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.
Through how many revolutions does the grindstone turn during the 4.0 s interval?

Options

A. 0.64
B. `4.0`
C. 3.8
D. 5.7

Answer

To solve the problem step by step, we need to determine how many revolutions the grindstone turns during the 4.0 seconds interval given its initial and final angular speeds and the time duration.

### Step 1: Identify the given values
- Initial angular speed (\( \omega_0 \)) = 8.0 rad/s
- Final angular speed (\( \omega_f \)) = 10.0 rad/s
- Time interval (\( t \)) = 4.0 s

### Step 2: Calculate the angular acceleration (\( \alpha \))
We can use the formula for angular acceleration:
\[ \alpha = \frac{\omega_f - \omega_0}{t} \]
Substituting the known values:
\[ \alpha = \frac{10.0 \, \text{rad/s} - 8.0 \, \text{rad/s}}{4.0 \, \text{s}} = \frac{2.0 \, \text{rad/s}}{4.0 \, \text{s}} = 0.5 \, \text{rad/s}^2 \]

### Step 3: Calculate the angular displacement (\( \theta \))
We can use the angular displacement formula:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Substituting the known values:
\[ \theta = (8.0 \, \text{rad/s})(4.0 \, \text{s}) + \frac{1}{2}(0.5 \, \text{rad/s}^2)(4.0 \, \text{s})^2 \]
Calculating each term:
\[ \theta = 32.0 \, \text{rad} + \frac{1}{2}(0.5)(16) = 32.0 \, \text{rad} + 4.0 \, \text{rad} = 36.0 \, \text{rad} \]

### Step 4: Convert angular displacement from radians to revolutions
To find the number of revolutions, we divide the angular displacement by \( 2\pi \):
\[ \text{Number of revolutions} = \frac{\theta}{2\pi} = \frac{36.0 \, \text{rad}}{2\pi} \]
Calculating this:
\[ \text{Number of revolutions} = \frac{36.0}{6.2832} \approx 5.73 \]

### Final Answer
The grindstone turns approximately **5.73 revolutions** during the 4.0 seconds interval.