A turn has a radius of `10m` if a vehicle goes round it at an average speed of `18km//h` , what should be the proper angle of banking?

Explanation

To find the proper angle of banking for a vehicle going around a turn, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The given speed is \( 18 \, \text{km/h} \). To convert this to meters per second, we use the conversion factor:
\[ 1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s} \]
So, we calculate:
\[ \text{Speed in m/s} = 18 \times \frac{5}{18} = 5 \, \text{m/s} \]

### Step 2: Identify the radius of the turn
The radius of the turn is given as \( R = 10 \, \text{m} \).

### Step 3: Set up the equations for forces
When a vehicle is on a banked curve, the forces acting on it include the gravitational force \( mg \) acting downwards and the normal force \( N \) acting perpendicular to the surface of the road. The angle of banking is \( \theta \).

- The horizontal component of the normal force provides the centripetal force required for circular motion:
\[ N \sin \theta = \frac{mv^2}{R} \]
- The vertical component of the normal force balances the weight of the vehicle:
\[ N \cos \theta = mg \]

### Step 4: Divide the two equations
To eliminate \( N \), we can divide the first equation by the second:
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{\frac{mv^2}{R}}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{v^2}{Rg} \]

### Step 5: Substitute the known values
Now we can substitute the values we have:
- \( v = 5 \, \text{m/s} \)
- \( R = 10 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating \( \tan \theta \):
\[ \tan \theta = \frac{(5)^2}{10 \times 10} = \frac{25}{100} = \frac{1}{4} \]

### Step 6: Calculate the angle \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{4}\right) \]

### Final Result
Using a calculator, we find:
\[ \theta \approx 14.04^\circ \]

Thus, the proper angle of banking is approximately \( 14.04^\circ \).

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