Albert stands on a frictionless turntable, holding a bike wheel. Both Albert and the wheel are initially stationary. Albert gives the bike wheel a good spin, and it begins rotating clockwise. If the bike wheel has a final angular momentum of one unit, clockwise, what is the final angular momentum had by the rest of the system (i. e., Albert and the turntable) ?

Albert and the turntable have a final angular momentum L₀₂ = 1 kgm²/s in the counterclockwise direction.

Explanation:

Let L₁ represent the initial angular momentum of the wheel and L₂ represent the initial angular momentum of Albert and the turntable. Let L₀₁ represent the final angular momentum of the wheel and L₀₂ represent the final angular momentum of Albert and the turntable.

From the law of conservation of angular momentum,

initial angular momentum = final angular momentum.

So, L₁ + L₂ = L₀₁ + L₀₂

Since Albert, the wheel and the turntable are initially at rest, L₁ = L₂ = 0 and L₀₁ = - 1 kgm²/s (the negative sign indicates that it is rotating in a clockwise direction). Substituting these values into the equation above,

L₁ + L₂ = L₀₁ + L₀₂

0 + 0 = - 1 + L₀₂

L₀₂ = + 1 kgm²/s

So, Albert and the turntable have a final angular momentum L₀₂ = 1 kgm²/s in the counterclockwise direction.

L₁ = 0

Explanation:

The total angular momentum of the system is conserved:

Lin = Lfin

If

Lin = Lwheel = - L₀ = - 1

Lwheel final = - L₀ = - 1

Lfin = LAlbert+turntable + Lwheel final = - L₀ + L₁ = - 1 + L₁ = ?

If Lin = Lfin ⇒

⇒ - 1 = - 1 + L₁ ⇒ L₁ = 0