A particle travels in a circle of radius 20 cm at a uniformly increasing speed. If the speed changes from `8 ms^(-1)` to `9 ms^(-1)` in 2s, what would be the angular acceleration in `"rad s"^(-2)` ?

Options

A. `1.5 "rad s"^(-2)`
B. `2.5 "rad s"^(-2)`
C. `3.5 "rad s"^(-2)`
D. `4.5 "rad s"^(-2)`

Explanation

To find the angular acceleration of a particle moving in a circle with a changing speed, we can follow these steps:

### Step 1: Identify the given values
- Initial speed (u) = 8 m/s
- Final speed (v) = 9 m/s
- Time interval (t) = 2 s
- Radius of the circle (r) = 20 cm = 0.2 m

### Step 2: Calculate the linear acceleration (a)
Linear acceleration can be calculated using the formula:
\[ a = \frac{\Delta v}{t} \]
where \(\Delta v = v - u\).

Substituting the values:
\[ \Delta v = 9 \, \text{m/s} - 8 \, \text{m/s} = 1 \, \text{m/s} \]
Now, substituting into the acceleration formula:
\[ a = \frac{1 \, \text{m/s}}{2 \, \text{s}} = 0.5 \, \text{m/s}^2 \]

### Step 3: Relate linear acceleration to angular acceleration
The relationship between linear acceleration (a) and angular acceleration (\(\alpha\)) is given by:
\[ \alpha = \frac{a}{r} \]
where \(r\) is the radius of the circle.

### Step 4: Substitute the values to find angular acceleration
Substituting the values we have:
\[ \alpha = \frac{0.5 \, \text{m/s}^2}{0.2 \, \text{m}} = 2.5 \, \text{rad/s}^2 \]

### Final Answer
The angular acceleration is:
\[ \alpha = 2.5 \, \text{rad/s}^2 \]

---