A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 10 rad/s over the next 4.0 s. Assume that the angular acceleration is constant.
What is the magnitude of the angular acceleration of the grindstone?

Options

A. `0.50rad//s^(2)`
B. `4.5rad//s^(2)`
C. `18rad//s^(2)`
D. `2.0rad//s^(2)`

Answer

To find the magnitude of the angular acceleration of the grindstone, we can use the formula that relates angular acceleration to the change in angular velocity over time:

\[ \alpha = \frac{\Delta \omega}{\Delta t} \]

Where:
- \(\alpha\) is the angular acceleration,
- \(\Delta \omega\) is the change in angular velocity,
- \(\Delta t\) is the change in time.

### Step-by-Step Solution:

1. **Identify the initial and final angular velocities**:
- Initial angular velocity (\(\omega_i\)) = 8.0 rad/s
- Final angular velocity (\(\omega_f\)) = 10.0 rad/s

2. **Calculate the change in angular velocity (\(\Delta \omega\))**:
\[ \Delta \omega = \omega_f - \omega_i = 10.0 \, \text{rad/s} - 8.0 \, \text{rad/s} = 2.0 \, \text{rad/s} \]

3. **Identify the time interval (\(\Delta t\))**:
- Time interval (\(\Delta t\)) = 4.0 s

4. **Substitute the values into the angular acceleration formula**:
\[ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{2.0 \, \text{rad/s}}{4.0 \, \text{s}} = 0.5 \, \text{rad/s}^2 \]

5. **Conclusion**:
The magnitude of the angular acceleration of the grindstone is \(0.5 \, \text{rad/s}^2\).