Bead Problem Part2

Example 3: A bead (or ring) sliding on a uniformly rotating wire in a force-free space • The wire is straight and is rotated uniformly about some fixed axis perpendicular to the wire. • Constraint is time dependent, with rotation axis along the z and the wire in the xy plane. The transformation equation explicitly contain the time, t: x = r cos t,

y = r sin t,

with

 - angular velocity of rotation

r – distance along the wire from rotation axis

Example 3: A bead (or ring) of mass m sliding on a uniformly rotating wire in a force-free space y

Euler-Lagrange Eq.: 

L=T–V

T = ½ mvr2 + ½ mv2 V=0 r



𝜃 = ω𝑡

x

𝑑𝑟 𝑣𝑟 = = 𝑟ሶ 𝑑𝑡

1 1 T = 𝑚𝑟 2ሶ + 𝑚𝑟 2 𝜃ሶ 2 2

2

(in polar coordinates)

𝑑𝜃 𝑣𝜃 = 𝑟𝜔 = 𝑟 = 𝑟𝜃ሶ = 𝑟𝜔 𝑑𝑡 𝑑𝜃 = 𝜃ሶ = 𝜔 𝑑𝑡

Example 3: A bead (or ring) of mass m sliding on a uniformly rotating wire in a force-free space y

Step 1: Construct the Lagrangian 1 1 L = 𝑚𝑟ሶ 2 + 𝑚𝑟 2 𝜃ሶ 2



2

r 

2

Step 2: Conjugate momenta x

𝜕𝐿 𝑝𝑟 = = 𝑚𝑟ሶ 𝜕𝑟ሶ

𝜕𝐿 𝑝𝜃 = = 𝑚𝑟 2 𝜃ሶ 𝜕𝜃ሶ





𝑝𝑟 𝑟ሶ = 𝑚

𝜃ሶ =

𝑝𝜃 𝑚𝑟 2

Example 3: A bead (or ring) of mass m sliding on a uniformly rotating wire in a force-free space y

Step 3: Form the Hamiltonian 

𝐻 = ෍ 𝑝𝑖 𝑞ሶ 1 − L 𝑖

𝐻 = 𝑝𝑟 𝑟ሶ + 𝑝𝜃 𝜃ሶ −

r 

x

𝐻 = 𝑝𝑟

1 1 𝑚𝑟 2ሶ + 𝑚𝑟 2 𝜃ሶ 2 2 2

𝑝𝑟 𝑝𝜃 1 𝑝𝑟 + 𝑝𝜃 − 𝑚 𝑚 𝑚𝑟 2 2 𝑚

𝑝𝑟2 𝑝𝜃2 𝑝𝑟2 𝑝𝜃2 𝐻= + − − 𝑚 𝑚𝑟 2 2𝑚 2𝑚𝑟 2

𝑝𝑟2 𝑝𝜃2 𝐻= + 2𝑚 2𝑚𝑟 2

2

1 𝑝𝜃 + 𝑚𝑟 2 2 𝑚𝑟 2

2

Example 3: A bead (or ring) of mass m sliding on a uniformly rotating wire in a force-free space y

Step 4: Obtain 𝑞𝑖ሶ as functions of (q, p, t) 

𝜕𝐻 𝑝𝑟 𝑟ሶ = = 𝜕𝑝𝑟 𝑚 r



x

2 𝜕𝐻 𝑝𝜃2 𝑝 𝑝𝑟ሶ = − =− −2𝑟 −2−1 = 𝜃3 𝜕𝑟 2𝑚 𝑚𝑟

𝜕𝐻 𝑝𝜃 𝜃ሶ = = 𝜕𝑝𝜃 𝑚𝑟 2 𝜕𝐻 𝑝𝜃ሶ = − =0 𝜕𝜃

Example 3: A bead (or ring) of mass m sliding on a uniformly rotating wire in a force-free space y

Equation of motion for r: 𝑑 𝜕𝐻 𝑚 = 𝑝𝑟ሶ 𝑑𝑡 𝜕𝑝𝑟



𝑝𝜃2 𝑚𝑟ሷ = 𝑚𝑟 3

r 

x

𝑚𝑟 2 𝜃ሶ 𝑚𝑟ሷ − 𝑚𝑟 3

2

=0

𝑚𝑟ሷ − 𝑚𝑟𝜃ሶ 2 = 0 𝑚 𝑟ሷ − 𝑟𝜃ሶ 2 = 0

Example 3: A bead (or ring) of mass m sliding on a uniformly rotating wire in a force-free space y

Equation of motion for :

𝑑 𝜕𝐻 𝑚 = 𝑝𝜃ሶ 𝑑𝑡 𝜕𝑝𝜃



𝜕𝐻 =0 𝜕𝑝𝜃

r 



𝑝𝜃ሶ = 0

x

Hence:

𝑝𝜃 = 𝑚𝑟 2 𝜃ሶ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

 Conservation of angular momentum arises because  is a cyclic coordinate.

Cyclic