Class 11 > Unit # 04: Rotational & Circular Motion > Banked Curve and Angle of Banking

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Unit No. 4 Rotational and Circular Motion - Banked Curve and Angle of Banking

Define Banked Curve. Describe the forces and motion of bodies on a banked curve.

Banked Curve

"A banked curve is a curve that has its surface inclined at an angle with respect to the horizontal ground on which the curve is positioned."

Figure: Free-body diagram of a vehicle navigating a banked curve.

Description of Forces and Motion

The primary reason for banking curves is to reduce the vehicle's reliance solely on static friction to make a safe turn. On a flat (unbanked) curve, a car traveling along the path relies entirely on the horizontal force of static friction directed toward the center of the circle to provide the necessary centripetal acceleration.

On a banked curve, however, the normal force ($N$) exerted by the road on the car acts perpendicularly to the inclined surface, at an angle $\theta$ with the vertical. This normal force can be resolved into two components:

• Vertical Component ($N \cos\theta$): Balances the downward weight of the vehicle ($mg$).
• Horizontal Component ($N \sin\theta$): Acts directly toward the center of the circular path, providing all or part of the necessary centripetal acceleration.

Consequently, for a specific design speed and angle of banking, no frictional force is required at all to negotiate the curve safely. This allows vehicles to turn successfully even under slippery conditions, such as on ice or wet pavement.

Prove that the angle of banking is independent of the mass of body.

Consider a vehicle of mass $m$ moving with speed $v$ along a banked road of radius $r$. Let $\theta$ be the angle of banking. For ideal banking (ignoring friction), the external forces acting on the car are its downward weight ($W = mg$) and the normal force ($N$) exerted by the road.

1. Horizontal Equilibrium (Centripetal Force):

The horizontal component of the normal force provides the required centripetal force:

$N \sin\theta = F_c = \frac{mv^2}{r} \quad \text{--- (i)}$

2. Vertical Equilibrium:

Since the car does not move vertically off the road, the vertical component of the normal force balances the weight of the car:

$N \cos\theta = mg \quad \text{--- (ii)}$

3. Deriving the Angle of Banking Formula:

Dividing equation (i) by equation (ii):

$\frac{N \sin\theta}{N \cos\theta} = \frac{\frac{mv^2}{r}}{mg}$

Simplifying terms ($N$ cancels out on the left side, and $m$ cancels out on the right side):

$\tan\theta = \frac{v^2}{rg}$

Conclusion:
Solving for the angle of banking $\theta$:

$\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)$

This final mathematical expression demonstrates that the angle of banking ($\theta$) depends only on the vehicle's speed ($v$), the radius of the curve ($r$), and acceleration due to gravity ($g$). The mass ($m$) of the vehicle cancels out entirely, proving that the angle of banking is independent of the mass of the body.

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