Classical Mechanics Quiz

Sample Quiz 1, Physics 5011, Fall, 2007 Instructions: This is a closed book exam. Calculators are not permitted (note that any numerical answers required can be done to order-of-magnitude accuracy). Do each problem on the sheets of paper provided. Make sure your name and ID number are on every sheet, so they can be identified if they become separated. After you are done, staple together all of your sheets. 1. (30 pts.) A sphere of mass m and radius a rolls without slipping inside a spherical shell of radius R (with R > a) under the influence of a uniform gravitational force. Assuming that all of the motion occurs in one vertical plane, set up appropriate generalized coordinates for this problem and determine the Lagrangian and any constraint conditions. Determine the equations of motion and find the frequency of small oscillations for this system. (Hint: recall that I = (2/5)ma2 for a sphere.) 2. (30 pts.) Consider a particle of mass m and angular momentum l that moves in a central force field that has the form F = –mkr. (a) Find the total energy and sketch the effective potential for this particle, and determine the radius and period of a circular orbit. (b) Suppose that the orbit is almost circular. Show that the particle reaches its closest distance to the center twice per orbit (or equivalently, that the frequency of small oscillations about the circular orbit is twice the orbital frequency). 3. (40 pts.) Consider a symmetrical top (I1 = I2 ≠ I3) with no torques acting upon it. (a) Write down the Lagrangian for this top in terms of the Euler angles and determine two constants of motion associated with the generalized momenta of φ and ψ. Find the equation of motion for θ, and show that steady precession (constant θ) is possible when ϕ = I 3ω3 / I1 cos θ . (b) Show that one of the two constants of motion in part (a) corresponds to the angular momentum around the body z′ axis and the other corresponds to the fixed z axis. (Hint: Determine the angular momentum vector in body coordinates, and perform the coordinate transformation to find the z component in the fixed coordinates.)

Physics 5011, Quiz 1, October 6, 2006 Important formulas: (Note that not all formulas are valid in all cases. Also, tensors are denoted by a double arrow above the symbol rather than a double underline) T=

1 2 mv 2

p = mv

L = r×p

M = ∑ mi

R cm =

i

d ⎛ ∂L ⎜ dt ⎜⎝ ∂q j

L = T −V pj =

⎞ ∂L = ∑ λ i aij ⎟⎟ − ∂ q i j ⎠

∂L ∂q j

V = − ∫ F ⋅ dr

F = −∇V

∑ miri

Vcm =

i

M

h = ∑ p j q j − L

l = mr 2 θ

j

mk [1 + e cos(θ − θ′)] l2

I v′ = A ⋅ v

M

j

d u m m dV ( u ) + u = − 2 2 f (u ) = − 2 2 du dθ l u l a=

b=

i

∑ aij dq j = 0

Constraints:

2

u=

∑ mi vi

u=

rmax + rmin k l 2 / mk =− = 2 2 E 1 − e2

Θ k cot 2E 2

I I A = A −1 vi′ = Aij v j I I I I B ′ = A ⋅ B ⋅ A −1

b db sin Θ d Θ I A ji = A −1

1 r e=

rmax − rmin 2 El 2 = 1+ rmax + rmin mk 2

σ=

( )

ij

Aik A jk = δij

Bij′ = Aik A jl Bkl

cos ψ sin ϕ + cos θ cos ϕ sin ψ sin ψ sin θ ⎞ ⎛ cos ψ cos ϕ − cos θ sin ϕ sin ψ I ⎜ ⎟ A = ⎜ − sin ψ cos ϕ − cos θ sin ϕ cos ψ − sin ψ sin ϕ + cos θ cos ϕ cos ψ cos ψ sin θ ⎟ ⎜ − sin θ cos ϕ sin θ sin ϕ cos θ ⎟⎠ ⎝ ⎛ cos ψ cos ϕ − cos θ sin ϕ sin ψ − sin ψ cos ϕ − cos θ sin ϕ cos ψ sin θ sin ϕ ⎞ I −1 ⎜ ⎟ A = ⎜ cos ψ sin ϕ + cos θ cos ϕ sin ψ − sin ψ sin ϕ + cos θ cos ϕ cos ψ − sin θ cos ϕ ⎟ ⎜ sin ψ sin θ cos ψ sin θ cos θ ⎟⎠ ⎝

ωx′ = ϕ sin θ sin ψ + θ cos ψ ω = ϕ sin θ cos ψ − θ sin ψ

 sin θ sin ϕ ωx = θ cos ϕ + ψ ω y = θ sin ϕ − ψ sin θ cos ϕ

ωz′ = ϕ cos θ + ψ

ωz = ψ cos θ + ϕ

y′

⎛d⎞ ⎛d⎞ + ω× ⎜ ⎟ =⎜ ⎟ ⎝ dt ⎠ fix ⎝ dt ⎠body

( u ) fix = ( u )body + ω × u

Feff = F − 2mω × v − mω × ( ω × r ) I I I 1 I I = ∫ d 3r ρ ( r ) ⎡⎣ r 2 1 − rr ⎤⎦ L = I ⋅ω T = ω⋅ I ⋅ω 2 I I I 2  i + εijk ω j ωk I k = N i I = Icm + M Rcm 1 − R cm R cm Ii ω

(

L=

(

1 I1 θ 2 + ϕ 2 sin 2 2

Math: Binomial Theorem: (1 + x ) = 1 + px + p

) 1 θ) + I 2

p ( p − 1)

3

( ψ + ϕ cos θ )

x 2 + ... +

2

− V ( cos θ )

p ( p − 1) ... ( p − n + 1)

x n + ...

2 n! 2 n x x ( n) Taylor expansion: f ( x ) = f ( 0 ) + xf ′ ( 0 ) + f ′′ ( 0 ) + ... + f ( 0 ) + ... 2 n! Stokes Theorem: ∫ da nˆ ⋅ ( ∇ × A ) = v∫ A ⋅ dl Divergence Theorem: ∫ d 3 x∇ ⋅ A = v∫ da nˆ ⋅ A

εijk

sin 2 θ + cos 2 θ = 1 1 cos 2 θ = (1 + cos 2θ ) 2 ⎧1 for ijk = 123, 231,312 ⎪ = ⎨−1 for ijk = 132, 213,321 ⎪0 otherwise ⎩

A × ( B × C) = B ( A ⋅ C) − C ( A ⋅ B )

sec 2 θ − tan 2 θ = 1 cosh 2 θ − sinh 2 θ = 1 1 sin 2 θ = (1 − cos 2θ ) sin 2θ = 2sin θ cos θ 2 ⎧1 for i = j δij = ⎨ ⎩0 for i ≠ j

εijk εilm = δ jl δkm − δ jm δkl

A ⋅ ( B × C) = B ⋅ (C × A ) = C ⋅ ( A × B )

( A × B ) ⋅ ( C × D ) = ( A ⋅ C )( B ⋅ D ) − ( A ⋅ D )( B ⋅ C ) ∇ × ∇ψ = 0 ∇ ⋅ (∇ × A ) = 0 ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇2 A ∇ ⋅ ( ψA ) = ∇ψ ⋅ A + ψ∇ ⋅ A ∇ × ( ψA ) = ∇ψ × A + ψ∇ × A ∇ ( A ⋅ B ) = A ⋅∇B + B ⋅∇A + A × ( ∇ × B ) + B × ( ∇ × A ) ∇ ⋅ ( A × B ) = B ⋅ (∇ × A ) − A ⋅ (∇ × B ) ∇ × ( A × B ) = A∇ ⋅ B − B∇ ⋅ A + B ⋅∇A − A ⋅∇B Div, Grad, Curl and all that: Cylindrical ∂ψ 1 ∂ψ ∂ψ ∇ψ = sˆ + ϕˆ + zˆ ∂s s ∂ϕ ∂z 1 ∂ 1 ∂Aϕ ∂Az ∇⋅A = + ( sAs ) + s ∂s s ∂ϕ ∂z

Spherical ∂ψ ˆ 1 ∂ψ 1 ∂ψ ∇ψ = rˆ +θ + ϕˆ r ∂θ r sin θ ∂ϕ ∂r ∇⋅A =

1 ∂ 2 1 ∂ 1 ∂Aϕ r Ar + ( sin θAθ ) + 2 r sin θ ∂θ r sin θ ∂ϕ r ∂r

(

)

⎛ 1 ∂Az ∂Aϕ ⎞ ⎛ ∂As ∂Az ⎞ ∇ × A = sˆ ⎜ − − ⎟ + ϕˆ ⎜ ∂s ⎟⎠ ⎝ ∂z ⎝ s ∂ϕ ∂z ⎠ ∂A ⎤ 1⎡ ∂ + zˆ ⎢ ( sAϕ ) − s ⎥ s ⎣ ∂s ∂ϕ ⎦ ∇ 2ψ =

1 ∂ ⎛ ∂ψ ⎞ 1 ∂ 2ψ ∂ 2 ψ + ⎜s ⎟+ s ∂s ⎝ ∂s ⎠ s 2 ∂ϕ2 ∂z 2

∇ × A = rˆ

∂A ⎤ 1 ⎡∂ sin θAϕ ) − θ ⎥ ( ⎢ ∂ϕ ⎦ r sin θ ⎣ ∂θ

⎡ 1 ∂Ar 1 ∂ ⎤ ∂A ⎤ 1⎡ ∂ + θˆ ⎢ − rAϕ ) ⎥ + ϕˆ ⎢ ( rAθ ) − r ⎥ ( ∂θ ⎦ r ⎣ ∂r ⎣ r sin θ ∂ϕ r ∂r ⎦

∇ 2ψ =

1 ∂2 1 ∂ ⎛ ∂ψ ⎞ 1 ∂ 2ψ r sin ψ + θ + ( ) ⎜ ⎟ r ∂r 2 ∂θ ⎠ r 2 sin 2 θ ∂ 2 ϕ r 2 sin θ ∂θ ⎝

dl 2 = ds 2 + s 2 d ϕ2 + dz 2

dl 2 = dr 2 + r 2 d θ2 + r 2 sin 2 θ d ϕ2

d 3 x = s ds d ϕ dz

d 3 x = r 2 sin θdr d θ d ϕ = r 2 dr d ( cos θ ) d ϕ

sˆ = cos ϕ xˆ + sin ϕ yˆ ϕˆ = − sin ϕ xˆ + cos ϕ yˆ zˆ = zˆ

rˆ = sin θ cos ϕ xˆ + sin θ sin ϕ yˆ + cos θ zˆ θˆ = cos θ cos ϕ xˆ + cos θ sin ϕ yˆ − sin θ zˆ ϕˆ = − sin ϕ xˆ + cos ϕ yˆ