A particle is moving in a circle of radius 1.5 m. Its speed is increasing by 180 rev/ minute in one mintue , what is its linear acceleration ?

Options

A. `0.25 m//s^(2)`
B. `0.30 m//s^(2)`
C. `0.39 m//s^(2)`
D. `0.47 m//s^(2)`

Explanation

To solve the problem, we need to find the linear acceleration of a particle moving in a circle with a given radius and an increase in speed. Here are the step-by-step calculations:

### Step 1: Understand the given data
- Radius of the circle (r) = 1.5 m
- Increase in speed = 180 revolutions per minute (rev/min)
- Time interval = 1 minute

### Step 2: Convert revolutions per minute to radians per second
1 revolution = \(2\pi\) radians.
To convert 180 revolutions per minute to radians per second:
\[ \text{Angular speed} (\omega) = 180 \text{ rev/min} \times \frac{2\pi \text{ radians}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ seconds}} \]
Calculating this gives:
\[ \omega = 180 \times \frac{2\pi}{60} = 6\pi \text{ radians/second} \]

### Step 3: Calculate angular acceleration
Since the speed is increasing by 180 rev/min in 1 minute, the angular acceleration (\(\alpha\)) can be calculated as:
\[ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{6\pi \text{ radians/second}}{60 \text{ seconds}} = \frac{\pi}{10} \text{ radians/second}^2 \]

### Step 4: Relate linear acceleration to angular acceleration
The linear acceleration (a) is related to angular acceleration (\(\alpha\)) by the formula:
\[ a = r \alpha \]
Substituting the values we have:
\[ a = 1.5 \text{ m} \times \frac{\pi}{10} \text{ radians/second}^2 \]

### Step 5: Calculate the linear acceleration
Now, substituting the value of \(\pi \approx 3.14\):
\[ a = 1.5 \times \frac{3.14}{10} = \frac{4.71}{10} = 0.471 \text{ m/s}^2 \]

### Final Answer
The linear acceleration of the particle is approximately \(0.471 \text{ m/s}^2\).

---