course outline 1.matlab tutorial 2.motion of systems that can be idealized as particles description...

Course Outline

1. MATLAB tutorial

2. Motion of systems that can be idealized as particles• Description of motion; Newton’s laws;

• Calculating forces required to induce prescribed motion;

• Deriving and solving equations of motion

3. Conservation laws for systems of particles• Work, power and energy;

• Linear impulse and momentum

• Angular momentum

4. Vibrations• Characteristics of vibrations; vibration of free 1 DOF systems

• Vibration of damped 1 DOF systems

• Forced Vibrations

5. Motion of systems that can be idealized as rigid bodies• Description of rotational motion; Euler’s laws; deriving and solving

equations of motion; motion of machines

Exam topics

Particle Dynamics: Concept Checklist• Understand the concept of an ‘inertial frame’• Be able to idealize an engineering design as a set of particles, and know when this

idealization will give accurate results• Describe the motion of a system of particles (eg components in a fixed coordinate system;

components in a polar coordinate system, etc)• Be able to differentiate position vectors (with proper use of the chain rule!) to determine

velocity and acceleration; and be able to integrate acceleration or velocity to determine position vector.

• Be able to describe motion in normal-tangential and polar coordinates (eg be able to write down vector components of velocity and acceleration in terms of speed, radius of curvature of path, or coordinates in the cylindrical-polar system).

• Be able to convert between Cartesian to normal-tangential or polar coordinate descriptions of motion

• Be able to draw a correct free body diagram showing forces acting on system idealized as particles

• Be able to write down Newton’s laws of motion in rectangular, normal-tangential, and polar coordinate systems

• Be able to obtain an additional moment balance equation for a rigid body moving without rotation or rotating about a fixed axis at constant rate.

• Be able to use Newton’s laws of motion to solve for unknown accelerations or forces in a system of particles

• Use Newton’s laws of motion to derive differential equations governing the motion of a system of particles

• Be able to re-write second order differential equations as a pair of first-order differential equations in a form that MATLAB can solve

Particle Kinematics

r i j k( ) ( ) ( ) ( )t x t y t z t r(t)

r(t+dt)

v(t) dt

i

j

path of particle

kO

( ) ( ) ( ) ( )

( ) ( ) ( )

x y z

x y z

t v t v t v t

d dx dy dzx y z

dt dt dt dtdx dy dz

v t v t v tdt dt dt

v i j k

i j k i j k

• Direction of velocity vector is parallel to path• Magnitude of velocity vector is distance traveled / time

Inertial frame – non accelerating, non rotating reference frameParticle – point mass at some position in space

Position Vector

Velocity Vector

Acceleration Vector

2 2 2

2 2 2

( ) ( ) ( ) ( )

( ) ( ) ( )

yx zx y z x y z

yx zx y z

dvdv dvdt a t a t a t v v v

dt dt dt dt

dvdv dvd x d y d za t a t a t

dt dt dtdt dt dt

a i j k i j k i j k

Particle Kinematics

• Straight line motion with constant acceleration

20 0 0

1

2X V t at V at a

r i v i a i

• Time/velocity/position dependent acceleration – use calculus

0

0

0

( )

0

( )( ) ( )

( )

( )( ) ( )

( )

v t

V

x t t

X

dv g ta f v dv g t dt

dt f v

dx g tv f x dv v t dt

dt f v

0 00 0

( ) ( )t t

X v t dt V a t dt

r i v i

0

( ) ( )

0

( )v t x t

V

vdv a x dx

Particle Kinematics

• General circular motion

R

i

j

Rcos

Rsin

tsin

cos

n

2

22

cos sin

sin cos

( sin cos ) (cos sin )

R

R V

R R

dV VR R

dt R

r i j

v i j t

a i j i j

t n t n

2 2/ / /

/

d dt d dt d dt

s R V ds dt R

• Circular Motion at const speed

22 2

cos sin

sin cos

(cos sin )

R

R V

VR R

R

r i j

v i j t

a i j n n

t s R V R

Particle Kinematics

• Polar Coordinates

r

i

j

ee r

22 2

2 22

r

r

dr dr

dt dt

d r d d dr dr r

dt dt dtdt dt

v e e

a e e

2

V

dV V

dt R

v t

a t n

2 2

2 2

3/222

1

d y dydx d x

d dd d

Rdydx

d d

( ) ( )x y r i j

t

R

t

n

• Arbitrary path

Newton’s laws

• For a particle

• For a rigid body in motion without rotation, or a particle on a massless frame

ij

NA NBTBW

mF a

c M 0

You MUST take momentsabout center of mass

Calculating forces required to cause prescribed motion of a particle

• Idealize system

• Free body diagram

• Kinematics

• F=ma for each particle.

• (for rigid bodies or frames only)

• Solve for unknown forces or accelerationsc M 0

Deriving Equations of Motion for particles

1. Idealize system

2. Introduce variables to describe motion(often x,y coords, but we will see other examples)

3. Write down r, differentiate to get a

4. Draw FBD

5.

6. If necessary, eliminate reaction forces

7. Result will be differential equations for coords defined in (2), e.g.

8. Identify initial conditions, and solve ODE

mF a

2

02sin

d x dxm kx kY t

dtdt

Motion of a projectile

i

j

kX0

V0

0 0 0

0 0 00

x y z

X Y Ztd

V V Vdt

r i j k

ri j k

20 0 0 0 0 0

0 0 0

1

2x y z

x y z

X V t Y V t Z V t gt

V V V gt

g

r i j k

v i j k

a k

Rearranging differential equations for MATLAB

22

22 0n n

d y dyy

dtdt • Example

/v dy dt• Introduce

22 n n

vyd

vdt v y

• Then

( , )yd

f tvdt

ww w• This has form

Conservation laws for particles: Concept Checklist• Know the definitions of power (or rate of work) of a force, and work done by a force• Know the definition of kinetic energy of a particle• Understand power-work-kinetic energy relations for a particle• Be able to use work/power/kinetic energy to solve problems involving particle motion• Be able to distinguish between conservative and non-conservative forces• Be able to calculate the potential energy of a conservative force• Be able to calculate the force associated with a potential energy function• Know the work-energy relation for a system of particles; (energy conservation for a closed

system)• Use energy conservation to analyze motion of conservative systems of particles

• Know the definition of the linear impulse of a force• Know the definition of linear momentum of a particle• Understand the impulse-momentum (and force-momentum) relations for a particle• Understand impulse-momentum relations for a system of particles (momentum

conservation for a closed system)• Be able to use impulse-momentum to analyze motion of particles and systems of particles• Know the definition of restitution coefficient for a collision• Predict changes in velocity of two colliding particles in 2D and 3D using momentum and the

restitution formula

• Know the definition of angular impulse of a force• Know the definition of angular momentum of a particle• Understand the angular impulse-momentum relation• Be able to use angular momentum to solve central force problems

Work and Energy relations for particles

Rate of work done by a force(power developed by force)

ij

k

O

PF

vP F v

Total work done by a force

ij

k

O

PF(t)

r0

r1

1

0

t

W dt F v1

0

W d r

r

F r

2 2 2 21 1

2 2 x y zT m m v v v vKinetic energy

ij

k

O

P

v

r0

r1Work-kinetic energy relation1

0

0W d T T r

r

F r

Power-kinetic energy relationdT

Pdt

Potential energy

ij

k

O

PF

r0

r1

0

( ) constantV d r

r

r F r

Potential energy of aconservative force (pair)

grad( )VF

Type of force Potential energy

Gravity acting on a particle near earths surface

V mgy

Fm

ji

y

Gravitational force exerted on mass m by mass M at the origin

GMmV

r

r

F

r m

Force exerted by a spring with stiffness k and unstretched length

0L

201

2V k r L F

i

j

Force acting between two charged particles

1 2

4

Q QV

r

r

+Q1+Q2

ij

1

F2

Force exerted by one molecule of a noble gas (e.g. He, Ar, etc) on another (Lennard Jones potential). a is the equilibrium spacing between molecules, and E is the energy of the bond.

12 6

2a a

Er r

ri

j

1

F2

m1

m2

m3

m4

R21

R12

R13 R31

R23

R32

F2ext

F3ext

F1ext

Energy relations forconservative systems subjected to external forces

ijRInternal Forces: (forces exerted by one part of the system on another)

External Forces: (any other forces) extiF

0

0

( ) ( )t t

extext i

forces t

W t t dt

F vWork done by external forces

0 0extW T V T V

System is conservative if all internal forces are conservative forces (or constraint forces)

Energy relation for a conservative system 0 0t t t t

Kinetic and potential energy at time 0t 0 0T V

Kinetic and potential energy at time t T V

Linear Impulse of a force

ij

k

O

F(t) v

m

1

0

( )t

t

t dtF

Linear momentum of a particle mp v

Impulse-momentum relations d

dt

pF 1 0 p p

Impulse-momentum relations

m1

m2

m3

m4

R21

R12

R13 R31

R23

R32

F2ext

F3ext

F1ext

0

0

( )t t

exttot i

particles t

t dt

FTotal external impulse

tot i iparticles

m p vTotal linear momentum

( )ext toti

particles

dt

dt p

FImpulse-momentum tot tot p

Impulse-momentum for a system of particles

Impulse-momentum for a single particle

0 0tot tot pSpecial case (closed system) (momentum conserved)

Collisions

*

A B

A B

vxB1vx

A1

vxB0

vxA0

1 1 0 0A B A BA x B x A x B xm v m v m v m v

1 1 0 0B A B Av v e v v

1 0 0 0

1 0 0 0

(1 )

(1 )

B B B AA

A B

A A B AB

A B

mv v e v v

m m

mv v e v v

m m

AB

A B

B0vA0v

v A1 B1v

n

1 1 0 0 0 0(1 )B A B A B Ae v v v v v v n n

1 1 0 0B A B AB A B Am m m m v v v v

1 0 0 0(1 )B B B AA

B A

me

m m

v v v v n n

1 0 0 0(1 )A A B AB

B A

me

m m

v v v v n n

Restitution formula

Momentum

Momentum

Restitution formula

Angular Impulse-Momentum Equations for a Particle

i

j

k

x

y

zO

F(t)

r(t)0

0

( ) ( )t t

t

t t dt

r F

m h r p r v

d

dt

hMImpulse-Momentum relations

1 0 h h

Useful for central force problems

Angular Momentum

Angular Impulse

Special Case 1 0 0 h h Angular momentum conserved

Free Vibrations – concept checklistYou should be able to:1.Understand simple harmonic motion (amplitude, period, frequency, phase)2.Identify # DOF (and hence # vibration modes) for a system3.Understand (qualitatively) meaning of ‘natural frequency’ and ‘Vibration mode’ of a system4.Calculate natural frequency of a 1DOF system (linear and nonlinear)

5.Write the EOM for simple spring-mass-damper systems by inspection

6.Understand natural frequency, damped natural frequency, and ‘Damping factor’ for a dissipative 1DOF vibrating system7.Know formulas for nat freq, damped nat freq and ‘damping factor’ for spring-mass system in terms of k,m,c8.Understand underdamped, critically damped, and overdamped motion of a dissipative 1DOF vibrating system9.Be able to determine damping factor and natural frequency from a measured free vibration response10.Be able to predict motion of a freely vibrating 1DOF system given its initial velocity and position, and apply this to design-type problems

Vibrations and simple harmonic motion

Period, T

Peak to PeakAmplitude A

time

Displacementor Acceleration

y(t)

Typical vibration response

• Period, frequency, angular frequencyamplitude

Simple Harmonic Motion 0( ) sin

( ) cos( )

( ) sin )

x t X X t

v t V t

a t A t

V X A V

Vibration of 1DOF conservative systems

Harmonic Oscillator

2

02

m d ss L

k dt Derive EOM (F=ma)

Compare with ‘standard’ differential equation

Solution

2 2 20 0 0 0( ) ( ) / sin( )n ns t L s L v t

0 0 0

2

1

n

x s C L x s

m

k

nk

m Natural Frequency

k

m mk k

x1 x2

Vibration modes and natural frequencies

•Vibration modes: special initial deflections that cause entire system to vibrate harmonically•Natural Frequencies are the corresponding vibration frequencies

Number of DOF (and vibration modes)

In 2D: # DOF = 2*# particles + 3*# rigid bodies - # constraintsIn 3D: # DOF = 3*# particles + 6*# rigid bodies - # constraints

Expected # vibration modes = # DOF - # rigid body modes

A ‘rigid body mode’ is steady rotation or translation of the entire system at constant speed. The maximum number of ‘rigid body’ modes (in 3D) is 6; in 2D it is 3. Usually only things like a vehicle or a molecule, which can move around freely, have rigid body modes.

k

m mk k

x1 x2

Calculating nat freqs for 1DOF systems – the basics

EOM for small vibration of any 1DOF undamped system has form

m

k,L0y

2

2 2

1

n

d yy C

dt

1. Get EOM (F=ma or energy)2. Linearize (sometimes)3. Arrange in standard form4. Read off nat freq.

n is the natural frequency

Tricks for calculating natural frequencies of 1DOF undamped systems

k,L0

m

L0+

n

g

• Nat freq is related to static deflection

• Using energy conservation to find EOM

s

k, dm

22

0

2

02

2

02

1 1( )

2 2

( ) ( ) 0

dsKE PE m k s L const

dt

d ds d s dsKE PE m k s L

dt dt dt dt

d sm ks kL

dt

Linearizing EOM

2

2( )

d yf y C

dt Sometimes EOM has form

We cant solve this in general… Instead, assume y is small

2

20

2

20

(0) ...

(0)

y

y

d y dfm f y C

dt dy

d y dfm y C f

dt dy

There are short-cuts to doing the Taylor expansion

Writing down EOM for spring-mass-damper systems

s=L0+xk, L0

mc

2

2

22

2

0

2 02

n n n

d x c dx km x

m dt mdt

d x dx k cx

dt m kmdt

F a

k1

k2

Commit this to memory! (or be able to derive it…)

x(t) is the ‘dynamic variable’ (deflection from static equilibrium)

Parallel: stiffness 1 2k k k

c2

c1

Parallel: coefficient 1 2c c c

k1 k2

Series: stiffness 1 2

1 1 1

k k k

c2c1

Parallel: coefficient1 2

1 1 1

c c c

s=L0+xk, L0

mc

2

02

d s dsm c ks kL

dtdt EOM

2

2 2

1 2

nn

d x dxx C

dtdt

Standard Form

02

nk c

C Lm km

21d n

Overdamped 1

Critically Damped 1

Underdamped 1

Canonical damped vibration problem

Overdamped 0 0 0 0( )( ) ( )( )( ) exp( ) exp( ) exp( )

2 2n d n d

n d dd d

v x C v x Cx t C t t t

1

Critically Damped 1 0 0 0( ) ( ) ( ) exp( )n nx t C x C v x C t t

0 00

( )( ) exp( ) ( )cos sinn

n d dd

v x Cx t C t x C t t

Underdamped 1

0 0 0dx

x x v tdt

with 0 0 0ds

s s v tdt

0 0x ss x

Calculating natural frequency and damping factor from a measured vibration response

Displacement

time

t0 t1 t2 t3

T

x(t0) x(t1) x(t2) x(t3)

t4

0( )1log

( )n

x t

n x t

Measure log decrement:

2 2

2 2

4

4n T

Measure period: T

Then

Forced Vibrations – concept checklist

You should be able to:1.Be able to derive equations of motion for spring-mass systems subjected to external forcing (several types) and solve EOM using complex vars,or by comparing to solution tables2.Understand (qualitatively) meaning of ‘transient’ and ‘steady-state’ response of a forced vibration system (see Java simulation on web)3.Understand the meaning of ‘Amplitude’ and ‘phase’ of steady-state response of a forced vibration system4.Understand amplitude-v-frequency formulas (or graphs), resonance, high and low frequency response for 3 systems 5.Determine the amplitude of steady-state vibration of forced spring-mass systems.6.Deduce damping coefficient and natural frequency from measured forced response of a vibrating system 7.Use forced vibration concepts to design engineering systems

x(t)k, L0

m

x(t)

m

m

F(t)=F0 sin t

External forcing

Base Excitation

x(t)

Rotor Excitation

y(t)=Y0 sint

m0

y(t)=Y0sint

L0

k, L0

L0

k, L0

L0

2

02 2

1 2sin

nn

d x dxx KF t

dtdt

1, ,

2n

kK

m kkm

EOM for forced vibrating systems

2

2 2

1 2 2

n nn

d x dx dyx K y

dt dtdt

, , 12

nk

Km km

22 20

2 2 2 2 2

1 2sin

nn n n

Yd x dx K d yx K t

dtdt dt

00

2n

mkK M m m

M MkM

2

02 2

1 2sin( )

nn

d x dxx C KF t

dtdt

0 0 0dx

x x v tdt

( ) ( ) ( )h px t C x t x t

Equation Initial Conditions

Full Solution

0( ) sinpx t X t Steady state part (particular integral)

100 1/2 2 22 22 2

2 /tan

1 /1 / 2 /

n

nn n

KFX

Transient part (complementary integral)

Overdamped0 0 0 0( ) ( )

( ) exp( ) exp( ) exp( )2 2

h h h hn d n d

h n d dd d

v x v xx t C t t t

1

Critically Damped 1 0 0 0( ) exp( )h h hh n nx t C x v x t t

0 00( ) exp( ) cos sin

h hh n

h n d dd

v xx t C t x t t

Underdamped 1

0 0 0 0 0 0 0 00

(0) sin cosph h

pt

dxx x C x x C X v v v X

dt

21d n

Steady-state and Transient solution to EOM

0( ) sinpx t X t

10 0 1/2 2 22 22 2

2 /1( / , ) tan

1 /1 / 2 /

nn

nn n

X KF M M

2

02 2

1 2sin( )

nn

d x dxx C KF t

dtdt

Steady state solution to 2 / T

s=L0+xk, L0

mc

0( ) sinF t F t

Canonical externally forced system (steady state solution)

1

2n

k cK

m kkm

0 0.5 1 1.5 2 2.5 310

-1

100

101

Frequency Ratio /n

Mag

nific

atio

n X

0/KY

0

= 0.01

= 0.05 = 0.1 = 0.2 = 0.4

= 1.0 = 0.6

2

2 2

1 2 2

n nn

d x dx dyx C K y

dt dtdt

0( ) sinpx t X t

1/223 3

10 0 1/2 2 2 22 22 2

1 2 / 2 /( , , ) tan

1 (1 4 ) /1 / 2 /

nn

nn

n n

X KY M M

Steady state solution to

Canonical base excited system (steady state solution)

12

nk c

Km km

0( ) sinpx t X t

0 0 ( , , )nX KY M

Steady state solution to2 2

2 2 2 2

1 2

nn n

d x dx K d yx C

dtdt dt

0 siny Y t

2 21

1/2 2 22 22 2

/ 2 /tan

1 /1 / 2 /

n n

nn n

M

c

0

0 002 ( )n

mk cK

m m m mk m m

Canonical rotor excited system (steady state solution)