Final answer:
To calculate the angular acceleration of a Ferris wheel coming to a stop during one quarter of a revolution, we determine the initial angular velocity and use the kinematic equation that relates angular displacement, initial and final angular velocities, and angular acceleration.
Step-by-step explanation:
The question at hand involves calculating the angular acceleration of a Ferris wheel when it comes to a stop during one quarter of a revolution. Given that the Ferris wheel completes one revolution in 13.5 seconds and has a radius of 9.07 meters, we need to first determine its initial angular velocity. To calculate the angular velocity (ω), we use the formula ω = 2π / T, where T is the period of one revolution. In this case, T = 13.5 s, so ω = 2π / 13.5 s⁻¹.
Next, we note that the wheel comes to a stop during a quarter revolution, which is equivalent to an angular displacement (Δθ) of π/2 radians. The uniform deceleration over this displacement implies that the final angular velocity (ω_f) is 0 rad/s. Using the kinematic equation ω_f² = ω² + 2αΔθ, where α is the angular acceleration and ω_f is the final angular velocity, we can solve for α. Given that ω_f = 0, we have 0 = ω² + 2α(π/2). Rearranging and solving for α gives us the angular acceleration needed for the Ferris wheel to stop.