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.

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step-by-step explanation:

a and b sometimes have a positive value