Question

A rope of negligible mass is wound round a hollow cylinder of mass $3 \,kg$ and radius $40 \,cm$. If the rope is pulled with a force of $30 \,N$, then the angular acceleration produced in the cylinder is

• A. $15\,rad\, s^{-2}$
• B. $20\,rad\, s^{-2}$
• C. $25\,rad\, s^{-2}$
• D. $30\,rad\, s^{-2}$

Answer 1

The Correct Option is C

To determine the angular acceleration produced in the hollow cylinder when the rope is pulled with a force, we begin by analyzing the situation using the following physics principles:

1. **Understand the Concept**: The problem involves a force applied to a cylindrical object causing it to rotate. The angular acceleration can be calculated using the relation between torque, moment of inertia, and angular acceleration.

2. **Identify Given Values**:

• Mass of the cylinder, m = 3 \, \text{kg}
• Radius of the cylinder, r = 40 \, \text{cm} = 0.4 \, \text{m}
• Force applied, F = 30 \, \text{N}

3. **Formula for Torque**: Torque (\tau) can be computed using the formula:

\tau = F \cdot r

Substituting the given values:

\tau = 30 \, \text{N} \times 0.4 \, \text{m} = 12 \, \text{Nm}

4. **Moment of Inertia for a Hollow Cylinder**: The moment of inertia (I) for a hollow cylinder about its central axis is given by:

I = m \cdot r^2

Substituting the known values:

I = 3 \, \text{kg} \times (0.4 \, \text{m})^2 = 0.48 \, \text{kg} \, \text{m}^2

5. **Calculating Angular Acceleration**: Using the relation between torque and angular acceleration (\alpha):

\tau = I \cdot \alpha

Substituting for torque and moment of inertia:

12 \, \text{Nm} = 0.48 \, \text{kg} \, \text{m}^2 \times \alpha

Solving for \alpha:

\alpha = \frac{12 \, \text{Nm}}{0.48 \, \text{kg} \, \text{m}^2} = 25 \, \text{rad} \, \text{s}^{-2}

Hence, the angular acceleration produced in the cylinder is 25 \, \text{rad} \, \text{s}^{-2}. Therefore, the correct answer is the option:

• 25 \, \text{rad} \, \text{s}^{-2}