the angular acceleration of flywheel having moment of inertia 50 kg `m^2` when a torque of 5 `N-m` is applied on the flywheel is

Options

A. `20.1 rad s^-2`
B. `250 rad s^-2`
C. `10 rad s^-2`
D. `1.25 rad s^-2 `

Answer

To find the angular acceleration of the flywheel, we can use the relationship between torque, moment of inertia, and angular acceleration. The formula we will use is:

\[ \tau = I \cdot \alpha \]

Where:
- \(\tau\) is the torque applied (in N·m),
- \(I\) is the moment of inertia (in kg·m²),
- \(\alpha\) is the angular acceleration (in rad/s²).

### Step-by-Step Solution:

1. **Identify the given values:**
- Moment of inertia, \(I = 50 \, \text{kg·m}^2\)
- Torque, \(\tau = 5 \, \text{N·m}\)

2. **Rearrange the formula to solve for angular acceleration (\(\alpha\)):**
\[ \alpha = \frac{\tau}{I} \]

3. **Substitute the known values into the equation:**
\[ \alpha = \frac{5 \, \text{N·m}}{50 \, \text{kg·m}^2} \]

4. **Perform the calculation:**
\[ \alpha = \frac{5}{50} = 0.1 \, \text{rad/s}^2 \]

5. **State the final answer:**
The angular acceleration of the flywheel is \(0.1 \, \text{rad/s}^2\).