**Continuation... **

It is also useful
to consider the unit vectors associated with polar coordinates in central force
problems: **n **pointing in direction of
increasing r, and **l**, in the direction
of increasing q .

The velocity components can then be written:

**v **_{r } = dr/
dt, **v** q
= r dq /dt

** **Using the polar
coordinate unit vectors (

**n**,,

**l**) this can be rewritten as:

**v
= **dr/ dt
**n** +
r dq /dt **l**

For the change in
the radial coordinate alone

**v
= **d**r**/ dt
= d(r** n**)/ dt = ( dr/ dt **n** + r
d**n** /dt)

d**n** /dt
= d**n** / dq (dq
/dt)

To evaluate d**n** / dq one makes use of the fact that by radian
measure of angles:

** **D **n **= D q
and in the limit** **as** **D q ** **® ** **0,

‖D
**n**
‖ / ‖D q ‖ = 1

So: d**n** / dq = **l**

i.e**. **As** **D q ** **® ** **0, D **n**
becomes perpendicular to **n **and
assumes the direction of ** l** .

In an analogous
way we may obtain the relation**:**

d**l** / dq = **-n**

Finally, we have: d**n**
/dt = (dq /dt) **l**

And: d**l**
/dt =
- (dq /dt ) **n **

**The Nature of
Constraints**

Consider the diagram below:

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**.

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}

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

**Use of 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 Problem: **Find the Lagrangian multiplier for the
system discussed above:

*Solution*: 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 of the double pendulum shown in the sketch below which includes two
masses, m1 and m2, at two angles to the vertical, f1 and f2.

**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} )

a)Obtain the Lagrangian then apply constraints and eliminate one coordinate (z).

b)Write the new Lagrange equations, viz.

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

d/dt** **(¶ L/¶ q** **)
- ¶ L/¶ q = 0

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