A rope 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`, what is the angualr acceleration of the cylinder ?

Options

A. `15 rad s^(-2)`
B. `20 rad s^(-2)`
C. `25 rad s^(-2)`
D. `30 rad s^(-2)`

Explanation

To find the angular acceleration of the hollow cylinder when a rope is pulled with a force, we can follow these steps:

### Step 1: Identify the given values
- Mass of the hollow cylinder, \( m = 3 \, \text{kg} \)
- Radius of the hollow cylinder, \( r = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Force applied on the rope, \( F = 30 \, \text{N} \)

### Step 2: Calculate the torque (\( \tau \))
The torque (\( \tau \)) exerted on the cylinder due to the force applied on the rope can be calculated using the formula:
\[ \tau = r \times F \]
Here, since the force is applied perpendicular to the radius, we can simplify this to:
\[ \tau = r \cdot F \]
Substituting the known values:
\[ \tau = 0.4 \, \text{m} \times 30 \, \text{N} = 12 \, \text{N m} \]

### Step 3: Determine the moment of inertia (\( I \))
The moment of inertia (\( I \)) for a hollow cylinder is given by the formula:
\[ I = m r^2 \]
Substituting the values:
\[ I = 3 \, \text{kg} \times (0.4 \, \text{m})^2 = 3 \, \text{kg} \times 0.16 \, \text{m}^2 = 0.48 \, \text{kg m}^2 \]

### Step 4: Relate torque to angular acceleration (\( \alpha \))
According to the rotational dynamics, the relationship between torque and angular acceleration is given by:
\[ \tau = I \alpha \]
From this, we can solve for angular acceleration (\( \alpha \)):
\[ \alpha = \frac{\tau}{I} \]
Substituting the values we found:
\[ \alpha = \frac{12 \, \text{N m}}{0.48 \, \text{kg m}^2} = 25 \, \text{rad/s}^2 \]

### Final Answer
The angular acceleration of the cylinder is:
\[ \alpha = 25 \, \text{rad/s}^2 \]

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