A car is moving in a horizontal level of circular track with uniform speed of `10 m/s`. If radius of circular path is 50 m then the minimum coefficient of friction to avoid over turning is

Options

A. 0.3
B. 0.5
C. 0.1
D. 0.2

Answer

To solve the problem of finding the minimum coefficient of friction required to avoid overturning for a car moving in a circular path, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Speed of the car, \( v = 10 \, \text{m/s} \)
- Radius of the circular path, \( r = 50 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (assuming standard value for simplicity)

2. **Understand the Forces Acting on the Car:**
- When a car moves in a circular path, it experiences centripetal acceleration directed towards the center of the circle.
- The net force required for this centripetal acceleration is provided by the frictional force between the tires and the road.

3. **Write the Equation for Centripetal Force:**
- The centripetal force \( F_c \) required to keep the car moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the car.

4. **Write the Expression for Frictional Force:**
- The frictional force \( F_f \) that prevents the car from sliding out is given by:
\[ F_f = \mu N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a car on a flat surface, the normal force \( N \) equals the weight of the car:
\[ N = mg \]
Therefore, the frictional force can be expressed as:
\[ F_f = \mu mg \]

5. **Set the Forces Equal:**
- For the car to avoid overturning, the frictional force must be equal to the centripetal force:
\[ \frac{mv^2}{r} = \mu mg \]

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

7. **Solve for the Coefficient of Friction \( \mu \):**
- Rearranging the equation gives:
\[ \mu = \frac{v^2}{rg} \]

8. **Substitute the Known Values:**
- Now, substituting the known values into the equation:
\[ \mu = \frac{(10 \, \text{m/s})^2}{(50 \, \text{m})(10 \, \text{m/s}^2)} \]
\[ \mu = \frac{100}{500} = 0.2 \]

### Final Answer:
The minimum coefficient of friction required to avoid overturning is \( \mu = 0.2 \).