A particle travels in a circle of radius 20 cm at a speed thast uniformly increases. If the speed changes from 5.0 m/s to 6.0 m/s in 2.0s, find the angular aceleration.

Options

A. `12.5rads^2`
B. `3.5rads^2`
C. `2.5rads^2`
D. `4.5rads^2`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Initial speed \( v_i = 5.0 \, \text{m/s} \)
- Final speed \( v_f = 6.0 \, \text{m/s} \)
- Time interval \( t = 2.0 \, \text{s} \)
- Radius of the circle \( r = 20 \, \text{cm} = 0.20 \, \text{m} \)

### Step 2: Calculate the change in velocity
The change in velocity \( \Delta v \) can be calculated as:
\[ \Delta v = v_f - v_i = 6.0 \, \text{m/s} - 5.0 \, \text{m/s} = 1.0 \, \text{m/s} \]

### Step 3: Calculate the tangential acceleration
Tangential acceleration \( a_t \) is defined as the change in velocity over time:
\[ a_t = \frac{\Delta v}{t} = \frac{1.0 \, \text{m/s}}{2.0 \, \text{s}} = 0.5 \, \text{m/s}^2 \]

### Step 4: Relate tangential acceleration to angular acceleration
Angular acceleration \( \alpha \) is related to tangential acceleration \( a_t \) by the formula:
\[ \alpha = \frac{a_t}{r} \]
Substituting the values:
\[ \alpha = \frac{0.5 \, \text{m/s}^2}{0.20 \, \text{m}} = 2.5 \, \text{rad/s}^2 \]

### Step 5: Final answer
The angular acceleration of the particle is:
\[ \alpha = 2.5 \, \text{rad/s}^2 \]

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