ME 416/516 Dynamics The Mathematics of Analytical Dynamics - - PowerPoint PPT Presentation

ME 416/516 Dynamics

The Mathematics of Analytical Dynamics (Integral Formulation)

Gregory P. Starr

starr@unm.edu

Department of Mechanical Engineering University of New Mexico Albuquerque, NM 87131

ME 416/516 Dynamics – p.1

Stationary Values of a Definite Integral

We wish to minimize the following integral I: I = x=b

x=a

F (x, y, y′) dx (y′ = dy dx) (1) where I is a functional, and y(x) is the unique function that results in the minimum for I... Consider the varied function (see text Fig. B.1) y(x) + ǫη(x), where η(a) = η(b) = 0 So I(ǫ) = x=b

x=a

F (x, y + ǫη, y′ + ǫη′) dx (2)

ME 416/516 Dynamics – p.2

For integral I to have a stationary value, dI dǫ = 0 when ǫ = 0 (3) This results in dI dǫ

• ǫ=0

= b

a

     ∂F ∂y η + ∂F ∂y′ η′

integrate by parts

     dx = 0 (4)

ME 416/516 Dynamics – p.3

Recall integration by parts is

• udv = uv −
• vdu

(5) (6) We want to integrate b

a

∂F ∂y′ η′dx. (7) and a good choice for u and dv is u = ∂F ∂y′ , dv = η′dx (8)

ME 416/516 Dynamics – p.4

Then since u = ∂F ∂y′ = ⇒ du = d dx ∂F ∂y′

• dx

(9) also dv = η′dx = dη dx dx = dη = ⇒ v = η (10) Our integral is thus b

a

[⋆] = ∂F ∂y′ η(x)

• x=b

x=a

− b

a

d dx ∂F ∂y′

• η(x)dx

(11)

ME 416/516 Dynamics – p.5

Getting back to the “stationary value” problem...

dI dǫ

• ǫ=0

= ∂F ∂y′ η(x)

• x=b

x=a

• η(a)=η(b)=0

+ x=b

x=a

∂F ∂y − d dx ∂F ∂y′

• η(x)dx = 0

(12) Since function η(x) is arbitrary, we must have

∂F ∂y − d dx ∂F ∂y′

• = 0

(13) This is the famous “Euler-Lagrange” equation!

ME 416/516 Dynamics – p.6

Application to Dynamics

Newton’s 2nd law is given by F = ˙ p (14) where p = mv (momentum) (15) D’Alembert’s principle is F − ˙ p = 0 = ⇒ (F − ˙ p) · δr = 0 (16) Equation (16) holds at each instant of time, so t2

t1

(F − ˙ p) · δr dt = 0 (17)

ME 416/516 Dynamics – p.7

Integrate the 2nd term by parts (1st term is simply δW): − t2

t1

˙ p · δr dt (18) Select the following for u and dv: u = δr = ⇒ du = d dtδr dt (19) dv = ˙ p dt = ⇒ v =

• ˙

p dt = p (20) Thus we have − t2

t1

˙ p · δr dt = −p · δr|t2

t1 +

t2

t1

• p · d

dtδr

• dt

(21)

ME 416/516 Dynamics – p.8

The 2nd term in (21) can be expanded t2

t1

• p · d

dtδr

• dt =

t2

t1

p · δ˙ r dt = (22) t2

t1

p · δv dt = t2

t1

mv · δv dt (23) But T = 1

2mv · v, so δT = mv · δv, and combining everything

we get t2

t1

(δW + δT) dt = −p · δr|t2

t1 = 0

(24) Like η(x), let δr(t1) = δr(t2) = 0, then we get...

ME 416/516 Dynamics – p.9

t2

t1

(δW + δT) = 0 (25) This is the “Extended Hamilton’s Principle!” For conservative forces, δW = −δV , so we have t2

t1

(δT − δV ) dt = 0 (26)

ME 416/516 Dynamics – p.10

Cleaning it Up...

Define L = T − V, (27) then we have t2

t1

δL dt = 0 = ⇒ δ t2

t1

Ldt = 0 (28) Equation (28) implies a STATIONARY VALUE of the integral, and we know how to do that! F(x, y, y′) = ⇒ L(q, ˙ q, t) (29)

ME 416/516 Dynamics – p.11

Equations of Motion

Using equation (13), the Euler-Lagrange equation, we have d dt ∂L ∂ ˙ q

• − ∂L

∂q = 0 (30) = Qnc (for nonconservative forces) (31)

Another formulation: d dt ∂T ∂ ˙ qk

• − ∂T

∂qk + ∂V ∂qk = Qknc, k = 1, 2, ...n (32)

ME 416/516 Dynamics – p.12

The End

I hope you enjoyed the show!

ME 416/516 Dynamics – p.13