lecture 2. a day of principles

39
Lecture 2. A Day of Principles The principle of virtual work d’Alembert’s principle Hamilton’s principle 1 (with an example that applies ‘em all at the end)

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Lecture 2. A Day of Principles. (with an example that applies ‘ em all at the end). The principle of virtual work. d’Alembert’s principle. Hamilton’s principle. Principle of virtual work says the work done by a virtual displacement from an equilibrium position must be zero . - PowerPoint PPT Presentation

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Page 1: Lecture 2. A Day of Principles

1

Lecture 2. A Day of Principles

The principle of virtual work

d’Alembert’s principle

Hamilton’s principle

(with an example that applies ‘em all at the end)

Page 2: Lecture 2. A Day of Principles

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Principle of virtual work says the work done by a virtual displacement from an equilibrium position must be zero.

δW = fi ⋅δrii=1

N

∑ = 0

This can be used in statics to find forcesWe need to move this up to dynamics

d’Alembert’s principle says that

−ma = −m˙ v can be treated as an

inertial force

Page 3: Lecture 2. A Day of Principles

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Combine these

δW = fi − m˙ v i( ) ⋅δrii=1

N

∑ = 0

We eliminated internal forces last time, so the fs are external

We suppose we know them, and since the displacement is tinythey don’t change during the virtual displacement

Page 4: Lecture 2. A Day of Principles

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The virtual displacements, if unconstrained, can be written

δri = δxi + δyj + δzk

What I’d like to do is gather all of this up into what I’ll call configuration space

q = x1 y1 z1 x2 y2 z2 L xK yK zK{ }T

The dimension of q is N = 3K

Page 5: Lecture 2. A Day of Principles

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That’s just an example of a configuration space.We’ll see more as we go along.

The general rule, which we’ll investigate as we go along,there are (at least) as many qs as there are degrees of freedom.

We’ll take up the matter of constraints — relations among the qs —later

Bottom line: we can write the rs in terms of the qs.

Page 6: Lecture 2. A Day of Principles

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m1 m2

y1 y2

k1 k2k3

Simple example of possible qs

q =y1

y2

⎧ ⎨ ⎩

⎫ ⎬ ⎭

, y1 + y2

y1 − y2

⎧ ⎨ ⎩

⎫ ⎬ ⎭

Page 7: Lecture 2. A Day of Principles

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Each r can depend on all the qs.

δri = ∂ri

∂qk δqk

i=1

N= 3K

∑ = ∂ri

∂qk δqk

repeatedindices

δW = fi − m˙ v i( ) ⋅ ∂ri

∂qk δqk

i=1

N

∑ = 0

summation convention does not apply to i in this formula

Page 8: Lecture 2. A Day of Principles

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fi ⋅∂ri

∂qk δqk

i=1

N

∑ = m˙ v i ⋅∂ri

∂qk δqk

i=1

N

Let’s do some rearranging

fi ⋅∂ri

∂qki=1

N

∑ ⎛

⎝ ⎜

⎠ ⎟δqk = m˙ v i ⋅

∂ri

∂qki=1

N

∑ ⎛

⎝ ⎜

⎠ ⎟δqk

Define the generalized force

Qk = fi ⋅∂ri

∂qki=1

N

∑ ⎛

⎝ ⎜

⎠ ⎟

This force goes with the kth generalized coordinate

Page 9: Lecture 2. A Day of Principles

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m˙ v i ⋅∂ri

∂qki=1

N

∑ ⎛

⎝ ⎜

⎠ ⎟δqk = Qkδqk

The left hand side looks a lot like a momentum changedotted into something. If you like manipulations

you’ll LOVE what comes next

First a bit of what I call “integration by parts”

ddt

fg( ) = ˙ f g + f ˙ g ⇒ ˙ f g = ddt

fg( ) − f ˙ g

This is a useful trick in deriving things

Page 10: Lecture 2. A Day of Principles

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mi ˙ v i ⋅∂ri

∂qk = middt

vi ⋅∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟− mivi ⋅

ddt

∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟

˙ q j is not a function of

qk

ddt

∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟= ∂

∂q j

∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟˙ q j

∂∂q j

∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟˙ q j = ∂

∂qk

∂ri

∂q j

⎛ ⎝ ⎜

⎞ ⎠ ⎟˙ q j

∂∂qk

∂ri

∂q j

⎛ ⎝ ⎜

⎞ ⎠ ⎟˙ q j = ∂

∂qk

∂ri

∂q j ˙ q j ⎛ ⎝ ⎜

⎞ ⎠ ⎟

∂∂qk

∂ri

∂q j ˙ q j ⎛ ⎝ ⎜

⎞ ⎠ ⎟= ∂

∂qk ˙ r i( ) = ∂vi

∂qk

ddt

( ) = ∂∂q j ( ) ˙ q j ⇒

Page 11: Lecture 2. A Day of Principles

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mi ˙ v i ⋅∂ri

∂qk = middt

vi ⋅∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟− mivi ⋅

∂∂q j

∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟˙ q j

mi ˙ v i ⋅∂ri

∂qk = middt

vi ⋅∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟− mivi ⋅

∂vi

∂qk

so

turns into

We still aren’t done. The second term on the right is cool

mivi ⋅∂vi

∂qk = ∂∂qk

12

mivi ⋅vi

⎛ ⎝ ⎜

⎞ ⎠ ⎟= ∂Ti

∂qk

but we need to play with the first term

Page 12: Lecture 2. A Day of Principles

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middt

vi ⋅∂ri

∂qk

⎛ ⎝ ⎜

⎞ ⎠ ⎟

vi = ∂ri

∂q j ˙ q j ⇒ ∂vi

∂˙ q j= ∂ri

∂q j

so

middt

vi ⋅∂vi

∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟= d

dt∂

∂ ˙ q k12

mivi ⋅vi

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛ ⎝ ⎜

⎞ ⎠ ⎟= d

dt∂

∂˙ q kTi( )

⎛ ⎝ ⎜

⎞ ⎠ ⎟

sort of a trick

Page 13: Lecture 2. A Day of Principles

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m˙ v i ⋅∂ri

∂qki=1

N

∑ ⎛

⎝ ⎜

⎠ ⎟δqk = Qkδqk

turns into

ddt

∂T∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂T

∂qk − Qk

⎝ ⎜

⎠ ⎟δqk = 0

Remember that there is a sum on k here!

I can split Qk into two pieces: the potential and nonpotential parts

Qk = − ∂V∂qk + QNP( )k

so

Page 14: Lecture 2. A Day of Principles

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ddt

∂T∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂T

∂qk + ∂V∂qk − QNP( )k

⎝ ⎜

⎠ ⎟δqk = 0

V = V qk( )⇒∂V∂˙ q k

= 0For our purposes

ddt

∂ T −V( )∂ ˙ q k

⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂ T −V( )∂qk − QNP( )k

⎝ ⎜

⎠ ⎟δqk = 0

ddt

∂ T −V( )∂ ˙ q k

⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂ T −V( )∂qk

⎝ ⎜

⎠ ⎟δqk = QNP( )k

δqk

Page 15: Lecture 2. A Day of Principles

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ddt

∂L∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂qk

⎝ ⎜

⎠ ⎟δqk = Qkδqk

The Lagrangian

L = T −V

If the qk are independent, then we have the Euler-Lagrange equations

From here on in Qk will be understood to be the nonpotential generalized forcesWe’ll have an algorithm for the calculation later.

ddt

∂L∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂qk

⎝ ⎜

⎠ ⎟= Qk

Page 16: Lecture 2. A Day of Principles

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What did we do and what did we assume?

Principle of virtual work

d’Alembert’s principle (inertial forces)

Independence of the generalized coordinates

We will spend a lot of time dealing with cases where the qs are not independent. . . . . but not today . . .

All the rest was clever manipulation

Page 17: Lecture 2. A Day of Principles

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??

Page 18: Lecture 2. A Day of Principles

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Hamilton’s Principle

This is a formalism that leads to the Euler-Lagrange equationsand it will help us when we need to consider constraints

(relations between the δqs).

This is generally posed in terms of the Lagrangianbut that eliminates the generalized forces

which I’d like to include

Let me define

L* = T + W

where W denotes the work, potential and nonpotential

Page 19: Lecture 2. A Day of Principles

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We write the action integral

I = L * qk, ˙ q k,t( )dtt1

t2∫

The action integral depends on the path between the two end points.

Page 20: Lecture 2. A Day of Principles

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Hamilton’s principle statesThe actual path will be the path that minimizes the action integral.

Suppose that

qk = qk η( ), qk 0( ) = q0k where the zero denotes the desired path

I η( ) = L * qk η ,t( ), ˙ q k η ,t( ), t( )dtt1

t2∫

dI η( )dη

= 0 = ∂L *∂qk

∂qk

∂η+ ∂L *

∂˙ q k∂˙ q k

∂η

⎛ ⎝ ⎜

⎞ ⎠ ⎟dt

t1

t2∫

The first piece is fine; we need to play with the second.

Page 21: Lecture 2. A Day of Principles

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∂L *∂ ˙ q k

∂˙ q k

∂η

⎛ ⎝ ⎜

⎞ ⎠ ⎟dt

t1

t2∫ = ∂L *∂˙ q k

ddt

∂qk

∂η

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎝ ⎜

⎠ ⎟dt

t1

t2∫

∂L *∂˙ q k

ddt

∂qk

∂η

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎝ ⎜

⎠ ⎟dt

t1

t2∫ = ddt

∂L *∂˙ q k

∂qk

∂η

⎛ ⎝ ⎜

⎞ ⎠ ⎟− d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟∂qk

∂η

⎝ ⎜

⎠ ⎟dt

t1

t2∫

I can integrate the first part. All the paths hit the end points, and the integral is zero.

∂qk

∂η= δqk → 0

I do my integration by parts trick again

qk = q0k + ∂qk

∂η η = 0

⎝ ⎜ ⎜

⎠ ⎟ ⎟η +L

Page 22: Lecture 2. A Day of Principles

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so I have

dI η( )dη

= 0 = ∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎝ ⎜

⎠ ⎟∂qk

∂ηdt

t1

t2∫ = ∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎝ ⎜

⎠ ⎟δqkdt

t1

t2∫

and this gives us the Euler-Lagrange equations

No matter how we look at it, we have a governing system

ddt

∂L∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂qk

⎝ ⎜

⎠ ⎟δqk = Qkδqk

Page 23: Lecture 2. A Day of Principles

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from a governing integral

Qk + ∂L∂qk − d

dt∂L∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎝ ⎜

⎠ ⎟δqkdt

t1

t2∫

We get the Euler-Lagrange equations if the δqk are independent

We will have issues regarding independenceand solutions for them.

We will have issues regarding the generalized forcesand we’ll develop techniques for finding them.

Page 24: Lecture 2. A Day of Principles

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??

Page 25: Lecture 2. A Day of Principles

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A special kind of friction: “viscous” friction, damper/dashpot

y€

friction force = −c˙ y

damping constant

Page 26: Lecture 2. A Day of Principles

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We can get the force by differentiating something called the Rayleigh dissipation function

F =12

ν ij ˙ q i ˙ q j

The (unconstrained) Euler-Lagrange equations become

ddt

∂L∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ ∂F

∂˙ q k− ∂L

∂qk = Qk

where Qk no longer includes the friction forces

note double summationthe coefficients are the most generalthey will usually be much simpler

Page 27: Lecture 2. A Day of Principles

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I’d like to put all this together in some sort of procedure.

??

We can look at some mechanical systems that can be viewed as collections of point masses

OK. Away we go . . .

Page 28: Lecture 2. A Day of Principles

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The Euler-Lagrange process

1. Find T and V as easily as you can

2. Apply geometric constraints to get to N coordinates

3. Assign generalized coordinates

4. Define the Lagrangian

Page 29: Lecture 2. A Day of Principles

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5. Differentiate the Lagrangian with respect to the derivative of the first generalized coordinate

6. Differentiate that result with respect to time

7. Differentiate the Lagrangian with respect to the same generalized coordinate

8. Subtract that and set the result equal to Q1

∂L∂˙ q 1

ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

∂L∂q1

ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q1 = Q1

Repeat until you have done all the coordinates

Page 30: Lecture 2. A Day of Principles

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M

y1, f1

OVERHEAD CRANE

(y2, z2)

m

q

Steps 1-4 lead us to

T = 12

M + m( ) ˙ q 12

+ 12

ml2 ˙ q 22

+ ml ˙ q 1˙ q 2 cosq2

V = −mglcosq2

L = 12

M + m( ) ˙ q 12

+ 12

ml2 ˙ q 22

+ ml ˙ q 1˙ q 2 cosq2 + mglcosq2

(start without the generalized forces)

y1 = q1, θ = q2

Page 31: Lecture 2. A Day of Principles

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For q1

5. ∂L∂˙ q 1

= M + m( ) ˙ q 1 + ml ˙ q 1 cosq1

6. ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟= M + m( )˙ ̇ q 1 + mlcosq1˙ ̇ q 2 − ml ˙ q 2

2sinq2

7. ∂L∂q1 = 0

8. ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q1 = M + m( )˙ ̇ q 1 + mlcosq2˙ ̇ q 2 − ml ˙ q 22sinq2 = 0

L = 12

M + m( ) ˙ q 12

+ 12

ml2 ˙ q 22

+ ml ˙ q 1˙ q 2 cosq2 + mglcosq2

Page 32: Lecture 2. A Day of Principles

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For q2

5. ∂L∂˙ q 2

= ml2 ˙ q 2 + ml ˙ q 1 cosq2

6. ddt

∂L∂˙ q 2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟= ml2˙ ̇ q 2 + ml˙ ̇ q 1 cosq2 − ml ˙ q 1˙ q 2 sinq2

7. ∂L∂q2 = −ml ˙ q 1˙ q 2 sinq2 − mglsinq2

8. ddt

∂L∂˙ q 2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q2 = ml2˙ ̇ q 2 + ml˙ ̇ q 1 cosq2 − ml ˙ q 1˙ q 2 sinq2 + ml ˙ q 1˙ q 2 + mgl( )sinq2 = 0

L = 12

M + m( ) ˙ q 12

+ 12

ml2 ˙ q 22

+ ml ˙ q 1˙ q 2 cosq2 + mglcosq2

Page 33: Lecture 2. A Day of Principles

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8a. M + m( )˙ ̇ q 1 + mlcosq2˙ ̇ q 2 − ml ˙ q 22sinq2 = 0

8b. ml2˙ ̇ q 2 + ml˙ ̇ q 1 cosq2 − ml ˙ q 1˙ q 2 sinq2 + ml ˙ q 1˙ q 2 + mgl( )sinq2 = 0

The governing equations are then

8a. M + m( )˙ ̇ y 1 + mlcosθ ˙ ̇ θ − ml ˙ θ 2 sinθ = 0

8b. ml2˙ ̇ θ + ml˙ ̇ y cosθ − ml˙ y ̇ θ sinθ + ml˙ y ˙ θ + mgl( )sinθ = 0

Put the physical variables back so it looks more familiar

This is all without forcing or damping — let’s add those

Page 34: Lecture 2. A Day of Principles

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M

y1, f1

OVERHEAD CRANE

(y2, z2)

m

q

If y1 changes, f1 does workQ1 = f1

If q changes, t1 does workQ2 = t1

add a torque, t1

If y1 changes, t1 does no workQ1 does not change

If q changes, f1 does no workQ2 = 0

(with forces)

Page 35: Lecture 2. A Day of Principles

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The governing equations were

8a. M + m( )˙ ̇ y + mlcosθ ˙ ̇ θ − ml ˙ θ 2 sinθ = 0

8b. ml2˙ ̇ θ + ml˙ ̇ y cosθ − ml˙ y ̇ θ sinθ + ml˙ y ˙ θ + mgl( )sinθ = 0

We added the generalized forces

8a. M + m( )˙ ̇ y + mlcosθ ˙ ̇ θ − ml ˙ θ 2 sinθ = Q1 = f1

8b. ml2˙ ̇ θ + ml˙ ̇ y cosθ − ml˙ y ̇ θ sinθ + ml˙ y ˙ θ + mgl( )sinθ = Q2 = τ 1

Now we need the Rayleigh dissipation function

Page 36: Lecture 2. A Day of Principles

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F =12

c ˙ θ 2

The damper works when the angle changes, but not when the cart movesSo, the Rayleigh dissipation function for this problem is

∂F∂ ˙ y

= 0, ∂ F∂ ˙ θ

= c ˙ θ

8a. M + m( )˙ ̇ y + mlcosθ ˙ ̇ θ − ml ˙ θ 2 sinθ = f1

8b. ml2˙ ̇ θ + ml˙ ̇ y cosθ − ml˙ y ˙ θ sinθ + ml˙ y ˙ θ + mgl( )sinθ + c ˙ θ = τ 1

Page 37: Lecture 2. A Day of Principles

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We aren’t really up to discussing solving these problems,but let me say a few things that we will revisit.

M + m( )˙ ̇ y + mlcosθ ˙ ̇ θ − ml ˙ θ 2 sinθ = f1

ml2˙ ̇ θ + ml˙ ̇ y cosθ − ml˙ y ̇ θ sinθ + ml˙ y ˙ θ + mgl( )sinθ + c ˙ θ = τ 1

a pair of coupled second order ordinary differential equations

It would be nice to have first order equations

There are lots of ways to do this, and we’ll look at many of thembut the simplest is to let

˙ q i = ui

Page 38: Lecture 2. A Day of Principles

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Then we’ll have

M + m( ) ˙ u 1 + mlcosq2 ˙ u 2 − mlu22sinq2 = f1

mlcosθ ˙ u 1 + ml2 ˙ u 2 − mlu1u2 sinq1 + mlu1u2 + mgl( )sinq2 + cu2 = τ 1

˙ q 1 = u1

˙ q 2 = u2

If you supply a force, a torque and initial conditionsyou can solve this set numerically.

Page 39: Lecture 2. A Day of Principles

39

You can solve for the variables, which I won’t do because it is pretty messy, and you’ll wind up with

ddt

q1

q2

u1

u2

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎬ ⎪ ⎪

⎭ ⎪ ⎪

=

u1

u2

rhs( )3

rhs( )4

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎬ ⎪ ⎪

⎭ ⎪ ⎪

Later on we’ll learn to call this a state vector