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

Options

A. `0.0015m//s^(2)`
B. `0.54m//s^(2)`
C. `0.057m//s^(2)`
D. 1.6 m/s

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the maximum angular speed (ω₀)
The Ferris wheel completes a full revolution (2π radians) every 2 minutes. We first convert the time into seconds:

\[ \text{Time} = 2 \text{ minutes} = 2 \times 60 = 120 \text{ seconds} \]

Now we can calculate the maximum angular speed (ω₀):

\[ \omega_0 = \frac{2\pi \text{ radians}}{120 \text{ seconds}} = \frac{\pi}{60} \text{ radians/second} \]

### Step 2: Determine the final angular speed (ω)
Since the Ferris wheel comes to a stop, the final angular speed (ω) is:

\[ \omega = 0 \text{ radians/second} \]

### Step 3: Calculate the angular deceleration (α)
The wheel is uniformly slowed to a stop in 35 seconds. We can use the equation of motion for angular motion:

\[ \omega = \omega_0 + \alpha t \]

Substituting the known values:

\[ 0 = \frac{\pi}{60} + \alpha \times 35 \]

Now, solving for α:

\[ \alpha = -\frac{\frac{\pi}{60}}{35} = -\frac{\pi}{2100} \text{ radians/second}^2 \]

### Step 4: Calculate the tangential acceleration (a_t)
The tangential acceleration (a_t) at the outer rim of the wheel can be calculated using the formula:

\[ a_t = r \cdot \alpha \]

Where:
- r = radius of the wheel = 38 m
- α = angular acceleration = -\frac{\pi}{2100} radians/second²

Substituting the values:

\[ a_t = 38 \cdot \left(-\frac{\pi}{2100}\right) \]

Calculating the magnitude:

\[ |a_t| = 38 \cdot \frac{\pi}{2100} \approx 0.0568 \text{ m/s}² \]

### Final Answer
The magnitude of the tangential acceleration at the outer rim of the wheel during its deceleration is approximately **0.0568 m/s²**.

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