Constraints and Generalized Coordinates

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PHYS 705: Classical Mechanics Constraints and Generalized Coordinates

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Constraints Problem Statement: In solving mechanical problems, we start with the 2nd law

F

(e)  F  miri ji i

(*)

j

In principle, one can solve for ri(t) (trajectory) for the ith particle by specifying all the external and internal forces acting on it . However, if constraints are present, these external forces in general are NOT known. Therefore, we need to understand the various constraints and know how to handle them.

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Holonomic Constraints Holonomic constraints can be expressed as a function in terms of the coordinates and time,

f  r1 , r2 ,; t   0 e.g. (a rigid body) 

r  r 

non-holonomic examples:

i

j

2

 cij2  0

-Gas in a container -Object rolling on a rough surface without slipping… more later

More quantifiers: - Rheonomous: depend on time explicitly - Scleronomous: not explicitly depend on time e.g. a bead constraints to move on a fixed vs. a moving wire

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Constraints and Generalized Coordinates Difficulties involving constraints: 1. Through f  r1 , r2 , ; t   0 , the individual coordinates ri are no longer independent  eqs of motion (*) for individual particles are now coupled (not independent) 2. Forces of constraints are not known a priori and must be solved as additional unknowns With holonomic constraints: Prob #1 can be handled by introducing a set of “proper” (independent) Generalized Coordinates Prob #2 can be treated with: D’Alembert’s Principle & Lagrange’s Equations (with Lagrange multipliers)

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Generalized Coordinates •

Without constraints, a system of N particles has 3N dof

•

With K constraint equations, the # dof reduces to 3N-K

•

With holonomic constraints, one can introduce (3N-K) independent

 q1 , q2 , , q3 N  K  such that: r1  r1  q1 , q2 , , q3 N  K , t 

(proper) generalized coordinates

 rN  rN  q1 , q2 , , q3 N  K , t 

a point transformation

 Generalized coordinates can be anything: angles, energy units, momentum units, or even amplitudes in the Fourier expansion of ri  But, they must completely specify the state of a given system The choice of a particular set of generalized coordinates is not unique. No specific rule in finding the most “suitable” (resulting in simplest EOM)

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Generalized Coordinates In regular Cartesian coord

Example: O l (x1, y1)

2

y

 x1 , y1 , x2 , y2 

x

1

l

(x2, y2)

ri : 4 dof

 x12  y12  l 2  0 2 constraints:  2 2 x x y y     l2  0      2 1 2 1 But, there are only 2 indep dof… In generalized coord

q :

1 , 2 

(Double Plane Pendulum)

j

2 indep dof

Coord Transformation: (constraints are implicitly encoded here)

1  tan 1  x1 y1 

 2  tan 1   x2  x1   y2  y1  

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Non-Holonomic Constraints -

can’t use constraint equations to eliminate dependent coordinates

-

in general, solution is problem specific. Example: Vertical Disk rolling without slipping on a horizontal plane z

y a

Described by 4 coordinates:



(x, y) of the contact point v

: orientation of diskangle of disk axis with x-axis

O

 : angle of rotation of the disk

 x

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Non-Holonomic Constraints (rolling disk exp) Now, consider the constraints:

1. No-slip condition: s  a  v  a y

top view

2. Disk rolling vertically  v  disk axis  x  v sin     y  v cos 

see graph

v sin



v



-v cos x

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Non-Holonomic Constraints (rolling disk exp) Putting them together, gives the following differential equations of constraint,

d  dx   dt  a sin   a sin  dt    dy  a cos   a cos  d  dt dt

or

 dx  a sin  d  0  dy  a cos  d  0

The point is that we can’t write this in Holonomic form:

  f  x, y ,    0

with f being a function!

(hw)

 Roll the disk in a circle with radius R. Upon completion of the circle, x, y and  will have returned to their original values  but,  will depend on R (can’t be specified by  x, y,   )

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How to deal with Constraints? Principle of Virtual Work Consider a system in equilibrium first, - The net force on each particle vanishes:

Fi  0 (note: i labels the particles)

Consider an arbitrary “virtual” infinitesimal change in the coordinates,  ri -

Virtual means that it is done with no change in time during which forces and constraints do not change.

-

These changes are done consistent with the constraints (we will be more specific later).

Since all the Fi are zero (equilibrium), obviously we have

 F  r  0 i

i

i

(virtual work)

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Principle of Virtual Work (a)

Separating the forces into applied Fi and constraint forces fi ,

Fi  Fi( a )  fi Then,

(a) F  i   ri   fi   ri  0

i

i

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Principle of Virtual Work For virtual displacements to be consistent with the constraints means that  the virtual work done by the constraint forces along the virtual displacement must be zero.

fi

Alternative geometric view (more a bit later)

 ri

For N particles with K constraints, motion is restricted on a (3N-K)-D surface and the constraint forces fi must be  to that surface.

fi   ri or fi   ri  0 For

 ri

to be consistent with constraints means that the net virtual work

of the forces of constraints is zero !

 f  r  0 i

i

i

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Principle of Virtual Work Back to our original equation for a constrained system in equilibrium, (a) F  i   ri   fi   ri  0

i

i

This leaves us with the statement, (a) F  i   ri  0

i

 The virtual work of the applied forces must also vanish! This is called the Principle of Virtual Work.

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Principle of Virtual Work (a) F  i   ri  0

i

Note: Since the coordinates (and the virtual variations) are not necessary independent. They are linked through the constraint equations. The Principle of Virtual Work does not implies,

Fi( a )  0 for all i independently. The trick is now to change variables to a set of proper (independent) generalized coordinates. Then, we can rewrite the equation as,

 ?

j

 q j  0

j

With qj being independent, we can then claim:

 ?  j  0 for all j.

As we will

see, this will give us expressions which will lead to the solution of the problem.

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D’Alembert’s Principle Now, we consider the more general case when the system is not necessary in equilibrium so that the net force on the particles is NOT zero. We continue to assume the constraints forces to be unknown a priori… Similar to our discussion on the Principle of Virtual Work, we would like to reformulate the mechanical problem to include the constraint forces such that they “disappear”  you solve the “new” problem using only the (given) applied forces. This is the basis for the D’Alembert’s Principle and by choosing a set of proper generalized coordinates, it will result in the EulerLagrange’s Equations.

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D’Alembert’s Principle In deriving the Principle of Virtual Work, the system was in equilibrium. In extending it to include dynamics , we will begin with the 2nd law,

Fi  p i

or Fi  p i  0

for the ith particle in the system.

We again consider a virtual infinitesimal displacement

 ri consistent with

the given constraint. Since we have Fi  p  i  0 for all particles, We have,

  F  p    r  0 i

i

i

i

Again, we separate out the applied and constraint forces, Fi  Fi( a )  fi This gives,

 F

(a) i

i

 p i    ri   fi   ri  0 i

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D’Alembert’s Principle Then, following a similar argument for the virtual displacement to be consistent with constraints, i.e, We can write down,

 F

(a) i

 f  r  0 i

i

(no virtual work for fi)

 p i    ri  0 i

i

This is the D’Alembert’s Principle. Again, since the coordinates (and the virtual variations) are not necessary



(a)

independent. This does NOT implies, Fi

 p i   0 for the individual i.

We then need to look into changing variables to a set of independent generalized coordinates so that we have

 ?

j

  q j  0 with the coefficients

in the sum independently equal to zero, i.e.,  ?   0 j j

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Geometric View of the D’Alembert’s Principle Consider a particle moving in 3D with one Holonomic constraint, z

equation of motion: m r  F(a)  f

r  ( x, y , z )

equation of constraint: g (r, t )  0 Here,

m

- F(a) is the known applied force

r (t) y

- And, we model the unknown constraint force by the vector f.

x Note: r(t) has 3 unknown

trajectory (red) is constraint to move in

components + 1 constraint

a 3-1=2 dimensional surface (blue).

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Geometric View of the D’Alembert’s Principle - There are three unknown components to the constraint force f. A scalar constraint does not specify the vector f completely. - There are multiple choices for f which satisfy g(r, t)=0 BUT there is an additional physical restriction on f that we should consider… z

z

f

f y

x

y x

Observation: For a given f, adding a component // to the surface will still keep the particle on the surface (satisfying g(r, t)=0) but will result with an additional acceleration along the surface).

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Geometric View of the D’Alembert’s Principle  A reasonable physical argument is to restrict the choice of f so that: Constraint Force f needs to lay  to the constraint surface Note that g(r,t) =0 is the equation for the constraint surface and

z

 g (r, t )  surface

f y x

g (r, t )  0 This gives,

So, the  condition can be stated as,

f   g (r, t )

where  can depend on t

m x  F ( a )   g (r, t )  4 unknowns r and   4 equations g (r, t )  0 

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Geometric View of the D’Alembert’s Principle m r  F ( a )   g (r, t )   g (r, t )  0 

4 unknowns r and  4 equations

This system is solvable but now we would like to solve the system w/o knowing the constraint forces explicitly … Note that g is  to the surface of constraint and we can project the dynamical equation onto the tangent plane of the constraint surface at (r, t): This gives two independent scalar equations,

 mr  F   e  mr  F   e

a

  g (r, t )  ea  0

b

  g (r, t )  eb  0

(a)

(a)

where e a and eb are two basis vectors spanning the tangent plane to the constraint surface at (r, t).

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Geometric View of the D’Alembert’s Principle Together with the constraint equation itself, we then have 3 eqs for the 3 unknown components of r.

 mr  F   e

 0   g (r, t )  0  (a)

a ,b

3 unknowns r 3 equations

So, now, in principle, we can solve for the dynamical equation (EOM), r(t), without knowing the constraint forces f explicitly.

 mr  F   e (a)

a ,b

0

 This is the D’Alembert’s Principle (for a single particle).

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Geometric View of the D’Alembert’s Principle We can generalize the argument to a system of N particles with K constraints (Holonomic): Note: The virtual  ri (a) mi ri  Fi  e k  0 displacements consistent i with the constraints are   (a) in the tangent space  p F  r 0    i i i   spanned by the basis e k   i 









Geometric Interpretation: The K constraints restrict the system to a (3N-K)-D surface within the 3N-D space. There are (3N-K) ek vectors spanning that tangent plane to the constraint surface so that the above expression gives (3N-K) equations that the problem can be solved without knowing the constraint forces explicitly.

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D’Alembert’s Principle

 F

(a) i

 p i    ri  0

i

To solve for EOM using the D’Alembert’s Principle … We need to look into changing variables to a set of independent generalized coordinates so that we have

 ?

j

 q j  0

j

Then, we can claim the coefficients  ?  j in the sum to be independently equal to zero. The Euler-Lagrange equation will give an explicit expression for the EOM as:

? j  0

d  T  dt  q j

 T  Qj  0    q j 

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An Aside: Constraint and Work Recall that we have from the EOM:

m r  F ( a )   g (r, t )

Let F(a) be a conservative force , i.e., F ( a )  U (r, t ) so that

mr  U  g Dotting

r

into both sides,

m r  r 

d  1 2  dT  mr   dt  2  dt

U  r  g  r

Consider the last term, from chain rule, we have,

dg  g x g y g z  g g        g  r     dt  x t y t z t  t t

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An Aside: Constraint and Work As the particle moves, it is constraint to stay on the g=0 surface, So,

g dg  0 and,  g  r    dt t

Similarly, from chain rule, we can write, U  r 

dU U  dt t

Putting everything together,

m r  r  U  r  g  r With E=T+U,

dT dU U g    dt dt t t

dE U g   dt t t

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An Aside: Constraint and Work dE U g   dt t t So, either U or g explicitly depends on time, the total energy changes with time. Since we typically do not consider time-dependent U potential functions, So, we can make the following assertions: Scleronomous (g not explicitly depends on t) Holonomic Constraints:

 g  r   

g 0 t

and constraint force won’t do work!

Rheonomous (g explicitly depends on t) Holonomic Constraints:

 g  r   

g 0 t

and constraint force can do work!