MEE 214 (Dynamics)

Tuesday 8.30-11.20

• Dr. Soratos Tantideeravit (สรทศ ตนตธรวทย) soratos@oaep.go.th

• Lecture Notes, Course updates, Extra problems, etc

•• No Homework

• Final Exam (Date & Time – TBD)

12/03/58 MEE214 – Dynamics 1

Course Overview

• Kinematics of a Particle

– Rectilinear and Curvilinear Motion

• Kinetics of a Particle

– Force and Acceleration– Force and Acceleration

– Work and Energy

– Impulse and Momentum

• Kinetics of a System of Particles

12/03/58 MEE214 – Dynamics 2

Introduction to Dynamics

EngineeringMechanics

Statics

12/03/58 MEE214 – Dynamics 3

Mechanics

Dynamics

Kinematics Kinetics

Kinematics of a particle

Objectives

• Concepts of position, displacement, velocity, and

acceleration.

• Particle motion along a straight line

• Particle motion along a curved path using • Particle motion along a curved path using

different coordinate systems.

• Analysis of dependent motion of two particles.

• Principles of relative motion of two

particles using translating axes.

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Rectilinear Kinematics

• Origin

– Define a fixed point in space

• Position

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• Position

– Defined by a position vector r

or an algebraic scalar s

Rectilinear Kinematics

• Displacement

– Change in position

• Velocity

s∆

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• Velocity

t

svavg

∆

∆=

dt

dsv =

Rectilinear Kinematics

• Acceleration

t

vaavg

∆

∆=

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2

2

dt

sd

dt

dva ==

dvvdsa =

Constant Acceleration

caa =tavv c+= 0

21tatvss ++=

12/03/58 MEE214 – Dynamics 9

2

002

1tatvss c++=

)(2 0

2

0

2 ssavv c −+=

Problem 12-31

The acceleration of a particle along a straight

line is defined by a=(2t-9) m/s2, where t is in

seconds. At t=0, s=1 m and v=10 m/s. When

t=9 s, determine (a) the particle’s position, (b)

12/03/58 MEE214 – Dynamics 10

t=9 s, determine (a) the particle’s position, (b) the total distance traveled and (c) the velocity.

General Curvilinear Motion

Curvilinear motion occurs when the particle moves

along a curved path

Position. The position of the particle, measured from a

fixed point O, is designated by the position vector

12/03/58 MEE214 – Dynamics 11

fixed point O, is designated by the position vectorr = r(t).

General Curvilinear Motion

Displacement. Suppose during a small time interval

Δt the particle moves a distance Δs along the curve to a

new position P`, defined by r` = r + Δr. The displacementΔr represents the change in the particle’s position.

12/03/58 MEE214 – Dynamics 12

Δr represents the change in the particle’s position.

srt ∆=∆→∆ ,0

General Curvilinear Motion

Velocity

t

rvavg

∆

∆=

vv

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dt

ds

dt

rdvins ==

vv

General Curvilinear Motion

Acceleration.

t

vaavg

∆

∆=

vv

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2

2

2

2

dt

sd

dt

rd

dt

vda ===

vvv

Curvilinear Motion:

Rectangular Components

kzjyixr ˆˆˆ ++=v

Position. Position vector is defined by

The magnitude of is always positive and defined asrv

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222 zyxr ++=

The direction of r is specified by

the components of the unit

vector rrur /ˆv

=

Curvilinear Motion:

Rectangular Components

Velocity.

zvyvxv

kvjvivdt

rdv

zyx

zyx

&&&

vv

===

++== ˆˆˆ

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vvuv /ˆv

=

The velocity has a magnitude defined as the positive value of

222

zyx vvvv ++=

Curvilinear Motion:

Rectangular Components

Acceleration.

yva

xva

kajaiadt

vda

xx

zyx

&&&

&&&

vv

==

==

++== ˆˆˆ

12/03/58 MEE214 – Dynamics 17

zva

yva

zz

yy

&&&

&&&

==

==

The acceleration has a magnitude defined as the

positive value of

222

zyx aaaa ++=

Curvilinear Motion:

Rectangular Components

• The acceleration has a direction specified by

the components of the unit vector .

• Since a represents the time rate of change in

aaua /ˆv

=

12/03/58 MEE214 – Dynamics 18

• Since a represents the time rate of change in velocity, a will not be tangent to the path.

Motion of Projectile

• Constant downward acceleration, no air

resistance

• Mathema<cal expressions, ↑ [=] +, → [=] +

−= x 0=&&

12/03/58 MEE214 – Dynamics 19

2

00

0

2

1gttvyy

gtvy

gy

y

y

−+=

−=

−=

&

&&

tvxx

vx

x

x

x

00

0

0

+=

=

=

&

&&

Example 12.12

The chipping machine is designed to eject wood chips

at vO = 25 ft/s. If the tube is oriented at 30° from the

horizontal, determine how high, h, the chips strike the

pile if they land on the pile 20 ft from the tube.

12/03/58 MEE214 – Dynamics 20

Curvilinear Motion:

Normal and Tangential Components

• When the path of motion of a particle is known,

describe the path using n and t coordinates which act

normal and tangent to the path

•Consider origin located at the particle

12/03/58 MEE214 – Dynamics 21

tu

nu

Tangential direction

Normal direction

Curvilinear Motion:

Normal and Tangential Components

Velocity.• Since the particle is moving, s is a function of time

• Particle’s velocity v has direction that is always tangent to the path and a magnitude that is determined

by taking the time derivative of the path function s = s(t)

12/03/58 MEE214 – Dynamics 22

by taking the time derivative of the path function s = s(t)

sv

uvv t

&

v

=

= ˆ

Curvilinear Motion:

Normal and Tangential Components

Acceleration• Acceleration of the particle is the time rate of change

of velocity

tt uvuvdt

uvdva &&&vv

ˆˆ)ˆ(

+===

12/03/58 MEE214 – Dynamics 23

tt uvuvdt

va & ˆˆ +===

2

2

dt

sd

dt

dvv ==&

Curvilinear Motion:

Normal and Tangential Components

Acceleration• Find tu

&'ˆˆˆttt uudu =+

nuntt uduud ˆˆ =

θθ dddut == )1(

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'ˆtu

tu tud ˆθd

t

nt udud ˆˆ θ=

nnnt uv

us

uu ˆˆˆˆρρ

θ ===&

&&

Curvilinear Motion:

Normal and Tangential Components

vdvdsava

uauaa

tt

nntt

==

+=

&

vˆˆ

2v

12/03/58 MEE214 – Dynamics 25

ρ

2van =

22

2/32

/

])/(1[

dxyd

dxdy+=ρ

Problem 12-120

The automobile is originally at rest at s=0. If it

then starts to increase its speed at

ft/s2 where t is in seconds, determine the

magnitudes of its velocity and acceleration at

)05.0( 2tv =&

12/03/58 MEE214 – Dynamics 26

magnitudes of its velocity and acceleration at s = 550 ft.

Problem 12-131

At a given instant the train engine at E has a

speed of 20 m/s and an acceleration of 14

m/s2 acting in the direction shown. Determine

the rate of increase in the train’s speed and

12/03/58 MEE214 – Dynamics 27

the rate of increase in the train’s speed and the radius of curvature of the path.

Problem 12-152

If the speed of the box at

point on the track is 30ft/s

which is increasing at the rate

of ft/s2 , determine the 5=v&

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of ft/s2 , determine the

magnitude of the acceleration of the box at this instant.

5=v&

Curvilinear Motion:

Cylindrical Components

• Fixed origin

u Radial direction

12/03/58 MEE214 – Dynamics 29

ru

θu

Radial direction

Transverse direction

Curvilinear Motion:

Cylindrical Components

Position

urr ˆ=v

12/03/58 MEE214 – Dynamics 30

rurr ˆ=v

Curvilinear Motion:

Cylindrical Components or Polar

Velocity

rr ururrv &&&vv

ˆˆ +==

θθuu ˆˆ && =

12/03/58 MEE214 – Dynamics 31

θθururv rˆˆ &&

v+=

θθuur ˆˆ && =

θθ uvuvv rrˆˆ +=

v

Curvilinear Motion:

Cylindrical Components

Acceleration

θθθ θθθ urururururva rr&&&&&&&&&&&

vvˆˆˆˆˆ ++++==

ruu ˆˆ θθ&& −=

12/03/58 MEE214 – Dynamics 32

ruu ˆˆ θθ −=

θθ uauaa rrˆˆ +=

v

2θ&&& rrar −=

θθθ&&&& rra 2+=

Example 12-19

The searchlight casts a spot of light along the face of a wall that is

located 100m from the searchlight. Determine the magnitudes of

the velocity and acceleration at which the spot appears to travel

across the wall at the instant θ = 45°. The searchlight is rotating at

a constant rate of 4 rad/s

12/03/58 MEE214 – Dynamics 33

Problem 12-184

The slotted arm AB drives pin C through the spiral groove

described by the equation r = (1.5Ө) ft, where Ө is in radians. If

the arm starts from rest when Ө = 60˚ and is driven at an angular

velocity of Ө) = (4t) rad/s, where t is in seconds, determine the

radial and transverse components of velocity and acceleration of

the pin C when t=1 s.

12/03/58 MEE214 – Dynamics 34

the pin C when t=1 s.

Absolute Dependent Motion

• Dependent motions of two particles are

normally associated with systems of

connected masses via inextensible cords and pulleys.

12/03/58 MEE214 – Dynamics 35

pulleys.

Absolute Dependent Motion

BA ssl 3+=

vv 30 +=

12/03/58 MEE214 – Dynamics 36

BA vv 30 +=

BA aa 30 +=

Problem 12-206

If the hydraulic cylinder at H draws in rod BC

by 200 mm at 2ft/s, determine how far the slider A moves and the speed of the slider.

12/03/58 MEE214 – Dynamics 37

Example 3

A man at A is hoisting a safe Sby walking to the right with a

constant velocity vA = 0.5m/s.

Determine the velocity and

12/03/58 MEE214 – Dynamics 38

Determine the velocity and

acceleration of the safe when it

reaches the elevation at E. The

rope is 30m long and passes

over a small pulley at D.

Problem 12-208

If block A is moving downward with a speed of

6 ft/s while C is moving down at 18 m/s, determine the speed of block B.

12/03/58 MEE214 – Dynamics 39

Relative Motion Analysis

The relative position of B with respect to A is given by

The relative velocity and acceleration of B with respect

ABAB rrrvvv

−=/

12/03/58 MEE214 – Dynamics 40

The relative velocity and acceleration of B with respect

to A are given by

ABAB vvvvvv

−=/

ABAB aaavvv

−=/

Example 4

A train, traveling at a constant speed of 60 mi/h, crosses

over a road. If automobile A is traveling t 45 mi/h along

the road, determine the magnitude and direction of

relative velocity of the train with respect to the

automobile.

12/03/58 MEE214 – Dynamics 41

automobile.

Problem 12-149

The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA=0.7 m/s and vB=1.5 m/s respectively.

12/03/58 MEE214 – Dynamics 42

Determine at t= 2s, (a) the displacement along the path of each particle, (b) the position vector to each particle, and (c) the magnitude of the acceleration of particle B.