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

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 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?

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Diễm Kiều 3 years 2021-07-27T11:34:53+00:00 1 Answers 114 views 0

Answers ( )

  1. Edana
    0
    2021-07-27T11:36:11+00:00
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    Answer:

    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

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