Final Answer:
The angular momentum of the spinning wheel in the air the moment the bike leaves the ground is (b) 4.9 kg⋅m²/s. This choice is determined by the conservation of angular momentum, where the initial angular momentum of the system is equal to the final angular momentum when no external torques act. Therefore, the correct answer is (b) 4.9 kg⋅m²/s.
Step-by-step explanation:
The angular momentum (L) of a rotating object is given by the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. In the scenario of a bike leaving the ground, the moment of inertia remains constant as no external torques are acting on the system. Initially, the wheel's angular velocity (ω) is non-zero, and as the bike takes off, it remains the same due to the absence of external torques.
The conservation of angular momentum is expressed as I₁ω₁ = I₂ω₂, where the subscripts 1 and 2 represent initial and final states. Since ω₂ (final angular velocity) is the same as ω₁ (initial angular velocity), the moment of inertia I₂ must be equal to I₁. The moment of inertia for a spinning wheel is given by I = 0.5 * m * r², where m is the mass of the wheel and r is its radius. Assuming no change in the wheel's configuration during takeoff, I₂ = I₁.
Now, considering the given choices, the angular momentum L is proportional to I, and the correct answer corresponds to the correct moment of inertia. Choice (b) 4.9 kg⋅m²/s aligns with the conservation of angular momentum principle, making it the accurate answer for the angular momentum of the spinning wheel the moment the bike leaves the ground.
Therefore, the correct answer is (b) 4.9 kg⋅m²/s.