A wheel is rotating at the rate of `33 "rev min"^(-1)`. If it comes to stop in 20 s. Then, the angular retardation will be

Options

A. `pi rad s^(-2)`
B. `11 pi rads^(-2)`
C. `(pi)/(200) rads^(-2)`
D. `(11 pi)/(200) rads^(-2)`

Explanation

To find the angular retardation of the wheel, we can follow these steps:

### Step 1: Convert the initial angular velocity from revolutions per minute to radians per second.
The initial angular velocity (ω₀) is given as 33 revolutions per minute (rev/min). We can convert this to radians per second using the conversion factor \(2\pi\) radians per revolution and the fact that there are 60 seconds in a minute.

\[ \omega_0 = 33 \, \text{rev/min} \times \frac{2\pi \, \text{radians}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{33 \times 2\pi}{60} \, \text{radians/s} \]

### Step 2: Set up the equation for angular motion.
Since the wheel comes to a stop, the final angular velocity (ω) is 0 rad/s. We can use the equation of motion for angular velocity:

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

Where:
- ω = final angular velocity = 0 rad/s
- ω₀ = initial angular velocity (calculated in Step 1)
- α = angular retardation (what we want to find)
- t = time = 20 s

### Step 3: Rearrange the equation to solve for angular retardation (α).
Substituting the values into the equation:

\[ 0 = \frac{33 \times 2\pi}{60} - \alpha \times 20 \]

Rearranging gives:

\[ \alpha \times 20 = \frac{33 \times 2\pi}{60} \]

\[ \alpha = \frac{33 \times 2\pi}{60 \times 20} \]

### Step 4: Simplify the expression for α.
Now we simplify the expression:

\[ \alpha = \frac{66\pi}{1200} \]

Dividing both the numerator and denominator by 6:

\[ \alpha = \frac{11\pi}{200} \, \text{radians/s}^2 \]

### Conclusion
Thus, the angular retardation of the wheel is:

\[ \alpha = \frac{11\pi}{200} \, \text{radians/s}^2 \]

### Final Answer
The correct option is 4: \( \frac{11\pi}{200} \, \text{radians/s}^2 \).

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