A body moves along a circular path of radius 1m & `mu = 0.4`. The maximum angular velocity in rad/sec of the body so that it does not slide is (g = 10`ms^(-2)`)

Options

A. 2
B. 30
C. 15
D. 4

Answer

To solve the problem, we need to find the maximum angular velocity of a body moving in a circular path without sliding, given the radius of the path, the coefficient of friction, and the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the circular path, \( r = 1 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.4 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Understand the Forces Involved:**
- For a body moving in a circular path, the centripetal force required to keep the body in circular motion is provided by the frictional force.
- The centripetal force \( F_c \) is given by:
\[ F_c = \frac{m v^2}{r} \]
- The frictional force \( F_f \) is given by:
\[ F_f = \mu m g \]

3. **Set Up the Equation:**
- For the body not to slide, the centripetal force must equal the frictional force:
\[ \frac{m v^2}{r} = \mu m g \]

4. **Cancel Out the Mass \( m \):**
- Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \mu g \]

5. **Substitute the Known Values:**
- Substitute \( r = 1 \, \text{m} \), \( \mu = 0.4 \), and \( g = 10 \, \text{m/s}^2 \):
\[ \frac{v^2}{1} = 0.4 \times 10 \]
\[ v^2 = 4 \]

6. **Find the Linear Velocity \( v \):**
- Taking the square root of both sides:
\[ v = \sqrt{4} = 2 \, \text{m/s} \]

7. **Relate Linear Velocity to Angular Velocity:**
- The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
- Since \( r = 1 \, \text{m} \):
\[ 2 = 1 \cdot \omega \]
- Therefore, \( \omega = 2 \, \text{rad/s} \).

### Final Answer:
The maximum angular velocity \( \omega \) of the body so that it does not slide is:
\[ \omega = 2 \, \text{rad/s} \]