I have been planning to buy a new car and one of the major considerations for me is the fastest way to make a U turn on a horizontal road at constant speed. Suppose to make a `180^(@)` turn on a circular path. The car's steering system does not allow a radius of less than 5m. The road is wide enough to allow a circle of maximum radius of 20m. If the acceleration is greater than `5 m//s^(2)`, the car skids.
What is coefficient of static friction between car tires and road?

Options

A. 0.3
B. 0.5
C. 0.8
D. `1.0`

Explanation

To find the coefficient of static friction between the car tires and the road, we can follow these steps:

### Step 1: Understand the forces acting on the car
When a car makes a U-turn, it experiences centripetal acceleration directed towards the center of the circular path. The frictional force between the tires and the road provides the necessary centripetal force to keep the car moving in a circular path.

### Step 2: Write the equation for centripetal force
The centripetal force \( F_c \) required to keep the car moving in a circle of radius \( r \) at speed \( v \) is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the car,
- \( v \) is the speed of the car,
- \( r \) is the radius of the turn.

### Step 3: Relate the frictional force to the centripetal force
The maximum frictional force \( F_f \) that can act on the car without skidding is given by:
\[ F_f = \mu_s N \]
where:
- \( \mu_s \) is the coefficient of static friction,
- \( N \) is the normal force, which equals \( mg \) (the weight of the car).

Thus, we can write:
\[ F_f = \mu_s mg \]

### Step 4: Set the centripetal force equal to the frictional force
For the car to make a turn without skidding, the centripetal force must equal the maximum frictional force:
\[ \frac{mv^2}{r} = \mu_s mg \]

### Step 5: Cancel the mass \( m \) from both sides
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \mu_s g \]

### Step 6: Solve for the coefficient of static friction \( \mu_s \)
Rearranging the equation gives:
\[ \mu_s = \frac{v^2}{rg} \]

### Step 7: Determine the maximum acceleration
The problem states that the acceleration should not exceed \( 5 \, \text{m/s}^2 \). The centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]
Setting \( a_c \) to \( 5 \, \text{m/s}^2 \):
\[ 5 = \frac{v^2}{r} \]

### Step 8: Substitute the maximum radius
The minimum radius for the turn is \( 5 \, \text{m} \) and the maximum radius is \( 20 \, \text{m} \). Using the maximum radius \( r = 20 \, \text{m} \):
\[ 5 = \frac{v^2}{20} \]
This gives:
\[ v^2 = 5 \times 20 = 100 \quad \Rightarrow \quad v = \sqrt{100} = 10 \, \text{m/s} \]

### Step 9: Substitute \( v \) and \( r \) into the equation for \( \mu_s \)
Now substituting \( v = 10 \, \text{m/s} \), \( r = 20 \, \text{m} \), and \( g = 10 \, \text{m/s}^2 \):
\[ \mu_s = \frac{v^2}{rg} = \frac{100}{20 \times 10} = \frac{100}{200} = 0.5 \]

### Final Answer
The coefficient of static friction \( \mu_s \) between the car tires and the road is \( 0.5 \).

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