Banking of Road — Motion on Circular Track | Safe Turning Speeds, Friction & Worked Problems

Banking of Road — Motion of a Car on a Circular Track, Maximum & Minimum Safe Speeds

Why do we slow down on sharp turns and in rain? How does banking of roads help? This post explains motion on horizontal & banked curves, derives safe speed formulas (frictionless and with friction), provides worked practice problems and answers common FAQs.

Table of Contents

• Intro: why speed matters on turns
• Turning on horizontal roads (with & without friction)
• Banking of road — frictionless case (ideal speed)
• Banking of road — with friction (range of safe speeds)
• Quick derivations & final formulas
• Practice problems (worked)
• FAQs

Intro — Why we slow down at turns

While taking a turn a vehicle needs a centripetal force directed toward the center of the curve. If the required centripetal force exceeds what friction (and/or the banking geometry) can supply, the tires slip and the vehicle skids outward. Banking the road—raising the outer edge—adds a horizontal component of the normal force to help provide centripetal acceleration even when friction is small (e.g., in rain).

Turning on horizontal roads

Case 1 — Frictionless horizontal surface (μ = 0)

If the road is perfectly smooth (no friction) there is no horizontal force to provide centripetal acceleration — the vehicle cannot negotiate the turn and will slip tangentially. Practically: impossible to turn without friction or banking.

Case 2 — Rough horizontal surface (static friction μ)

Static friction provides the centripetal force. For a vehicle of mass m traveling at speed v on a curve of radius R:

Vertical equilibrium: \(N = mg\). Horizontal (centripetal): \(f_{\text{max}} = \mu_s N = \dfrac{mv^2}{R}\).

Maximum safe speed on a horizontal rough track:

\[ v_{\max} = \sqrt{\mu_s g R}. \]

Above this speed the required centripetal force \(mv^2/R\) exceeds available friction and the vehicle will skid outward.

Banking of road (without friction)

Consider a banked curve with banking angle \(\theta\) (road tilted so outer edge is higher). On a frictionless bank, the horizontal component of the normal reaction provides the centripetal force.

Figure: Normal force N resolves into horizontal \(N\sin\theta\) (centripetal) and vertical \(N\cos\theta\) components.

Balance forces:

Vertical: \(N\cos\theta = mg\). Horizontal (centripetal): \(N\sin\theta = \dfrac{m v^2}{R}\).

Divide to eliminate N:

\[ \tan\theta = \frac{v^2}{R g}. \] Hence the ideal (design) speed for a frictionless bank is \[ \boxed{v_{\text{design}} = \sqrt{R g \tan\theta}.} \]

If v > v_design the car tends to skid outward; if v < v_design it tends to skid inward (on a frictionless bank).

Banking of road (with friction) — range of safe speeds

If friction is present, it may act either up or down the slope depending on whether the car tends to skid inward or outward. This gives a range of safe speeds \([v_{\min}, v_{\max}]\) for which no slipping occurs.

Set-up and sign convention

Let N be the normal reaction (perpendicular to surface), f the frictional force (|f| ≤ μN) acting along the surface. Resolve forces into vertical and horizontal (toward center) components.

Maximum safe speed (tendency to skid outward — friction points down the slope)

Vertical: \(N\cos\theta - f\sin\theta = mg\). Horizontal: \(N\sin\theta + f\cos\theta = \dfrac{m v_{\max}^2}{R}\).

At limiting condition take \(f = \mu N\) (down the slope). Solving yields:

\[ v_{\max} = \sqrt{ \; R g \; \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} \; }. \]

Minimum safe speed (tendency to skid inward — friction points up the slope)

Vertical: \(N\cos\theta + f\sin\theta = mg\). Horizontal: \(N\sin\theta - f\cos\theta = \dfrac{m v_{\min}^2}{R}\).

At limiting condition take \(f = \mu N\) (up the slope). Solving yields:

\[ v_{\min} = \sqrt{ \; R g \; \frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta} \; }. \]

These formulas assume the denominators and numerators are positive. If the numerator for \(v_{\min}^2\) becomes negative, it means that static friction can hold the vehicle even at \(v=0\) (no minimum positive speed required) for that combination of θ and μ.

Quick derivations & final formulas (boxed)

Situation	Formula
Horizontal road (rough, max safe speed)	\(\displaystyle v_{\max}=\sqrt{\mu_s g R}\)
Banked, frictionless (design speed)	\(\displaystyle v=\sqrt{R g \tan\theta}\)
Banked with friction (max speed)	\(\displaystyle v_{\max}=\sqrt{R g \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}}\)
Banked with friction (min speed)	\(\displaystyle v_{\min}=\sqrt{R g \frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta}}\)

Practice problems (with worked solutions)

Q1. Required banking angle

What should be the angle of banking for a circular track of radius \(R\) designed for cars with average speed \(v_0\)?

Use frictionless design relation: \(\tan\theta = v_0^2/(Rg)\). So the banking angle is \(\displaystyle \theta = \arctan\!\left(\frac{v_0^2}{R g}\right)\).

Q2. Speed for no-friction condition

A circular road of radius \(R\) is banked at angle \(\theta\). At what speed should a vehicle go so that no friction is needed?

For frictionless motion: \(\displaystyle v = \sqrt{R g \tan\theta}.\)

Q3. Maximum safe speed (numerical)

A circular road of radius \(R=50\ \text{m}\) is banked at \(\theta=12^\circ\). The coefficient of static friction is \(\mu=0.25\). What is the maximum safe speed?

Compute: \[ v_{\max}=\sqrt{R g \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}}. \] Using \(g=9.8\ \text{m/s}^2\), \(\theta=12^\circ\): \[ \sin12^\circ\approx0.2079,\quad \cos12^\circ\approx0.9782. \] Numerator = \(\sin\theta + \mu\cos\theta \approx 0.2079 + 0.25\times0.9782 = 0.2079 + 0.2446 = 0.4525.\) Denominator = \(\cos\theta - \mu\sin\theta \approx 0.9782 - 0.25\times0.2079 = 0.9782 - 0.0520 = 0.9262.\) So factor ≈ \(0.4525/0.9262 = 0.4886.\) Then \[ v_{\max}=\sqrt{50\times9.8\times0.4886}\approx\sqrt{239.3}\approx15.47\ \mathrm{m/s}\ (\approx55.7\ \mathrm{km/h}). \] Answer: ≈ 15.5 m/s.

Q4. Minimum safe speed (numerical)

Same road as Q3 (\(R=50\) m, \(\theta=12^\circ, \mu=0.25\)). What is the minimum safe speed?

Use: \[ v_{\min}=\sqrt{ R g \frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta} }. \] Numerator = \(\sin\theta - \mu\cos\theta = 0.2079 - 0.25\times0.9782 = 0.2079 - 0.2446 = -0.0367\) → negative. This negative numerator indicates the formula gives a non-physical real value — in practice it means static friction can hold the car at rest (v=0) without slipping inward for these θ and μ. So the minimum safe speed is effectively 0 (no positive lower bound).

FAQs

Q: Why is banking used on highways & curves?

A: Banking provides a horizontal component of the normal force to supply centripetal acceleration so vehicles can make turns more safely even with low friction (wet roads). It allows higher safe speeds and reduces reliance on tire friction alone.

Q: What happens if you exceed the maximum safe speed?

A: If v > v_max the required centripetal force exceeds what friction + banking can supply and the vehicle tends to skid outward (toward the outer edge of the curve).

Q: Can a steeply banked curve have no minimum speed?

A: Yes — if the combination of θ and μ makes the numerator \(\sin\theta-\mu\cos\theta\) negative, then the mathematical minimum speed becomes non-physical (≤0). Physically that means the car will not skid inward even at very low speeds—static friction and component of normal force are enough to hold it.

Q: Is the design speed the same as speed limit?

A: Not exactly. The design (or ideal) speed for a frictionless bank gives the speed at which no friction is necessary. Engineers choose banking, friction allowance, and an appropriate posted speed limit that factors safety margins, vehicle types and road conditions.

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