Final answer:
The magnitude of the angular momentum at full capacity of the Singapore Flyer is 164.04 x 10^5 kg*m^2/s. The average net external torque applied to the spindle (axle) when the Singapore Flyer comes to a complete stop in 15 min is 172.0425 x 10^5 kg*m^2/s^2.
Step-by-step explanation:
To calculate the magnitude of the angular momentum at full capacity, we need to find the total moment of inertia of the Singapore Flyer. The moment of inertia, I, for a wheel with mass m and radius r is given by the equation:
I = 0.5 * m * r^2
Since the wheel consists of 28 capsules, each weighing 1 x 10^4 kg, the total mass of the capsules is m = 28 * 1 x 10^4 kg = 2.8 x 10^5 kg. The radius of the wheel, r, is half the diameter, so it is r = 150 m / 2 = 75 m. Plugging in these values into the moment of inertia equation, we get:
I = 0.5 * 2.8 x 10^5 kg * (75 m)^2 = 787.5 x 10^5 kg*m^2
The magnitude of the angular momentum, L, is given by the equation:
L = I * ω
where ω is the angular velocity. The entire Singapore Flyer rotates one complete revolution in 30 min, which is equal to 2π rad. So the angular velocity, ω, is:
ω = 2π rad / 30 min = 0.209 rad/s
Plugging in the values of I and ω, we get:
L = (787.5 x 10^5 kg*m^2) * (0.209 rad/s) = 164.04 x 10^5 kg*m^2/s
Therefore, the magnitude of the angular momentum at full capacity of the Singapore Flyer is 164.04 x 10^5 kg*m^2/s.
To calculate the average net external torque applied to the spindle (axle) when the Singapore Flyer comes to a complete stop in 15 min, we can use the equation:
τ = I * α
where α is the angular acceleration. The Singapore Flyer rotates one complete revolution in 30 min, which is equal to 2π rad. So the angular acceleration, α, is:
α = 2π rad / (30 min * 60 s/min) = 0.0219 rad/s^2
Plugging in the value of I and α, we get:
τ = (787.5 x 10^5 kg*m^2) * (0.0219 rad/s^2) = 172.0425 x 10^5 kg*m^2/s^2
Therefore, the average net external torque applied to the spindle when the Singapore Flyer comes to a complete stop in 15 min is 172.0425 x 10^5 kg*m^2/s^2.