A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after `2pi` revolutions is :

Options

A. `2xx10^(6)Nm`
B. `2xx10^(-6)Nm`
C. `2xx10^(-3)Nm`
D. `12xx10^(-4)Nm`

Explanation

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

### Step 1: Convert the initial angular velocity from rpm to rad/s
Given:
- Initial speed = 3 rpm

To convert this to radians per second:
\[ \omega_0 = 3 \text{ rpm} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{3 \times 2\pi}{60} = \frac{6\pi}{60} = \frac{\pi}{10} \text{ rad/s} \]

### Step 2: Determine the angular displacement in radians
We need to stop the cylinder after \(2\pi\) revolutions. The angular displacement (\(\theta\)) in radians is:
\[ \theta = 2\pi \text{ rev} \times 2\pi \text{ rad/rev} = 4\pi^2 \text{ rad} \]

### Step 3: Use the equation of motion for angular motion
We will use the equation:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Where:
- \(\omega\) (final angular velocity) = 0 (since we want to stop the cylinder)
- \(\omega_0 = \frac{\pi}{10} \text{ rad/s}\)
- \(\theta = 4\pi^2 \text{ rad}\)

Substituting the values:
\[ 0 = \left(\frac{\pi}{10}\right)^2 + 2\alpha(4\pi^2) \]
\[ 0 = \frac{\pi^2}{100} + 8\pi^2\alpha \]
Rearranging gives:
\[ 8\pi^2\alpha = -\frac{\pi^2}{100} \]
\[ \alpha = -\frac{1}{800} \text{ rad/s}^2 \]

### Step 4: Calculate the moment of inertia (I) of the solid cylinder
The moment of inertia of a solid cylinder rotating about its axis is given by:
\[ I = \frac{1}{2} m r^2 \]
Where:
- \(m = 2 \text{ kg}\)
- \(r = 4 \text{ cm} = 0.04 \text{ m}\)

Calculating \(I\):
\[ I = \frac{1}{2} \times 2 \times (0.04)^2 = \frac{1}{2} \times 2 \times 0.0016 = 0.0016 \text{ kg m}^2 \]

### Step 5: Calculate the torque (\(\tau\))
Using the formula for torque:
\[ \tau = I \alpha \]
Substituting the values:
\[ \tau = 0.0016 \times \left(-\frac{1}{800}\right) = -\frac{0.0016}{800} = -2 \times 10^{-6} \text{ N m} \]

The negative sign indicates that the torque is applied in the opposite direction of the rotation.

### Final Answer
The torque required to stop the cylinder after \(2\pi\) revolutions is:
\[ \tau = 2 \times 10^{-6} \text{ N m} \]

---