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

Options

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

Explanation

To solve the problem, we need to determine the torque required to stop a rotating wheel with a given moment of inertia in a specified time. Here's the step-by-step solution:

### Step 1: Convert the angular speed from RPM to radians per second.
The wheel rotates at a rate of 60 revolutions per minute (rpm). We need to convert this to radians per second.

\[ \text{Angular speed} (\omega) = 60 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = 2\pi \, \text{radians/second} \]

**Hint:** Remember that 1 revolution is equal to \(2\pi\) radians and there are 60 seconds in a minute.

### Step 2: Determine the time in seconds.
The time given for the wheel to stop is 1 minute, which we convert to seconds.

\[ t = 1 \, \text{minute} = 60 \, \text{seconds} \]

**Hint:** Always convert time to seconds when dealing with angular motion equations.

### Step 3: Use the angular motion equation to find angular acceleration (\(\alpha\)).
We know the initial angular speed (\(\omega_0\)) is \(2\pi\) radians/second, and the final angular speed (\(\omega\)) when the wheel stops is 0 radians/second. We can use the equation:

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

Substituting the known values:

\[ 0 = 2\pi + \alpha(60) \]

Rearranging gives:

\[ \alpha = \frac{-2\pi}{60} = -\frac{\pi}{30} \, \text{radians/second}^2 \]

**Hint:** The negative sign indicates that the wheel is decelerating.

### Step 4: Calculate the torque (\(T\)) using the formula \(T = I \alpha\).
The moment of inertia (\(I\)) is given as \(2 \, \text{kg m}^2\). Now we can substitute \(I\) and \(\alpha\) into the torque equation:

\[ T = I \alpha = 2 \, \text{kg m}^2 \times \left(-\frac{\pi}{30} \, \text{radians/second}^2\right) = -\frac{2\pi}{30} = -\frac{\pi}{15} \, \text{Nm} \]

**Hint:** The torque is negative because it acts in the opposite direction to the rotation.

### Step 5: Determine the magnitude of the torque.
Since we are interested in the magnitude of the torque, we take the absolute value:

\[ |T| = \frac{\pi}{15} \, \text{Nm} \]

### Conclusion
Thus, the torque required to stop the wheel's rotation in one minute is:

\[ \text{Torque} = \frac{\pi}{15} \, \text{Nm} \]

The correct answer is option 1: \(\frac{\pi}{15} \, \text{Nm}\).

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