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)|
x = xtranslational + xrotational
To find xtrans:
Conservation of linear momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v'
m(2v) + m(−v) = 2mv'
→ v' = 0.5v
→ xtrans = v't = 0.5vt
→ A, B and E are FALSE.
Check answer C. x = 0.5vt + 0.5b sin (3vt/b)
→ xrot = 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