A flywheel of moment of inertia 0.4 kg `m^(2)` and radius 0.2 m is free to rotate about a central axis. If a string is wrapped around it and it is pulled with a force of 10N. Then its angular velocity after 4s will be

Options

A. `10 rad s^(-1)`
B. `5rad s^(-1)`
C. `20 rad s^(-1)`
D. None of these

Answer

To solve the problem step by step, we will use the concepts of torque, moment of inertia, and angular acceleration.

### Step 1: Identify the given values
- Moment of inertia (I) = 0.4 kg·m²
- Radius (r) = 0.2 m
- Force (F) = 10 N
- Time (t) = 4 s

### Step 2: Calculate the torque (τ) applied to the flywheel
The torque exerted by the force on the flywheel can be calculated using the formula:
\[ \tau = F \cdot r \]
Substituting the values:
\[ \tau = 10 \, \text{N} \cdot 0.2 \, \text{m} = 2 \, \text{N·m} \]

### Step 3: Relate torque to angular acceleration (α)
The torque is also related to the moment of inertia and angular acceleration by the formula:
\[ \tau = I \cdot \alpha \]
We can rearrange this to find angular acceleration (α):
\[ \alpha = \frac{\tau}{I} \]
Substituting the values:
\[ \alpha = \frac{2 \, \text{N·m}}{0.4 \, \text{kg·m²}} = 5 \, \text{rad/s²} \]

### Step 4: Calculate the final angular velocity (ω) after 4 seconds
Since the flywheel starts from rest, the initial angular velocity (ω₁) is 0. We can use the formula for angular velocity:
\[ \omega_2 = \omega_1 + \alpha \cdot t \]
Substituting the values:
\[ \omega_2 = 0 + (5 \, \text{rad/s²}) \cdot (4 \, \text{s}) = 20 \, \text{rad/s} \]

### Final Answer
The angular velocity after 4 seconds is:
\[ \omega_2 = 20 \, \text{rad/s} \]

---