Taylor j r Classical Mechanics Solutions1

Contents

1

Newton’ Newton’ss Laws Laws of of Motion Motion

2

2

Projecti Projectiles les and Charge Charged d Particl Particles es

3

3

Moment Momentum um and Angula Angular r Momentum Momentum

4

4

Ener En ergy gy

5

5

Oscill Oscillati ations ons

6

6

Calcul Calculus us of Vari Variati ations ons

7

7

Lagrange Lagrange’s ’s Equ Equati ations ons

8

8

Two-Body Two-Body CentralCentral-F Force Problems Problems

15

9

Mechanics Mechanics in in Non-inertia Non-inertiall Frames Frames

18

10 Rotational Motion of Rigid Bodies

19

11 Coupled Oscillators and Normal Modes

35

12 Nonlinear Mechanics and Chaos

39

13 Hamiltonian Mechanics

40

14 Collision Theory

53

15 Special Relativity

54

16 Continuum Mechanics

55

1

Chapter 1

Newton’s Laws of Motion 1.1 No plans to do this chapter yet. yet.

2

Chapter 2

Projectiles and Charged Particles 2.1 No plans to do this chapter yet. yet.

3

Chapter 3

Momentum and Angular Momentum 3.1 No plans to do this chapter yet. yet.

4

Chapter 4

Energy 4.1 No plans to do this chapter yet. yet.

5

Chapter 5

Oscillations 5.1 No plans to do this chapter yet. yet.

6

Chapter 6

Calculus of Variations 6.1 No plans to do this chapter yet. yet.

7

Chapter 7

Lagrange’s Equations 7.1 Write down the Lagrangian Lagrangian for a projectile projectile (subject to no air resistance) resistance) in terms of its Cartesian Cartesian coordinates (x,y,z (x,y,z), ), with z   measured measured vertically vertically upward. upward. Find the three Lagrange Lagrange equations equations and show that they are exactly what you would expect for the equations of motion. Solution.  The

Lagrangian can be easily identified and written as such:

L = T  −  − U 

1 = m(x˙ 2 + y˙ 2 + z˙ 2 ) 2

(7.1)

− mgz

A direct direct application application of the Euler-Lagra Euler-Lagrange nge equations equations yields: ∂  d ∂  = ∂x dt ∂  ˙ x˙ ∂  d ∂  = ∂y dt ∂  ˙ y˙ ∂  d ∂  = ∂z dt ∂  ˙ z˙

L L

L ⇒ L ⇒ L ⇒

L

mx ¨=0

(7.2)

my¨ = 0

(7.3)

mz¨ =

−mg

 

(7.4)

All three of which are expected for a projectile in free fall. Note the negative sign on the  z  direction is due to the chosen sign convention.

7.8 (a) Write Write down the Lagrangian Lagrangian (x1 , x2 , x˙1 , x˙2 ) for two particles of equal masses, m1 = m2 = m,

 L

confined to the x  axis and connected by a spring with potential energy U  = 12 kx2 . [Her [Heree 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.] (b) Rewrite L in terms of the new variables

X  = 12 (x1 + x2 ) (the CM position) position) and x and x (the  (the extension), and write down the two Lagrange equations for X  for  X  and x and  x.. (c) Solve for X  for  X ((t) and x and  x((t) and describe the motion. 8

Classical Mechanics [Taylor, J.R.] Solution Manual Solution.  (a)

The Lagrangian is simply:

L = 21 m(x˙12 + x˙22) − 21 k(x1 − x2 − l)2

(7.5)

(b) We note that we can manipulate the new variables X  = 12 (x1  + x  +  x2 ) and x  = (x1

− x2 − l), as

such: ˙ = 1 (x˙1  + x˙2 ) X  2

and

x˙  = x˙1

 1 ˙  + x˙1  = X   +  ˙x 2  1 ˙ x˙2  = X   ˙x 2

− x˙2

 

(7.6)

(7.7)

−

We can then rewrite the Lagrangian from Equation (7.5 (7.5)) as:

L

 

1  1 1 2 ˙  + ˙ = m (X   +  ˙x)2 + (X   ˙x) 2 2 2 1 1 2 ˙ 2 +  1  ˙x2 = m 2X  kx 2 2 2

−

−

−

 1 2 kx 2

(7.8)

The Lagrange equations can be written fairly easily, one for each coordinate. (1) :

∂  d ∂  = ∂x dt ∂  ˙ x˙ 1 kx = kx  = mx ¨ 2

L

L

−

∂  d ∂  = ˙ ∂X  dt ∂ X   ¨ 0 = 2mX 

L

(2) :

 

⇒

x ¨  =

(7.9)

− 2mk x

 

(7.10)

L

(7.11)

⇒

¨ =0 X 

(7.12)

(c) Now we are required to solve for  X (  X (t) and x and  x((t). We tackle the easier of the two first: ¨ =0 X 

X (t) = v 0 t + X   + X 0

⇒

 

(7.13)

where we have taken the liberty to introduce two integration constants that depend upon the initial conditions. As for the other coordinate  x,  x , we invoke the familiar SHM solution: x ¨  =

−

2k x m

⇒

x(t) =  A cos

   2k + δ  m

 

(7.14)

where we have introduced two other integration constants  A  and  δ . The center of mass moves with constant constant velocity velocity since no external forces forces are acting on it. The extension extension of the spring undergoes undergoes simple harmonic motion, which just means that the two masses are oscillating with respect to each other. 9

Classical Mechanics [Taylor, J.R.] Solution Manual

7.33 A bar of soap soap (mass m (mass m)) is at rest on a frictionless rectangular plate that rests on a horizontal table. At time t   = 0, I start raising one edge of the plate so that the plate pivots about the opposite edge with constant angular velocity ω velocity  ω , and the soap starts to slide toward the downhill edge. Show that the equation of motion for the soap has the form x¨

ωt, where x   is the soap’s − ω2x = −g sin ωt,

distance from the downhill edge. Solve this for  x(  x (t), given that x that  x(0) (0) = x = x 0 . [You can easily solve the homogeneou homogeneouss equation; equation; for a particular particular solution try x try  x  = B  =  B sin  sin ωt and ωt  and solve for B for  B .] Solution.

To obtain obtain the equation of motion, we shall utilise utilise the Lagrange Lagrange equations. equations. First, First, we write

the Lagrangian, noting that the motion of the soap consists of both translational and rotational motion:

L = 21 mx˙ 2 + 21 I ω2 − mgx sin ωt 1 1 = mx˙ 2 + mx2 ω 2 − mgx sin ωt 2 2

 

(7.15)

Applying the Lagrange equations, we get d ∂  ∂  = dt ∂  ˙ x˙ ∂x

L

L ⇒ x ¨

¨  = mω  =  mω 2 x mx

− mg sin ωt

− ω2x = −g sin ωt

(shown)

(7.16)

Note that Equation Equation (7.16  (7.16)) is a second second order, order, inhomogene inhomogeneous ous differential differential equation. equation. We first consider consider the homogeneous equation, which has a simple harmonic motion solution. xH (t) = A 1 eωt + A2 e−ωt

(7.17)

Now we consider the particular solution. We make a trigonometric ansatz  x P  =  B sin  B  sin ωt. ωt. Substituting into Equation (7.16 (7.16), ),

−ω2B sin ωt − ω2B sin ωt = ωt  = −g sin ωt ⇒ B  = 2ωg 2

   

(7.18) (7.19)

We can then compose our general solution for x(t): x(t) =  x H  + x  +  xP  =  A1 eωt + A2 e−ωt +

10

g  sin ωt 2ω 2

 

(7.20)

Classical Mechanics [Taylor, J.R.] Solution Manual We have the initial conditions: x(0) = x = x 0   and x˙ (0) = 0, which yield us: A1 + A  +  A2  =  x 0

and

ω(A1

− A2) + 2gω = 0

(7.21)

Using these conditions, we can solve for the integration constants A constants  A 1 and A and  A 2 : x0 g 2 4ω 2 x0 g A2  = + 2 2 4ω A1  =

−

 

(7.22)

 

(7.23)

Finally, we are able to construct our general equation as: x0 x(t) = 2

g x0 g −ωt + g  sin ωt ωt e + + e 4ω 2 2 4ω 2 2ω 2 g g =  x 0 cosh ωt  sinh ωt + ωt  +  sin ωt   2ω 2 2ω 2



−

 



 

(7.24) (7.25)

−

7.40 The ”spherical ”spherical pendulum” pendulum” is just a simple simple pendulum that is free to move in any sideways sideways direction. direction. (By contra contrast st a ”simpl ”simplee pendulu pendulum” m” is confined confined to a single single vertic vertical al plane.) plane.) The bob of a spheri spherical cal pendulum moves on a sphere, centered on the point of support with radius  r  = R  =  R,, the length of the pendulum. A convenient choice of coordinates is spherical polars,  r,  r, θ,φ, θ,φ, with the origin at the point of support and the polar axis (z ( z axis) pointing straight straight down. The two variables variables θ and φ  make a

−

good choice choice of generalize generalized d coordinates coordinates.. (a) Find the Lagrangian Lagrangian and the two Lagrange Lagrange equations. (b) Explain what the φ  equation tells us about the z   component of angular momentum lz . (c) For the special case that φ   =const, describe what the θ   equati equation on tells tells us. (d) Use the φ  equation to replace φ˙ by lz   in the θ   equation and discuss the existence of an angle θ0   at which θ  can remain constant. Why is this motion called a conical pendulum? (e) Show that if  θ  θ  = θ  =  θ 0  +   +  ,, with   small, then θ then  θ   oscillates about θ about  θ 0  in harmonic motion. Describe the motion of the pendulum’s bob.

Solution.  (a)

With reference to Figure  7.1, we  7.1,  we can write the Lagrangian Lagrangian in spherical spherical coordinates coordinates as: + mgR cos θ L = 21 m(R2 ˙θ2 + R2 ˙φ sin2 θ) + mgR 11

 

(7.26)

Classical Mechanics [Taylor, J.R.] Solution Manual

Figure 7.1: Sketch of the problem

Then, applying Lagrange equation to the two generalized coordinates: d ∂  ∂  = dt ∂  ˙θ ∂θ

L

⇒

L

mR2 θ¨ = mR  =  mR 2  ˙φ2 sin θ cos θ

− mgR sin θ

 

(7.27)

d ∂  ∂  = dt ∂  ˙φ ∂φ

L

⇒

L

mR2  ˙φ sin2 θ  = l  =  lz = const

(7.28)

We leave it to the reader to simplify Equation (7.27 ( 7.27). ). As for the latter equation, we have noted that

L the Lagrangian is independent of  φ,  φ , such that ∂  ∂φ = 0. (b) Equation (7.28 (7.28)) tells us that the z the  z   component of the angular momentum is constant. (c) If  φ  φ  = const, then φ˙  = 0, and Equation (7.27 ( 7.27)) reduces to: θ¨ =

− Rg  sin θ

 

(7.29)

That is, we can describe the motion of the pendulum like a simple pendulum, at a fixed plane characterized by φ = φ  =  φ 0 , for some fixed  φ 0 . (d) We want to first express φ˙  in terms of  lz  (using Equation (7.28 (7.28), ), then substitute it into Equa12

Classical Mechanics [Taylor, J.R.] Solution Manual tion (7.27 (7.27). ). φ˙  =



lz 2 mR sin2 θ

lz θ¨ = mR2 sin2 θ lz2 cos θ = 2 4 3 m R sin θ



(7.30)

2

−

sin θ cos θ

− Rg  sin θ

(7.31)

g  sin θ R

To explore a possible stationary solution of  θ, we consider θ¨ = 0. Following ollowing from Equation Equation (7.31 ( 7.31), ), we obtain: lz2 cos θ g =  sin θ 3 R m2 R4 sin θ cos θ  = k  =  k sin4 θ  = k  =  k(1 (1

 

(7.32)

− cos2 θ)2

(7.33)

where we have introduced the constant k = gm2 R3 /lz2 . To analyz analyzee the solutio solution, n, we substitu substitute te x  = cos θ, into Equation (7.33 (7.33)) where x where  x

∈ [0, [0, 1]. WolframAlpha gives us the following plot: We note

Figure 7.2: Plot of Equation (7.33 Equation  (7.33))

that there is only one solution for which Equation (7.33 (7.33)) is satisfied. This would correspond to the solution θ solution  θ 0 , as specified in the question. (e) Consider a perturbation, θ =  θ 0  +   +  .. We can Taylor aylor expand the following following quantities quantities about the equilibrium position x position  x 0 , and also write down the derivatives of  θ: cos θ

≈ cos θ0 −  sin θ0 sin θ ≈ sin θ0  +   +   cos θ0 θ¨ = ¨

 

 

(7.34)

 

(7.35) (7.36)

Before we substitute these equations into Equation (7.31 ( 7.31), ), we recall that at equilibrium, Equation (7.32 (7.32)) is satisfied. That is, we have: lz2 g sin4 θ0 = m2 R4 R cos θ0 13

(7.37)

Classical Mechanics [Taylor, J.R.] Solution Manual which which we can use to simplify simplify our expression expression slightly slightly later on. With the last 4 equations from above, above, Equation Equation (7.31 7.31)) gives us: 4

θ0 g (cos θ0 −  sin θ0 )(sin θ0  +   +   cos θ0 )−3 −  (sin θ0  +   +   cos θ0 )  ≈ Rg sin cos θ0 R

¨

=

g sin4 θ0 (cos θ0 R cos θ0

 +   cos θ0 ) −  sin θ0)(sin−3 θ0 − 3 sin−4 θ0 cos θ0) − Rg  (sin θ0 + 

(7.38) (7.39)

After some simplification and ignoring terms in   2 , we obtain: ¨ =

−

 g  R



1 + 3cos2 θ0 cos θ0



 

(7.40)

This shows that the pendulum will oscillate in a simple harmonic motion, with angular frequency g 1 + 3cos2 θ0 2 approximately ω approximately ω = , about the equilibrium position θ position  θ 0 . R cos θ0





14

Chapter 8

Two-Body Central-Force Problems 8.13 Two particles particles whose reduced reduced mass is µ   interact via a potential energy U  =

1 2 2 kr ,

where r   is the

distance between them. (a) Make a sketch showing U  showing  U ((r ), the centrifugal potential energy U  energy  U cf  cf (r), and the effective potential energy U eff  (Treat the angular angular momentum momentum l l  as a known, fixed constant.) eff (r ). (Treat (b) Find the “ equili equilibri brium um ” separa separatio tion n r0 , the distance at which the two particles can circle each other with constant r constant r.. [Hint: This requires that dU  that  dU eff  /dr  be zero.] (c) By making a Taylor expansion eff /dr be of  U eff  (r eff (r ) about the equilibrium point r0  and neglecting all terms in (r

− r0)3 and higher, find the

frequency of small oscillations about the circular orbit if the particles are disturbed a little from the separation r separation  r 0 .

Solution.  (a)

A sketch of all three graphs are shown in the figure below.

Figure 8.1: Graphs of the potentials  U (r ),  U cf  cf (r ) and  U eff  eff (r )

15

Classical Mechanics [Taylor, J.R.] Solution Manual We can write the potentials as such: 1 U ( U (r ) = kr 2 2 l2 U cf  cf (r ) = 2µr2

(8.1)  

(8.2)

1 2 l2 U eff  ( r ) =  U (  U  ( r ) + U  +  U  ( r ) = kr + eff  cf  cf  2 2µr2 (b) The equilibrium separation by setting

 

(8.3)

dU eff  eff  = 0. We obtain: obtain: dr

dU eff  eff  =  kr 0 dr

−

1/4

   l 2 µk

r0  =

⇒

l2 =0 µr03 (8.4)

(c) Let us define r = r0  +  .   . The Taylo Taylorr expansion can be p erformed erformed about the equilibriu equilibrium m r0 , noting that the first derivative was calculated before, and is equal to 0 at  r 0 :

      

d2 U eff  eff (r ) U eff  ( r ) =  U  ( r ) +  +   eff  eff  eff  0 dr2 2

 ≈  ≈  ≈

1 2 l2 kr  + 2 0 2µr02 1 2 l2 kr  + 2 0 2µr02 1 2 l2 kr  + 2 0 2µr02

+ 

2

+ 

2

+ 

2

+ . . .

r =r0 +



3l2 k  + µ(r0  +   +  ))4  3l  3 l2 −4 k  + (r0 +  r0−5 ) µ 3l2  3 l2 3  3l k  + 4 +  µr0 µr05



 (8.5)

For small oscillations , we can ignore terms of  3 and above, and note that the effective potential takes on the form of a harmonic oscillator potential (e.g. one of an elastic spring). That is, it takes the form of  U (  U (x) = U (0)  U (0) + 12 k  x2 , and the frequency of oscillation for such a system is ω is  ω  =

 

k /m. /m.

Making Making such such an observation, observation, we can deduce deduce the ‘ spring spring constant’ constant’ k  and oscillation frequency  ω :



    

3l2 k  = 2 k + 4  = 2k  + 6k 6 k  = 8k µr0

ω  =

k = µ

8k µ

 

(8.6)

where we have used the result from Equation (8.4 ( 8.4)) to rewrite r0   in terms of  k, the central force constant. Do not confuse this with the spring constant k constant  k   used in the analogy earlier.

16

Classical Mechanics [Taylor, J.R.] Solution Manual

Figure 8.2: Transfer orbit for Problem 8.34 8.34

8.34 Suppose that we decide to send a spacecraft to Neptune, using the simple transfer described in Example 8.6 (c.f. textbook pg 318). The craft starts in a circular orbit close to the Earth (radius 1 AU) and is to end up in a circular orbit near Neptune (radius about 30 AU). Use Kepler’s third law to show that the transfer will take about 31 years. Solution.  The

transfer orbit in question is depicted in Figure 8.2, Figure  8.2, with  with the radius of the Earth orbit

as r as  r E  and the radius of the Neptune orbit as  r N . a3 GM  Kepler’s third law is stated as: 3 =  . τ  4π 2 We know (from the figure) that the sum of the radii equals twice the semi-major axis a   of the elliptical orbit. Explicitly, Explicitly, we have rE  + r  +  rN  = 2a

⇒

a  = 15 15..5 AU

(8.7)

And so to get from P  to P  to  P  : 1 1 Time of transfer = τ  = 2 2

 

4π 2 a3 GM 

= 1.926

× 109 s  31 years

Note Note that that I used used the follow following ing constan constants: ts: 1 AU AU = 1.496 M Sun Sun  = 1.989

× 1030 kg.

17

(shown)

(8.8)

× 1011 m, G = 6.67 × 10−11 Nm2kg-2,

Chapter 9

Mechanics in Non-inertial Frames 9.1 No plans to do this chapter yet. yet.

18

Chapter 10

Rotational Motion of Rigid Bodies 10.1 The result result that that



mα rα  = 0, can be paraphrased to say that the position vector of the CM relative

to the CM is zero, and, in this form, is nearly obvious. Nevertheless, to be sure that you understand the result, prove it by solving We Solution.   We

 for r   and substituting into the sum concerned. α

rα  =  R  + rα

start start with with the equati equation on given given in the question, question, and substituti substituting ng into into the sum as

suggested:





rα  =  R  + rα



mα rα  =

⇒ (mα rα ) − R 



rα  =  r α



 

− R 

mα

 

(10.1) (10.2)

where the sum is taken implicitly over  α.  α . We also have the following expression for the position of  the center of mass:  = R  =

1 M 

Using this relation, and noting that the term becomes:



 

(mα rα )

(10.3)

mα  is simply the total mass M , M , Equation (10.2 (10.2))

mα rα  =  R M 

M  = 0  −  − R M 

(10.4)

We have thus shown that the position of the CM relative to itself is zero.

10.3 Five Five equal poin p ointt masses masses are placed placed at the five corners of a square square pyramid whose square square base is centered on the origin in the xy the  xy  plane, with side L side  L,, and whose apex is on the z the  z  axis at a height H  height  H  above the origin. Find the CM of the five mass system. 19

Classical Mechanics [Taylor, J.R.] Solution Manual Solution.  The

(x,y,z (x,y,z)) coordinates of the 4 corners are (L/ ( L/22, L/2 L/2, 0), (L/ (L/22, L/2 L/2, 0), ( L/2 L/2, L/2 L/2, 0),

−

−

( L/2 (0 , 0, H ). ). Let m Let  m  be the mass of one L/2, L/2 L/2, 0), while the coordinates of the apex is given by (0,

−

−

point mass. To calculate the CM of the system, we perform the following: m

  

 

1 i xi xCM  = = 5m 5 m i yi 1 yCM  = = 5m 5 m i zi H  zCM  = = 5m 5

L  L + 2 2 L  L + 2 2

L 2 L 2

 −  −

 −  −

L 2 L 2

 

=0

(10.5)

=0

(10.6)

 

(10.7)

The coordinates of the CM is given by (0, (0 , 0, H/ H/5). 5).

10.4 The calculation calculation of centers centers of mass or moments moments of inertia inertia usually involves involves doing an integral, integral, most often a volume integral, and such integrals are often best done in spherical polar coordinates. Prove that:

 

dV f (r ) =

    dr r

2

dθ sin θ

 

dφ f (r,θ,φ) r,θ,φ).

[Think about the small volume dV  volume dV  enclosed  enclosed between r between  r  and  r + dr, dr,  θ  and  θ + dθ, dθ, and φ and  φ and  and φ  φ + dφ.] dφ.] If the volume integral on the left runs over all space, what are the limits of the three integrals on the right?

Figure 10.1: Taken from   http://astro-learned.blogspot.sg/

20

Classical Mechanics [Taylor, J.R.] Solution Manual Solution.  Referring

to the figure, a small volume element dV  element  dV  can be approximated approximated to a cuboid with

volume: dV  = (dr)( dr)(rdθ rdθ)( )(rr sin θ dφ) dφ) =  r 2 sin θ drdθdφ drdθdφ

 

(10.8)

Substituting this into the integral, taking note of the independence (or dependence) of the integrand with respect to the 3 variables, we obtain, trivially, the result:

 

dV f (r) =

    dr r

2

dθ sin θ

 

dφ f (r,θ,φ) r,θ,φ).

 

(10.9)

Anothe Anotherr approa approach ch to this question question is by consid consideri ering ng the Jacobi Jacobian. an. We expres expresss the Cartes Cartesian ian coordinates in terms of the spherical polar coordinates: x  = r  =  r sin θ cos φ,

y  = r  =  r sin θ sin φ,

z  = r  =  r cos θ.

 

(10.10)

We then have the volume element as dV  as  dV  =  dxdydz =  dxdydz  = J   J ((r,θ,φ) r,θ,φ) drdθdφ, drdθdφ, with J  with  J ((r,θ,φ) r,θ,φ) being:

  

∂x ∂r ∂y J (r,θ,φ) r,θ,φ) = ∂r ∂z ∂r

∂x ∂φ ∂y ∂φ ∂z ∂φ

  

∂x ∂θ sin θ cos φ ∂y = sin θ sin φ ∂θ cos θ ∂z ∂θ

= r 2 sin θ

 

−r sin θ sin φ r sin θ cos φ 0

 

r cos θ cos φ r cos θ sin φ r sin θ

−

 

(10.11)

This yields us the same result as above. The integration limits for a volume integral over all space are: 0

≤ r ≤ ∞, 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ π.

10.5 A uniform solid hemisphere hemisphere of radius radius R   has its flat base in the xy  plane, with its center at the origin. Use the result of Problem 10.4 to find the center of mass. Solution.  Due

to the inherent spherical symmetry of the object, one might expect the x and y

coordinates of the center of mass to be zero. Nevertheless, they can be computed by: xCM  =

  

x  dm  d m =0 dm

yCM  =

21

  

y dm =0 dm

(10.12)

Classical Mechanics [Taylor, J.R.] Solution Manual Let ρ Let  ρ  be the density of the hemisphere. The  z  coordinate can be computed in a similar fashion: zCM  = = =

         ·      

z dm ρ r cos θ dV  = dm ρ dV 

R 0

 

π/ π/ 2 2π sin θ cos θ dθ 0 0 R 2 π/ π/ 2 2π r dr 0 sin θ dθ 0 dφ 0

r r2 dr

(R4 /4) (1/ (1/2) (2π (2π ) 3R = 3 (R /3) (1) (2π (2π) 8

 

dφ

 

(10.13)

10.66 (a) Find the CM of a unifor 10. uniform m hemisp hemispher herica icall shell shell of inner inner and outer outer radii a and b   and mass M  positioned as in Problem 10.5. (b) What becomes of your answer when a when  a  = 0? (c) What if  b  b Solution.  (a)

→ a?

Once again, by symmetry we have xCM =  y CM   = 0. We set up a similar similar integral integral to

compute z compute  z CM, but with modified limits: zCM  = = = =

     ·   −  −−

            

z dm ρ r cos θdV  = dm ρ dV 

b  r a

π/ π/ 2 2π sin θ cos θ dθ 0 0 b 2 π/ π/ 2 2π  r dr 0 sin θ dθ 0 dφ a 1 4 a4 ) (1/ (1/2) (2π (2π ) 4 (b 1 3 3 a ) (1) (2π (2π ) 3 (b 3 b4 a4

8

r2 dr

b3

− a3

dφ

 

(10.14)

(b) Note that when a   = 0, we are simply computing the CM of a uniform hemisphere (as in Problem 10.4), with b = R. It is then then no surpris surprisee that our answe answerr here is

3 8 b,

which corresponds

with our answer from the previous problem. (c) When b

 a , the hemispherical shell becomes infinitely thin, with radius a. We would expect expect  → a,

that z that  z CM  approaches a/  approaches  a/2. 2. Now let us compute it. Note that here, a here,  a  is simply a number, and so we are only treating b  as a variable as it tends towards a. Since Since we have have the limit limit of [0]/[0 [0]/[0], ], we can use L’Hopital’s rule:

−  −  

3 zCM  = lim b→a 8 3 = lim 8 b→a 3 4a3 = 8 3a2

22

b4 a4 b3 a3 4b3 3b2 a  = 2

 

(10.15)

Classical Mechanics [Taylor, J.R.] Solution Manual

10.9 The moment moment of inertia of a continu continuous ous mass distribution distribution with density  density   is obtained by converting the sum I z =



mα ρ2α  into the volume integral

 

ρ2 dm  =

 

ρ2  dV  .   . Find Find the moment moment of inerti inertiaa

of a uniform circular cylinder of radius  R and mass  M  for  for rotation about its axis. Explain why the  R  and mass M  products of inertia are zero.

Solution.  For

continuous masses, we have: I z = 

 

2π

R

ρ dV  =  

0

=  (    (h h)(2π )(2π)

Since the density  density    =

        h

2

0

ρ2 ρ dρ dθ dz

·

0

R4 4

 

(10.16)

M  , we finally obtain the moment of inertia I  inertia  I z : πR2 h M R2 I z = 2

 

(10.17)

The products of inertia inertia are zero because of cylindrical cylindrical symmetry. symmetry. More explicitly explicitly,, we have, have, for example:

 −     −

I xy xy =  

xy dV 

h

=



R

dz

0

2

ρ

0

       2π

· ρ dρ

sin θ cos θ dθ

0

=0

=0

10.15 (a) Write Write down the integral (as in Problem Problem 10.9) for the moment moment of inertia inertia of a uniform cube of  side a  and mass M , M , rotating about an edge, and show that it is equal to

2 2 3Ma .

(b) (b) If I bala balanc ncee

the cube on an edge in unstable equilibrium on a rough table, it will eventually topple and rotate until it hits the table. By considering considering the energy of the cube, find its angular angular velocity velocity just before it hits the table. (Assume the edge does not slide on the table.) 23

Classical Mechanics [Taylor, J.R.] Solution Manual Solution.   (a)

The detail detailss of this integra integrall can be found found in the textbook’ textbook’ss Exampl Examplee 10. 10.2. 2. I shall shall

present the final answer here without proof:

   − − − −  − −  − −   −  −− −  −− − −  y 2 + z 2 yx zx

  a,a,a

I  =  ρ

0,0,0

=

M  a3

2 5 a 3 1 5 a 4 1 5 a 4

M a2 = 12

8 3 3

xy 2 x + z 2 zy

1 5 a 4 2 5 a 3 1 5 a 4

3 8 3

−xz −yz 2 2

x + y

1 5 a 4 1 5 a 4 2 5 a 3

 

dV 

3 3 8

 

(10.18)

Hence we can directly read off the moment of inertia about an edge: I  = 23 M a2 . (b) When the cube is resting stably on the table, the height of the CM above the table is a/2. a/2. When it is held at an unstable equilibrium (assuming balanced on an edge at a 45◦   angle), the a a height of the CM above the table is = . Hence Hence we can calculat calculatee the change change in the ◦ 2tan45 2 potential energy of the cube as it falls:

√ 

∆P.E. = P.E.  = M  M g

 √  −  a 2

a  = M  =  M ga 2

 √  −  1 2

 1 2

 

(10.19)

1 2 I ω , where ω  is the angular velocity just 2 before it hits the table and I  and  I  is  is the moment of inertia about an edge (calculated in (a)). Equating The corresponding gain in kinetic energy is ∆K.E. ∆K.E. =

the two energies together, and solving for ω: ω2 =

√ 

3g ( 2 2a

− 1)

(10.20)

10.23 Consider Consider a rigid plane body or ”lamina”, ”lamina”, such as a flat piece of sheet metal, rotating rotating about a point O   in the body. body. If we choose choose axes so that the lamina lamina lies in the xy  plane, which elements of the inertia tensor I  tensor  I  are  are automatically zero? Prove that I zz  +  I yy zz =  I xx xx  + I  yy .

Solution.   If

the lamina is on the xy  plane, then z   = 0 automatical automatically ly.. Hence, Hence, products of inertia inertia

containing z containing z   is thus automatically zero. Explicitly, I  Explicitly,  I xz  =  I zy xz =  I zx zx  = I  zy =  I yz yz  = 0. We will not be able to deduce anything about I  about  I yx and  I xy yx and I  xy . 24

Classical Mechanics [Taylor, J.R.] Solution Manual Consider: I zz zz =  ρ = ρ

   

x2 + y 2 dV  2

2

x + z dV  + ρ

= I yy  +  I xx yy  + I  xx

 

y 2 + z 2 dV   

(10.21)

We can perform the second step because  z 2 = 0.

10.25 (a) Find all nine elements of the moment of inertia tensor with respect to the CM of a uniform cuboid (a rectangular brick shape) whose sides are 2a, 2 a, 2b  and 2c 2c  in the x the  x,, y,z directions y,z  directions and whose mass is M . M . Explain Explain clearly why you could write down the off-diagona off-diagonall elements elements without doing any any integ integrat ration ion.. (b) Combine Combine the results results of part part (a) and Proble Problem m 10. 10.24 24 to find the mom momen entt of  inertia tensor of the same cuboid with respect to the corner  A  at (a,b,c). a,b,c). (c) What is the angular momentum about A  if the cuboid is spinning with angular velocity ω  around the edge through A and parallel to the x the  x  axis? Solution.

(a)

     − c, b, a

I  =  ρ

−c,−b,−a Identify the density to be ρ be  ρ  =

M  . 8abc

y 2 + z 2 yx zx

−xy 2 −xz 2 x + z −yz 2 2 −zy x + y

−

 

dV 

 

(10.22)

We compute the diagonal elements first.

              c

I xx xx  =  ρ

b

a

y 2 + z 2 dxdydz

−c −b −a

c b M  = 2a(y 2 + z 2 )  dydz 8abc −c −b M  c 2b3 = + 2bz 2bz2 dz 4bc −c 3

M  2b2 c 2c3 M  2 +  = (b + c2 ) 2c 3 3 3 M  2 I yy = (a + c2 ) yy = 3 M  2 I zz = (a + b2 ) zz = 3 =

··· ···

where the other two moments of inertia can be directly calculated, or inferred from  I xx xx  (due to the ’symmetry’ present). 25

Classical Mechanics [Taylor, J.R.] Solution Manual The off-diagonal elements are zero because:

−xy, −yz, yz , −xz   etc. etc.

are all all odd functions functions in  in   x,y,z. x,y,z.

Integrating from

 −a  to  t o a of  x, for example, will yield 0.

(b) We shift the ‘origin’ by

= (a,b,c). a,b,c). Using Using the result resultss from from Proble Problem m 10. 10.24, 24, the modified modified

d

moment of inertia is given by: I   =

M  3

M  = 3

    −−

b2 + c2 0 0

0 2 a + c2 0

4b2 + 4c 4 c2 3ba 3ca

(c) If the angular velocity

ω   is

    −− −  − 

b2 + c2 + M  ba ca

0 0 2 a + b2

3ab 4a2 + 4c 4c2 3cb

−ab 2 −ac 2 −bc 2 a + c 2 −cb a + b

3ac 3bc 2 4a + 4b 4 b2

− −

 

parallel to the x   axis, we can write

ω

(10.23)

=

momentum is then given as the matrix multiplication:  =  I ω  = L  = I 

Mω 3

 

 

4(b 4(b2 + c2 ) 3ba 3ca

− −

 

 

ω 0 . The The ang angul ular ar 0

 

(10.24)

10.26 (a) Prove Prove that in cylindrical cylindrical polar coordinates, coordinates, a volume integral integral takes the form

 

dV f ( f (r) =

      ρ dρ

dφ

dzf ( zf (ρ,φ,z) ρ,φ,z ).

(b) Show that the moment of inertia of a uniform solid cone pivoted at its tip and rotating about its axis is given by the integral: I zz zz =  

 

2

      2π

h

dV ρ =  

dz

0

V 

r

dφ

0

ρ dρ ρ2 ,

0

explaining explaining clearly the limits of integrati integration. on. Show that the integral integral evaluate evaluatess to also that I  that  I xx xx  = Solution.  (a)

3 M ( M (R2 20

3 M R2 . 10

(c) Prove

+ 4h 4h2 ).

Using cylindrical polar coordinates, we have the following coordinate transformations:

x  = ρ  =  ρ cos φ,  y  = ρ  =  ρ sin φ. Hence consider the Jacobian for a small volume element:

  

∂x ∂ρ ∂y J (ρ,φ,z) ρ,φ,z ) = ∂ρ ∂z ∂ρ

∂x ∂φ ∂y ∂φ ∂z ∂φ

  

∂x ∂z cos φ ∂y = sin φ ∂z 0 ∂z ∂z 26

 

−ρ sin φ ρ cos φ 0

 

0 0 =  ρ 1

 

(10.25)

Classical Mechanics [Taylor, J.R.] Solution Manual And so we have have dV  =  ρ  dρ  d ρ dφ dz , which then yields us the integral as required. (b) If we were to use cylindrical coordinates,  I zz zz  will be calculated as: I zz zz =  

 

2

2

x + y dV  = 

V 

 

ρ2 dV 

V 

Utilising the result from part (a), we then have:

      h

I zz zz =  

2π

dz

0

r

dφ

0

ρ3 dρ

 

(10.26)

0

M  . The volume could have have been b een obtained by considering considering 1/3  π R2 h dV  ,   , but here we quote a direct result.

where we have the density:   = the triple integral

 

The limits of integration are as such:

• h  is the height of the cone. • 2π  is taken due to the cylindrical symmetry. • r =  R · hz  is derived from considering a set of similar triangles in Figure  10.2,  10.2, with R   being the radius of the base of the cone. Let us now evaluate this integral.

      h

I zz zz =  

2π

dz

0

dφ

0

R4 z 4 4h 4

h

πR4 z 4 =   dz 2h4 0 3M  πR4 h = πR 2 h 10 3 = M R2 (shown) 10

 ·

(c) Using the same notations as before, I  before,  I xx xx  is calculated as: I xx xx  =  

                y 2 + z 2 dV  = 

V  h

=  

V 

2π

dz

0

0

0

h

=  

0

=

2π

dz

0

r

dφ

h

=  

ρ2 sin2 φ + z  + z 2 dV 

0 4 4 R z

4h4

ρ3 sin2 φ + ρz  + ρz 2 dρ

sin2 φ +

 R 2 z 4 dφ 2h2

πR4 z 4 πR 2 z 4 + dz 4h4 h2

3 M (R2 + 4h 4 h2 ) 20 27

(shown)

(10.27)

Classical Mechanics [Taylor, J.R.] Solution Manual

Figure 10.2: Similar triangles in a cone for Problems 10.26 and 10.27

10.27 10. 27 Find Find the inertia inertia tensor tensor for a unifor uniform, m, thin thin hollo hollow w cone, cone, such such as an ice-crea ice-cream m cone, cone, of mass mass M , height h height  h,, and base radius R radius  R,, spinning about its pointed end.

Solution.  At

any particular z particular  z , we have the corresponding radius  r  = R  =  R

· hz . We obtain this ratio by

considering a set of similar triangles, as seen in Figure  10.2.   Using cylindrical coordinates (r,θ,z (r,θ,z), ), we first calculate the density  density    = M  =  M/A /A,, where A where  A  is the curved surface area of the cone.

     

2π

h

A =

dA  =

0

r dθ dz

0

=  πRh

⇒

  =

 

M  πRh

(10.28)

 

(10.29)

The inertia tensor is then given by:

I  =  

   − −     − h

=  

 

y 2 + z 2 yx zx

2π

0

0

−xy 2 −xz 2 −yz dA x + z −zy x2 + y2 r2 sin2 θ + z  +  z 2 −r2 sin θ cos θ −rz cos θ 2 r sin θ cos θ r2 cos2 θ  + z  + z 2 −rz sin θ r2 −rz cos θ −rz sin θ

 

Note that off-diagonal elements equal to zero because

 

2π

0

 

2π

sin θ dθ  =

 

2π

cos θ dθ  =

0

0

28

sin θ cos θ dθ  = 0

 r dθ dz

 

(10.30)

Classical Mechanics [Taylor, J.R.] Solution Manual

 

2π

Using the fact that

sin θ dθ =

0

inertia tensor to be:

 

2π

2

cos2 θ dθ = π , we have the diagonal components of the

0

 

h

I xx xx  =  I yy yy =  

0

R3 z 3 Rz 3 π 3 + 2π 2π dz h h

M  2 = (R + 2h 2 h2 ) 4

 

h

I zz zz =  

2π

0

=

(10.31a)

R3 z 3 dz h3

M R2 2

 

(10.31b)

Finally we obtain the inertia tensor for a hollow cone to be: I  =

M  4

 

R2 + 2h 2h2 0 0

0 2 R + 2h 2 h2 0

0 0 2R2

 

 

(10.32)

10.35 A rigid body consists consists of three masses fastened fastened as follows: m at (a, 0, 0), 2m 2m  at (0, (0, a , a) a) and 3m 3m at (0, (0, a, a). (a) Find the inertia inertia tensor tensor I . (b) Find the principal principal moments and a set of orthogonal orthogonal

−

principle axes.

Solution.  (a)

I  =

  − − − −       mi

i

=

The inertia tensor of a rigid body with discrete masses is given by: yi2 + zi2 yi xi zi xi

xi yi x2i + zi2 zi yi

−xi zi −2 yizi2

xi + yi

 

m(0) + 2m 2m(2a (2a2 ) + 3m 3 m(2a (2a2 ) 0 0 2 2 2 2 0 m(a ) + 2m 2 m(a ) + 3m 3 m(a ) 2m(a ) 3m( a2 ) 0 2m(a2 ) 3m( a2 ) m(a2 ) + 2m 2 m(a2 ) + 3m 3 m(a2 )

= ma

2

10 0 0 0 6 1 0 1 6

−

−

−

−

−

−

 

(b) We have to find directions of  ω  such that I ω = λω , where λ   is the moment of inertia about the principal axes. This eigenvalue problem has a non-trivial solution when det(I  det( I 

 

− λ1) = 0.

1 Firstly, by inspection, we know that ω1 = 0   is an eigenvec eigenvector, tor, or equivalen equivalently tly,, one of the 0 principal axes directions, with λ with  λ 1  = 10 10ma ma2 . To check that this is true, one only needs to substitute substitute 29

Classical Mechanics [Taylor, J.R.] Solution Manual into  I ω , ω1   into I 

and verify that one obtains  λ 1 ω .

For the other two solutions, we just need to solve:



 −

6ma2 λ ma2 =0 ma2 6ma2 λ

−

(6ma (6ma2

⇒

− λ2)2 − (ma2)2 = 0

And so the eigenvalues are: λ2  = 5ma2

λ3  = 7ma2

and

When λ When  λ 2  = 5ma2, we have: 2

ma

⇒

     1 1 1 1

ωy  = ωz

0 0

ωy = 1, ωz =

−1

Normalizing, we obtain

 √    − −      0 1 1

1 ω2  = 2

 

(10.33)

Then when λ when  λ 3  = 7ma2, we have, instead: ma2

1 1

⇒

1 1

ωy  = ωz

−

0 0

ωy = 1, 1 , ωz = 1

Normalizing, we obtain

 √   

1 ω3  = 2

0 1 1

Summarizing, with the set of orthogonal principle axes

 

(10.34)

 {ω1, ω2, ω3}, the inertia tensor consisting

of the principal moments is given by: I   =

 

10 10ma ma2 0 0

0 5ma2 0

0 0 7ma2

 

 

(10.35)

10.36 10. 36 A rigid rigid body consis consists ts of three equal equal masses masses (m ( m) fastened at the positions (a, (a, 0, 0), (0, (0, a, 2a), and (0, (0, 2a, a). (a) Find Find the inerti inertia a tensor tensor I . (b) Find the princi principal pal moment momentss and a set of orthog orthogona onall principal axes. 30

Classical Mechanics [Taylor, J.R.] Solution Manual Solution.   (a)

We take take refere reference nce from the previous previous problem problem (Probl (Problem em 10. 10.35) 35).. A simila similar, r, trivia triviall

calculation can be performed to obtain the inertia tensor as: I  =

  − − − −    − − yi2 + zi2 yi xi zi xi

mi

i

= ma

10 0 0

2

xi yi 2 xi + zi2 zi yi

0 6 4

−xi zi −2 yizi2

xi + yi

0 4 6

 

(b) Similar to the previous problem, we have to find directions of  ω   (principal axes), such that we have I ω  =  λ ω , with λ  being the moment moment of inertia about each each principal axis. This eigenva eigenvalue lue equation has non-trivial solutions if and only if det(I  det(I 

− λ1) = 0.

From the form of the inertia tensor above, we observe by inspection that the first eigenvector is 1 0 , with an eigenvalue of  λ  λ 1  = 10 10ma ma2 . ω1  = 0 For the other 2 axes, we consider the determinant:

 

 −

 −

6ma2 λ 4ma2 =0 4ma2 6ma2 λ

(6ma (6ma2

⇒

−

−

(4ma2 )2 = 0 − λ2)2 − (4ma

And so the eigenvalues are: λ2  = 10 10ma ma2

λ3  = 2ma2

and

When λ2  = 10 10ma ma2 , we have: ma2

− −      4 4

4 4

ωy  = ωz

0 0

ωy = 1, ωz =

−1

− −

⇒ Normalizing, we obtain ω2  =

Then when λ3  = 2ma2 , we have, instead: 2

ma



⇒

4 4

−

 √    − 1 2

ωy  = ωz

ωy = 1, ωz = 1 31

 

   

−4 4

0 1 1

0 0

(10.36)

Classical Mechanics [Taylor, J.R.] Solution Manual (0, 1, 0)

dy

(0, 0, 0)

(1, 0, 0)

Figure Figure 10.3: Metal Triangl Trianglee for Problem 10.37 10.37

Normalizing, we obtain ω3  =

 √    0 1 1

1 2

Summarizing, with the set of orthogonal principle axes

 

(10.37)

 {ω1, ω2, ω3}, the inertia tensor consisting

of the principal moments is given by: I   =

 

10 10ma ma2 0 0

0 10 10ma ma2 0

0 0 2ma2

 

 

(10.38)

10.37 A thin, flat, uniform metal metal triangle lies lies in the xy the  xy  plane with its corners at (1, (1, 0, 0), (0, (0, 1, 0) and the origin. Its surface density (mass/area) is σ is  σ  = 24. (a) Find the triangle’s inertia tensor I  tensor  I .. (b) What are its principal moments and the corresponding axes? Solution.  (a)

The inertia tensor is given by: I  = σ

   − −    − A

(z =0)

=

y2 + z 2 yx zx

σ

A

−xy −xz −yz 2 x2 + z 2 2 −zy x + y 0 −xy2

y2 yx 0

x 0

x2

0 + y 2

 

 

dA

dA

 

(10.39)

We shall compute compute the elements elements of the inertia inertia tensor component component wise. Take reference reference from Figure 10.3, ure  10.3, noting  noting that the hypotenuse is characterized as: y  = 1

      −       − 1

I xx xx  =  σ

1 y

−

0

y 2 dx dy

0

1

=  σ

− x:

y 2 (1

y ) dy = 2

(10.40)

0

1 x

1

I yy yy =  σ

−

0

0

1

=  σ

x2 dy dx

x2 (1

0

32

x) dx = 2

(10.41)

Classical Mechanics [Taylor, J.R.] Solution Manual

          −

1 y

1

I zz zz =  σ

0

−

x2 + y 2 dx dy

0

=  I xx  +  I yy xx  + I  yy = 4 1

I xy xy =  I yx yx  =  σ

(10.42)

1 y

−

−xy dx dy  (1 − y )2 y dy  = −1

0

0

1

=  σ

(c.f. Problem 10.23) .23)

0

(10.43)

2

Putting all the numbers together, we get the inertia tensor to be: I  =

(b) For the set of principal axes

 −

2 1 0

 

−1

0 0 4

2 0

 

(10.44)

have  I ω  =  λ i ω {ω , ω , ω }, we necessarily have I  1

2

i

3

i

for i for  i =  = 1, 2, 3. Note

that λi   is the moment of inertia about the principal axis in the direction of  ωi . The eigenv eigenvalue alue equation has non-trivial solutions if and only if det(I  det(I  λ1) = 0. 0 By inspection, we have that ω3  = 0  is an eigenvector, with corresponding eigenvalue/ moment 1 of inertia λ3  = 4. Verify this by ensuring the eigenvalue equation stated above is satisfied for this

−

 

choice of  ω3 . As for the remaining two principal axes, we shall consider:

 − − 2

⇒



λ 1 =0 1 2 λ

− − (2 − λ)2 − 1 = 0

And so the eigenvalues are: λ1 = 1

and

λ2  = 3

When λ1  = 1, we have:



1 1

−

⇒

   

−1 1

ωx  = ωy

0 0

ωx  = 1, ωy = 1

Normalizing, we obtain

 √   

1 ω1  = 2

33

1 1 0

 

(10.45)

Classical Mechanics [Taylor, J.R.] Solution Manual Then when λ when  λ 2  = 3, we have:

− −      1 1

⇒

1 1

ωx  = ωy

0 0

ωx  = 1, ωy =

−1

− −

Normalizing, we obtain

 √  −  1 1 0

1 ω2  = 2

Summarizing, with the set of orthogonal principle axes

 

(10.46)

 {ω1, ω2, ω3}, the inertia tensor consisting

of the principal moments is given by: I   =

 

 

1 0 0 0 3 0 0 0 4

34

 

(10.47)

Chapter 11

Coupled Oscillators and Normal Modes 11.5 (a) Find the normal normal frequencies, frequencies, ω  ω 1 and ω and  ω 2 , for the two carts shown in Figure 11.1 Figure  11.1,, assuming that m1  = m  =  m 2 and k and  k 1  =  k 2 . (b) Find and describe the motion for each of the normal modes in turn. Solution.   (a)

Let m1  =  m2  =  m,  m ,  k 1  =  k 2  = k  =  k,, and rightwards be positive. We can write down the

forces acting on each cart: F 1  =  k 2 x2  + ( k1

− − k2)x1 =  kx  k x2 − 2kx1 F 2  = −k2 x2  + k  +  k2 x1  = −kx2  + kx  +  kx1

   

(11.1) (11.2)

Using Newton’s second law, we express the above two equations as a single matrix equation:

  − mx  ¨1  = mx  ¨2

2k k

k k

−

  x1 x2

We look for the normal mode solutions that take the form

  iωt

x  =  A e

, where

(11.3)

x  and A  are

column

vectors. vectors. Substituting Substituting into Equation (11.3 (11.3), ),

−mω

2

  − A1  = A2

2k k

k k

−

  A1 A2

Figure 11.1: Two carts for Problems 11.5 11.5 and  and 11.6 11.6

35

 

(11.4)

Classical Mechanics [Taylor, J.R.] Solution Manual For non-trivial solutions, we have:

−

2k + mω  +  mω 2 k

⇒

k2



k =0 k + mω  +  mω 2

−

(mω2 )2 = 0 − 3kmω2 + (mω

We obtain the eigenfrequencies to be: 2

ω =

(3

± √ 5) ω0 ,

 

2

(11.5)

where ω where  ω 0  =  k/m.  k/m . This works out to be ω be  ω 1  = 0.62 62ω ω0 and ω and  ω 2  = 1.62 62ω ω0   approximately. (b) When ω When  ω  = ω  =  ω 1 , we have:



−1−√ 5 k 2

k

k −1−√ 5 k 2

⇒ A1 = 1

    A1  = A2

0 0

√ 

1+ 5 A2  = 2

and

 

(11.6)

And so we have the amplitudes (the reader can normalize the expression if he so wishes) to be:

   A1  = A2

1√ 

 

1+ 5 2

(11.7)

This means that both masses oscillate in phase, with the amplitude of mass 2 to be approximately 1.62 times that of mass 1. When ω When  ω  = ω  =  ω 2 , we then have:



−1+√ 5 k 2

k

k −1+√ 5 k 2

⇒ A1 = 1

    A1  = A2

and

A2  =

1

0 0

− √ 5

 

2

(11.8)

The amplitudes, when written in vector form is:

   A1  = A2

1 −√ 5

1

 

(11.9)

2

In this situation, both masses oscillate out of phase, with the amplitude of mass 2 being approximately 0.62 times that of mass 1.

11.6 Answer Answer the same questions as in Problem Problem 11.5 11.5  but for the case that m1 = m2 and k1 = 3k2 /2. (Write k (Write  k 1  = 3k and k and  k 2  = 2k.) Explain the motion in the two normal modes. 36

Classical Mechanics [Taylor, J.R.] Solution Manual Solution.  We

use the same sign conventions as Problem 11. 5,  5,   with m1 = m2 = m and k1 = 3k ,

k2  = 2k . The forces on each mass is then:  =  k 2 x2  + ( k1 F 1  = k

− − k2)x1  = 2kx2 − 5kx1 F 2  = −k2 x2  + k  +  k2 x1  = −2kx2  + 2kx 2 kx1

   

(11.10) (11.11)

We then use Newton’s second law to express the two equations above in a single matrix equation:

  − mx  ¨1  = mx  ¨2

Normal mode solutions:

iωt

x  =  A e

5k 2k

2k 2k

−

  x1 x2

 

(11.12)

. Substituting into the above equation, we obtain:

−mω

2

  − A1  = A2

5k 2k

2k 2k

−

  A1 A2

 

(11.13)

For non-trivial solutions, we have the determinant vanishing:

⇒

−

5k + mω  +  mω 2 2k 6k2



2k =0 2k  + mω  + mω 2

−

(mω2 )2 = 0 − 7kmω 2 + (mω

We obtain the eigenfrequencies to be: ω 2 = 6ω 6 ω0 where ω where  ω 0  = k  =  k/m /m.. And so ω so  ω 1  = When ω1  =

√ 6ω

0

or

ω0 ,

 

(11.14)

and ω2  = ω  =  ω 0 .

√ 6ω , we have: 0

      k 2k

⇒

2k 4k

A1 = 2

A1  = A2

and

0 0

A2  =

−1

(11.15)

In vector form,

  A1  = A2

2 1

 

−

(11.16)

With this normal mode, mass 1 and mass 2 are oscillating out of phase, with the amplitude of mass 1 being twice that of mass 2. We do the same for  ω 2  = ω  =  ω 0 . We get:

−

4k 2k

⇒

A1 = 1 37

2k k

−

   

and

A1  = A2

0 0

A2 = 2

(11.17)

Classical Mechanics [Taylor, J.R.] Solution Manual In vector form,

   A1  = A2

1 2

 

(11.18)

With this normal mode, mass 1 and mass 2 are oscillating in phase, with the amplitude of mass 1 being half that of mass 2.

38

Chapter 12

Nonlinear Mechanics and Chaos 12.1 No plans to do this chapter yet. yet.

39

Chapter 13

Hamiltonian Mechanics 13.3 Consider Consider the Atwood Atwood Machine of Fig 13.1, Fig  13.1, but  but suppose that the pulley is a uniform disc of mass  M  and radius R radius  R.. Using x Using  x  as your generalized coordinate, write down the Lagrangian, the generalized momentum p, and the Hamiltonian

 H = px˙ − L. Find Hamilton’s equations and use them to find

the acceleration x ¨. R M  x y

m1

m2

Figure 13.1: The Atwood Machine

M R2 , and angular velocit velocity y of the disc is given by by Solution.  The moment of inertia of the disc is 2 x˙ ω = . Since Since the length length of the string is fixed fixed (say l), we can write the following equation to R incorporate the constraint:

y + x  +  x +  + πR  πR =  = l l

⇒ 40

y  =

−x + const

(13.1)

Classical Mechanics [Taylor, J.R.] Solution Manual Now we can express the kinetic and potential energies in terms of a single coordinate  x.  x . 1  1 M R2 T  = (m1  + m  +  m2 )x˙ 2 + 2 2 2 1 1 = m1  + m  +  m2  + M   ˙x2 2 2



U  =



2

 x˙ R

(13.2)

−m2gy − m1gx = (m2 − m1 )gx + const

(13.3)

The Lagrangian is therefore:

L = T  −  − U  1 = 2





1 m1  + m  +  m2 + M   ˙x2 + (m (m1 2

− m2)gx + const

(13.4)

The generalized momentum p momentum  p  can be written as such:





∂  1  p =  p  = = m1  + m  +  m2  + M  x˙ ∂  ˙ x˙ 2

L

 

(13.5)

The Hamiltonian is then given by:

H = px˙ − L

  − −

p2 = m1 + m  +  m2  + 12 M  1 = 2

We note that



 1 2

p2 m1  + m  +  m2  + 12 M 

p2 m1 + m  +  m2  + 12 M 

(m1

−

(m1

− m2)gx − const

− m2)gx − const

(13.6)

 T  + U . U . The Hamilton’s equations H is just the total energy of the system, that is, H = T +

are given by: x˙  =

∂  ∂p

H

p m1  + m  +  m2 + 12 M  ∂   p =  p˙  = ∂x =

 

(13.7)

H

= (m2

− m1)g

 

(13.8)

We can obtain the accelerat acceleration ion rather rather trivially trivially from Hamilton’s equations equations (Eqns 13.7 (Eqns 13.7 and 13.8 and  13.8). ).  p˙ m1  + m  +  m2 + 12 M  (m2 m1 )g = m1  + m  +  m2 + 12 M 

x ¨  =

−

41

 

(13.9)

Classical Mechanics [Taylor, J.R.] Solution Manual

13.55 A bead of mass 13. mass  m  is threaded on a frictionless wire that is bent into a helix with cylindrical polar coordinates (ρ,φ,z (ρ,φ,z)) satisfying z = cφ and ρ = R, with c and R   consta constant nts. s. The z   axis points vertically up and gravity vertically down. Using φ  as your generalized coordinate, write down the kinetic and potential energies, and hence the Hamiltonian

 H  as a function of  φ  and its conjugate

momentum p. Write down Hamilton’s equations and solve for φ¨ and hence z¨. Explain your result in terms of Newtonian mechanics and discuss the special case that  R  = 0. Solution.   In

cylindrical coordinates, we have, in general, the kinetic energy  T  to  T  to be: T  =

1 m ρ˙ 2 + ρ2  ˙φ2 + z˙ 2 2





 

(13.10)

Substituting in the constraints z constraints  z  = cφ  =  cφ  and  ρ  = R  =  R,, we obtain

 

 

 ˙  = 1 m R2  ˙φ2 + cφ  ˙ T  φ 2

2

 

(13.11)

The potential energy is simply given by the gravitational potential energy: U  (φ  ( φ) = mgz  =  mgcφ We can write the Lagrangian

 

(13.12)

 L as: L = T  −  − U  = 21 m





R2 + c2 φ˙ 2

− mgcφ

 

(13.13)

The next step is to find the canonical momentum p  and substitute it into the general expression for the Hamiltonian. ∂  =  m R2 + c2 φ˙  ˙ ∂ φ  ˙ =  p φ

 p =  p  =

H

L

−L



   −

p2 p2 mR2 (R2 + c2 ) 2m (R2 + c2 ) p2 = + mgcφ 2m (R2 + c2 ) =

 

(13.14)

 

(13.15)



 − mgcφ

 

 

(13.16) (13.17)

Now, Now, we turn to Hamilton’s Hamilton’s equations by taking appropriate appropriate derivative derivativess of the Hamiltonian Hamiltonian.. ∂  p φ˙  = = 2 ∂p m (R + c2 ) ∂   p =  p˙  = = mgc ∂φ

H

− H − 42

   

(13.18) (13.19)

Classical Mechanics [Taylor, J.R.] Solution Manual We are able to solve for φ¨  and z¨ as such:  p˙ = m (R2 + c2 ) gc2 ¨ z¨ = c  =  c φ = R2 + c2

φ¨ =

− R2gc+ c2

   

−

(13.20) (13.21)

Let us unwrap unwrap the helix into into a 2-D plane. plane. When When go round round the helix an angle of ∆φ ∆φ = 2π , the vertical height we have gained is  z  = 2πc, πc , while the ’horizontal’ distance travelled is 2πR 2πR.. That is, we have the following: −g

−g sin α

2πc α

2πR Figure 13.2: Unwrapped helix into a 2-D plane

where the angle tan α  is given by c/R by  c/R.. From this, we note that we can write Eqn  13.21 as:  13.21  as: z¨ =

−g sin2 α

 

(13.22)

An object sliding down a slope of angle α  has a component of acceleration down the slope given by atang = g sin α. So the bead has a tangen tangential tial accele accelerat ration ion atang   along the wire. wire. The vertical vertical component of this acceleration (along the z the  z  axis) is z¨ = a  =  a tang sin α  = g  =  g sin2 α. This is precisely the acceleration calculated in Equation (13.22 ( 13.22). ). If  R =  R  = 0, then by Equation (13.21 ( 13.21), ), z¨ =

−g, as expected, since α = π  =  π//2.

13.6 In discussing the oscillation of a cart on the end of a spring, we almost always always ignore the mass of the spring. spring. Set up the Hamiltonian Hamiltonian

mass m on  on a spring (force constant k constant  k)) whose mass M  mass M  H for a cart of mass m

is  not  negligible, using the extension x extension  x of  of the spring as the generalized coordinate. Solve Hamilton’s equations and show that the mass oscillates with angular frequency ω frequency  ω  =

 

k/( k/(m + M/  + M/3). 3). That is,

the effect of the spring’s mass is to add M/3 M/ 3 to m. (Assume (Assume that the spring’s spring’s mass is distributed distributed uniformly and that it stretches uniformly.) Solution.  With

reference to the figure, let L  be the length of the spring, while ρ  be the mass per

unit length of spring. 43

Classical Mechanics [Taylor, J.R.] Solution Manual m

y (s)

s

y(L) =  x

ds

L

Consider an element A of spring of length  ds and  ds  and mass ρds mass  ρds,, at a distance s distance  s from  from the fixed end. Let y (s) be the displacement of the element A, and y and  y((L) = x  be the displacement of mass m mass  m  at  s =  s  = L  L.. Assume that the spring stretches uniformly, i.e. y (s) =  xs/L,  xs/L, with y(0) = 0, y (L) = x. We can can then express the velocity and kinetic energy of element A as: xs ˙ L 1 ρ x˙ 2 s2 2 T A  = (ρ ds) ds)y˙ = ds 2 2L2 y˙  =

 

(13.23)

Thus, the kinetic energy of the spring is given by:

 

ρ  ˙x2 L 2 T spring s ds spring  = 2L2 0 1 1 = ρ Lx˙ 2 = M  ˙ x˙ 2 6 6

(13.24)

where in the last line we have identified the total mass of the spring as  M  =  ρL.  ρL. 1 With the kinetic energy of the cart as T cart ˙ 2 , we have the total KE: cart  = 2 mx

T  = T spring  +  T cart spring + T  cart  =

1  1 M  ˙ x˙ 2 + mx˙ 2 6 2

1 = meff  ˙x2 2

(13.25)

where m where  m eff  = m  =  m +  + 13 M  is M  is the effective mass. The potential energy can be identified to be  U  = 12 kx2 , and thus the Lagrangian is:

L = 21 meff  ˙x2 − 21 kx2

(13.26)

We use the Lagrangian to derive the momentum in terms of the velocities, and then use the inverse relation to obtain the Hamiltonian

 H( px, x), as such: ∂ L  px  = =  meff  ˙x

 

∂  ˙ x˙

(13.27)

2

H = px ˙x − L = 2mpxeff  +  12 kx2 44

(13.28)

Classical Mechanics [Taylor, J.R.] Solution Manual The Hamilton’s equations of motions give us: x˙  =

∂  px = ∂p x meff 

H

 p˙x  =

− ∂ ∂xH = −kx

 

(13.29)

Differentiating the first of the two equations with respect to time, and using the second equation, we obtain: x ¨  =

 p˙x = meff 

− mkeff x = −ω2x

 

(13.30)

where ω where  ω 2 =  k/m eff . More explicitly explicitly,, this is the equation of simple simple harmonic harmonic motion with frequency frequency ω  =

    k = meff 

k m + M/  + M/33

 

(13.31)

13.17 Consider Consider the mass confined to the surface of a cone described described in Example 13.4 (page 533). We saw that there are solutions for which the mass remains at the fixed height  z =  z 0 , with fixed angular velocity φ˙ 0 , say say. (a) For For any chosen chosen value value of  pφ , use (13.34) to get an equation that gives the corresponding value of the height z0 . (b) Use the equations equations of motion to show that this motion motion is stable. stable. That That is, show show that if the orbit orbit has z = z0  +   +  ,, with   small, then   will oscillate about zero. (c) Show that the angular frequency frequency of these oscillatio oscillations ns is ω =

√ 3φ ˙  sin α, where α  is the 0

half angle of the cone (tan α  = c  =  c where  where c  c  is the constant in ρ in  ρ =  = cz  cz). ). (d) Find the angle α angle  α for  for which the frequency of oscillation ω  is equal to the orbital angular velocity φ˙ 0 , and describe the motion in this case. Solution.  The

two equations in (13.34) from the textbook (pg 534) are: pz m(c2 + 1)  pφ2  p˙z = mg mc2 z 3 z˙  =

 −

 

(13.32)  

(13.33)

The above equations were obtained from constructing the Lagrangian, and writing the Lagrange equations. For completeness, the Hamiltonian of the system can be written as:

H





 pφ2 1 pz2 = + + mgz 2m c2 + 1 c2 z 2 45

 

(13.34)

Classical Mechanics [Taylor, J.R.] Solution Manual

Figure 13.3: A mass  m  constrained to move on the surface of a cone. Taken from textbook pg 533.

(a) To maintain maintain at a fixed height, height, we require z˙  = 0, which from Equation (13.32 (13.32), ), we infer that it is necessary that p that  p z   = 0. And And so so  p˙z  = 0. From Equation Equation (13.33 (13.33), ), we see that the height z0   which corresponds to a particular value of  p of  p φ is: z0  =

  pφ2 m 2 c2 g

1 /3

(13.35)

(b) From Equation (13.32 (13.32), ), we differentiate once with respect to time to get an equation of motion for z for  z as:  p˙z m(c2 + 1)

z¨ = =

1 + 1)

m(c2



pφ2 mc2 z 3

 − mg



 

(13.36)

We consider a perturbation of  z  z  about the equilibrium z equilibrium  z 0 , that is, z is,  z  =  z 0 + , for small  small  .. We have the following approximations: z¨ = ¨

 

− 

z −3 =  z 0−3 1

3

 z0

(13.37)  

(13.38)

Substituting Substituting these relations relations into Equation Equation (13.36 13.36), ), we obtain: ¨ =

−

 pφ2 3  m(c2 + 1) mc2 z04

 

(13.39)

 pφ2 =  mg (from  mg  (from Equation (13.35 (13.35)) in an intermediate step. To mc2 z03 simplify Equation (13.39 (13.39)) further, we recall a relation between φ˙  and p  and  pφ  derived from the Hamilton’s where we have used the fact that

46

Classical Mechanics [Taylor, J.R.] Solution Manual equations. ∂  pφ φ˙  = = ∂p φ mc2 z 2

H

 

(13.40)

We use this to replace  p φ  in Equation (13.39 (13.39): ): ¨ =

 ˙ 2

2

− c32φ+c1  

 

(13.41)

This takes on the form of a SHM equation. (shown) (c) The angular frequency of the oscillation can be simply obtained from the SHM equation: ω  =

 

 ˙ 2 c2 3φ = c2 + 1

√  ˙ √  c 3φ c2 + 1

(13.42)

Finally, we can note that the fraction above is none other than sin α, where α  is defined in the following way (c.f. Figure  13.3):  13.3): tan α  = cz/z  =  cz/z.. Hence, we have shown that ω that  ω  =

√ 3φ ˙  sin α. 0

√ 

(d) In order for ω for  ω  = φ˙ 0 , we require sin α  = 1/ 3, or: α  = 35 35..26◦

 

(13.43)

In this case, the mass’s mass’s rotational frequency frequency equals its oscillation oscillation frequency frequency.. As a result, result, its orbit is closed. closed.

13.23 Consider Consider the modified Atwood Atwood mac machine hine shown in Figure Figure  13.4.   The two weights on the left have equal masses m masses  m and  and are connected by a massless spring of force constant k constant  k.. The weight on the right has mass M  = 2m, and the pulley is massless and frictionless frictionless.. The coordinate coordinate x  is the extension of the spring from its equilibrium length; that is, the length of the spring is le  +  x   where le is the equilibrium length (with all the weights in position and M   M   held stationa stationary) ry).. (a) Show Show that the total potential energy (spring + gravitational) is just U  = 12 kx2 (plus a constant that we can take to be zero). (b) Find the two momenta momenta conjugate conjugate to x and y . Solv Solvee for for x˙   and y˙ , and write down the Hamiltonian. Hamiltonian. Show that the coordinate coordinate y  y  is ignorable. (c) Write down the four Hamilton equations and solve them for the following initial conditions: You hold the mass  M  fixed  M  fixed with the whole system in equilibrium and  y  = y  =  y 0 . Still holding M  fixed, M  fixed, you pull the lower mass m  down a distance x0  and at time t  = 0 you let go of b oth masses. masses. Describe Describe the motion. In particular, particular, find the frequency frequency with which which  x  oscillates. 47

Classical Mechanics [Taylor, J.R.] Solution Manual

Figure 13.4: Modified Atwood machine for Problem 13.23

Solution.   (a)

Let string length be L. We take downwar downwards ds to be b e positive, positive, and the origin to be the

centre centre of the pulley pulley wheel. wheel. We also also assume assume the radius radius of the pulley pulley to be 0, without without any loss in generalit generality y. In addition, let us define the original, original, unstretched unstretched length of the spring (without load) to be l be  l 0 . That is, l is,  l e  =  l 0  + mg/k  +  mg/k,, where k where  k  is the spring constant. The two components of the potential energy can be written as: U grav grav  =

 +  le  + x  +  x)) − 2m(L − y )g −mgy − mg( mg(y + l = −mgle − 2mLg − mgx  

U spring spring  =

(13.44)

1 k(x + mg/k  + mg/k))2 2

(13.45)

Summing the two potential energies give us: U  = ( mgle

−

=

mgx) + − 2mLg − mgx)

1 2 kx + const 2



1 2 (mg) mg)2 kx + mgx +  mgx  + 2 2k

(shown)

 (13.46)

Note that we can ignore the constant, constant, and take it to b e equal to zero. Henceforth, Henceforth, the potential potential energy shall be computed only with U  with  U  = 12 kx2 . (b) We first complete the expression for the Lagrangian (

U ) by considering the kinetic L = T  −  − U )

energy T  energy  T  of  of the system: T  =

1  1 1 my˙ 2 + (2m (2m)y˙ 2 + m(y˙  + x˙ )2 2 2 2 48

(13.47)

Classical Mechanics [Taylor, J.R.] Solution Manual Then, we can write the momentum conjugates as: ∂  ∂ T  = = m(  m (y˙  + x˙ ) ∂  ˙ x˙ ∂  ˙ x˙ ∂  ∂ T   py  = = = 3m 3 my˙  + m  + m((y˙  + x˙ ) =  m(4  m(4y˙  + x˙ ) ∂  ˙ y˙ ∂  ˙ y˙

 px  =

L L

(13.48)

Solving Solving the above equation equation for x˙   and y˙ , we find their expressions: 4 px  py 3m py  px y˙  = 3m

x˙  =

− −

(13.49)

The Hamiltonian can thus be written as: ˙ x  + yp ˙ y −L H = xp 1 1 = ( py − px )2 + px2 2m 3





 1 + kx2 2

(13.50)

after some simplificati simplification. on. Note that this is simply the sum of the kinetic energies energies and potential potential energy. energy. Thus

 H  represents the total energy of the system. ∂ H Sinc Sincee  p˙y = − = 0,  p y  is constant and thus y thus  y  must be an ignorable coordinate. (shown) ∂y (c) The four Hamilton equations are:

− ∂ ∂xH = −kx ∂ H 4 px − py x˙  = =

− ∂ ∂yH = 0 ∂ H py − px y˙  = =

 p˙x  =

∂p x

 p˙y =

3m

∂p y

3m

(13.51)  

(13.52)

noting that the expressions in Equation (13.52 ( 13.52)) matches with those from Equation (13.49 (13.49). ).

To solve solve these equations, equations, we shall use the following following initial initial conditions: conditions: x(0) = x0 , y(0) = y0 , x˙ (0) = y˙ (0) = 0. Furthermore, with Equation (13.48 ( 13.48), ), we also have  p x (0) = p = p y (0) = 0. For the coordinate in  x,  x , we differentiate differentiate the expression expression for x˙  to get: x ¨  =

4 p˙x  p˙y = 3m

−

− 34mk x,

 

(13.53)

where we we have substitute substituted d the expressions expressions for  p˙x   an and  p˙y   from Equation (13.51 (13.51). ). One can can easily easily recognise this taking the form of a simple harmonic motion, where the position and velocity is given by:

        −  4k  4 k t + δ   + δ  3m

x(t) = A cos x˙ (t) =

 4k  4 k A sin 3m

49

 4k  4 k t + δ   + δ  3m

 

(13.54)  

(13.55)

Classical Mechanics [Taylor, J.R.] Solution Manual where A and δ  can δ  can be found found by consid consideri ering ng the initia initiall condit condition ions. s. It turns turns out that δ   δ   = 0 and A = x0 , after after a quick quick substitu substitutio tion n of the initia initiall condit condition ions. s. As such, such, the mass attached attached to the spring spring is undergoing undergoing simple harmonic motion described described by: x(t) =  x 0 cos

    4k  4 k t 3m

 

(13.56)

 

 4k  4 k t   . As for the motion character characterized ized by the coordinate coordinate y , since 3m we know that (i) p (i)  p y  is constant in time, and (ii) p (ii)  p y (0) = 0, we have, necessarily, that p that  p y  = 0 for all

with angular frequency ω  =

time t time  t.. From the second equation in Equation (13.48 (13.48), ), we have: y˙  =

− x4˙

y˙  =

⇒

1 4

 

 4k  4 k x0 sin 3m

    4k  4 k t 3m

 

(13.57)

A quick integration (and applying initial conditions) yields us:  1 y (t) = (y0  + x0 ) 4

−

1 x0 cos 4

    4k  4 k t 3m

 

(13.58)

It is not surprising to see that y that  y  is executing simple harmonic motion as well.

13.25 Here is another example example of a canonical transforma transformation, tion, which which is still too simple to be of any real use, use, but does nevert neverthel heless ess illustra illustrate te the power power of these these chang changes es of coordin coordinate ates. s. (a) Consider Consider a system with one degree of freedom and Hamiltonian

 H = H(q, p) and a new pair of coordinates  Q √  √  and P   P   defined so that q  = 2P  sin  sin Q and p = 2P  cos  cos Q. Prove Prove that that if  ∂ H = − p˙ and ∂ H = q ˙, ∂q

∂p

 −P  ˙ and ∂ ∂P H = Q˙ . In other words, the Hamiltonian Hamiltonian formalism formalism

H it automatically follows that ∂  ∂Q =

applie appliess just as well well to the new coordin coordinate atess as to the old. (b) Show Show that the Ham Hamilto iltonia nian n of a one-dimensional harmonic oscillator with mass  m =  m  = 1 and force constant k constant  k  = 1 is

 H = 12 (q 2 + p2).

(c) Show that if you rewrite this Hamiltonian in terms of the coordinates  Q and P   P   defined above, then Q then  Q is  is ignorable. What is P  is  P ?? (d) Solve the Hamiltonian Hamiltonian equation equation for Q for Q((t) and verify that, when rewritten for q  for  q , your solution gives the expected behaviour. Solution.  (a)

We have the following relation given in the question: q  =  =  p =  p =

√  2P  sin  sin Q √ 

2P  cos  cos Q

50

(13.59)

Classical Mechanics [Taylor, J.R.] Solution Manual We compute the relevant derivatives below: ∂ q  sin Q = ∂P  2P  ∂p cos Q = ∂P  2P 

√ 

∂ q  = ∂Q ∂p = ∂Q

√ 

2P  cos  cos Q

√   sin Q − 2P  sin ∂ H ∂ H Consider the Hamiltonian equation for Q for  Q,, recalling that = q ˙ and = − p˙ : √ 

∂p

H

(13.60)

 

(13.61)

∂q 

∂  ∂  ∂ p ∂  ∂ q  = + ∂Q ∂p ∂Q ∂q  ∂Q  ∂ p  ∂ q  = q ˙  p˙ ∂Q ∂Q  ∂ q  ˙ ∂ q  ˙ ∂ p  ∂ p ˙ ∂ p ˙ ∂ q  = Q + P  Q + P  ∂Q ∂P  ∂Q ∂Q ∂P  ∂Q  ∂ q  ∂ p ∂ p ∂ q  ∂ p ∂ q  ˙  ∂ q  ∂ p = Q˙ + P  ∂Q ∂Q ∂Q ∂Q ∂P  ∂Q ∂P  ∂Q ∂ p ∂ q  ˙  ∂ q  ∂ p = 0 + P  ∂P  ∂Q ∂P  ∂Q ˙  ˙   (shown) = P  sin2 Q cos2 Q = P 

H

 

H

 −



  −    −     −  − −  −



 −



Similarly, we can consider the Hamiltonian equation for P : P : ∂  ∂  ∂ p  ∂  ∂ q  = + ∂P  ∂p ∂P  ∂q  ∂P   ∂ p  ∂ q  = q ˙  p˙ ∂P  ∂P   ∂ q  ˙ ∂ q  ˙ ∂ p  ∂ p ˙ ∂ p ˙ ∂ q  = Q + P  Q + P  ∂Q ∂P  ∂P  ∂Q ∂P  ∂P   ∂ q  ∂ p ∂ p ∂ q  ∂ p ∂ q  ˙  ∂ q  ∂ p = Q˙ + P  ∂Q ∂P  ∂Q ∂P  ∂P  ∂P  ∂P  ∂P   ∂ q  ∂ p ∂ p ∂ q  = Q˙ +0 ∂Q ∂P  ∂Q ∂P  = Q˙ cos2 Q + sin2 Q = Q˙   (shown)

H

H



H

−

 −   −  − 

  



−



(b) The Lagrangian for the 1-D harmonic oscillator is simply given by the quantity  T 

U :  −  − U :

L = T  −  − U  = 21 mq ˙2 − 21 kq 2

(13.62)

We can express the velocity as a function of the momentum:  p =  p  =

∂  =  mq ˙ ∂  ˙ q ˙

L

⇒

q ˙  =  =

p m

 

(13.63)

Thus the Hamiltonian can be written as such:

H

1  p 2  1 2 = T  + U  = m + kq  2 m 2 1 2 = (q  + p2 ) (shown) 2



51

(13.64)

Classical Mechanics [Taylor, J.R.] Solution Manual where the last line was obtained by setting  k  = m  =  m =  = 1, as given in the question. (c) Using the coordinate transformations defined in Equation (13.59 (13.59), ), we get:

H = 21





2P  sin  sin 2 Q + 2P  2P  cos  cos 2 Q  = P   =  P 

 

(13.65)

And so P  so  P  is  is the Hamiltonian of the system. It can also be observed that  Q  is no longer present in the Hamiltonian, and so, is an ignorable coordinate. (d) We have from Hamilton’s equations: ∂  Q˙  = =1 ∂P 

H

(13.66)

As such, we can easily find  Q(  Q (t) by integrating, then solving for q: Q  = t  =  t +  + δ   δ  q  =  =

√ 

2sin t + δ   + δ 

 

(13.67)  

(13.68)

where δ  is   is a constant of integration. integration. This gives the expected behaviour behaviour (a simple harmonic harmonic oscillation).

52

Chapter 14

Collision Theory 14.1 No plans to do this chapter yet. yet.

53

Chapter 15

Special Relativity 15.1 No plans to do this chapter yet. yet.

54

Chapter 16

Continuum Mechanics 16.1 No plans to do this chapter yet. yet.

55