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

To solve this problem, we will use the concepts of rotational motion to calculate the torque required to stop the rotating cylinder. The initial given data includes:

• Mass of the cylinder, m = 2 \, \text{kg}
• Radius of the cylinder, r = 4 \, \text{cm} = 0.04 \, \text{m}
• Initial angular velocity, \omega_i = 3 \, \text{rpm}
• Number of revolutions before stopping, \theta = 2\pi \, \text{revolutions}

Step 1: Convert Angular Velocity

The initial angular velocity in radians per second is obtained by converting the rotations per minute:

\omega_i = 3 \times 2\pi \times \frac{1}{60} \, \text{rad/s} = \frac{1}{10} \, \pi \, \text{rad/s}

Step 2: Calculate Moment of Inertia

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

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

Step 3: Use Angular Kinematic Equation

We need the final angular velocity \omega_f to be zero because the cylinder is to stop. Using the kinematic equation for angular motion:

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

Since \omega_f = 0, solve for angular acceleration \alpha:

0 = \left(\frac{1}{10} \pi\right)^2 + 2 \cdot \alpha \cdot (2\pi)

Solving for \alpha:

\alpha = -\frac{\left(\frac{1}{10} \pi\right)^2}{4\pi} = -\frac{\pi}{400} \, \text{rad/s}^2

Step 4: Calculate Torque

The torque \tau needed can be calculated using the relation:

\tau = I \cdot \alpha

Substitute I and \alpha:

\tau = 0.0016 \, \text{kg} \cdot \text{m}^2 \cdot \left(-\frac{\pi}{400} \, \text{rad/s}^2\right) = -0.00000126\pi \, \text{Nm}

Approximate the numerical value:

\tau \approx -2 \times 10^{-6} \, \text{Nm}

The negative sign indicates that the torque is in the direction opposite to the rotation. The magnitude of the torque required is 2 \times 10^{-6} \, \text{Nm}.

Therefore, the correct answer is: 2 \times 10^{-6} \, \text{Nm}