Motion of particle in Plane(3)

FIRST PART:
Speed = r'R + rθ'P
Acceleration = [r"– r(θ')^2]R + (2r'θ' + rθ")P

where
R is unit vector in radial direction,
P is unit vector in angular direction

Magnitude of speed v
v^2 = (r') ^2 + (rθ')^ 2

Since
r = a(1 + cosθ)
r' = –a(sinθ)(θ')
r" = –a(cosθ)(θ')^2 – a(sinθ)(θ")

Therefore
v^2 = a^2sin^2θ(θ')^2 + a^2 (θ')^2(1 + cosθ)^2
v^2 = 2a^2 (1 + cosθ)(θ')^ 2 …………………(1)

Rewrite (1) gives (1 + cosθ)(θ')^ 2 = v^2/2a^2………………(2)

Differentiate (2) on both side gives
(1 + cosθ)2(θ')(θ") + (θ')^2[–sinθ(θ')] = 0
Or θ" = sinθ(θ')^2/[2(1 + cosθ)]……………(3)

Radial component of the acceleration is r"– r(θ')^2 which is equal to

–a(cosθ)(θ')^2 – a(sinθ)(θ") – a(1+cosθ)(θ')^2
= –a[cosθ(θ')^2 + sinθ(θ") + (θ')^2 + cosθ(θ')^2]
= –a[(θ')^2(1 + 2cosθ) + sinθ(θ")]

Use (3), = –a{(θ')^2(1 + 2cosθ) + sinθ.sinθ(θ')^2/[2(1 + cosθ)]}
= –a(θ')^2{1 + 2cosθ + sin^2θ/[2(1 + cosθ)]}
= –a(θ')^2 (2 + 6cosθ + 4cos^2θ + sin^2θ)/[2(1 + cosθ)]
= –a(θ')^2 (3 + 6cosθ + 3cos^2θ)/[2(1 + cosθ)]
= –(3/2)a(θ')^ 2 (1 + cosθ)……………(4)

Sub (2) into (4) gives –(3/2)a(v^2/2a^2) = –3v^2/4a which is a constant

2009-05-21 11:39:03 補充：
SECOND PART:
Angular component of acceleration = 2r'θ' + rθ"
= –2a sinθ(θ') ^2 + a(1 + cosθ)θ"
Use (3)
= –2a sinθ(θ') ^2 + a(1+cosθ)sinθ(θ') ^2/[2(1 + cosθ)]
=–2a sinθ(θ') ^2 + a sinθ(θ') ^2/2
=–3/2a sinθ(θ') ^2……………(5)

2009-05-21 11:39:39 補充：
Square of magnitude of acceleration A = Square of (4) + Square of (5)
A^2 = 9/4a^2(θ')^4(1 + cosθ)^2 + 9/4a^2sin^2θ(θ')^4
A^2 = 9/4a^2(θ')^4(1 + 2cosθ + cos^2θ + sin^2θ)
A^2 = 9/4a^2(θ')^4(2 + 2cosθ)
A^2 = 9/2a^2(θ')^4(1 + cosθ)

2009-05-21 11:39:56 補充：
Use (2)
A^2 = 9/2a^2(1 + cosθ)v^4/[4a^4(1 + cosθ)^2]
A^2 = 9v^4/[8a.a(1 + cosθ))
A^2 = 9v^4/8ar
A = Sqrt(9v^4/8a).Sqrt(1/r)

You probably have to use the formula:
acceleration = [r" - r( x')^2]R + (1/r)d(r^2 d x/dt)/dt P.
x stands for theta, R and P are the 2 unit vectors.