Classical Mechanics solutionsFull description...

Author:
Hope Hamamoto

172 downloads 662 Views 410KB Size

Assignment #6 Solutions #1 (5 points) JRT Prob. 7.4 Consider a mass m moving in a frictionless plane that slopes at an angle α angle α with the horizontal. Write down the Lagrangian in terms of coordinates x, measured horizontally across the slope, and y, measured down the slope (treat the problem as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected. Solution The kinetic and potential energies are 1 T = m(x˙ 2 + y˙ 2 ) and U = mgh = mgh = 2 Therefore,

L = T − − U = 12 m(x˙

2

−mgy sin α.

+ y˙ 2 ) + mgy + mgy sin α.

(1)

(2)

The two Lagrange equations are then 0 = m = m m¨ x¨ and mg sin α = m¨ y¨, y,

(3)

which imply that the acceleration across the slope is zero, while that down the slope is g is g sin α, as expected (this goes to zero for a ﬂat surface and g for a vertical surface.)

#2 (10 points) JRT Prob. 7.8 (a) Write down the Lagrangian (x1 , x2 , ˙x1 , x˙ 2 ) for two particles of equal masses, m1 = m2 = m, conﬁned to the x axis and connected by a spring with potential potential energy U energy U = 21 kx2 (here, x (here, x is the extension of the spring, x = (x1 x2 ℓ), where ℓ is the spring’s unstretched length, and I assume that mass 1 remains to the right of mass 2 at all times).

L

− −

1

(b) Rewrite in terms of the new variables X = 21 (x1 + x 2 ) (the CM position) and x (the extension), and write down the two Lagrange equations for X and x.

L

(c) Solve for X (t) and x(t) and describe the motion. Solution (a)

L = T − U =

1 2

m(x˙ 21 + x˙ 22 )

−

1 2

k(x1

2

− x − ℓ) . 2

(b) Solving for x 1 and x2 in terms of X and x, we ﬁnd 1 1 x1 = X + x + ℓ, 2 2

and x2 = X

Diﬀerentiating these, and substituting into

L

1 = m 2

˙

2

− 12 x − 12 ℓ.

(4)

L, we wind

2

+ X ˙ − 1 ˙x − 1 kx = mX ˙ + 1 mx˙ − 1 kx .

1 X + ˙x 2

2

2

2

2

2

4

2

2

(5)

The two Lagrange equations are: ∂ d ∂ = ˙ ∂X dt ∂ X

L

L −→ 0 = 2mX ¨

(6)

L −→ −kx = 1 m¨x. 2

(7)

∂ d ∂ = ∂x dt ∂ x˙

L

˙ (t) is a constant (call it V 0 ), and hence (c) The X equation implies that X that X (t) = V 0 t+X 0 . That is, the CM moves like a free particle, which is to be expected since there are no external forces. The x equation has the general solution x(t) = A cos(ωt δ ) (simple harmonic motion), with ω = 2k/m. This tells us that the spring is compressed (or stretched) by twice the amount that either separate mass moves. Thus, the force on either mass is as if the spring had force constant 2 k.

−

#3 (5 points) JRT Prob. 7.18 A mass m is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius R and moment of inertia I , which is free to rotate about a ﬁxed horizontal axle. Using 2

a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and ﬁnd the acceleration of the mass m. Solution We will choose as our generalized coordinate the height x of m, below the axis of the cylinder. The angular velocity of the cylinder is ω = x/R, ˙ so the total KE is 1 1 1 I (8) T = mx˙ 2 + Iω2 = m + 2 x˙ 2 , 2 2 2 R while the PE is U = mgx. Therefore,

−

L = T −

1 I U = m + 2 ˙x2 + mgx, 2 R

(9)

and the Lagrange equation is

I mg = m + x¨,

R2

which implies that x¨ =

mg . m + RI 2

(10)

(11)

#4 (10 points) JRT Prob. 7.20 A smooth wire is bent into the shape of a helix, with cylindrical polar coordinates ρ = R and z = λφ, where R and λ are constants and the z axis is vertically up. Using z as your single generalized coordinate, write down the Lagrangian for a bead of mass m threaded on the wire. Find the Lagrange equation and hence the bead’s vertical acceleration ¨z . In the limit that R 0, what is z¨ ? Does this make sense?

→

Solution The helix is determined by the equations ρ = R and z = λφ. This helps us express the bead’s velocity in cylindrical coordinates, in terms of the single parameter z and the constants R and λ: ˙ z ˙ ) = (0, Rφ, ˙ z ˙ ) = z ˙ (0,R/λ, 1). v = (ρ, ˙ ρφ, The KE of the bead is thus 3

(12)

1 2 1 2 R 2 T = mv = mz ˙ 1 + 2 , 2 2 λ

(13)

and the PE is simply U = mgz . The Lagrangian is

L = T −

1 2 R 2 U = mz ˙ 1 + 2 2 λ

− mgz.

(14)

The Lagrange equation is

L ⇒ −mg = m¨z 1 + R λ

∂ d ∂ = = ∂z dt ∂ z ˙

L

which rearranges to z¨ =

2

2

1 +−g . R2 λ2

,

(15)

(16)

When R 0, this answer reduces to z¨ = g, which is correct because in this limit, the helix reduces to a vertical frictionless wire, on which the acceleration is just g, vertically downward.

→

−

#5 (10 points) JRT Prob. 7.32 Consider the cube balanced on a cylinder as described in Example 4.7. Assuming that b < r , use the Lagrangian approach to ﬁnd the angular frequency of small oscillations about the top. The simplest procedure is to make the small-angle approximations to before you diﬀerentiate to get Lagrange’s equation. As usual, be careful in writing down the kinetic energy; this is 1 (mv2 + I ˙θ2 ), where v is the speed of the CM and I is the moment of inertia 2 about the CM (equal to 2mb2/3 in this geometry - you don’t have to prove this). The safe way to ﬁnd v is to write down the coordinates of the CM and then diﬀerentiate.

L

Solution The position of the CM of the block - see Fig. 4.14 - is r = (x, y) = ([r + b]sin θ + rθ cos θ, [r + b]cos θ + rθ sin θ),

(17)

and so its velocity is v = (x, ˙ y) ˙ = (b cos θ + rθ sin θ,

4

−b sin θ + rθ cos θ)θ.˙

(18)

Thus,

v 2 = v v = (b2 + r 2 θ2 )θ˙2 ,

·

(19)

and the Lagrangian is

L

1 2 1 2 mv + Iω mgy 2 2 1 1 = m (b2 + r 2 θ2 ) + b2 θ˙2 2 3

=

−

(20)

− mg ([r + b]cos θ + rθ sin θ) . (21)

Using the small-angle approximation as suggested (sin θ θ, cos θ simpliﬁes to 5 2 ˙ 2 1 mb θ mg(r b)θ2 + const. 6 2 With this approximation, the Lagrange equation in θ reads

≈

L≈

−

−

5 2¨ mb θ = 3

Provided that b < r, this has the form ¨ θ = motion with angular velocity ω =

−mg(r − b)θ.

≈ 1), this (22)

(23)

2

−ω θ - indicating simple harmonic

3g(r − b) 5b2

.

(24)

#6 (10 points) JRT Prob. 7.34 Consider the well-known problem of a cart of mass m moving along the x axis attached to a spring (force constant k), whose other end is held ﬁxed. If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency ω = k/m. Using the Lagrangian approach, you can ﬁnd the eﬀect of the spring’s mass M , as follows:

(a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is 61 M x˙ 2 (as usual, x is the extension of the spring from its equilibrium length). Write down the Lagrangian for the system of cart plus spring.

5

(b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency ω = k/(m + M/3); that is, the eﬀect of the spring’s mass M is just to add M/3 to the mass of the cart.

Solution (a) Let the unstretched length of the spring be ℓ and consider a short segment of spring a distance ξ from the ﬁxed end and of length dξ . Since the spring is uniform, the mass of this segment is M dξ/ℓ, and since the spring stretches uniformly its velocity (when the cart has velocity x) ˙ is xξ/ℓ. ˙ Therefore, the KE of this segment is 21 M x˙ 2 ξ 2 dξ/ℓ3 , and the total KE of the spring is 1 M x˙ 2 T spr = 2 ℓ3

ξ dξ = 1 M x˙ . ℓ

2

2

6

0

(25)

Therefore, the Lagrangian for the system of spring and cart is

L = 12 (m + M/3) x˙ − 12 kx . 2

2

(26)

(b) The Lagrange equation is kx = (m + M/3)¨x, which is the same as for the usual massless spring except that m, the mass of the cart, has been replaced by m + M/3.

−

#7 (10 points) Refer to the ﬁgure below. A mass M is attached to a massless hoop

of radius R that lies in a vertical plane. The hoop is free to rotate about its ﬁxed center. M is tied to a string which winds part way around the hoop, then rises vertically up and over a massless pulley. A mass m hangs on the other end of the string. Find the equation of motion for the angle of rotation of the hoop (θ). What is the frequency of small oscillations? Assume that m moves only vertically, and that M > m. Solution As always, we need to ﬁnd T , U , and then . We will simplify matters a bit by using diﬀerent origins for the two masses, but this won’t aﬀect the ﬁnal

L

6

Fig. 1: Geometry of Problem #7

answer since constant potential oﬀsets are irrelevant when taking derivatives. For mass M , let the origin be at the center of the hoop, with θ = 0 when the mass is at the bottom of the hoop, as shown in the ﬁgure. Then, the position of M is R(sin θ, cos θ). As for m, its vertical position is determined by θ as well; an increase in θ will unwrap an arc length of Rθ from the string, lowering m by the same amount. Therefore, the height of m, relative to its ˙ while height when θ = 0, is y = Rθ. As for the velocities, vm = Rθ, ˙ latter is a purely tangential velocity...hopefully it’s clear that vM = R θ (the both masses must have the same speed, since they’re attached by a string). Thus, we can write

−

−

T = T M + T m =

−

1 1 1 1 1 2 2 + mvm = M R2 ˙θ2 + mR2 ˙θ2 = (M + m)R2 ˙θ2 M vM 2 2 2 2 2 (27)

7

and U = U M + U m =

−MgR(1 − cos θ) − mgRθ = −MgR cos θ − mgRθ + const., (28)

where we will ignore the constant. Then,

L = 12 (M + m)R ˙θ 2

2

+ M gR cos θ + mgRθ.

(29)

The equation of motion is then ∂ d ∂ = ∂θ dt ∂ ˙θ

L

L → (M + m)Rθ = g(m ¨ − M sin θ).

(30)

This simply represents F = ma along the direction of the string (since M g sin θ is the tangential component of the gravitational force on M ). For small oscillations, we ﬁrst need to ﬁnd the equilibrium point, θ0 . This occurs when dU/dθ = 0, giving us θ0 = sin−1 (m/M ). Now, let θ = θ0 + δ , and expand the equation of motion to ﬁrst order in δ . This gives ¨ = g [m (M + m)Rδ = g [m

− M sin(θ + δ )] − M (sin θ cos δ + sin δ cos θ)] m = g m − M cos δ + sin δ cos θ M = g(m − m cos δ − M sin δ cos θ ) ≈ g(m − m − M δ cos θ ) = −M g cos θ δ,

(31) (32)

0 0

0

0

(33) (34) (35) (36)

0

0

0

and therefore, the equation of motion for small oscillations reads ¨ + δ

Mg cos θ 0

(M + m)R

δ = 0.

(37)

The frequency of small oscillations is therefore

M cos θ g M (1 − sin θ ) g = M + m R M (1 − (m/M )) g M + m R 2

ω =

0

0

1/2

(38)

1/2

=

M + m

R

8

(39)

(M − m ) g M + m R [(M + m)(M − m)] g M + m R [(M + m)(M − m)] g R [(M + m)(M + m)] M − m g 2 1/2

2

=

(40)

1/2

=

(41)

(42)

1/2

=

1/2

1/4

=

M + m

R

.

(43)

#8 (10 points) JRT Prob. 13.10 Consider a particle of mass m moving in two dimensions, subject to x + K ˆ y, where k and K are positive constants. Write down a force F = kxˆ the Hamiltonian and Hamilton’s equations, using x and y as generalized coordinates. Solve the latter and describe the motion.

−

Solution The KE is T = 21 m (x˙ 2 + y˙ 2 ). If we choose the PE to be zero at the origin, r F dr = 21 kx2 + Ky. The generalized momenta are then we have U = 0 given by p = mr˙ and the Hamiltonian is therefore

− · H

1 1 = p2x + p2y + kx2 + Ky. 2m 2

(44)

The two Hamilton’s equations for x are ˙ x =

∂ px = and p˙ x = ∂p x m

H

− ∂ ∂xH = −kx,

(45)

which combine to give x ¨ = (k/m)x. Thus, x oscillates in SHM: x = A cos(ωt δ ), with ω = k/m. Meanwhile, the two y equations are

−

˙ y =

−

∂ py = and p˙ y = ∂p y m

H

− ∂ ∂yH = −K,

(46)

which combine to give y¨ = K/m. Thus, y accelerates in the negative y direction, y = 12 (K/m)t2 + vy0 t + y0 , with constant acceleration K/m.

−

−

−

9

#9 (5 points) JRT Prob. 13.14 Consider the mass conﬁned to the surface of a cone described in Example 13.4. We saw that there have to be maximum and minimum heights z max and z min, beyond which the mass cannot stray. When z is a maximum or minimum, in must be that z ˙ = 0. Show that this can happen if and only if the conjugate momentum pz = 0, and use the equation = E , where is the Hamiltonian function, to show that, for a given energy E , this occurs at exactly two values of z . Describe the motion of the mass.

H

H

Solution According to Eq. (13.32) of the text, z ˙ = p z /m(c2 + 1), which means that z ˙ can vanish if and only if p z = 0. According to Eq. (13.33), p2φ 1 p2z + + mgz = E, 2m (c2 + 1) c2 z 2

(47)

which indicates that if p z = 0, then p2φ + mgz = E. 2mc2 z 2

(48)

All of the terms in this equation are positive, so it is clear that the LHS of the equation approaches + as z 0 and z . Furthermore, by diﬀerentiating the LHS, we see that its derivative vanishes exactly once. Therefore, the LHS has a single minimum value, which we call E min. If E < E min, then no motion is possible. If E > E min, the equation must have exactly two solutions, the turning points z max and z min.

∞

→

→ ∞

#10 (10 BONUS points) JRT Prob. 7.38 A particle is conﬁned to move on the surface of a circular cone with its axis on the vertical z axis, vertex at origin (pointing down), and half-angle α. (a) Write down the Lagrangian nates r and φ.

L in terms of the spherical polar coordi-

(b) Find the two equations of motion. Interpret the φ equation in terms of the angular momentum ℓz , and use it to eliminate φ˙ from the r 10

equation in favour of the constant ℓ z . Does your r equation make sense in the case that ℓz = 0? Find the value r0 of r at which the particle can remain in a horizontal circular path. (c) Suppose that the particle is given a small radial kick, so that r(t) = r0 + ǫ(t), where ǫ(t) is small. Use the r equation to decide whether the circular path is stable. If so, with what frequency does r oscillate about r 0 ? Solution (a) In spherical polar coordinates, the angle θ is ﬁxed at θ = α. Thus, all time derivatives of θ are zero, and the KE is 1 ˙ 2 ). T = m(r˙ 2 + r 2 sin2 αφ 2

(49)

Since the PE is just U = mgz = mgr cos α, we have

L = 12 m(r˙

2

˙ 2 ) + r2 sin2 αφ

− mgr cos α.

(50)

(b) Since does not depend on φ, the φ equation simply indicates that ∂ L ˙ constant. That is, ℓ z is conserved. The r equation = mr 2 sin2 αφ is ˙ ∂ φ is

L

˙ 2 m¨ r = mr sin αφ 2

−

ℓ2z mg cos α, or r¨ = 2 3 2 m r sin α

− g cos α.

(51)

Note that - as expected - if ℓz = 0, the particle simply slides in the radial direction with the well-known acceleration g cos α (remember that α is the angle of the incline with respect to the vertical ). The value of r0 that permits the particle to remain in a horizontal circular path is that which gives r¨ = 0. Thus, ℓ2z r0 = m2 g sin2 α cos α

11

1/3

.

(52)

(c) If we write r = r 0 + ǫ, the r equation becomes ℓ2z ǫ −3 1 + ǫ¨ = g cos α r0 m2 r03 sin2 α ℓ2z ǫ 1 3 g cos α 2 3 r0 m2 r0 sin α 3ℓ2z = ǫ. m2 r04 sin2 α

− − −

≈

(53)

(54)

−

(55)

This represents SHO with ω =

√ 3ℓ

z

mr02

=

√ mg 3

ℓz

2

1/3

sin α cos α 2

2/3

.

(56)

Don’t worry about seeing a mass dependence in ω. There is a corresponding mass dependence buried in ℓz - these cancel out. Changing m will have no eﬀect on ω.

12

Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close