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## 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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Physics
3 years
2021-07-27T11:34:53+00:00
2021-07-27T11:34:53+00:00 1 Answers
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## Answers ( )

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/secUsing 1st equation of motion for angular motion:

ωf = ωi + α twhere,

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 secα = – 1.49 x 10⁻³ rad/sec²Since,

a = rαwhere,

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