Question

When cars travel around a banked (curved) road at the optimum angle,the normal reaction force (n) can provide the necessary centripetal force without the need for a friction force. (a)Describe whar would happen to Optimum banking angle when radius doubles? (b)Describe what would happen to optimum angle when speed doubles? (c)A car negotiate a turn of radius 80cm.What is the optimum banking angle for this curve if the speed is to be equal to 12m/s?​

Answer 1

Answer:

(a) The optimum banking Decreases

(b) The optimum banking Increases

(c) The optimum banking is approximately 86.88°

Explanation:

(a) The equation of motion on a banked road is given as follows;

$v = \sqrt{R \cdot g \times \left(\dfrac{tan (\theta) + \mu}{1 - \mu \cdot tan (\theta) }\right) }$

For no friction, we have;

v = √(R·g·tan(θ))

Where;

R₁ = The radius of the road

g = The acceleration due to gravity ≈ 9.81 m/s² = Constant

θ₁ = The bank angle

μ = The coefficient pf friction = Constant

v = The vehicle's speed

If the radius doubles, for no friction, we have;

v² = R·g·(tan(θ))

tan(θ) = v²/(R·g)

Therefore, when the radius doubles, tan(θ) becomes smaller and therefore, the optimum banking angle θ decreases (becomes smaller)

(b) When the speed doubles, we have;

v₁ = 2·v

∴ tan(θ₁) = (v₁)²/(R·g) = 4·(v)²/(R·g) = 4·tan(θ)

When the speed doubles, tan(θ) increases and therefore, the optimum banking angle θ increases increases

(c) The radius negotiated by the car, R = 80 cm = 0.8 m

The speed of the car, v = 12 m/s

From tan(θ) = v²/(R·g), we have;

tan(θ) = 12²/(0.8 × 9.81) ≈ 18.349

θ ≈ arctan(18.349°) ≈ 86.88°