## Motion of a car on a Banked Road

Friction is not always reliable at circular turns, if high speed and sharp turns are involved.

To avoid dependence on friction, the roads are banked by an angle (θ) at the turn, so that the outer part of the road is somewhat raised compared to the inner part.

Applying Newton’s second law along the radius and the first law in the vertical direction for the motion of a car on the banked road having mass m.

We get, N sin θ = mv²/r and N cos θ = mg

Here, r = radius of circular turn, θ = banking angle. From these two equations, we get

**tan θ = v²/rg **or **v = √(rg tan θ)**

## Maximum velocity (V_{max}) at which a car can navigate a banked curve without slipping.

**-> V _{max} = (Rg(μ_{s} + tan θ) / (1 – μ_{s} tan θ))^{1/2}**; where μ

_{s}= coefficient of static friction, θ = angle of the bank, R = radius of the curve, g = acceleration due to gravity (9.8 m/s²), v = velocity of the car.

-> **Banked roads:** The edges of curved roads are banked (tilted inward) to provide the centripetal force required for vehicles to turn safely. The angle at which the road is inclined is called the **bank angle**.**-> Driving at V _{0}:** When driving at a speed of

**V**, there is little wear and tear on the tires.

_{0}= (Rgtan θ)^{1/2}**Related Question:** A circular race track of a radius of 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race car and the road is 0.2, then what will be the maximum permissible speed to avoid slipping? (Take, tan 15° = 0.26)

**Solution:** Given:

r = radius of the track = 300 m

θ = banking angle = 15°

g = acceleration due to gravity = 9.8 m/s²

μs = coefficient of friction = 0.2

tan 15° = 0.26

**Solving:** 300 × 9.8 (0.2 + tan 15°) / (1 – 0.2 × tan 15°)]^{1/2}

=> [300 × 9.8 (0.2 + 0.26) / (1 – 0.2 × 0.26)]^{1/2}

= [1353.6 / 0.948]^{1/2}

V_{max} = [1427.85]^{1/2} = 37.8 m/s

The maximum permissible speed to avoid slipping is **37.8 m/s**.