A rope is wound around a hollow cylinder of mass `3 kg` and radius `40 cm`. What is the angular acceleration of the cylinder if the rope is pulled with a force of `30 N` ?

Options

A. `25 m//s^(2)`
B. `0.25 rad//s^(2)`
C. `25 rad//s^(2)`
D. `5 m//s^(2)`

Answer

To find the angular acceleration of the hollow cylinder when a force is applied to the rope, 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 (\( \tau \)) applied on the cylinder
The torque is given by the formula:
\[ \tau = F \times r \]
Substituting the values:
\[ \tau = 30 \, \text{N} \times 0.4 \, \text{m} = 12 \, \text{N m} \]

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

### Step 4: Use the relationship between torque, moment of inertia, and angular acceleration
The relationship is given by:
\[ \tau = I \alpha \]
Where \( \alpha \) is the angular acceleration. Rearranging the formula to solve for \( \alpha \):
\[ \alpha = \frac{\tau}{I} \]
Substituting the values we calculated:
\[ \alpha = \frac{12 \, \text{N m}}{0.48 \, \text{kg m}^2} = 25 \, \text{rad/s}^2 \]

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

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