Ask Question
26 April, 07:09

The original Ferris wheel had a radius of 38 m and completed a full revolution (2π radians) every two minutes when operating at its maximum speed. If the wheel were uniformly slowed from its maximum speed to a stop in 35 seconds, what would be the magnitude of the tangential acceleration at the outer rim of the wheel during its deceleration?

+5
Answers (1)
  1. 26 April, 07:17
    0
    a = 0.0568 m/s²

    Explanation:

    First we find the initial angular velocity of the wheel:

    Initial Angular Velocity = ωi = (2π rad/2 min) (1 min/60 sec)

    ωi = 0.0523 rad/sec

    Using 1st equation of motion for angular motion:

    ωf = ωi + α t

    where,

    ωi = initial angular velocity = 0.0523 rad/sec

    ωf = final angular velocity = 0 rad/sec (Since, wheel finally stops)

    α = angular deceleration

    t = time to stop = 35 sec

    Therefore,

    0 rad/sec = 0.0523 rad/sec + α (35 sec)

    α = (-0.0523 rad/sec) / 35 sec

    α = - 1.49 x 10⁻³ rad/sec²

    Since,

    a = rα

    where,

    a = tangential deceleration

    r = radius of wheel = 38 m

    Therefore,

    a = (38 m) (1.49 x 10⁻³ rad/sec²)

    a = 0.0568 m/s²
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “The original Ferris wheel had a radius of 38 m and completed a full revolution (2π radians) every two minutes when operating at its maximum ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers