A particle is moving along a circle and its linear velocity is 0.5ms when its angular velocity is `2.5rads^(-1)`. Its tangential acceleration is `0.2ms^(-2)` then its angular acceleration is

Options

A. `1rads^(-2)`
B. `2 rads^(-2)`
C. `2.5 rads^(-2)`
D. `3 rads^(-2)`

Answer

To find the angular acceleration of a particle moving in a circular path, we can follow these steps:

### Step 1: Identify the given values
- Linear velocity (v) = 0.5 m/s
- Angular velocity (ω) = 2.5 rad/s
- Tangential acceleration (a_t) = 0.2 m/s²

### Step 2: Relate linear velocity to angular velocity
The relationship between linear velocity (v) and angular velocity (ω) is given by the formula:
\[ v = r \cdot \omega \]
where r is the radius of the circular path.

### Step 3: Calculate the radius (r)
We can rearrange the formula to find the radius:
\[ r = \frac{v}{\omega} \]
Substituting the values we have:
\[ r = \frac{0.5 \, \text{m/s}}{2.5 \, \text{rad/s}} = 0.2 \, \text{m} \]

### Step 4: Relate tangential acceleration to angular acceleration
The tangential acceleration (a_t) is related to angular acceleration (α) by the formula:
\[ a_t = \alpha \cdot r \]
We can rearrange this to find angular acceleration:
\[ \alpha = \frac{a_t}{r} \]

### Step 5: Substitute the values to find angular acceleration (α)
Now we substitute the values of tangential acceleration and radius into the equation:
\[ \alpha = \frac{0.2 \, \text{m/s}^2}{0.2 \, \text{m}} = 1 \, \text{rad/s}^2 \]

### Conclusion
The angular acceleration (α) of the particle is:
\[ \alpha = 1 \, \text{rad/s}^2 \]

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