The WindSeeker ride at Canada's Wonderland lifts riders up to 300 feet into the air while spinning them around in a 15.9 m radius circle. The WindSeeker starts its circular motion from rest with a constant angular acceleration of 0.0187 rad/s2. It maintains that angular acceleration for 47.0 s, after which the angular speed remains constant for the rest of the ride - another 138 s. During the entire time on the WindSeeker, what angle does the rider pass through?
Question
The WindSeeker ride at Canada's Wonderland lifts riders up to 300 feet into the air while spinning them around in a 15.9 m radius circle. The WindSeeker starts its circular motion from rest with a constant angular acceleration of 0.0187 rad/s2. It maintains that angular acceleration for 47.0 s, after which the angular speed remains constant for the rest of the ride - another 138 s. During the entire time on the WindSeeker, what angle does the rider pass through?
Solution
To solve this problem, we need to find the total angle the rider passes through during both the acceleration and constant speed phases of the ride.
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First, let's find the final angular speed after the acceleration phase. We can use the formula for final angular speed (ω) which is ω = ω0 + αt, where ω0 is the initial angular speed, α is the angular acceleration, and t is the time. Since the ride starts from rest, ω0 = 0. So, ω = 0 + 0.0187 rad/s² * 47.0 s = 0.8799 rad/s.
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Next, let's find the angle passed through during the acceleration phase. We can use the formula for angular displacement (θ) which is θ = ω0t + 0.5αt². Again, since the ride starts from rest, ω0 = 0. So, θ = 0 + 0.5 * 0.0187 rad/s² * (47.0 s)² = 20.8 rad.
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Now, let's find the angle passed through during the constant speed phase. The formula for angular displacement is θ = ωt. So, θ = 0.8799 rad/s * 138 s = 121.4 rad.
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Finally, let's add the angles from the acceleration and constant speed phases to find the total angle passed through. So, the total angle is 20.8 rad + 121.4 rad = 142.2 rad.
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