A wheel having moment of inertia 4 kg `m^(2)` about its axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is :-

Options

A. `(5pi)/(7)Nm`
B. `(8pi)/(15)Nm`
C. `(2pi)/(9)Nm`
D. `(3pi)/(7)Nm`

Answer

To solve the problem of finding the torque required to stop the rotation of a wheel in one minute, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Information:**
- Moment of Inertia (I) = 4 kg·m²
- Initial angular velocity (ω_initial) = 240 revolutions per minute (rpm)
- Time (t) = 1 minute = 60 seconds

2. **Convert Angular Velocity to Radians per Second:**
- We know that 1 revolution = 2π radians.
- Therefore, to convert rpm to radians per second:
\[ \omega_{\text{initial}} = 240 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = 240 \times \frac{2\pi}{60} = 8\pi \, \text{rad/s} \]

3. **Use the Angular Deceleration Formula:**
- The final angular velocity (ω_final) when the wheel stops is 0 rad/s.
- We can use the equation:
\[ \omega_{\text{final}} = \omega_{\text{initial}} - \alpha t \]
- Rearranging gives:
\[ \alpha = \frac{\omega_{\text{initial}} - \omega_{\text{final}}}{t} \]
- Substituting the values:
\[ \alpha = \frac{8\pi - 0}{60} = \frac{8\pi}{60} = \frac{2\pi}{15} \, \text{rad/s}^2 \]

4. **Calculate the Torque Required to Stop the Wheel:**
- The torque (τ) is given by the formula:
\[ \tau = I \cdot \alpha \]
- Substituting the known values:
\[ \tau = 4 \, \text{kg·m}^2 \cdot \frac{2\pi}{15} \, \text{rad/s}^2 = \frac{8\pi}{15} \, \text{N·m} \]

5. **Final Answer:**
- The torque required to stop the rotation of the wheel in one minute is:
\[ \tau = \frac{8\pi}{15} \, \text{N·m} \]