A wheel having moment of interia `2 kgm^(-2)` about its axis, rotates at 50 rpm about this axis. The angular retardation that can stop the wheel in one minute is

Options

A. `(pi)/(36)"rad s"^(-2)`
B. `(pi)/(18)"rad s"^(-2)`
C. `(pi)/(72)"rad s"^(-2)`
D. `(pi)/(9)"rad s"^(-2)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Convert RPM to Radians per Second
The initial angular velocity (ω) is given in revolutions per minute (rpm). We need to convert this to radians per second.

1. **Given**:
- Initial angular velocity (ω) = 50 rpm
2. **Conversion**:
- 1 revolution = 2π radians
- 1 minute = 60 seconds
- Therefore, to convert rpm to rad/s:
\[ \omega = 50 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = \frac{100\pi}{60} = \frac{5\pi}{3} \, \text{rad/s} \]

### Step 2: Use the Angular Retardation Formula
We know that the final angular velocity (ω_f) will be 0 when the wheel stops. We can use the equation of motion for rotational motion to find the angular retardation (α).

1. **Given**:
- Final angular velocity (ω_f) = 0 rad/s
- Initial angular velocity (ω_i) = \( \frac{5\pi}{3} \, \text{rad/s} \)
- Time (t) = 1 minute = 60 seconds

2. **Using the equation**:
\[ \omega_f = \omega_i + \alpha t \]
Rearranging gives:
\[ \alpha = \frac{\omega_f - \omega_i}{t} \]

### Step 3: Substitute the Values
Now we can substitute the values into the equation to find α.

1. **Substituting**:
\[ \alpha = \frac{0 - \frac{5\pi}{3}}{60} \]
\[ \alpha = \frac{-\frac{5\pi}{3}}{60} = -\frac{5\pi}{1800} = -\frac{\pi}{360} \, \text{rad/s}^2 \]

### Step 4: Final Result
The angular retardation that can stop the wheel in one minute is:
\[ \alpha = \frac{\pi}{360} \, \text{rad/s}^2 \]

### Summary
The angular retardation required to stop the wheel in one minute is \( \frac{\pi}{360} \, \text{rad/s}^2 \).

---