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ME 407 Advanced Dynamics

We will learn to model systems that can be viewed as collections of rigid bodies

Common mechanical systems

Robots

Various wheeled vehicles

The focus will be on engineering applications

Divers and gymnasts

I’m open to applications you all care about

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I expect you to be comfortable with mathematics and abstract thinking in general

even though our applications will be concrete

I expect you to be familiar with

geometry

trigonometry

linear algebra

systems of ordinary differential equations

vectors

Prerequisites

and some basic physics

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YOU NEEDTO INTERRUPT MEIF YOU DON’T KNOW WHAT IS GOING ON

THIS IS IMPORTANT

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BoilerplateThere’s a web site: www.me.rochester.edu/courses/ME407

(NOT UP TO DATE — STAY TUNED)

My email, which I read regularly: gans@me.rochester.edu

Text: Engineering Dynamics: From the Lagrangian to Simulationavailable in preprint form from Jill in the department office.

Weekly problems sets

Probably two midterms

Meirovitch and/or Goldstein will be useful at the beginningboth on two hour reserve in Carlson

Office hours Tuesday-Thursday 2 – 4 or by appointment.

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We will go from very fundamental to very applied

conservation of momentum and angular momentum

What is a rigid body?

Moments of inertia

internal and external forces and torques

work and energy

geometry of three dimensional motion

angular velocity and angular momentum

coordinate systems

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We will go from very fundamental to very applied

Hamilton’s principle

The Euler-Lagrange equations

Hamilton’s equations

Kane’s method

The null-space method

Computational tricks: the method of Zs

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We will go from very fundamental to very applied

engineering mechanisms: linkages, gears, etc.

robots and their relatives

wheeled vehicles of different sorts

I’m open to applications you all care about

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Let me show you a couple of hard problemsso you can see where we are going

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10

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We will also need mathematical and computational tools

We need notation to understand ourselves better

Most of the interesting problems are wildly nonlinearand we’ll need to integrate differential equations numerically

I’m perfectly happy to use commercial code to do thisbut you do need to have an idea of what to expect

so you can figure out if it’s right.

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You will find Mathematica very useful. It’s available on many UR computers.

We can take part of a class to deal with this if necessary.The following link will get you to more information than you need.

http://www.me.rochester.edu/courses/ME201/websoft/softw.html

Mathematica

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A little bit about notation

vectors will be lower case bold face

matrices will be upper case bold face

“Vector notation”

Matrix/linear algebra notation

vectors will be column vectors, their transposes row vectors

Indicial notation

vectors have one superscript, their transposes have one subscript

“real matrices” have one superscript and one subscriptdenoting row and column respectively

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€

a = ai =

a1

a2

MaN

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

, aT = ai = a1 a2 L aN{ }

€

A = A. ji =

A11 A2

1 L AN1

A21 A2

2 M MM M O M

AN1 L L AN

N

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

Matrices do not have to be square.

Examples of the notations

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Vector-matrix multiplication

€

Ax = A. ji x j =

A11 A2

1 L AN1

A21 A2

2 M MM M O M

AN1 L L AN

N

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

x1

x 2

Mx N

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

€

a ⋅b = aibii=1

N

∑ = aibi

i=1

N

∑

€

ab = abT = aib j =

a1b1 a1b2 L a1bN

a1b2 a2b2 M MM M O M

a1bN L L aNbN

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

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Summation convention

€

a ⋅b = aibii=1

N

∑ = aibi

i=1

N

∑ ⇒ aibi

“Metric tensor”

€

gij =1,i = j0,i ≠ j ⎧ ⎨ ⎩

, gij =1,i = j0,i ≠ j ⎧ ⎨ ⎩

⇒ a ⋅b = gijaib j , a ⋅b = gijaib j

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??

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The inertial coordinate system: coordinates x, y, z; unit vectors i, j, k

i

j

k

€

r = xi + yj+ zk

We will also have body coordinates, but not today

We have to do physics in the inertial coordinate system

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Start from the very basic: “f = ma” and consider a single particle/point mass — moments of inertia all zero

€

v = ˙ r ⇔ v i = ˙ r i, a = ˙ v = ˙ ̇ r ⇔ ai = ˙ v i = ˙ ̇ r i

€

f = m˙ ̇ r ⇔ f i = m˙ ̇ r i

Conservation of momentum

€

˙ x = dxdt

, ˙ r = drdt

, ˙ A = dAdt

L

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Angular momentum

This doesn’t mean much for a particle, but we might as well start here

€

l = r × p = mr × v

This angular momentum is defined wrt the inertial origin, but any reference will do —

different reference, different angular momentum

€

l* = r − r0( ) × p = mr *×v

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Its rate of change

€

˙ l * = m ˙ r − ˙ r 0( ) × v + m r − r0( ) × ˙ v

€

˙ r 0 = 0 = ˙ r × v⇒ ˙ l * = m r − r0( ) × ˙ v

which we call the torque.

The torque depends on the point of reference — remember this

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Example: a particle falling under gravity

€

x = x0, y = v0t, z = w0t − 12

gt 2

i

j

k

€

p = mv0j+ m w0 − gt( )k

€

l = x0i + v0tj + w0t − 12

gt 2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟k

⎛ ⎝ ⎜

⎞ ⎠ ⎟× mv0j+ m w0 − gt( )k( )

€

l = − 12

gt 2mv0i − m w0 − gt( )x0j+ mx0v0k

€

τ =˙ l = −gtmv0i + mgx0j

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??

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WORK AND ENERGY

work = force times distance, so

€

dW = f ⋅ds = f ⋅dr

€

dWdt

= f ⋅ drdt

= f ⋅v

€

f = m˙ v ⇒ dWdt

= m˙ v ⋅v = ddt

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mv ⋅v ⎛ ⎝ ⎜

⎞ ⎠ ⎟= dT

dt

€

T = 12

mv ⋅v

The kinetic energy of a particle

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€

dWdt

dt = W2 −W1 =t1

t2∫ dTdt

dt =t1

t2∫ T2 − T1 ⇔ ΔW = ΔT

and we can go back to the beginning and note that

€

ΔT = f ⋅dss1

s2∫

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i

j

k

1

2

€

ΔT = f ⋅dss1

s2∫

In general the integral

will be different for the red path and the blue path

€

ΔT = f ⋅ds = 0∫

If the integral is the same for all paths, we’ll have

and the force is conservative

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Conservative forces come from potentials

A force is conservative iff

€

f = −∇V r( )

Potentials can be time-dependent; we will not deal with time-dependent potentials

There’s a discussion of potentials in the text, and I’ll do a little on the board

Bottom line

The total energy, T + V, is conserved for a single particle under conservative forces

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An aside regarding potentials

M

m

€

f = −G mMr2 er

€

V = −G mMr

⇒ f = −∇V = −G mMr2 er

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For celestial mechanics we do not include the m in the potentialWe associate the potential with the gravitating body

€

V = −G Mr⇒ f = −m∇V = −G mM

r2 er

There are several simple orbital examples in the text.

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SYSTEMS OF PARTICLES

€

f1 = m˙ ̇ r 1

€

f4 = m˙ ̇ r 4

€

f5 = m˙ ̇ r 5

€

f2 = m˙ ̇ r 2

€

f3 = m˙ ̇ r 3

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The particles can interact — including action at a distance

€

f1 = f1(e ) + f21 + f31 + f41 + f51

f2 = f2(e ) + f12 + f32 + f43 + f53

M

Split each force into an external part and an interaction part, within the system

momentum of the system

€

p = p1 + p2 + p3 + p4 + p5

the rate of change is equal to the force, so we have

€

˙ p = ˙ p 1 + ˙ p 2 + ˙ p 3 + ˙ p 4 + ˙ p 5= f1

(e ) + f21 + f31 + f41 + f51 + f2(e ) + f12 + f32 + f43 + f53 +L

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€

˙ p = ˙ p 1 + ˙ p 2 + ˙ p 3 + ˙ p 4 + ˙ p 5= f1

(e ) + f21 + f31 + f41 + f51 + f2(e ) + f12 + f32 + f43 + f53 +L

cancel

All such pairs cancel by Newton’s third law of action and reaction

This is called

The weak law of action and reaction

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from which we deduce

€

˙ p = f1(e ) + f2

(e ) + f3(e ) + f4

(e ) + f5(e )

or, more generally

€

˙ p = fi(e )

i−1

N

∑

Only the external forces change the momentum of a systemunder the weak law of action and reaction

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What is the momentum of a system?

€

p = ddt

mirii=1

N

∑

write

€

M = mii=1

N

∑ then

€

p = M ddt

miri

Mi=1

N

∑ = M drCM

dt⇒ rCM = miri

Mi=1

N

∑

€

M˙ ̇ r CM = fi(e )

i=1

N

∑ = F

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If the sum of the external forces acting on a system is zero, the momentum of the system is conserved

For example: the contents of a shotgun shell fired in a vacuum

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We can do the same thing for torque and angular momentum,and we’ll find we need a new law

€

l = r1 × p1 + r2 × p2 + r3 × p3 + r4 × p4 + r5 × p5

€

˙ l = τ = m2r1 × ˙ v 1 + m2r2 × ˙ v 2 + m3r3 × ˙ v 3 + m4r4 × ˙ v 4 + m5r5 × ˙ v 5 = r1 × f1 + r2 × f2 + r3 × f3 + r4 × f4 + r5 × f5

Look at a pair for simplicity’s sake

€

r1 × f1(e ) + f12( ) + r2 × f2

(e ) + f21( )

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€

r1 × f1(e ) + f12( ) + r2 × f2

(e ) + f21( )

€

r1 × f1(e ) + r2 × f2

(e ) + r1 × f12 + r2 × f21

The internal torques will cancel if the forces are parallel to a line connecting the two particles€

f21 = −f12

€

r1 − r2( ) × f12

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reference point

r2

r1

r1 – r2

€

r1 − r2( ) × f12 = 0

if f12 is parallel to r1 – r2

Gravity works this way, as does electrostatics

Not all internal forces work this way, but all the ones we care about do

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That is the strong law of action and reaction

I will assume that throughout.

We have the following for systems

€

˙ p = fi(e )

i−1

N

∑

€

˙ l = ri × fi(e )

i−1

N

∑

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The angular momentum of a system can be written

€

l = MrCM × vCM + mi ′ r i × ′ v ii−1

N

∑

where

€

′ r i = ri − rCM , ′ v i = ˙ r i − ˙ r CM = vi − vCM

You can establish this for homework. It’s not hard and it’s a good exercise.

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€

l = MrCM × vCM + mi ′ r i × ′ v ii−1

N

∑

angular momentum of the system wrt the reference

angular momentum of the system wrt the CM

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??

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Kinetic Energy

€

T = 12

mivi ⋅vii−1

N

∑

€

vi = vCM + ′ v i

€

T = 12

mi vCM + ′ v i( ) ⋅ vCM + ′ v i( )i−1

N

∑ = 12

M mi

MvCM + ′ v i( ) ⋅ vCM + ′ v i( )

i−1

N

∑

€

T = 12

M mi

MvCM ⋅vCM

i−1

N

∑ + M mi

MvCM ⋅ ′ v i

i−1

N

∑ + 12

M mi

M′ v i ⋅ ′ v i

i−1

N

∑

€

T = 12

MvCM ⋅vCM + 12

mi ′ v i ⋅ ′ v ii−1

N

∑

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€

M mi

MvCM ⋅ ′ v i

i−1

N

∑ = MvCM ⋅ mi

M′ v i

i−1

N

∑ = MvCM ⋅ ddt

mi

M′ r i

i−1

N

∑

€

rCM = mi

Mri

i−1

N

∑ = mi

MrCM + ′ r i( )

i−1

N

∑ = rCM

Mmi

i−1

N

∑ + 1M

mi ′ r ii−1

N

∑

these are equal thisis

zero

so the kinetic energy is as on the previous slide

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€

T = 12

MvCM ⋅vCM + 12

mi ′ v i ⋅ ′ v ii−1

N

∑

kinetic energy of the center of mass

internal kinetic energy

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Let’s try to summarize today’s beginning

€

rCM = miri

Mi−1

N

∑

€

p = M˙ r CM

€

˙ p = M˙ ̇ r CM = fi(e )

i=1

N

∑ = f

€

l = ri × pii−1

N

∑

€

˙ l = ri × ˙ p ii−1

N

∑ = ri × fi(e )

i−1

N

∑ = τ

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??