A 1.0 m roulette wheel reaches a maximum angular speed of 18 rad/s before it begins decelerating. After reaching this maximum angular speed, it turns through 35 rev (220 rad) before it stops. How long did it take the wheel to stop after reaching its maximum angular speed?

Options

A. 12 s
B. 24 s
C. 48 s
D. 88 s

Answer

To solve the problem step by step, we will use the equations of motion for rotational dynamics.

### Given Data:
- Initial angular speed, \( \omega_0 = 18 \, \text{rad/s} \)
- Final angular speed, \( \omega = 0 \, \text{rad/s} \) (since the wheel stops)
- Angular displacement, \( \theta = 220 \, \text{rad} \)

### Step 1: Use the equation of motion for rotational dynamics
We will use the third equation of motion for rotational motion, which is:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
where:
- \( \omega \) is the final angular velocity,
- \( \omega_0 \) is the initial angular velocity,
- \( \alpha \) is the angular acceleration,
- \( \theta \) is the angular displacement.

### Step 2: Substitute the known values
Substituting the known values into the equation:
\[ 0 = (18)^2 + 2\alpha(220) \]
This simplifies to:
\[ 0 = 324 + 440\alpha \]

### Step 3: Solve for angular acceleration \( \alpha \)
Rearranging the equation gives:
\[ 440\alpha = -324 \]
\[ \alpha = -\frac{324}{440} = -0.738 \, \text{rad/s}^2 \]

### Step 4: Use the angular acceleration to find the time
Now we can use the angular acceleration to find the time taken to stop. We can use the equation:
\[ \omega = \omega_0 + \alpha t \]
Substituting the known values:
\[ 0 = 18 + (-0.738)t \]

### Step 5: Solve for time \( t \)
Rearranging gives:
\[ 0.738t = 18 \]
\[ t = \frac{18}{0.738} \approx 24.4 \, \text{seconds} \]

### Final Answer:
The time taken for the wheel to stop after reaching its maximum angular speed is approximately **24.4 seconds**.

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