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, we need to calculate the torque required to stop a solid cylinder that is rotating. Here’s a step-by-step solution:

### Step 1: Convert the rotation speed from rpm to rad/s
The cylinder is rotating at 3 revolutions per minute (rpm). We need to convert this to radians per second.

\[ \text{Angular velocity} \, (\omega) = 3 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} \]

Calculating this gives:

\[ \omega = 3 \times \frac{2\pi}{60} = \frac{6\pi}{60} = \frac{\pi}{10} \, \text{rad/s} \]

### Step 2: Calculate the total angle in radians for 2π revolutions
The cylinder needs to stop after 2π revolutions. We convert this to radians:

\[ \theta = 2\pi \, \text{revolutions} \times 2\pi \, \text{radians/revolution} = 4\pi^2 \, \text{radians} \]

### Step 3: Use the angular kinematic equation to find angular acceleration (α)
We can use the equation:

\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]

Where:
- \(\omega = 0\) (final angular velocity since the cylinder stops)
- \(\omega_0 = \frac{\pi}{10}\) (initial angular velocity)
- \(\theta = 4\pi^2\) (total angle in radians)

Substituting the values:

\[ 0 = \left(\frac{\pi}{10}\right)^2 + 2\alpha(4\pi^2) \]

This simplifies to:

\[ 0 = \frac{\pi^2}{100} + 8\pi^2\alpha \]

Rearranging gives:

\[ 8\pi^2\alpha = -\frac{\pi^2}{100} \]

Dividing both sides by \(8\pi^2\):

\[ \alpha = -\frac{1}{800} \, \text{rad/s}^2 \]

### Step 4: Calculate the moment of inertia (I) of the solid cylinder
The moment of inertia of a solid cylinder 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 (τ)
Torque is given by the formula:

\[ \tau = I \alpha \]

Substituting the values we have:

\[ \tau = 0.0016 \times \left(-\frac{1}{800}\right) \]

Calculating this gives:

\[ \tau = -\frac{0.0016}{800} = -2 \times 10^{-6} \, \text{N m} \]

Since we are interested in the magnitude of torque, we can state:

\[ \tau = 2 \times 10^{-6} \, \text{N m} \]

### Final Answer
The torque required to stop the cylinder after \(2\pi\) revolutions is:

\[ \boxed{2 \times 10^{-6} \, \text{N m}} \]