If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficeint so that a scooter going at 18 km/hr does not skid?

Explanation

To find the minimum coefficient of friction required for a scooter to navigate a turn without skidding, we can follow these steps:

### Step 1: Identify the Forces Acting on the Scooter
When the scooter is taking a turn, the following forces are acting on it:
- The weight of the scooter and rider, \( mg \), acting downwards.
- The normal force, \( N \), acting upwards.
- The centrifugal force, which is the inertial force acting outward, given by \( \frac{mv^2}{R} \), where \( m \) is the mass of the scooter and rider, \( v \) is the velocity, and \( R \) is the radius of the turn.
- The frictional force, which acts inward and is given by \( \mu N \), where \( \mu \) is the coefficient of friction.

### Step 2: Set Up the Equations
From the forces acting on the scooter, we can establish the following equations:
1. The normal force \( N \) balances the weight:
\[ N = mg \]
2. The frictional force must provide the necessary centripetal force to keep the scooter moving in a circle:
\[ \mu N = \frac{mv^2}{R} \]

### Step 3: Substitute the Normal Force
Substituting the expression for \( N \) from the first equation into the second equation gives:
\[ \mu mg = \frac{mv^2}{R} \]
Here, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = \frac{v^2}{R} \]

### Step 4: Solve for the Coefficient of Friction
Rearranging the equation to solve for \( \mu \):
\[ \mu = \frac{v^2}{gR} \]

### Step 5: Convert the Speed to Meters per Second
The speed given is 18 km/hr. We need to convert this to meters per second:
\[ v = 18 \text{ km/hr} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}} = 5 \text{ m/s} \]

### Step 6: Plug in the Values
Now we can substitute \( v = 5 \text{ m/s} \), \( g = 10 \text{ m/s}^2 \) (approximating the acceleration due to gravity), and \( R = 30 \text{ m} \) into the equation for \( \mu \):
\[ \mu = \frac{(5)^2}{10 \times 30} \]
Calculating this gives:
\[ \mu = \frac{25}{300} = \frac{1}{12} \approx 0.0833 \]

### Final Answer
Thus, the minimum coefficient of friction required is approximately:
\[ \mu \approx 0.084 \]