Understanding Banking of Roads
When a vehicle rounds a circular turn, it requires a centripetal force to maintain its curved path. While friction between the tires and the road can provide some of this force, relying solely on friction is dangerous, especially in wet or icy conditions. This is why engineers 'bank' roads—sloping them at an angle towards the center of the curve.
The Physics Formula
For a vehicle moving at velocity $v$ on a turn of radius $r$, the optimal banking angle $\theta$ (where friction is not needed) is given by:
$$\tan(\theta) = \frac{v^2}{rg}$$
Where:
- $v$ is the velocity in meters per second (m/s).
- $r$ is the radius of the turn in meters (m).
- $g$ is the acceleration due to gravity (approx. $9.8 \text{ m/s}^2$).
Step-by-Step Solution
1. Identify the given values:
- Velocity $v = 50 \text{ km/hr}$
- Radius $r = 200 \text{ m}$
- Gravity $g = 9.8 \text{ m/s}^2$
2. Convert units: Velocity must be in SI units (m/s). $v = 50 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 13.89 \text{ m/s}$
3. Apply the formula: $\tan(\theta) = \frac{(13.89)^2}{200 \times 9.8}$ $\tan(\theta) = \frac{192.93}{1960}$ $\tan(\theta) \approx 0.0984$
4. Find the angle: $\theta = \arctan(0.0984) \approx 5.62^{\circ}$
Thus, the road should be banked at an angle of approximately $5.62^{\circ}$ for a car to safely navigate the turn at $50 \text{ km/hr}$.