A car has to take a safe circular turn on horizontal flat rough road of radius 200 m. If the coefficient of friction between the tyres and road is 0.1 ,then maximum speed of car would be(All are in SI units)

Options

A. `sqrt(20g)`
B. `sqrt(10g)`
C. `sqrt(g)`
D. `g/sqrt(20)`

Answer

To find the maximum speed of a car taking a safe circular turn on a horizontal flat rough road, we can use the concept of centripetal force and friction. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Car
When a car turns in a circular path, the following forces are acting on it:
- Weight (mg) acting downwards.
- Normal force (N) acting upwards.
- Frictional force (f) acting towards the center of the circular path.

### Step 2: Identify the Condition for Circular Motion
For the car to safely navigate the turn without slipping, the frictional force must provide the necessary centripetal force required for circular motion. This can be expressed as:
\[ f \geq \frac{mv^2}{R} \]
where:
- \( f \) is the frictional force,
- \( m \) is the mass of the car,
- \( v \) is the speed of the car,
- \( R \) is the radius of the turn.

### Step 3: Express the Frictional Force
The maximum frictional force can be expressed in terms of the coefficient of friction (\( \mu \)) and the normal force (\( N \)):
\[ f = \mu N \]
Since the car is on a flat road, the normal force is equal to the weight of the car:
\[ N = mg \]
Thus, the frictional force becomes:
\[ f = \mu mg \]

### Step 4: Set Up the Inequality
Substituting the expression for friction into the inequality for circular motion gives:
\[ \mu mg \geq \frac{mv^2}{R} \]

### Step 5: Cancel the Mass
Since the mass \( m \) appears on both sides of the inequality, we can cancel it (assuming \( m \neq 0 \)):
\[ \mu g \geq \frac{v^2}{R} \]

### Step 6: Solve for Maximum Speed
Rearranging the inequality to solve for \( v \) gives:
\[ v^2 \leq \mu g R \]
Taking the square root of both sides:
\[ v \leq \sqrt{\mu g R} \]

### Step 7: Substitute the Given Values
Now, we can substitute the values provided in the question:
- Coefficient of friction (\( \mu \)) = 0.1
- Radius of the turn (\( R \)) = 200 m
- Acceleration due to gravity (\( g \)) = 10 m/s² (approximately)

Substituting these values into the equation:
\[ v \leq \sqrt{0.1 \times 10 \times 200} \]
\[ v \leq \sqrt{200} \]
\[ v \leq 14.14 \, \text{m/s} \]

### Step 8: Conclusion
Thus, the maximum speed of the car while taking the turn safely is approximately:
\[ v_{\text{max}} \approx 14.14 \, \text{m/s} \]

---