−m ˙θ2 = mg cos θ− FT ∴ FT = mg cos θ + m ˙θ2 ar

(Dated: April 23, 2005)

θ O

l

θ^ˆ FrT

m rˆ⊥

g mr Example of 2-dimensional motion: Simple pendulum

approach 1:

eq. of motion ma =X

F = mg + F|{z}T tension

ˆ

r_⊥ component

r_⊥ = = const. ∴ −m ˙θ² = mg cos θ− F^T ∴ FT = mg cos θ + m ˙θ² ar_⊥ = m¨r_⊥− mr⊥˙θ² , v = v ˆθ , v = ˙θ

θˆcomponent

m ¨θ = −mg sin θ i.e. m d²θ

dt² =−mg sin θ Die hard in t-domain to get θ (t) if θ ¿ 1 sin θ≈ θ =⇒ S.H.M.

d²θ dt² =−g

θ

∴ θ = θ (t) = θ|{z}^A sin (2πνt + δ)

→ Amplitude

→ ν = 1 2π

rg

→ frequency

approach 2.

conservation of mechanical energy K.E. = 1

2mv² = 1

2m ²˙θ² , P.E. = mg (1 − cos θ) = U (θ) E = 1

2m ²θ˙²+ mg (1− cos θ) = mg (1 − cos θ^A)

∴ ˙θ = dθ dt =

∙2g

(cos θ− cos θ^A)

¸¹₂

= f (θ)

dt = dθ

f (θ) = dg (θ)

∴ t − t⁰ = g (θ)− g (θ⁰) , θ (0) ≡ θ⁰ g (θ) =

Z dθ

∙2g

(cos θ− cos θ^A)

¸¹₂ = 1 2

µ g

¶¹₂ Z dθ

£sin²(θA/2)− sin²(θ/2)¤¹₂

→ θ = θ (t) elliptic integral of the 1-st kind. → Look into table.

set t0 = 0 , θ0 = 0 , θ = θA→ t = 1

4T , T = 1

ν period.

∴ 1 ν = 2

µ g

¶¹₂ Z θA

0

£ dθ

sin²(θA/2)− sin²(θ/2)¤¹₂

"

reference: 1 ν ≈ 2π

µ g

¶¹₂ µ 1 + 1

16θ²_A+ 11

3072θ⁴_A+· · ·

¶

if θA< π

#

if θ0 ¿ 1 small oscillation U (θ)≈ ¹₂mg θ² , sin θ ≈ θ

g (θ)≈ µ

g

¶¹₂ Z

dθ (θ²_A− θ²)¹²

→ θ (t) = θ^Asin∙³g ´¹2

t + δ

¸

, δ = sin⁻¹ θ0

θA

, ν = 1 2π

³g´¹2

φ^ˆ _rˆ

φ rr

O M kˆ

m Orbital motion of planet

F =−GM m

r² rˆ → two-dim.

r = r· ˆr v = ˙rˆr + r ˙φ ˆφ mdv

dt = m

³

¨

rˆr + 2 ˙r ˙φ ˆφ + r ¨φ ˆφ− r ˙φ²rˆ

´

∴

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ m³

¨

r− r ˙φ²´

=−GMm/r² m³

2 ˙r ˙φ + r ¨φ´

= 0 → L = mr × v = mr²φˆ˙k dL/dt =³

2r ˙r ˙φ + r²φ¨´ k = 0ˆ

angular momentum conservation of L

∴ r²φ = const.˙ → Kepler’s 2nd law

area sweeping rate: dA = ¹₂rr ˙φdt ∴ dA/dt = L/2m = const.

direction of L fixed → plane of motion E = 1

2mv²+ U (r) = 1 2m

³

˙r²+ r²φ˙²

´

−GM m r φ =˙ L

mr²

∴ E = 1

2m ˙r²+ µ L

2mr² −GM m r

¶

= 1

2m ˙r²+ Uef f(r) for a given L,

Ueff

energy

O r

r1 r₀ r₂ r3

Ueff

energy

O r

r

−GMm

2 2

2mr L

case 1 E < U (r0) = min (Uef f) no way!

case 2 E = U (r0)

˙r = 0→ r = r⁰ = const.

dUef f (r) dr

¯¯

¯¯

r=r0

= 0

circular orbit

case 3 0 > E > Uef f(r0)→ E = U^{ef f} (r1) = Uef f (r2) at r = r1, r2 ˙r = 0

∴ E = L²

2mr² −GM m

r =⇒ 2mEr²+ 2GM m²r− L² = 0 r1 , r2 roots of the eq.

r1+ r2 =−2GM m²

2mE ∴ E = −GM m

r1+ r2

elliptical orbit

case 4 E = 0 = U (r3) = U (∞) ∴ ˙r = 0 at r³ and ∞ parabolic orbit

case 5 E > 0 E = U (r4) , ˙r = 0 at r4 , ˙r > 0 as r → ∞ hyperbolic orbit

eq. of orbit

m³

¨

r− r ˙φ²´

=−GM m r² r²φ =˙ L

m = const.

set u≡ 1

r → ˙φ = L mr² = L

mu²

˙r = dr

dφφ =˙ L mr²

dr

dφ =−L m

du

dφ (by chain rule)

¨

r =−L m

d²u

dφ²φ =˙ −L² m²u²d²u

dφ²

→ d²u

dφ² + u = GM m² L²

∴ u = u (φ) = GM m²

L² − η cos φ = 1 r

l φ

r

F

focus

directrix

x y

eq. of polar coordinates of a conic section

r

+ r cos φ = → eccentricity or 1

r = 1

− 1 cos φ

∴ = η L²

GM m² →

"

1 +2E m

µ L

GM m

¶2#¹₂

case 1. < 1 ellipse

φ = 0 , π → r = r¹, r2 , r1+ r2 = 2a

→ 1

r = 1− cos φ

a (1− ²) E =−GM m r1+ r2

case 2. = 1 parabola → comet case 3. > 1 hyperbola → scattering case 4. = 0 circle

E =−G²M²m³ 2L²

another approach by conservation of energy

E = 1

2m ˙r²+ µ L²

2mr² − GM m r

¶

& mr²φ = L˙

→ dr dφ = 1

L

¡2mEr⁴− L²r²+ 2GM m²r³¢¹₂

set u = 1

r · · · ·

→ dφ =

d µ

u−GM m² L²

¶

"

2mE L² +

µGM m² L²

¶2

− µ

u− Gm²M L²

¶2#¹₂ = dβ (α² − β²)¹²

· · · → 1

r = GM m² L²

"

1− µ

1 + 2EL² G²M²m³

¶¹₂ cos φ

#