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 angular speed and the linear speed of the motion ?
What is the magnitude of the centripetal acceleration in above problem ?

Answer

To solve the problem step by step, we will calculate the angular speed, linear speed, and centripetal acceleration of the insect moving in a circular groove.

### Step 1: Calculate the Distance Covered
The insect completes 7 revolutions. The distance covered in one revolution is given by the circumference of the circle, which is calculated using the formula:
\[ \text{Circumference} = 2\pi r \]
where \( r \) is the radius of the groove.

Given:
- Radius \( r = 12 \) cm = \( 12 \times 10^{-2} \) m = \( 0.12 \) m

So, the distance covered in 7 revolutions is:
\[ \text{Distance} = 7 \times 2\pi r = 7 \times 2\pi \times 0.12 \]
Calculating this:
\[ \text{Distance} = 7 \times 2 \times 3.14 \times 0.12 \approx 5.305 \text{ m} \]

### Step 2: Calculate the Linear Speed
The linear speed \( v \) can be calculated using the formula:
\[ v = \frac{\text{Distance}}{\text{Time}} \]
Given that the time taken is 100 seconds, we have:
\[ v = \frac{5.305}{100} = 0.05305 \text{ m/s} \]

### Step 3: Calculate the Angular Speed
The angular speed \( \omega \) can be calculated using the relationship between linear speed and angular speed:
\[ v = \omega r \]
Rearranging gives:
\[ \omega = \frac{v}{r} \]
Substituting the values we found:
\[ \omega = \frac{0.05305}{0.12} \approx 0.4421 \text{ rad/s} \]

### Step 4: Calculate the Centripetal Acceleration
The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \omega^2 r \]
Substituting the values:
\[ a_c = (0.4421)^2 \times 0.12 \]
Calculating this:
\[ a_c \approx 0.0234 \text{ m/s}^2 \]

### Final Results
1. **Linear Speed**: \( v \approx 0.05305 \text{ m/s} \)
2. **Angular Speed**: \( \omega \approx 0.4421 \text{ rad/s} \)
3. **Centripetal Acceleration**: \( a_c \approx 0.0234 \text{ m/s}^2 \)

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