed
of 27 m/s. When the brakes are applied, the
motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it stops.
Plan: Establish the positive coordinate s in the direction the
motorcycle is traveling. Since the acceleration is given
as a function of time, integrate it once to calculate the
velocity and again to calculate the position.
EXAMPLE (continued)
Solution:
1) Integrate acceleration to determine the vvelocity.
t
a = dv / dt => dv = a dt =>  dv =  (-6t )dt
vo
o
=> v – vo = -3t2 => v = -3t2 + vo
2) We can now determine the amount of time required for
the motorcycle to stop (v = 0). Use vo = 27 m/s.
0 = -3t2 + 27 => t = 3 s
3) Now calculate the distance traveled in 3s by integrating the
velocity using so = 0:
s
t
v = ds / dt => ds = v dt =>  ds =  (-3t 2 + vo)dt
so
o
=> s – so = -t3 + vot
=> s – 0 = (3)3 + (27)(3) => s = 54 m
1.3 Rectilinear Kinematics: Erratic Motion
Today’s Objectives:
Students will be able to
determine position, velocity,
and acceleration of a particle
using graphs.
In-Class Activities:
• Applications
• s-t, v-t, a-t, v-s, and a-s
diagrams
APPLICATION
In many experiments, a
velocity versus position (v-s)
profile is obtained.
If we have a v-s graph for
the rocket sled, can we
determine its acceleration at
position s = 300 meters ?
How?
GRAPHING
Graphing provides a good way to handle complex
motions that would be difficult to describe with
formulas. Graphs also provide a visual description of
motion and reinforce the calculus concepts of
differentiation and integration as used in dynamics.
The approach builds on the facts that slope and
differentiation are linked and that integration can be
thought of as finding the area under a curve.
S-T GRAPH
Plots of position vs. time can be
used to find velocity vs. time
curves. Finding the slope of the
line tangent to the motion curve at
any point is the velocity at that
point (or v = ds/dt).
Therefore, the v-t graph can be
constructed by finding the slope at
various points along the s-t graph.
V-T GRAPH
Plots of velocity vs. time can be used to
find acceleration vs. time curves.
Finding the slope of the line tangent to
the velocity curve at any point is the
acceleration at that point (or a = dv/dt).
Therefore, the a-t graph can be
constructed by finding the slope at
various points along the v-t graph.
Also, the distance moved
(displacement) of the particle is the
area under the v-t graph during time t.
A-T GRAPH
Given the a-t curve, the change
in velocity (v) during a time
period is the area under the a-t
curve.
So we can construct a v-t graph
from an a-t graph if we know the
initial velocity of the particle.
A-S GRAPH
A more complex case is presented by
the a-s graph. The area under the
acceleration versus position curve
represents the change in velocity
(recall  a ds =  v dv ).
s2
½ (v1² – vo²) =  a ds = area under the
s1
a-s graph
This equation can be solved for v1, allowing you to solve for
the velocity at a point. By doing this repeatedly, you can
create a plot of velocity versus distance.
V-S GRAPH
Another complex case is presented
by the v-s graph. By reading the
velocity v at a point on the curve
and multiplying it by the slope of
the curve (dv/ds) at this same point,
we can obtain the acceleration at
that point.
a = v (dv/ds)
Thus, we can obtain a plot of a vs. s
from the v-s curve.
EXAMPLE
Given: v-t graph for a train moving between two stations
Find: a-t graph and s-t graph over this time interval
Think about your plan of attack for the problem!
EXAMPLE (continued)
Solution: For the first 30 seconds the slope is constant
and is equal to:
a0-30 = dv/dt = 40/30 = 4/3 ft/s2
Similarly, a30-90 = 0
and a90-120 = -4/3 ft/s2
a(ft/s2)
4
3
t(s)
-4
3
EXAMPLE (continued)
s(ft)
The area under the v-t graph
represents displacement.
3600
3000
s0-30 = ½ (40)(30) = 600 ft
s30-90 = (60)(40) = 2400 ft
s90-120 = ½ (40)(30) = 600 ft
600
30
90
t(s)
120
1.4 General Curvilinear Motion & 1.5 Curvilinear Motion:
Rectangular Components
Today’s Objectives:
Students will be able to:
a) Describe the motion of a
particle traveling along
a curved path.
b) Relate kinematic
quantities in terms of
the rectangular
components of the
vectors.
In-Class Activities:
• Applications
• General curvilinear motion
• Rectangular components of
kinematic vectors
APPLICATIONS
The path of motion of each plane in
this formation can be tracked with
radar and their x, y, and z coordinates
(relative to a point on earth) recorded
as a function of time.
How can we determine the velocity
or acceleration of each plane at any
instant?
Should they be the same for each
aircraft?
APPLICATIONS (continued)
A roller coaster car travels down
a fixed, helical path at a constant
speed.
How can we determine its
position or acceleration at any
instant?
If you are designing the track, why is it important to be
able to predict the acceleration of the car?
POSITION AND DISPLACEMENT
A particle moving along a curved path undergoes curvilinear motion.
Since the motion is often three-dimensional, vectors are used to
describe the motion.
A particle moves along a curve
defined by the path function, s.
The position of the particle at any instant is designated by the vector
r = r(t). Both the magnitude and direction of r may vary with time.
If the particle moves a distance s along the
curve during time interval t, the
displacement is determined by vector
subtraction:  r = r’ - r
VELOCITY
Velocity represents the rate of change in the position of a
particle.
The average velocity of the particle
during the time increment t is
vavg = r/t .
The instantaneous velocity is the
time-derivative of position
v = dr/dt .
The velocity vector, v, is always
tangent to the path of motion.
The magnitude of v is called the speed. Since the arc length s
approaches the magnitude of r as t→0, the speed can be
obtained by differentiating the path function (v = ds/dt). Note
that this is not a vector!
ACCELERATION
Acceleration represents the rate of change in the
velocity of a particle.
If a particle’s velocity changes from v to v’ over a
time increment t, the average acceleration during
that increment is:
aavg = v/t = (v - v’)/t
The instantaneous acceleration is the timederivative of velocity:
a = dv/dt = d2r/dt2
A plot of the locus of points defined by the arrowhead
of the velocity vector is called a hodograph. The
acceleration vector is tangent to the hodograph, but
not, in general, tangent to the path function.
RECTANGULAR COMPONENTS: POSITION
It is often convenient to describe the motion of a particle in
terms of its x, y, z or rectangular components, relative to a fixed
frame of reference.
The position of the particle can be
defined at any instant by the
position vector
r=xi+yj+zk .
The x, y, z components may all be
functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
The magnitude of the position vector is: r = (x2 + y2 + z2)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
RECTANGULAR COMPONENTS: VELOCITY
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt
Since the unit vectors i, j, k are constant in magnitude and
direction, this equation reduces to v = vxi + vyj + vzk
•
•
•
x
y
z
where vx =
= dx/dt, vy =
= dy/dt, vz =
= dz/dt
The magnitude of the velocity
vector is
v = [(vx)2 + (vy)2 + (vz)2]0.5
The direction of v is tangent
to the path of motion.
RECTANGULAR COMPONENTS: ACCELERATION
The acceleration vector is the time derivative of the
velocity vector (second derivative of the position vector):
a = dv/dt = d2r/dt2 = axi + ayj + azk
where
•
•
•
••
ax = vx = x = dvx /dt, ay = vy = y
= dvy /dt,
••
••
az = vz = z = dvz /dt
The magnitude of the acceleration vector is
a = [(ax)2 + (ay)2 + (az)2 ]0.5
The direction of a is usually
not tangent to the path of the
particle.
EXAMPLE
Given:The motion of two particles (A and B) is described by
the position vectors
rA = [3t i + 9t(2 – t) j] m
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
Find: The point at which the particles collide and their
speeds just before the collision.
Plan: 1) The particles will collide when their position
vectors are equal, or rA = rB .
2) Their speeds can be determined by differentiating
the position vectors.
EXAMPLE (continued)
Solution:
1) The point of collision requires that rA = rB, so xA =
xB and yA = yB .
x-components: 3t = 3(t2 – 2t + 2)
Simplifying: t2 – 3t + 2 = 0
Solving:
t = {3  [32 – 4(1)(2)]0.5}/2(1)
=> t = 2 or 1 s
y-components: 9t(2 – t) = 3(t – 2)
Simplifying:
3t2 – 5t – 2 = 0
Solving: t = {5  [52 – 4(3)(–2)]0.5}/2(3)
=> t = 2 or – 1/3 s
So, the particles collide when t = 2 s. Substituting this
value into rA or rB yields
xA = xB = 6 m
and yA = yB = 0
EXAMPLE (continued)
2) Differentiate rA and rB to get the velocity vectors.
.
.
vA = drA/dt = .xA i + yA j = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s
•
•
vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
Speed is the magnitude of the velocity vector.
vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5 = 6.71 m/s
1.6 MOTION OF A PROJECTILE
Today’s Objectives:
In-Class Activities:
Students will be able to
• Check homework, if any
analyze the free-flight motion of
• Applications
a projectile.
• Kinematic equations for
projectile motion
APPLICATIONS
A kicker should know at what angle, q, and initial velocity, vo, he
must kick the ball to make a field goal.
For a given kick “strength”, at what angle should the ball be
kicked to get the maximum distance?
APPLICATIONS (continued)
A fireman wishes to know the maximum height on the wall he can
project water from the hose. At what angle, q, should he hold the
hose?
CONCEPT OF PROJECTILE MOTION
Projectile motion can be treated as two rectilinear motions, one in
the horizontal direction experiencing zero acceleration and the other
in the vertical direction experiencing constant acceleration (i.e.,
gravity).
For illustration, consider the two balls on the
left. The red ball falls from rest, whereas the
yellow ball is given a horizontal velocity. Each
picture in this sequence is taken after the same
time interval. Notice both balls are subjected to
the same downward acceleration since they
remain at the same elevation at any instant.
Also, note that the horizontal distance between
successive photos of the yellow ball is constant
since the velocity in the horizontal direction is
constant.
KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains
constant (vx = vox) and the position in the x direction can be
determined by:
x = xo + (vox)(t)
Why is ax equal to zero (assuming movement through the air)?
KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = -g. Application of
the constant acceleration equations yields:
vy = voy – g(t)
y = yo + (voy)(t) – ½g(t)2
vy2 = voy2 – 2g(y – yo)
For any given problem, only two of these three
equations can be used. Why?
Example 1
Given: vo and θ
Find: The equation that defines
y as a function of x.
Plan: Eliminate time from the
kinematic equations.
vx = vo cos θ
Solution: Using
We can write: x = (vo cos θ)t
or
vy = vo sin θ
and
t =
x
vo cos θ
y = (vo sin θ)t – ½ g(t)2
By substituting for t:
y = (vo sin θ)
(
x
vo cos θ
) ( )(
–
g
2
2
x
vo cos θ
)
1.7 CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS
Today’s Objectives:
Students will be able to
determine the normal and
tangential components of
velocity and acceleration of a
particle traveling along a
curved path.
In-Class Activities:
• Applications
• Normal and tangential
components of velocity
and acceleration
• Special cases of motion
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, how can we determine the
magnitude and direction of its total
acceleration?
Why would you care about the total acceleration of the car?
APPLICATIONS (continued)
A motorcycle travels up a
hill for which the path can
be approximated by a
function y = f(x).
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?
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 (n) and tangential (t) coordinates are
often used.
In the n-t coordinate system, the
origin is located on the particle
(the origin moves with the
particle).
The t-axis is tangent to the path (curve) at the instant considered,
positive in the direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction
toward the center of curvature of the curve.
NORMAL AND TANGENTIAL COMPONENTS (continued)
The positive n and t directions are
defined by the unit vectors un and ut,
respectively.
The center of curvature, O’, always
lies on the concave side of the curve.
The radius of curvature, r, is defined
as the perpendicular distance from
the curve to the center of curvature at
that point.
The position of the particle at any instant is defined by the
distance, s, along the curve from a fixed reference point.
VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always
tangent to the path of motion
(t-direction).
The magnitude is determined by taking the time derivative of
the path function, s(t).
.
v = vut where v = s = ds/dt
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change
of velocity:
.
.
a = dv/dt = d(vut)/dt = vut + vut
.
Here v represents the change in
.
the magnitude of velocity and ut
represents the rate of change in
the direction of ut.
After mathematical manipulation,
the acceleration vector can be
expressed as:
.
a = vut + (v2/r)un = atut + anun.
ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
There are two components to the
acceleration vector:
a = at ut + an un
• The tangential component is tangent to the curve and in the
direction of increasing or decreasing velocity.
.
at = v or at ds = v dv
• The normal or centripetal component is always directed
toward the center of curvature of the curve. an = v2/r
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5
SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
.
2
r
 => an = v /r = 0 => a = at = v
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.
.
at = v = 0 => a = an = v2/r
The normal component represents the time rate of change
in the direction of the velocity.
SPECIAL CASES OF MOTION (continued)
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vot + (1/2)(at)ct2
v = vo + (at)ct
v2 = (vo)2 + 2(at)c(s – so)
As before, so and vo are the initial position and velocity of the
particle at t = 0. How are these equations related to projectile
motion equations? Why?
4) The particle moves along a path expressed as y = f(x).
The radius of curvature, r, at any point on the path can be
calculated from
2 ]3/2
[
1
(dy/dx)
+
r = ________________
d2y/dx 2
THREE-DIMENSIONAL MOTION
If a particle moves along a space
curve, the n and t axes are defined as
before. At any point, the t-axis is
tangent to the path and the n-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, b. The
binomial unit vector, ub, is directed perpendicular to the osculating
plane, and its sense is defined by the cross product ub = ut x un.
There is no motion, thus no velocity or acceleration, in the
binomial direction.
EXAMPLE PROBLEM
Given: Starting from rest, a motorboat
travels around a circular path of
r = 50 m at a speed that
increases with time,
v = (0.2 t2) m/s.
Find:
The magnitudes of the boat’s
velocity and acceleration at
the instant t = 3 s.
Plan: The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 3s using v(t).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.