Classical Mechanics - Rotational Motion

One ice skater of mass m moves with speed 2v, while another of the same mass m moves with speed v toward the left, as shown in Figure I. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in Figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2?

A.	x = 2vt	y = b/2
B.	x = vt + 0.5b sin (3vt/b)	y = 0.5b cos (3vt/b)
C.	x = 0.5vt + 0.5b sin (3vt/b)	y = 0.5b cos (3vt/b)
D.	x = vt + 0.5b sin (6vt/b)	y = 0.5b cos (6vt/b)
E.	x = 0.5vt + 0.5b sin (6vt/b)	y = 0.5b cos (6vt/b)

(GR9277 #78)

Solution:
x = x_{translational} + x_rotational

To find  x_trans:
Conservation of linear momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v'
m(2v) + m(−v) = 2mv'
→ v' = 0.5v
→ x_trans = v't = 0.5vt
→ A, B and E are FALSE.

Check answer C. x = 0.5vt + 0.5b sin (3vt/b)
→ x_rot = a sin ωt = 0.5b sin (3vt/b) → ω = 3v/b

To check if ω = 3v/b:

At t = 0,
L = L₁ + L₂ = r₁m₁v₁ + r₂m₂v₂ = ¹/₂bm2v + ¹/₂bmv = ³/₂bmv

At t > 0,  
L' = Iω
Moment inertia of the system: I = ½mb²
→ L' = ½mb²ω

Conservation of angular momentum:  L = L' 
→ ³/₂bmv = ¹/₂mb²ω → ω = 3v/b

Answer: C