Solutions to Homework #3

Solutions to Homework #3

Ben Shlaer

February 25, 2005

1

E=−∇Φ−A˙ B=∇ ×A L(x, ˙x, t) = 1 2m˙x 2_{−qΦ(x, t) +q˙x·A(x, t) (1)} a.) We compute d dt ∂L ∂x˙i − ∂L ∂xi = 0, (2) remembering that d dt = ∂ ∂t + ˙x· ∇. (3) Using ∂L ∂x˙i =mx˙ i +qAi (4) leads to m¨x = q(−∇Φ−A) +˙ q[∇(˙x·A)−(˙x· ∇)A] (5) = qE+q˙x×₍∇ ×_{A) (6)} = qE+q˙x×B (7)

Note that we use the notations xi _{= (x}1_{, x}2_{, x}3_{) and x}

i = (x1, x2, x3) interchangeably; there is no

significance to the upper or lower position of the index i. b.) We define the Hamiltonian

H = 3 X i=1 pix˙i− L (8) where pi = ∂L ∂x˙i =mx˙ i_+qAi ₍₉₎ ⇒ H = 1 2m˙x 2 _{+qΦ. (10)}

2

gα[r1(t) +δr1, ...,rn(t) +δrn(t)] = 0 1≤α≤f (11)

a.) By approximating the above equation with a Taylor expansion (to first order) we get

gα[r1(t) +δr1, ...,rn(t) +δrn(t)] = gα[r1(t), ...,rn(t)] + N X n=1 ∇ngα[r1(t), ...,rn(t)]·δrn (12) = 0 (13)

and since the zeroth order term in the expansion vanishes, so too must the first order term. b.) The net work done by all of the constraint forces for the virtual displacement is given by

W = N X n=1 FC n ·δrn = N X n=1 f X α=1 −λα(t)∇ngα·δrn (14) = − f X α=1 λα(t) N X n=1 ∇_ngα·_{δrn (15)} = 0 (16)

3

Let (r, θ, ϕ) be the spherical polar coordinates of the location of the particle. The constraints of the system imply that

r=a, ϕ =ωt+ϕ0 (17)

for some constant ϕ0. Therefore we can take the remaining coordinate, θ, to be the generalized

coordinate for the system. The relation between this generalized coordinate and the location of the particle is thus

x = asinθcos(ωt+ϕ0) (18)

y = asinθsin(ωt+ϕ0) (19)

z = acosθ. (20)

We now compute the Lagrangian L as a function of θ and ˙θ. First, taking time derivatives of Eqs. (18) – (20) gives

˙

x = aθ˙cosθcos(ωt+ϕ0)−aωsinθsin(ωt+ϕ0) (21)

˙

y = aθ˙cosθsin(ωt+ϕ0) +aωsinθcos(ωt+ϕ0) (22)

˙

z = −aθ˙sinθ. (23)

Squaring and summing gives for the kinetic energy

T = 1 2m( ˙x 2_{+ ˙y}2_{+ ˙z}2_{) (24)} = 1 2ma 2_θ˙2 ₊1 2ma 2_ω2_sin2_{θ. (25)}

The potential energy is

U =mgz=mgacosθ. (26)

Therefore the Lagrangian is L=T −U, giving

L= 1 2ma

2_θ˙2 ₊1

2ma

2_ω2_sin2_{θ−mgacosθ. (27)}

The Euler-Lagrange equation of motion is

d dt ∂L ∂θ˙ ! = ∂L ∂θ. (28) This gives d dt(ma

2_{θ˙) =ma}2_ω2_{sinθcosθ+mgasinθ (29)}

or

¨

θ =ω2sinθcosθ+g

asinθ. (30)

To obtain a constant of the motion, we use the fact that since the Lagrangian is not explicitly dependent on time (∂L/∂t= 0), the HamiltonianHis conserved. [The Hamiltonian does not coincide with T +U for this system since the constraints are time dependent.] The Hamiltonian is given by the formula

H = ˙θpθ− L, (31)

where pθ is the generalized momentum given by

pθ = ∂L ∂θ˙ =ma 2_{θ.˙ (32)} This gives H = 1 2ma 2_θ˙2₋ 1 2ma

2_ω2_sin2_{θ+mgacosθ. (33)}

This is the only constant of the motion.

Finally, we look for solutions of the form θ(t) =θ0 = constant. From Eq. (30) we find that such

a solution requires

ω2sinθ0cosθ0+

g

asinθ0 = 0, (34)

which implies that either

sinθ0 = 0 (35)

or

ω2cosθ0+

g

a = 0. (36)

Equation (35) always has solutions θ0 = 0 (the particle at the top of the hoop), and θ0 = π (the

particle at the bottom of the hoop). Equation (36) has the solution

θ0 = cos−1 − g ω2_a ; (37)

this solution exists only if g/(ω2_{a)≤1, or equivalently ω ≥ω}

0, where

ω0 =

rg

a. (38)

Lastly, although this was not asked in the question, we can figure out which of these stationary solutions are stable, as follows. We can write the energy conservation equation (33) as

H= 1 2ma

2_θ˙2_+U

eff(θ), (39)

where the effective potential for θ motion is

Ueff(θ) =−

1 2ma

2_ω2_sin2_{θ+mgacosθ. (40)}

The solutions we found are points θ0 at whichUeff0 (θ0) = 0. As is usual for 1-dimensional potentials,

the solutions are stable if U00

eff(θ0) >0 and unstable if Ueff00 (θ0)<0. Taking the second derivative we

find

U00

eff(θ0) = −ma2ω2cos(2θ0)−mgacosθ0. (41)

This is negative for θ0 = 0 (unstable), while for θ0 =π we get

U00

eff(π) =−ma2(ω2−ω02).

Motion at the bottom of the hoop is therefore unstable whenω > ω0, ie when the additional stationary

solution exists. Finally, evaluating the expression (41) using Eq. (36) gives that for the solution (37) we have U00 eff(θ0) = ma2 ω2− ω4 0 ω2 ! , (42)

which is positive forω > ω0. Thus the solution away from the bottom of the hoop is stable whenever

it exists.

4

F=mgˆey y=f(x) (43)

a.) The constraint equation is given by

gc_{(x, y) = y−f(x) = 0 (44)}

The Lagrange equations are then

m¨x=−mgˆey +λ(t)∇gc(x, y) (45)

with

∇gc_{(x, y) = −f}0_(x)ˆe

The two components of the Lagrange equations are therefore

mx¨ = −_λ(t)f0_{(x), (47)}

my¨ = −_{mg+λ(t). (48)}

Next, we differentiate the constraint (44) twice with respect to time. This gives ˙

y−f0_{(x) ˙x= 0 (49)}

and

¨

y−f0_(x)¨x−f00_{(x) ˙x}2 _{= 0. (50)}

We now substitute into this the expressions for ¨x and ¨y obtained from the equations of motion (47) and (48). This gives

−g+ λ m +f 0_(x) λf0(x) m ! −f00_{(x) ˙x}2 _{= 0. (51)}

Solving this for λ(t) gives

λ= m[g+f

00_{(x) ˙x}2_]

1 +f0_(x)2 . (52)

The constraint force is therefore

Fc _{= λ(t)∇g}c_{(x, y) (53)} = mf 00_{(x) ˙x}2_+g 1 +f0_(x)2 [−f 0_(x)ˆe x+ˆey] (54)

where we have used Eq. (46). To show that this force is orthogonal to the curve y=f(x), note that the tangent vector to the curve

vtan =ˆexdx+ˆeydy = (ˆex+f0(x)ˆey)dx (55)

has vanishing dot product with Fc_.

b.) Since the constraints do not depend explicitly on time, the total mechanical energy T +U is conserved, as proved in lecture. [ This can be understood from the fact that the force of constraint is always perpendicular to the velocity ˙x, and hence does no work, while the force of gravity is conservative.]

The total mechanical energy is

E = T +U (56)

= 1 2( ˙x

2_{+ ˙y}2_{) +mgy. (57)}

Eliminating y using the constraint (44) and eliminating ˙y using the time derivative (49) of the constraint gives E = 1 2mx˙ 2h 1 +f0_(x)2i +mgf(x). (58)

Solving this for ˙x2 _{and substituting into the expression (54) for the constraint force gives} Fc = f 00_(x)2(E−mgf(x)) 1+f0_(x)2 +mg 1 +f0_(x)2 [−f 0_(x)ˆe x+ˆey]. (59)

c.) The skier will leave the slope when the constraint force changes sign, that is, it no longer is pushing up from below the skier. The departure point xd is then defined by

Fc(xd) = 0 (60)

We know that if the skier starts at x = 0 with zero velocity, then E = 0, from Eq. (58) and using

f(x) =−axn_{. The prefactor in Eq. (59) then becomes}

−2f(x)f00_{(x) + 1 +f}0_(x)2

[1 +f0_(x)2_]2 . (61)

Simplifying this using f(x) =−axn _gives

n(2−_n)a2_x2n−2_{+ 1}

1 +a2_n2_x2n−2 , (62)

and setting this to zero gives yields

xd= [

1

a2_n(n−2)]

1

2(n₋1) (63)

In order for this expression to give a real number we must haven >2 (otherwise we obtain a negative number raised to a fractional power). So the skis leave the ground for n >2, at the point (63).