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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/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