An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100s.What is the magnitude of the centripetal acceleration.

Answer

To find the magnitude of the centripetal acceleration of the insect moving in a circular groove, we can follow these steps:

### Step 1: Identify the given values
- Radius of the circular groove, \( r = 12 \, \text{cm} = 0.12 \, \text{m} \) (convert to meters)
- Number of revolutions completed, \( n = 7 \)
- Time taken for these revolutions, \( t = 100 \, \text{s} \)

### Step 2: Calculate the angular displacement
The angular displacement in radians for \( n \) revolutions is given by:
\[ \theta = n \times 2\pi = 7 \times 2\pi \, \text{radians} \]

### Step 3: Calculate the angular speed (\( \omega \))
The angular speed is defined as the angular displacement divided by the time taken:
\[ \omega = \frac{\theta}{t} = \frac{7 \times 2\pi}{100} \]
Calculating this gives:
\[ \omega = \frac{14\pi}{100} = 0.14\pi \, \text{rad/s} \]

### Step 4: Calculate the centripetal acceleration (\( a_c \))
The formula for centripetal acceleration is:
\[ a_c = \omega^2 \times r \]
Substituting the values we have:
\[ a_c = (0.14\pi)^2 \times 0.12 \]
Calculating \( (0.14\pi)^2 \):
\[ (0.14\pi)^2 = 0.0196\pi^2 \]
Now substituting this into the centripetal acceleration formula:
\[ a_c = 0.0196\pi^2 \times 0.12 \]
Calculating this gives:
\[ a_c \approx 0.000235 \, \text{m/s}^2 \]
To express it in scientific notation:
\[ a_c \approx 2.35 \times 10^{-2} \, \text{m/s}^2 \]

### Final Answer
The magnitude of the centripetal acceleration is approximately:
\[ \boxed{2.35 \times 10^{-2} \, \text{m/s}^2} \]