Question

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

• A. \(2\times 10{-6}Nm \)
• B. \(2\times 10{-3}Nm \)
• C. \(12\times 10{-4}Nm \)
• D. \(2\times 10{6}Nm \)

Answer 1

The Correct Option is A

The problem involves calculating the torque required to stop a rotating solid cylinder. We'll start by understanding the physical principles and then go step-by-step through solving the problem.

Step 1: Calculate the moment of inertia (\(I\)) of the cylinder.

The moment of inertia for a solid cylinder rotating about its axis is given by the formula:

\( I = \frac{1}{2} m r^2 \)

where:

• \( m = 2 \, \text{kg} \) (mass of the cylinder)
• \( r = 0.04 \, \text{m} \) (radius of the cylinder converted to meters)

Substitute the values:

\( I = \frac{1}{2} \times 2 \times (0.04)^2 = \frac{1}{2} \times 2 \times 0.0016 = 0.0016 \, \text{kg} \, \text{m}^2 \)

Step 2: Calculate the initial angular velocity (\( \omega_i \)).

The initial angular velocity can be found from the given rotations per minute (rpm):

\( \omega_i = 2\pi \times \frac{\text{rpm}}{60} = 2\pi \times \frac{3}{60} = \frac{\pi}{10} \, \text{rad/s} \)

Step 3: Determine the angular displacement (\( \theta \)).

The problem states the cylinder makes \(2\pi\) revolutions before stopping:

\( \theta = 2\pi \times 2\pi = 4\pi^2 \, \text{rad} \)

Step 4: Use the equation of rotational motion.

We use the rotational motion equation to relate initial angular velocity, angular displacement, angular acceleration (\( \alpha \)), and final angular velocity:

\( \omega_f^2 = \omega_i^2 + 2\alpha\theta \)

Since the cylinder comes to a stop, \( \omega_f = 0 \):

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

Solve for \( \alpha \):

\( \alpha = -\frac{\left(\frac{\pi}{10}\right)^2}{2 \times 4\pi^2} = -\frac{\pi^2}{100 \times 8\pi^2} = -\frac{1}{800} \, \text{rad/s}^2 \)

Step 5: Calculate the torque (\( \tau \)).

The torque required to produce this angular acceleration is given by:

\( \tau = I \alpha \)

Substitute the known values:

\( \tau = 0.0016 \times -\frac{1}{800} = -2 \times 10^{-6} \, \text{Nm} \)

Since torque is a vector, we take the magnitude for the answer: \( \tau = 2 \times 10^{-6} \text{Nm} \).

Conclusion: The torque required to stop the rotating cylinder after \(2\pi\) revolutions is \(2 \times 10^{-6} \, \text{Nm}\).