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

Options

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

Explanation

To solve the problem, we need to calculate the torque required to stop a wheel with a given moment of inertia in a specified time.

### Step-by-Step Solution:

1. **Identify the given data:**
- Moment of inertia (I) = 2 kg m²
- Rotational speed (N) = 60 rpm
- Time (t) to stop the wheel = 1 minute = 60 seconds

2. **Convert the rotational speed from rpm to radians per second:**
- The formula to convert rpm to radians per second is:
\[ \omega = N \times \frac{2\pi}{60} \]
- Substituting the value of N:
\[ \omega = 60 \times \frac{2\pi}{60} = 2\pi \text{ rad/s} \]

3. **Calculate the angular acceleration (α):**
- Angular acceleration can be calculated using the formula:
\[ \alpha = \frac{\Delta \omega}{\Delta t} \]
- Here, the change in angular velocity (Δω) is from 2π rad/s to 0 rad/s, so:
\[ \Delta \omega = 0 - 2\pi = -2\pi \text{ rad/s} \]
- The time interval (Δt) is 60 seconds, thus:
\[ \alpha = \frac{-2\pi}{60} = -\frac{\pi}{30} \text{ rad/s}^2 \]

4. **Use the torque formula:**
- The torque (τ) is related to moment of inertia and angular acceleration by:
\[ \tau = I \cdot \alpha \]
- Substituting the values:
\[ \tau = 2 \cdot \left(-\frac{\pi}{30}\right) = -\frac{2\pi}{30} = -\frac{\pi}{15} \text{ N m} \]
- The negative sign indicates that the torque is acting in the opposite direction of the rotation.

5. **Final answer:**
- The magnitude of the torque required to stop the wheel is:
\[ \tau = \frac{\pi}{15} \text{ N m} \]