A wheel with a 0.10 m radius is rotating at 35 rev/s. It then slows uniformly to 15 rev/s over a 3.0 s interval. What is the angular acceleration of a point on the wheel?

Options

A. `-2.0rev//s^(2)`
B. `-6.7rev//s^(2)`
C. `-17rev//s^(2)`
D. `0.67rev//s^(2)`

Answer

To find the angular acceleration of the wheel, we can use the formula for angular acceleration, which is defined as the change in angular velocity over time. The formula is given by:

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

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

### Step 1: Identify the initial and final angular velocities
- The initial angular velocity (\(\omega_i\)) is given as 35 revolutions per second (rev/s).
- The final angular velocity (\(\omega_f\)) is given as 15 revolutions per second (rev/s).

### Step 2: Calculate the change in angular velocity
\[ \Delta \omega = \omega_f - \omega_i \]
Substituting the values:
\[ \Delta \omega = 15 \, \text{rev/s} - 35 \, \text{rev/s} = -20 \, \text{rev/s} \]

### Step 3: Identify the time interval
- The time interval (\(\Delta t\)) is given as 3.0 seconds.

### Step 4: Calculate the angular acceleration
Now, we can substitute the values into the angular acceleration formula:
\[ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{-20 \, \text{rev/s}}{3.0 \, \text{s}} \approx -6.67 \, \text{rev/s}^2 \]

### Final Answer
The angular acceleration of a point on the wheel is approximately \(-6.67 \, \text{rev/s}^2\).

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