Many Mechanical Degrees of Freedom

Classical Mechanics

1.5 Hamilton’s Principle

1.5.3 Many Mechanical Degrees of Freedom

Since the pendulum is confined to thexy-plane, we need thex- andy-components of the equation:

mx¨=−Tsinθ and my¨=Tcosθ₋mg, (1.85) the solution of which must satisfy the constraint that the lengthlof the pendulum does not change:

x²+y²=l²=const. (1.86)

In (1.85),T=||T||is the length of the unknown force vectorTrequired to enforce the constraint. Thus,Tmust be found as a part of the solution. In contrast,Tsim- ply does not arise in Lagrange’s equation of motion, because the constraint was automatically taken care of by choosingθ as a single variable to specify the config- uration of the system. That is,xandygiven by (1.73) automatically satisfy (1.86).

It is also worth pointing out that, in Newtonian mechanics, we work directly with vectors such as force, position, and velocity. In contrast, we deal primarily with scalars, such as kinetic and potential energies and Lagrangian in Lagrangian mechanics. This often leads to a simpler treatment of the same problem.

A variable needed to specify the configuration of the system, such asθ, is called ageneralized coordinate. We refer to its time derivative, ˙θ here, as ageneralized velocity.

Exercise 1.5.Show that (1.85) reduces to (1.82). ///

1.5 Hamilton’s Principle 21

. ......................

. ....... .............................

x y

1 l1

m1

g

2

l2

m2

Fig. 1.8 A coplanar double pendulum oscillating in thexy-plane. Bothl₁andl₂are constant.

The explicit time dependence ofLdoes not play any role when considering the variations ofq₁, . . . ,q_f at each instant of time between givent₁andt₂because the field is common to both actual and varied paths.

When applying the stationarity condition ofS, that is,δ^S=0, the variations of q₁, . . . ,q_f are entirely arbitrary except that the variations must vanish att=t₁and t₂. As one such variation, we may consider a variation, in which onlyδq_iis nonzero andδq_j≡0 for j=i. Even for this rather special variation, Hamilton’s principle demands thatδ^S =0. This gives

d dt

∂L

∂q˙_i

−∂L

∂^qi

=0. (1.88)

Lettingivary from 1 to f, we arrive at d

dt ∂^L

∂q˙_i

−∂^L

∂qi

=0, i=1, . . . ,f . (1.89) What we have shown is that (1.89) is necessary in order forδ^S to vanish with respect to arbitrary variations ofq₁, . . . ,q_f. Equation (1.89) is also sufficient for δ^S =0. (Take a moment to think about this.)

In arriving at (1.89), we made use of the fact thatq_i’s are all capable of indepen- dent variations. If the number ofq_i’s exceeds f, they are not independent. Thus, by demanding thatδq_j≡0 forj=i, we may be imposing a constraint on the possible values ofδq_i. In this case, Hamilton’s principle does not necessarily lead to (1.89).

Example 1.4. Coplanar double pendulum: Let us find a Lagrangian of a copla- nar double pendulum placed in a uniform gravitational field as depicted in Fig.1.8.

The configuration of the system is uniquely determined if we specify the position of the particles(x1,y₁)and(x2,y₂). In terms of these variables, we have

L=1

2m1(˙x12+y˙₁²)−m1gy1+1

2m2(x˙22+y˙₂²)−m2gy2. (1.90) However, since the particles are attached to the rods,x1,y1,x2, andy2are not independent and we cannot derive Lagrange’s equations of motion directly from the Lagrangian given by (1.90).

Instead, we note thatx₁,y₁,x₂, andy₂are completely determined by spec- ifying two variables φ1 andφ2. In other words, the mechanical degrees of freedom of the system is two and we should be able to find two Lagrange’s equations of motion from a Lagrangian expressed in terms ofφ1,φ2, ˙φ₁, and φ˙₂. From Fig.1.8, we note the following relations:

x₁=l₁sinφ1, y₁=−l₁cosφ1, x2=l1sinφ1+l2sinφ2,

y₂=−l₁cosφ1−l₂cosφ2. (1.91) Taking the time derivative,

˙

x₁=l₁φ˙₁cosφ1,

˙

y₁=l₁φ˙₁sinφ1,

˙

x2=l1φ˙₁cosφ1+l2φ˙₂cosφ2,

˙

y₂=l₁φ˙1sinφ1+l₂φ˙2sinφ2. (1.92) Using these relations in (1.90), we find

L= 1

2(m1+m2)l12φ˙₁²+1

2m2l22φ˙₂²+m2l1l2φ˙₁φ^˙₂cos(φ1−φ2) +(m1+m₂)gl1cosφ1+m2gl₂cosφ2, (1.93) where we used the identity:

cos(α+β) =cosαcosβ₋sinαsinβ. (1.94) ThisLcan be used in (1.89) to find Lagrange’s equations of motion forφ1(t) andφ2(t).

Exercise 1.6.Consider a pendulum consisting of a massless harmonic spring and a particle of massmattached at the end of the spring as shown in Fig.1.9. The spring constant and the natural length of the spring arekandl0, respectively.

1.5 Hamilton’s Principle 23

Fig. 1.9 A pendulum suspended by a harmonic spring.

a. Leter denote a unit vector pointing from the origin to the particle. Whether we are stretching the spring (r>l₀) or compressing it (r<l₀), the force exerted by the spring on the particle is given by

F=−k(r−l₀)er. (1.95)

(Draw a diagram to convince yourself of this point.) Use this expression to show that the potential energy of the harmonic spring is given by

ψ(r) =1

2k(r−l0)². (1.96)

b. Usingθandras your generalized coordinates, find the Lagrangian of the system.

You may assume that the motion of the spring and the particle is confined to the

xy-plane. ///

Exercise 1.7.A system consists of two particles, one atr1and the other atr2. They interact via a potential energyφ which depends only on their relative positionr:=

r1−r2. Using the position vectorRof the center of mass R=m₁r1+m₂r2

m₁+m₂ (1.97)

andras your generalized coordinates, show that L=1

2MR˙²+1

2µr˙²−φ(r), (1.98)

whereM:=m1+m2, ˙R²:=||R˙||²=R˙·R, and likewise for ˙˙ r². The quantity µ:= m₁m₂

m1+m2

(1.99) is the so-calledreduced mass. Why is it more convenient to work withRandrthan

withr1andr2? ///

Given a mechanical system, its Lagrangian is determined only to an additive function dF/dt. In fact, ifL₁denotes a Lagrangian of a mechanical system with f degrees of freedom, the new function

L2:=L1+dF(q1, . . . ,q_f,t)

dt (1.100)

also serves as a Lagrangian of the same system.

The validity of this claim is immediately obvious if we integrate (1.100) with respect to time. In fact, if

S_i[q1, . . . ,qf]:=

_t2 t1

Li(q1, . . . ,qf,q˙₁, . . . ,q˙_f,t)dt, (1.101) then

S₂[q1, . . . ,q_f] =S₁[q1, . . . ,q_f] +F(q1(t2), . . . ,q_f(t2),t₂)

−F(q1(t1), . . . ,q_f(t1),t₁). (1.102) Sinceδ^qi(t1) =δ^qi(t2) =0 for alli=1, . . . , f, we see thatδ^S2=δ^S1. Thus, if δ^S1=0, thenδ^S2=0 forthe same set of functions q₁(t), . . . ,q_f(t), which is then the solution of Lagrange’s equations of motion whether they are derived fromL₁or L₂. Alternatively, you can validate the claim through a more explicit computation of various derivatives as illustrated in the next exercise.

Exercise 1.8.Use (1.100) to compute d dt

∂L2

∂q˙_i

−∂L2

∂q_i and show that it is zero if and only if

d dt

∂L1

∂q˙_i

−∂L1

∂q_i =0.

This being the case fori=1, . . . ,f, the same set of functionsq1(t), . . . ,q_f(t)found by solving Lagrange’s equations of motion obtained from L1 also satisfies those

derived fromL2. ///