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)