Final answer:
The minimum angular momentum of a roller coaster at the bottom of the loop to safely make it through is calculated based on the mass and radius of the loop, resulting in 1.8×10⁵ kg·m²/s.
Step-by-step explanation:
To calculate the minimum angular momentum of a roller coaster at the bottom of a loop to make it safely through, we must consider the forces acting on it at the top of the loop. For the roller coaster not to fall off the track at the top of the loop, the centripetal force must be at least equal to the force of gravity (weight of the coaster). The centripetal force (Fc) is given by the equation Fc = m * v2 / r, where m is the mass of the roller coaster, v is the velocity, and r is the radius of the loop.
At the top of the loop, the minimum velocity (vmin) can be found using vmin = sqrt(r * g), where g is the acceleration due to gravity. Using the given mass of the coaster (3000.0 kg) and the radius of the loop (50.0 m), we find that vmin = sqrt(50.0 m * 9.8 m/s2). Subsequently, the minimum angular momentum (Lmin) at the bottom of the loop can be calculated using Lmin = m * r * vmin.
By calculating, we get Lmin = 3000.0 kg * 50.0 m * sqrt(50.0 m * 9.8 m/s2), which is approximately equal to 1.8 x 105 kg·m2/s.
Therefore, the correct answer is: a) 1.8×105 kg·m2/s.