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?

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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