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Advanced Dynamics

Paolo Tiso

Spring Semester 2017

ETH Zürich

- Lecture 4 –

Lagrange Equations

LECTURE OBJECTIVES

1. Derive the Lagrange equations of a system of particles;

2. Show that the equation of motion of the 2D rigid body could

be obtained by d’Alembert’s principle by projecting onto

virtual displacements;

3. Derive the equation of motion of the 2D rigid body via

Lagrange equations as well.

Lagrange equations are obtained by rewriting of the d’Alembert’s

principle by using energies. As a result, they are algebraically less

involved and can be automatized in a symbolic code.

N X

i= 1



d

dt

✓

@T

@r˙ i

◆

• F i

·

@r i

@q k

= 0 k = 1,... , p

Rewrite this differently

d

dt

✓

@T

@r˙ i

·

@r i

@q k

◆

=

d

dt

✓

@T

@r˙ i

◆

·

@r i

@q k

@T

@r˙ i

·

d

dt

✓

@r i

@q k

◆

d

dt

✓

@T

@r˙ i

◆

·

@r i

@q k

= -

@T

@r˙ i

·

d

dt

✓

@r i

@q k

◆

d

dt

✓

@T

@r˙ i

·

@r i

@q k

◆

Note that:

and therefore:

(5)

(6)

(7)

(8)

Substitute (7) into (5):

@T

@r˙ i

·

d

dt

✓

@r i

@qk

◆

=

@T

@r˙ i

·

@r˙ i

@qk

=

@T

@qk

We would like to work only with q k

, and eliminate r i

. Consider the red-boxed term:

(9)

N X

i= 1



@T

@r˙ i

·

d

dt

✓

@r i

@q k

◆

d

dt

✓

@T

@r˙ i

·

@r i

@q k

◆

• F i

·

@r i

@q k

= 0 k = 1,... , p

r˙

i

p

X

k= 1

@r

i

@q

k

q^ ˙

k

@r

i

@t

@r˙

i

@ q˙

k

@r

i

@q

k

@T

@r˙ i

@r i

@q k

@T

@r˙ i

@r˙ i

@ q˙ k

@T

@q k

Note also that, by taking the partial derivative with the velocity w.r.t. q˙k, we get

and therefore, the blue-boxed term could be simply written as

(10)

(11)

d

dt

@T

@ q˙ k

@T

@q k

- Q

k

= 0, k = 1,... , p (12)

The set of equations (12) is called Lagrange equations (LEs).

Ø LEs are nothing else than d’Alembert’s principle, written in terms of kinetic

energies.

Ø As such, the chosen generalized coordinates are supposed to identically satisfy

the constraints (i.e. they are unconstrained coordinates)

Ø By working with energy, the inertial terms involve less algebra than the

d’Alembert’s principle.

Giuseppe Ludovico Lagrangia

(1736-1713)

✓

x

y

m 1

m 2

L

g

EXAMPLE 4.

Find the equation of motion using Lagrange Equations.

Use y and (^) ✓as generalized coordinates.

Write kinematics to satisfy constraints identically: R s

R b

• R b

x 1

= 0

y 1

= y

x 2 = L cos ✓

y 2 = y + L sin ✓

x^ ˙ 1

= 0

y^ ˙ 1 =^ y˙

x ˙ 2

= - L sin ✓

˙ ✓

y ˙ 2

= y˙ + L cos ✓

˙ ✓

We need the kinetic and potential energy:

T =

1

2

m 1

( x˙ 1

2

• y˙ 1

2 ) +

1

2

m 2

( x˙ 2

2

• y˙ 2

2 ) V = - m 2

gx 2

(1) (2)

By using (1) and (2):

T =

1

2

m 1 y˙

2

1

2

m 2 [ y˙

2

• L

2 ˙ ✓

2

• 2 L y˙

˙ ✓ cos ✓] (^) V = - m 2 gL^ cos^ ✓

d

dt

✓

@T

@ y˙

◆

@T

@y

@V

@y

= 0

d

dt

✓

@T

@

˙ ✓

◆

@T

@✓

@V

@✓

= 0

Lagrange Equations

m 2

L

2 ¨ ✓ + m 2

Ly¨ cos ✓ + m

2 gL sin ✓ = 0

(m 1

• m 2

)y¨ + m 2

L

¨ ✓ cos ✓ - m 2

L

˙ ✓

2 sin ✓ = 0

e y

e x

P

k

r P

r Pk

✓

EQUATION OF MOTION OF 2D RIGID BODY USING D’ALEMBERT’S PRINCIPLE

O

B

Therefore, the position of a generic point k is

given by

r k

= r P

• r Pk

r k

= r P

• r Pk

r

Pk

· r

Pk

= const ) 2 r

Pk

· r

Pk

r k

The corresponding virtual displacement is

In order to obtain the virtual displacement , we can take the variation of

the rigid body constraint:

B

The position of the body is fully specified

by the position of an arbitrary point P

belonging to and one angle.

B

✓

r

Pk

Which yields, as expected, that has to be orthogonal to. (see rigid link

example). This can be expressed by posing

(1)

r

Pk

r Pk

(1) Alternatively, we could use the velocity transfer formula and use the kinematical method

v k

= v P

˙ ✓e z

⇥ r Pk

! r k

= r P

• ✓e z

⇥ r Pk

r

Pk

= ✓e

z

⇥ r

Pk

(1)

(2)

(3)

(4)

(2) Recall the identity a^ ·^ (b^ ⇥^ c) =^ b^ ·^ (c^ ⇥^ a) =^ c^ ·^ (a^ ⇥^ b)

N X

k= 1

r Pk

⇥ (m k

¨r k

- F

k

) · ✓e z

Equation (10) could be rewritten

2 as

N X

k= 1

m k

¨r k

N X

k= 1

F

k

N X

k= 1

m k

r Pk

⇥ ¨r k

N X

k= 1

r Pk

⇥ F

k

The resulting equations are

(11)

(12)

(13)

We still need to express (12) and (13) in terms of r P

and ; we need to find the

accelerations, by using the velocity transfer formula and differentiate in time:

r^ ˙ k

= r˙ P

✓e z

⇥ r Pk

(14)

(15)

¨r k

= ¨r P

✓e z

⇥ r Pk

✓e z

⇥ r˙ Pk

= ¨r P

✓e z

⇥ r Pk

✓e z

✓e z

⇥ r Pk

= ¨r P

✓e z

⇥ r Pk

2

r Pk

Substituting (15) and (16) in (12) and (13):

r PC

m tot

N X

k= 1

m k

r Pk

, m tot

N X

k= 1

m k

Recall the definition of the the center of mass

e y

e x

P

k

r P

r Pk

O

B

r k

C

r PC

Center of mass

(16)

m tot

(¨r P

✓e z

⇥ r PC

2

r PC

N X

k= 1

F

k

= F

m tot

r PC

⇥ ¨r P

N X

i= 1

m k

r Pk

⇥ (e z

⇥ r Pk

N X

i= 1

r Pk

⇥ F

k

= M

P

Note that:

(17)

(18)

(19)

m tot

(¨r P

✓e z

⇥ r PC

2

r PC

) = F

Finally, the equation of motion for the 2D rigid body become:

N X

k= 1

m k

r Pk

⇥ (e z

⇥ r Pk

N X

k= 1

m k

|r Pk

2

e z

= I

P

e z

m tot

r PC

⇥ ¨r P

+ I

P

✓e z

= M

P

(20)

(21)

EQUATION OF MOTION OF 2D RIGID BODY LAGRANGE EQUATIONS

We first set up the kinetic energy for the body according to its definition and the

kinematic relations across arbitrary points of the body:

where we used the definition of the moment of inertia and the center of mass

introduced in the previous section.

We now just need to evaluate the different terms of the Lagrange equations.

Coupling term for P not

being the center of mass

T =

1

2

N X

k= 1

m k

r˙ k

· r˙ k

=

1

2

N X

k= 1

mk(r˙ P

˙ ✓e z

⇥ r Pk

) · (r˙ P

˙ ✓e z

⇥ r Pk

)

=

1

2

N X

k= 1

m k

r˙ P

· r˙ P

1

2

N X

k= 1

m k

(

˙ ✓e z

⇥ r Pk

) · (

˙ ✓e z

⇥ r Pk

) +

N X

k= 1

m k

r˙ P

· (

˙ ✓e z

⇥ r Pk

)

=

1

2

m tot

r˙ P

· r˙ P

1

2

I P

˙ ✓

2

• m tot

˙ ✓r˙ P

· (e z

⇥ r PC

)

@T

@r˙ P

= m tot

r˙ P

• m tot

✓e z

⇥ r PC

d

dt

@T

@r˙ P

= m tot

¨r P

• m tot

✓e z

⇥ r PC

• m tot

2

e z

⇥ (e z

⇥ r PC

@T

@r P

For the translation:

For the rotation:

@T

= I

P

✓ + m tot

r˙ P

· (e z

⇥ r PC

d

dt

@T

= I

P

✓ + m tot

¨r P

· (e z

⇥ r PC

) + m tot

✓r˙ P

· (e z

⇥ (e z

⇥ r PC

@T

= m tot

✓r˙ P

e z

@r PC

= m tot

✓r˙ P

· (e z

⇥ (e z

⇥ r PC

m, I C

g

r

✓

C

O

k

R

A disk of radius r, mass m and centroidal moment of

inertia I C

rolls without slipping on a circular profile of

radius R. The center of the disk is connected to the

vertical axis by a spring of stiffness k and zero

unstretched length. The spring force always acts in

the horizontal direction, while gravity acts in the

vertical direction, as indicated. Denote with ψ the

rotation angle of the disk about its center C, and with

θ the angular position of the center of the disk with

respect to the center of the circular profile.

Find the equation of motion using Lagrange Equations.

Use θ as generalized coordinate.

EXAMPLE 4.

Rolling without slipping, integrable: ✓R^ =^ r

Spring is unstretched when θ=0.

We need to write the kinetic and potential energies of the body.

e x

e y

The kinetic energy w.r.t. the center of mass C writes:

T =

1

2

mr˙ C

· r˙ C

1

2

I C

|!|

2

r˙ C

= r˙ 0

+! ⇥ r OC

r ˙ 0

= 0

! = - (

˙ ✓ +

˙ )e z

=

˙ ✓

✓

1 +

R

r

◆

e z

r OC

= r sin ✓e x

• r cos ✓e y

r ˙ C

=

˙ ✓(r + R)(- sin ✓e x

• cos ✓e y

)

Use the velocity transfer formula where:

The velocity of the center of mass is therefore

T =

1

2

N X

k= 1

mk r˙ C

· r˙ C

1

2

IC|!|

2

1

2

m

˙ ✓

2 (r + R)

2

1

2

IC

(r + R)

2

r

2

˙ ✓

2

and the kinetic energy results in

The potential energy is given by the contributions on the gravity and the spring:

V = V g

• V s

=

1

2

k[(R + r) sin ✓]

2

• mg(R + r) cos ✓

d

dt

✓

@T

@

˙ ✓

◆

@T

@✓

@V

@✓

= 0

The Lagrange equation is given by (no non-conservative forces are present)

resulting in

EXAMPLE 4.2 cont’d



m(r + R)

2

• IC

(r + R)

2

r

2

¨ ✓ + k(R + r)

2 sin ✓ cos ✓ - mg(R + r) sin ✓ = 0

Ø Some active forces depend on constraint forces:

Ø We simply want to find some

constraint forces right away.

In the system on the left, the friction force f on

the collar C depends on the constraint force N C

:

f = - sgn( x˙C)μNC

The equation of motion must include the reaction

force N C

in order to determine the friction force

Ø Constraints are non-holonomic:

x

y

z

˙

↵

P

e

C

B

R

x^ ˙ C -^ R^

˙ sin ↵ = 0

y ˙ C

• R

˙ cos ↵ = 0

Some GCs cannot be eliminated since they are

related by non-integrable differential equations

This roller coaster is a complex multibody system. For the

design of the rail structure, the reactions between the

train and the rail might be the only ones of immediate

interest.

Idea: choose generalized coordinates that do not satisfy (some of the)

constraints to let constraint forces appear in the EoM

n

f(q, t) = 0

ˆe

2

, q

2

ˆe

1

, q

1

ˆe

n

, q

n

n =

rf

|rf|

trajectory

ˆr

Virtual displacement not compatible

with the constraint

ˆr · n 6 = 0

In this case, the virtual displacement does not lay on the tangent to the

constraint surface!

r = r(q, t)

In other words, the mapping between GCs and positions does not identically

satisfy the m constraints applied to the system.

Constraints have to be enforced

f i

(q, t) = 0, i = 1,... , m