Solutions To Theoretical Mechanics Problems (Holonomic Constraints)

1)Let:

x1 =   L1  sin f1

x1’ = L1  cos  f1  f1’

y1 =   L1  cos f1

y1’ = - L1  sin  f1 f1’

V =  mgL1 (1 -   cos  f1)

T1  =  ½  m( r q’ ² )  =    ½  m1( L1  f1’ ² )  and   T2 =   ½  m2( x2’ ²   + y2’ ²)

Where:

x2 =   L1  sin f1  +   L2  sin f2

y2 =   L1  cos f1  +   L2  cos f2

x2’   = L1  cos  f1  f1’    + L2 cos  f2  f2’   

y2’   = L1  sin  f1  f1’    -   L2 sin  f2  f2’   

Then:

T2  =    ½  m2(L1 ²   cos ²   f1  f1’ ²   + L2 ² cos ²  f2  f2’    +

 2L1 cos  f1  f1’  L2 cos  f2  f2’    +  L1 ²   sin ²    f1  f1’ ²   +

  L2 ² sin ²  f2  f2’   + 2L1 L2 sin  f1 sin  f2 f1’ f2’ )

T2 =   ½  m2(L1 ²  f1’ ²   +  L2 ²  f2’ ²   + 2L1 L2 cos (  f2  - f1)  f1’ f2’

Therefore:

L =  ½  m1 L1 ²   f1’ ²    +  ½  m2( L1 ²   f1’ ²   +  L2 ²  f2’ ² )  

+ m2 L1 L2 cos (  f2  - f1)  f1’ f2’ – m1 g L1 (1 -  cos  f1  )  -

m2 g [L1 (1 -  cos  f1 )  +   L2 ((1 -  cos  f2 ) 

2)  a)     The Lagrangian is L = T – V

L  =  ½  m( r” ²   +  r q’ ²   +  z’ ²  )  -  mg z

Applying constraints and eliminating one coordinate (z)  [Rem: z = ar]

L  =  ½  m( r’ ²   (1  + a ²  )+  r ² q’ ² )  -  mg (ar)

b)     The new Lagrange’s equations are then:

i) m r’’ ²  (1  + a ²  ) - m r ² q’ ²   +  mg a =  0

And:

ii) m r’’  (1  + a ²  ) -   ℓ / mr²   +  mg a =  0

Rem:  angular momentum:  ℓ =   mr²    dq/ dt   =   mr²   q’  

c)Using  undetermined multiplier  l:.

Write:  m r’’  -  m r ² q’ ² =       l a

Where    ¶ f  / ¶ r   =   a

d/dt [m r ² q’ ² ] =        ℓ

m z’’  +   m g   =  l  

We see   l   is the generalized force associated with z-component

But in terms of radial coordinate r:

z = ar,  so that:    

F _r    +     F _z    =  const.    

(Normal force exerted by cone’s side requires:  F _r    =  -  F _z )

z = ar,  so    z’’  =    ar”    then:

m z’’  +   m g      =   m (ar”)   +   mg  =   m(ar”  + g)

l   =  F _z     = m z’’  +   m g     = m(ar”  + g)

Then force acting along the radial direction is :

F _r    =    - m(ar”   +  g)

More Theoretical Mechanics - Holonomic Constraints and Lagrangian Multipliers

More complex aspects of a kind of general "circular" motion can now be examined along  with computing their Lagrangians. (See e.g. the previous two posts on theoretical mechanics and circular motion. )  Consider the diagram below for a body rolling without slipping:

We may write:  dS/ dt  -  R (dq /dt)  =   0

Or:  S’   -  R q’    =   0

Which implies:    å ^c _a   q’ _a    =   0

Then:  

ò  [dS/dt     -   R  (dq /dt ) ]  = const.

Or:   S   -  R q    = const. =  0 

This condition is what defines a holonomic constraint. If such integration is not possible the constraint is non-holonomic.  By looking at such constraints the aim is always to reduce the number of coordinates needed to describe the system.

Next, consider a disk rolling down an inclined plane as shown in the diagram below:

We want to calculate the constraints assuming no slipping, where the moment of inertia of the disk is given by:

I   =   ½  m R ²  

And we have the differential equation:

 x’   -  R q’    =   0

Which when integrated yields:

x   -  R q    =  const.   =    0

The kinetic energy of the system can then be written:

   T   =   ½  m x’ ²   +  ½  I q’ ²  

Or:  T =    ½  m x’ ²   +  ¼  m R ²  q’ ²  

The potential energy will be expressed:

V  =  mg (L – x) sin f

So the Lagrangian of the system can be written:

L  =  T – V  =

½  m x’ ²   +  ¼  m R ²  q’ ²    -  mg (L – x) sin f

Using the angular –linear relation seen earlier, e.g.

x’   =  R q’    

We can simplify further:

L  =   ¾  m x’ ²   -  mg (L – x) sin f

Then it can be shown:

d/dt (¶ L/¶ x’ )  -   ¶ L/¶ x  =  0

Working:

3/2 m x”   -  m g sin f    =   0

And:  q”    =  2/3  g sin f  /R

Lagrangian Multipliers:

The Lagrange equations can be rewritten in terms of multipliers, i.e.

d/dt (¶ L/¶ q’ _k)   -   (¶ L/¶ q _k)  =

l ₁ (¶f1/¶ q _k   )   -  l ₂   (¶f2 /¶ q _k)  +  …..

Where  l ₁  and   l ₂  denote Lagrange multipliers, with one multiplier per each equation of constraint.

Example: Find the Lagrangian multiplier for the system  discussed above:

We have for the relevant partial differential equations:

i) d/dt (¶ L/¶ x )  -   ¶ L/¶ x  -  l ((¶f/¶x  ) =  0

ii) d/dt (¶ L/¶ q’ )  -   ¶ L/¶ q  -  l ((¶f/¶q  ) =  0

From which we obtain:

i)                  mx” -  m g sin f    -  l (1)  =  0

ii)               ½  m R ²  q”     +  l R =  0

We find from the preceding equations:

l  =  -  ½  m R ²  q”    =  - ½  mx”

But:   x”   =   2/3   g sin f   

And:   q”    =  2/3  g sin f  /R

Hence:  l  =  - ½  m(2/3  g sin f )

Or:   l  =  -  m g sin f / 3

Problems:

1). Find the Lagrangian for the double pendulum shown below for the least number of coordinates possible.

The system includes two masses, m1 and m2, at two angles to the vertical, f1 and  f2,   respectively.

 2)  Consider a cone and particle situated on its inside surface:

 with a force :  F  =  - mg k  exerted:

Let the potential energy V  =  mg z

And:  z = ar

The equation for the total energy is given by:

T  =  ½  m( r” ²   +  r q’ ²   +  z ' ²  )

Find:

a)The Lagrangian  after applying constraints and eliminating one coordinate (z).

b)The new Lagrange equations, viz.

d/dt (¶ L/¶ r’ )  -   ¶ L/¶ r    =   0

  d/dt (¶ L/¶ q )  -   ¶ L/¶ q  =  0

c) The force acting along the radial direction.

Show how this is obtained from the preceding equations and an undetermined multiplier  l.