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

Answer

To solve the problem of finding the angular acceleration of a hollow cylinder when a rope is pulled with a force, we will follow these steps:

### Step 1: Understand the Given Data
- Mass of the hollow cylinder (m) = 3 kg
- Radius of the hollow cylinder (r) = 40 cm = 0.4 m
- Force applied on the rope (F) = 30 N

### Step 2: Calculate the Torque (τ)
The torque (τ) exerted on the cylinder by the force applied on the rope can be calculated using the formula:
\[ \tau = F \times r \]
Substituting the values:
\[ \tau = 30 \, \text{N} \times 0.4 \, \text{m} = 12 \, \text{N m} \]

### Step 3: Calculate the Moment of Inertia (I)
For a hollow cylinder, the moment of inertia (I) about its central axis 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: Calculate the Angular Acceleration (α)
The angular acceleration (α) can be calculated using Newton's second law for rotation:
\[ \tau = I \times \alpha \]
Rearranging the formula to find α gives:
\[ \alpha = \frac{\tau}{I} \]
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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