A wheel having moment of inertia `2 kg m^(2)` about its vertical axis, rotates at the rate of `60 rom` about this axis. The torque which can stop the wheel's rotation in one minuted woould be

Options

A. `(-pi)/(15)N-m`
B. `(pi)/(18)N-m`
C. `(2pi)/(15)N-m`
D. `(pi)/(12)N-m`

Explanation

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

### Step 1: Understand the given data
- Moment of inertia (I) = 2 kg m²
- Initial angular velocity (ω_initial) = 60 rotations per minute (rpm)
- Time to stop (t) = 1 minute

### Step 2: Convert the angular velocity to radians per second
1. We know that 1 rotation = 2π radians.
2. To convert rpm to radians per second:
\[ \omega_{\text{initial}} = 60 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{rotation}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = 2\pi \, \text{radians/second} \]

### Step 3: Determine the final angular velocity
- Since we want to stop the wheel, the final angular velocity (ω_final) = 0 radians/second.

### Step 4: Calculate the angular deceleration (α)
- We can use the equation of motion for angular motion:
\[ \omega_{\text{final}} = \omega_{\text{initial}} + \alpha t \]
Rearranging gives:
\[ \alpha = \frac{\omega_{\text{final}} - \omega_{\text{initial}}}{t} \]
Substituting the values:
\[ \alpha = \frac{0 - 2\pi}{60 \, \text{seconds}} = -\frac{2\pi}{60} = -\frac{\pi}{30} \, \text{radians/second}^2 \]

### Step 5: Calculate the torque (τ)
- The torque is related to angular acceleration by the equation:
\[ \tau = I \alpha \]
Substituting the known values:
\[ \tau = 2 \, \text{kg m}^2 \times -\frac{\pi}{30} \, \text{radians/second}^2 = -\frac{2\pi}{30} = -\frac{\pi}{15} \, \text{N m} \]

### Step 6: Interpret the result
- The negative sign indicates that the torque is acting in the opposite direction to the rotation, which is expected since it is a retarding torque.

### Final Answer
The torque required to stop the wheel's rotation in one minute is:
\[ \tau = -\frac{\pi}{15} \, \text{N m} \]

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