A circular disk of radius 0.010 m rotates with a constant angular speed of 5.0 rev/s. What is the acceleration of a point on the edge of the disk?

Options

A. `0.31m//s^(2)`
B. `9.9m//s^(2)`
C. zero `m//s^(2)`
D. `1.6m//s^(2)`

Answer

To solve the problem, we need to find the acceleration of a point on the edge of a circular disk that rotates with a constant angular speed. The acceleration of a point on the edge of the disk consists of two components: tangential acceleration and radial (centripetal) acceleration. Let's go through the steps to find the total acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the disk, \( r = 0.010 \, \text{m} \)
- Angular speed, \( \omega = 5.0 \, \text{rev/s} \)

2. **Convert Angular Speed to Radians per Second:**
- Since \( 1 \, \text{rev} = 2\pi \, \text{radians} \),
\[ \omega = 5.0 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 10\pi \, \text{rad/s} \]

3. **Calculate Tangential Acceleration:**
- Tangential acceleration \( a_t \) is given by the formula:
\[ a_t = r \alpha \]
- Here, \( \alpha \) (angular acceleration) is zero because the disk rotates with a constant angular speed.
\[ a_t = 0.010 \, \text{m} \times 0 = 0 \, \text{m/s}^2 \]

4. **Calculate Radial (Centripetal) Acceleration:**
- Radial acceleration \( a_r \) is given by the formula:
\[ a_r = \omega^2 r \]
- Substitute the values:
\[ a_r = (10\pi)^2 \times 0.010 \]
\[ a_r = 100\pi^2 \times 0.010 = 1.0\pi^2 \, \text{m/s}^2 \]
- Calculate \( \pi^2 \):
\[ \pi^2 \approx 9.87 \Rightarrow a_r \approx 1.0 \times 9.87 \approx 9.87 \, \text{m/s}^2 \]

5. **Determine Total Acceleration:**
- The total acceleration \( a \) is the vector sum of tangential and radial accelerations. Since \( a_t = 0 \):
\[ a = \sqrt{a_t^2 + a_r^2} = \sqrt{0^2 + (9.87)^2} = 9.87 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of a point on the edge of the disk is approximately \( 9.87 \, \text{m/s}^2 \).