A wheel of radius 0.5 m rotates with a constant angular speed about an axis perpendicular to its center. A point on the wheel that is 0.2 m from the center has a tangential speed of 2 m/s.
Determine the tangential acceleration of the point that is 0.2 m from the center.

Options

A. `0.4m//s^(2)`
B. `4.0m//s^(2)`
C. `"zero "m//s^(2)`
D. `2.0m//s^(2)`

Answer

To determine the tangential acceleration of a point on a wheel that is rotating with a constant angular speed, we can follow these steps:

### Step 1: Understand the Definitions
- **Tangential Speed (v)**: The speed of a point on the edge of the wheel as it moves along its circular path.
- **Tangential Acceleration (a_t)**: The rate of change of tangential speed. It is given by the formula:
\[ a_t = R \cdot \alpha \]
where \( R \) is the radius from the center to the point in question, and \( \alpha \) is the angular acceleration.

### Step 2: Identify Given Values
- The radius of the wheel \( R = 0.5 \, \text{m} \) (but we are interested in the point that is \( 0.2 \, \text{m} \) from the center).
- The distance from the center to the point \( r = 0.2 \, \text{m} \).
- The tangential speed \( v = 2 \, \text{m/s} \).
- The wheel rotates with a **constant angular speed**, which implies that the angular acceleration \( \alpha = 0 \).

### Step 3: Calculate Tangential Acceleration
Since the wheel is rotating with a constant angular speed, the angular acceleration \( \alpha \) is zero. Therefore, we can substitute \( \alpha = 0 \) into the formula for tangential acceleration:
\[ a_t = R \cdot \alpha = 0.2 \, \text{m} \cdot 0 = 0 \, \text{m/s}^2 \]

### Conclusion
The tangential acceleration of the point that is \( 0.2 \, \text{m} \) from the center of the wheel is:
\[ \text{Tangential Acceleration} = 0 \, \text{m/s}^2 \]