A rope is wound round a hollow cylinder of mass 5 kg and radius 0.5m. What is the angular acceleration of the cylinder if the rope is pulled with a force of 20 Ngt

Options

A. `"4 rad/s"^(2)`
B. `"5 rad/s"^(2)`
C. `"6 rad/s"^(2)`
D. `"8 rad/s"^(2)`

Explanation

To solve the problem of finding the angular acceleration of a 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) = 5 kg
- Radius of the hollow cylinder (r) = 0.5 m
- Force applied on the rope (F) = 20 N

### Step 2: Calculate the torque (τ) exerted by the force
The torque (τ) is given by the formula:
\[ \tau = F \times r \]
Substituting the values:
\[ \tau = 20 \, \text{N} \times 0.5 \, \text{m} = 10 \, \text{N m} \]

### Step 3: Determine the moment of inertia (I) of the hollow cylinder
For a hollow cylinder (or thin-walled cylinder), the moment of inertia (I) about its central axis is given by:
\[ I = m \times r^2 \]
Substituting the values:
\[ I = 5 \, \text{kg} \times (0.5 \, \text{m})^2 = 5 \, \text{kg} \times 0.25 \, \text{m}^2 = 1.25 \, \text{kg m}^2 \]

### Step 4: Relate torque to angular acceleration (α)
The relationship between torque and angular acceleration is given by:
\[ \tau = I \times \alpha \]
Rearranging this equation to solve for angular acceleration (α):
\[ \alpha = \frac{\tau}{I} \]

### Step 5: Substitute the values to find angular acceleration
Substituting the values of torque and moment of inertia:
\[ \alpha = \frac{10 \, \text{N m}}{1.25 \, \text{kg m}^2} = 8 \, \text{radians/s}^2 \]

### Final Answer
The angular acceleration of the hollow cylinder is:
\[ \alpha = 8 \, \text{radians/s}^2 \]

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