ENT142 ENGINEERING DYNAMICS

Programme: B.Eng.(Hons) (Mechatronic Eng.)

School of Mechatronic Engineering

Universiti Malaysia Perlis

UniMAP

Course Description

Unit Name: ENT142 Engineering Dynamics.

Lecturer: Dr. Khairul Salleh Basaruddin

Contact Hours: 3 hours or 4 hours/week.

Tuition Pattern: 3 hours lecture or 2 hours lecture + 2 hours tutorial.

Credits: 3 credits

Pre-Requisites : None

Course Outcomes

CO1:

Ability to analyze problems related to rectilinear kinematics, law of motions, and also concepts mechanics and vector mechanics.

CO2:

Ability to evaluate problems related to kinematics of particle, involving force and acceleration, work and energy, and also impulse and momentum.

CO3:

Ability to evaluate problems related to planar kinetics or a rigid body, involving force and acceleration, work and energy, and also impulse and momentum.

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 Newtons 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 Textbook :

R. C. Hibbler, Engineering Mechanics: Dynamics, Latest Ed., Prentice Hall, 2010.

Recommended Books:

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.

An Overview of Mechanics

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

Mechanics: the study of how bodies

react to forces acting on them

CHAPTER 1

KINEMATICS OF A PARTICLE

1.1 Introduction & 1.2 Rectilinear Kinetics: Continuous

Motion

Todays 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?

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 = s/t

The instantaneous velocity is the time-derivative of position.

v = ds/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.

Note that so and vo represent the initial position and

velocity of the particle at t = 0.

Velocity:

= t

o

v

v o

dt a dv = s

s

v

v o o

ds a dv v or = t

o

s

s o

dt v ds

Position:

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:

t a v v c o + = yields = t

o

c

v

v

dt a dv o

2 c o o

s

t (1/2)a t v s s + + = yields = t

o s

dt v ds o

) s - (s 2a ) (v v o c 2

o 2 + = yields =

s

s

c

v

v o o

ds a dv v

EXAMPLE

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.

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.

EXAMPLE (continued)

Solution:

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

1) Integrate acceleration to determine the velocity.

a = dv / dt => dv = a dt =>

=> v vo = -3t2 => v = -3t2 + vo

- = t

o

v

v

dt t dv o

) 6 (

3) Now calculate the distance traveled in 3s by integrating the

velocity using so = 0:

v = ds / dt => ds = v dt =>

=> s so = -t3 + vot

=> s 0 = (3)3 + (27)(3) => s = 54 m

+ - = t

o

o

s

s

dt v t ds o

) 3 ( 2

1.3 Rectilinear Kinematics: Erratic Motion

Todays 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

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.

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 ).

a-s graph

(v1 vo) = = area under the s2

s1 a ds

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

4

-4 3

3

a(ft/s2)

t(s)

Similarly, a30-90 = 0 and a90-120 = -4/3 ft/s2

EXAMPLE (continued)

The area under the v-t graph

represents displacement.

s0-30 = (40)(30) = 600 ft

s30-90 = (60)(40) = 2400 ft

s90-120 = (40)(30) = 600 ft 600

3000

3600

30 90 120

t(s)

s(ft)

1.4 General Curvilinear Motion & 1.5 Curvilinear Motion:

Rectangular Components

Todays 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 t0, 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 particles 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 time-

derivative 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 = x i + y j + z k .

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 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.

Since the unit vectors i, j, k are constant in magnitude and

direction, this equation reduces to v = vxi + vyj + vzk

where vx = = dx/dt, vy = = dy/dt, vz = = dz/dt x y z

The velocity vector is the time derivative of the position vector:

v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt

RECTANGULAR COMPONENTS: ACCELERATION

The direction of a is usually

not tangent to the path of the

particle.

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 = = = dvx /dt, ay = = = dvy /dt,

az = = = dvz /dt

vx x vy y

vz z

The magnitude of the acceleration vector is

a = [(ax)2 + (ay)

2 + (az)2 ]0.5

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)

1) The point of collision requires that rA = rB, so xA =

xB and yA = yB .

Solution:

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.

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

vA = drA/dt = = [3i + (18 18t)j] m/s

At t = 2 s: vA = [3i 18j] m/s

j yA i xA . +

vB = drB/dt = xBi + yBj = [(6t 6)i + 3j] m/s

At t = 2 s: vB = [6i + 3j] m/s

. .

1.6 MOTION OF A PROJECTILE

Todays Objectives:

Students will be able to

analyze the free-flight motion of

a projectile.

In-Class Activities:

Check homework, if any

Applications

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) g(t)2

vy2 = voy

2 2g(y yo)

For any given problem, only two of these three

equations can be used. Why?

Example 1

Given: vo and

Find: The equation that defines

y as a function of x.

Plan: Eliminate time from the

kinematic equations.

Solution: Using vx = vo cos and vy = vo sin

We can write: x = (vo cos )t or

y = (vo sin )t g(t)2

t = x

vo cos

y = (vo sin ) x g x

vo cos 2 vo cos

2

( ) ( ) ( ) By substituting for t:

1.7 CURVILINEAR MOTION:

NORMAL AND TANGENTIAL COMPONENTS

Todays 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 cars 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 particles 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.

.

.

. a = vut + (v

2/r)un = atut + anun.

After mathematical manipulation,

the acceleration vector can be

expressed as:

ACCELERATION IN THE n-t COORDINATE SYSTEM

(continued)

There are two components to the

acceleration vector:

a = at ut + an un

The normal or centripetal component is always directed

toward the center of curvature of the curve. an = v2/r

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 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.

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.

1) The particle moves along a straight line.

r => an = v2/r = 0 => a = at = v

.

The tangential component represents the time rate of

change in the magnitude 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

r = ________________ ]3/2 (dy/dx)2 1 [ +

2 d2y/dx

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 boats

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.