ESM 3124 Intermediate Dynamics 2012, HW6 Solutions

Roller coaster. A bead of mass m can slide without friction,under the action of gravity, on a smooth rigid wire which hasthe form y = f(x).

(a) Find the equation for x as a function of x and x.

(b) Find the expression for the Lagrange multiplier λ1, re-lated to the force maintaining the sliding constraint, asa function of x and x.

(c) Find the expression for the magnitude fc of the force ofconstraint, as a function of x and x.

Hint: the answers to (a) and (b) are

x = −f′(x)(f ′′(x)x2 + g)

(1 + f ′(x)2), λ1 =

m(f ′′(x)x2 + g)(1 + f ′(x)2)

m

g y=f (x)y

x

Solution

Lagrange Multiplier Method. We can first write the constraint y = f(x) in the form of a constraintequation equal to zero,

ψ1(x, y) = −f(x) + y = 0. (1)

We can use the Lagrange multiplier method to write the equations of motion. This involves writingthe Lagrangian function

L = T − V =12m(x2 + y2

)−mgy (2)

and then writing the Lagrange’s equation of motion in the presence of one constraint,

d

dt

(∂L

∂qi

)− ∂L

∂qi= λ1a1i, (i = 1, 2), (3)

where q1 = x, q2 = y, and λ1 is the Lagrange multiplier corresponding to the one constraint. Usingthe coefficients a11 = ∂ψ1

∂x = −f ′(x) and a12 = ∂ψ1

∂y = 1, we get two equations of motion,

mx = −f ′(x)λ1, my +mg = λ1. (4)

Using these two equations to eliminate λ1, we end up with

x+ f ′(x)y + f ′(x)g = 0. (5)

where we have divided through by −m and rearranged the order of the terms.

Getting x equation. To get an equation for x in terms of x and x, we need to write y in terms ofonly x and its time derivatives. To do this we start with the constraint y = f(x) and take two timederivatives so we have an equation for y,

y = f(x) ⇒ y = f ′(x)x ⇒ y = f ′′(x)x2 + f ′(x)x. (6)

Plugging this expression for y into eq. (5), we get

x = −f′(x)(f ′′(x)x2 + g)

(1 + f ′(x)2)(7)

1

Notice the right hand side is just a function of x and x.Alternate method for getting x. Going back to the Lagrangian function (2), we can directly write

this in terms of only x and x, so x is the only generalized coordinate and only degree of freedom.

L =12m(x2 + y2

)−mgy

=12m

((dx

dt

)2

+(dy

dt

)2)−mgf(x)

=12m

(dx

dt

)2(

1 +(dy

dx

)2)−mgf(x)

L(x, x) =12mx2

(1 + (f ′(x))2

)−mgf(x)

(8)

and then write the Lagrangian equation for x, as usual, to recover (7).This method is more direct for getting the equation for x since it skips the need for the Lagrange

multiplier.

Solving for the Lagrange multiplier and constraint force. Using (7) in the left-side equationof (4), we can solve for λ1 and we get

λ1 =m(f ′′(x)x2 + g)

(1 + f ′(x)2)(9)

If we want to know the magnitude of the constraint force, we proceed as follows. The constraint forceis Fc = λ1∇ψ1 = λ1

(∂ψ1

∂x n1 + ∂ψ1

∂y n2

). Taking the norm of this vector, we get

Fc = |Fc| = λ1

√√√√([∂ψ1

∂x

]2

+[∂ψ1

∂y

]2)

= λ1

√(f ′(x)2 + 1)

=(m(f ′′(x)x2 + g)

(1 + f ′(x)2)

)√(f ′(x)2 + 1)

(10)

so Fc is

Fc =m(f ′′(x)x2 + g)√

1 + f ′(x)2(11)

2

Problem 2Find the equation of motion for a situation addressed in anearlier homework, but now include Coulomb friction;i.e., the walls of the tube are rough so there’s a frictioncoefficient µ, where 0 < µ < 1, between the bead and thetube walls. Make no assumptions about the value of θ,although you can still assume θ is fixed.

Hint: You should find that the equation of motion for r is

mr −mrΩ2 sin2 θ +mg cos θ = −µNsgn(r)

where sgn(r) = |r|/r and the normal force N of the tubeacting on the bead has magnitude

N =√N2θ +N2

φ = m[4Ω2r2 + (rΩ2 cos θ + g)2]1/2 sin θ

106 Lagrange’s and Hamilton’s equations

and acts circumferentially on each circular strip with a moment arm equal to R sin θ . Thus,we obtain the integral

Q′φ = −

∫ π/2

02πσ f R

3 sin2 θ dθ

= −∫ π/2

03µFR sin2 θ cos θ dθ = −µFR (2.247)

The negative sign indicates that the frictional moment Q′φ opposes the rotation φ.

Example 2.10 A particle of mass m can slide inside a straight tube which is inclined atan angle θ from the vertical and is rigidly attached to a vertical shaft which rotates at aconstant rate (Fig. 2.9). There is a Coulomb friction coefficient µ between the particleand the tube. We wish to find the differential equation of motion.

Let us use Lagrange’s equation in the general form

d

dt

(∂T

∂q i

)− ∂T

∂qi+ ∂V

∂qi=

m∑j=1

λ j a ji + Q′i (2.248)

Choose the spherical coordinates (r, θ, φ) as generalized coordinates with the constraints

θ = 0, φ − = 0 (2.249)

g

m

er

eφ

eθ

µ

θ

r

Figure 2.9.

Solution

Write this as a constrained Lagrangian problem, where our coordinates are the usual n = 3 sphericalcoordinates (r, θ, φ), but we have two constraints, i.e., number of constraints, nc, is 2,

θ = 0

φ− Ω = 0(12)

which are in the nonholonomic form,∑n

i=1 aji(q, t)qi + bj(q, t) = 0, so we can read off the a and bcoefficients,

θ = 0 =⇒ a1r = 0, a1θ = 1, a1φ = 0, b1 = 0

φ− Ω = 0 =⇒ a2r = 0, a2θ = 0, a2φ = 1, b2 = −Ω

We need to write the unconstrained Lagrangian. The unconstrained velocity is what one expects forspherical coordinates,

R = rer + rθeθ + rφ sin θeφ (13)

so the unconstrained kinetic energy of the particle is,

T = 12m|R|2 = 1

2m(r2 + r2θ2 + r2φ2 sin2 θ). (14)

The potential energy is due to gravity alone. Using the origin of the spherical coordinate systemas the reference height, we have

V = mgr cos θ

The Lagrangian is:L = 1

2m(r2 + +r2θ2 + r2φ2 sin2 θ)−mgr cos θ

The Lagrangian equations of motion for a constrained system are,

d

dt

(dL

dqi

)− dL

dqi= Qnc,i + Ci, i = 1, 2, 3 (15)

where equivalently, we can view the indices i as being in the range r, θ, φ, and where the generalizedforces of constraint, Ci, can be written in terms of Lagrange multipliers

Ci =nc∑j=1

λjaji (16)

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and also in terms of the Newtonian (i.e., physical) forces of constraint on the particle, fc,

Ci = fc · ∂R∂qi , (17)

where, for this system, the force of constraint is equal to the reaction force of the tube walls on theparticle, so fc = N, which has components Nθ, and Nφ, i.e., N = Nθeθ +Nφeφ.

The non-conservative Newtonian force applied to the mass is Coulomb friction,

fnc = −µN sgn(r)er,

where N = |N|, so the generalized non-conservative forces are,

Qnc,r = fnc ·∂R∂r

= fnc · er = −µN sgn(r)

Qnc,θ = fnc ·∂R

∂θ= fnc · reθ = 0

Qnc,φ = fnc ·∂R

∂φ= fnc · r sin θeφ = 0

At this point, most of the hard work is done. Now we just work out the derivatives and do somealgebra. Remember, we do not yet apply the constraints. The Lagrange’s equations (15) give us,

mr −mrθ2 −mr sin2 θφ2 +mg cos θ = −µN sgn(r)

mr2θ −mr2 sin θ cos θφ2 −mgr sin θ = λ1

mr2 sin2 θφ+ 2mrr sin2 θφ+ 2mr2 sin θ cos θθφ = λ2

Applying the constraints (12), this reduces to

mr −mr sin2 θΩ2 +mg cos θ = −µN sgn(r)

−mr2 sin θ cos θΩ2 −mgr sin θ = λ1

2mrr sin2 θΩ = λ2

(18)

From (16)-(17), we can get the relationship between the λ’s and the components Nθ, and Nφ of theconstraint force N. Using (16), we have

Cθ = λ1

Cφ = λ2(19)

but using (17), we haveCθ = N · ∂R

∂θ= (Nθeθ +Nφeφ) · reθ = rNθ

Cφ = N · ∂R∂φ

= (Nθeθ +Nφeφ) · r sin θeφ = r sin θNφ

(20)

So, the magnitude of N can be written in terms of the Lagrange multipliers

N =√N2θ +N2

φ =

√(λ1r

)2+(

λ2r sin θ

)2(21)

and using the last two equations of (18), this gives

N = [(mr sin θ cos θΩ2 +mg sin θ)2 + (2mr sin θΩ)2]1/2

= m sin θ[4Ω2r2 + (rΩ2 cos θ + g)2]1/2(22)

We therefore obtain the r equation of motion from the first equation of (18)

mr −mrΩ2 sin2 θ +mg cos θ = −µNsgn(r)

where N is written completely in terms of r and r via (22).

4

Problem 3A dumbbell consists of two particles, each of mass m,connected by a massless rigid rod of length l, and movingin the horizontal plane. Assume there is a knife-edgeconstraint at particle 1, which constrains the velocityof particle 1 to be along the instantaneous axis of thedumbbell. Think of it this way; there’s a wheel or iceskate underneath particle 1 which is along the dumbbellaxis.

Find the differential equations of motion.

There are 4 unknowns (x, y, θ, λ1), where λ1 is the La-grange multiplier related to the knife-edge constraint. Soyou should get 4 equations.

1

2

[Hint: begin by writing the velocity constraint equation in the standard form. ]

Solution. The positions of particles 1 and 2 are, respectively,

~r1 = xn1 + yn2

~r2 = (x+ l cos θ)n1 + (y + l sin θ)n2

Their velocities are:

~r1 = xn1 + yn2

~r2 = (x− lθ sin θ)n1 + (y + lθ cos θ)n2

The unconstrained Lagrangian is:

L = T − V

= T − 0

= m(x2 + y2 + yθl cos θ − xθl sin θ + 12 l

2θ2)

The constraint imposed by the knife-edge is:

~r1 · eθ = 0 =⇒ −x cos θ + y sin θ = 0 (23)

This is a nonholonomic constraint. Writing it in the standard form, we see thatn∑i=1

aji(q, t)qi + bj(q, t) = 0, j = 1, . . . , nc =⇒ a1x = − sin θ, a1y = cos θ, a1θ = 0, b1 = 0,

since the number of constraints, nc = 1. The Lagrange’s equations of motion are

d

dt

(∂L

∂x

)− ∂L

∂x= λ1a1x

d

dt

(∂L

∂y

)− ∂L

∂y= λ1a1y

d

dt

(∂L

∂θ

)− ∂L

∂θ= λ1a1θ

(24)

Equations (24) and (23) give:

2x− θ sin θ − lθ2 cos θ = −λ1m sin θ

2y + lθ cos θ − lθ2 sin θ = −λ1m cos θ

lθ − x sin θ + y cos θ = 0−x cos θ + y sin θ = 0

5

OPTIONAL, EXTRA CREDIT

Two masses, m1 and m2, are connected by a string: m1 moveson the horizontal table, while m2 hangs suspended by thestring below the table. The string passes through a hole inthe center of the table. (See sketch). You may assume thatm2 moves only in the vertical direction, and that neither m1

or m2 pass through the hole. We’ve seen this problem in classbefore, but now I want you to tackle it using a Lagrangianformulation of the problem.

(a) Reduce the equations of motion to a single second-orderdifferential equation in the variable r (distance from holeto m1).

(b) This second-order equation can be interpreted as anequivalent 1-D system with an effective potential Veff(r).Find Veff(r), and make a qualitative sketch of Veff(r).

(c) What is the condition for m1 to move in a circle? Hint:think in terms of Veff(r).

m1

g

m2

SolutionThe Lagrangian for this system is:

L =12m1(r2 + r2θ2) +

12m2r

2 +m2g(l − r)

Since ∂L∂θ = 0, we see that ∂L

∂θ= const. = m1r

2θ = H. Hence the required EoM is:

(m1 +m2)r = m1rθ2 −m2g

The effective potential is defined as Veff = −∫rdr. Hence,

Veff = gr − 12r

2θ2

The condition for the mass m1 to move in a circle is that both r = 0 and r = 0. Thus,

r3 =H2

m1m2g

6