Why bank roads?
For a vehicle of mass \(m\) negotiating a horizontal curve of radius \(R\) at speed \(v\), the centripetal requirement is \(F_c = m v^2 / R\). On a level road this force has to come entirely from static friction between tyres and tarmac, so the upper limit is
\[ v_{\max}^{\text{flat}} = \sqrt{\mu_s R g}. \]
For a highway curve of \(R=300\) m with \(\mu_s = 0.2\) and \(g=10\) m/s², this caps the safe speed at \(v_{\max}^{\text{flat}}=\sqrt{600}\approx 24.5\) m/s, only about 88 km/h. Anything faster and the tyres skid outward. Worse, in rain or on icy patches \(\mu_s\) can drop to 0.05 — the safe limit collapses to 12 m/s.
Engineers solve this by tilting the entire road surface inward at a small angle \(\theta\). With the road tilted, the normal force is no longer vertical: it tilts inward too, and its horizontal component contributes to the centripetal force. Friction is now a backup rather than the primary actor, so the safe speed rises sharply and the design becomes far less sensitive to road condition. This is exactly the geometry NCERT presents in §4.10, "Motion of a car on a banked road".
Make sure you can write \(F_c=mv^2/R\) and \(v_{\max}^{\text{flat}}=\sqrt{\mu_s R g}\) without thinking. Brush up on the flat-road derivation in Circular Motion & Centripetal Force before continuing.
The geometry of a banked turn
Picture a circular curve viewed from above as a section of a horizontal circle of radius \(R\). Now imagine the road surface along this curve tilted up on the outer edge by an angle \(\theta\) above the horizontal. The "outside" of the bend is higher than the "inside". The angle \(\theta\) is the angle of banking.
Because the surface is tilted, the normal force \(\vec{N}\) — always perpendicular to the surface — points along a line tilted by \(\theta\) from the vertical, leaning toward the centre of the circle. Resolve it into two components:
- Vertical: \(N\cos\theta\), opposing gravity.
- Horizontal: \(N\sin\theta\), directed toward the centre of the curve. This is the new piece — on a flat road it would be zero.
Gravity \(m\vec{g}\) still points straight down. Friction \(\vec{f}\) lies along the surface (up or down the slope), and its direction depends on which way the car is about to slip, which in turn depends on whether the speed is above or below the optimum.
Optimum speed (no-friction case)
Start with the cleanest scenario: imagine the car is travelling at exactly the right speed so that no friction is needed. Then the only two forces are gravity and the normal force.
Vertical equilibrium (no vertical acceleration, because the circular motion is horizontal):
\[ N\cos\theta = m g. \]
Horizontal equation (centripetal):
\[ N\sin\theta = \frac{m v_0^{2}}{R}. \]
Divide the second by the first to eliminate \(N\):
\[ \tan\theta = \frac{v_0^{2}}{R g} \quad\Longrightarrow\quad \boxed{\,v_0=\sqrt{R\,g\,\tan\theta\,}.\,} \]
This is the optimum speed (NCERT Eq. 4.22). At \(v=v_0\) the horizontal component of \(N\) alone supplies the centripetal force, the vertical component balances gravity, and the tyres are skidding-free — friction is zero. Highway engineers therefore set \(\theta\) for the expected average traffic speed on that bend, so the typical driver experiences minimal tyre wear and the smoothest ride.
Notice that \(v_0\) does not depend on \(m\). Both the required centripetal force and the available normal force scale with mass, so they cancel — a two-wheeler, a sedan and a truck all share the same optimum speed for the same bend.
Forces when v > v0 (car tends to slide up the slope)
If you drive faster than \(v_0\), the horizontal component of \(N\) alone is no longer enough to keep the car on the circle. The car tends to slide outward — which, on a banked road, means up the slope. Static friction therefore acts down the slope, opposing the impending slip.
Now resolve all three forces along vertical and horizontal directions. Let \(f\) be the magnitude of friction.
Vertical (no acceleration):
\[ N\cos\theta = m g + f\sin\theta. \tag{4.19a} \]
Horizontal (centripetal):
\[ N\sin\theta + f\cos\theta = \frac{m v^{2}}{R}. \tag{4.19b} \]
These are NCERT's equations (4.19a) and (4.19b). Note the signs: friction now adds to \(N\sin\theta\) in the horizontal equation (it points partly inward) and adds to gravity in the vertical equation (its vertical component points downward).
Maximum safe speed
Static friction can supply at most \(f_{\max} = \mu_s N\). At this limit the speed is the maximum safe speed \(v_{\max}\). Substitute \(f=\mu_s N\) into the two equations above.
From the vertical equation:
\[ N\cos\theta - \mu_s N\sin\theta = m g \;\;\Longrightarrow\;\; N(\cos\theta - \mu_s\sin\theta) = m g, \]
so
\[ N = \frac{m g}{\cos\theta - \mu_s\sin\theta}. \]
From the horizontal equation:
\[ N\sin\theta + \mu_s N\cos\theta = \frac{m v_{\max}^{2}}{R} \;\;\Longrightarrow\;\; N(\sin\theta + \mu_s\cos\theta) = \frac{m v_{\max}^{2}}{R}. \]
Divide the second by the first to eliminate \(N\) and \(m\):
\[ \frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} = \frac{v_{\max}^{2}}{R g}. \]
Multiply numerator and denominator by \(1/\cos\theta\):
\[ \frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta} = \frac{v_{\max}^{2}}{R g}. \]
Hence the maximum safe speed on a banked road is
\[ \boxed{\,v_{\max} = \sqrt{\,\dfrac{R\,g\,(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta}\,}.\,} \tag{4.21} \]
Compare with the flat-road result \(v_{\max}^{\text{flat}}=\sqrt{\mu_s R g}\). Banking adds the \(\tan\theta\) term in the numerator and the \((1-\mu_s\tan\theta)\) factor in the denominator — both increase the safe speed. Even modest banking (\(\theta\) of a few degrees) gives a substantial gain in safety margin, especially when \(\mu_s\) is small (wet or icy roads).
Forces when v < v0 (car tends to slide down the slope)
Drive slower than \(v_0\) and the horizontal component of \(N\) over-supplies the centripetal requirement; the excess pushes the car inward. Geometrically this means the car tends to slide down the slope, so static friction now acts up the slope.
The equations are the mirror image of the high-speed case (flip the sign of \(f\) in both equations 4.19a and 4.19b). At the limit \(f=\mu_s N\), the algebra gives the minimum safe speed:
\[ v_{\min} = \sqrt{\,\dfrac{R\,g\,(\tan\theta - \mu_s)}{1 + \mu_s\tan\theta}\,}, \quad\text{valid only when}\quad \tan\theta > \mu_s. \]
If \(\tan\theta \le \mu_s\), friction alone is enough to hold the car even at rest — there is no lower speed limit; \(v_{\min}\) is zero. This is the case for most highway curves, where \(\theta\) is modest. On a steeply banked race track where \(\tan\theta > \mu_s\), a car driving too slowly will slide down toward the inner edge.
Parking on a banked road
Setting \(v=0\) in the low-speed equations gives the equilibrium condition for a stationary car:
\[ N\sin\theta = f\cos\theta \quad\text{and}\quad N\cos\theta + f\sin\theta = m g. \]
Solving for \(f\), the friction needed to keep the car at rest is \(f = m g \sin\theta / \cos\theta \cdot \cos\theta = m g \sin\theta\) (along the slope), and the normal force is \(N = m g \cos\theta\). The static friction required is therefore \(f/N = \tan\theta\). Since the available limit is \(\mu_s\), the parking condition is
\[ \boxed{\,\tan\theta \le \mu_s.\,} \]
For a 10° bank, \(\tan 10°\approx 0.18\), so any tyre–road combination with \(\mu_s \ge 0.2\) (which is most dry surfaces) is safe to park on. A steep 30° race-track bank (\(\tan 30°\approx 0.58\)) would slide a stationary car with \(\mu_s=0.4\).
Numerical comparison: flat versus banked
Take a curve of radius \(R=300\) m, \(\mu_s = 0.2\), \(g=10\) m/s², and bank it at \(\theta = 15°\) (\(\tan 15°\approx 0.268\)).
| Road type | Formula | Safe speed | In km/h |
|---|---|---|---|
| Flat | \(\sqrt{\mu_s R g}\) | \(\sqrt{0.2\cdot 300\cdot 10}=\sqrt{600}\approx 24.5\) m/s | ≈ 88 km/h |
| Banked, optimum | \(\sqrt{R g\tan\theta}\) | \(\sqrt{300\cdot 10\cdot 0.268}=\sqrt{804}\approx 28.4\) m/s | ≈ 102 km/h |
| Banked, maximum | \(\sqrt{Rg(\mu_s+\tan\theta)/(1-\mu_s\tan\theta)}\) | \(\sqrt{3000\cdot 0.468/0.946}=\sqrt{1484}\approx 38.5\) m/s | ≈ 139 km/h |
Banking at just 15° increases the maximum safe speed from 24.5 m/s to 38.5 m/s — a gain of about 57%. That is the engineering payoff: at unchanged \(\mu_s\) the bend is safer at higher speeds, and at unchanged speed the design tolerates a much lower \(\mu_s\) (wet road conditions).
Worked example — NCERT 4.11 (race-track at 15°)
A circular race-track of radius 300 m is banked at an angle of 15°. The coefficient of friction between wheels of a race-car and the road is 0.2. What is (a) the optimum speed of the race-car to avoid wear and tear on its tyres, and (b) the maximum permissible speed to avoid slipping? (Use \(g=10\) m/s².)
Given: \(R = 300\) m, \(\theta = 15°\), \(\mu_s = 0.2\). Trigonometry: \(\sin 15°\approx 0.259\), \(\cos 15°\approx 0.966\), \(\tan 15°\approx 0.268\).
(a) Optimum speed. At optimum speed no friction acts. Use \(v_0=\sqrt{R g\tan\theta}\):
\[ v_0 = \sqrt{300\cdot 10\cdot 0.268} = \sqrt{804}\approx 28.4\;\text{m/s}. \]
Converting, \(v_0 \approx 28.4 \times 3.6 \approx 102\) km/h. This is the speed for which the bank was engineered.
(b) Maximum permissible speed. Apply equation (4.21):
\[ v_{\max} = \sqrt{\dfrac{Rg(\mu_s+\tan\theta)}{1-\mu_s\tan\theta}}. \]
Plug in:
\[ v_{\max} = \sqrt{\dfrac{300\cdot 10\cdot(0.2+0.268)}{1-0.2\cdot 0.268}} = \sqrt{\dfrac{3000\cdot 0.468}{0.946}} = \sqrt{\dfrac{1404}{0.946}}=\sqrt{1484}\approx 38.5\;\text{m/s}. \]
So \(v_{\max}\approx 38.5\) m/s, or about 139 km/h. Driving faster than this, the static-friction limit is exceeded and the car slides outward off the curve.
Three speeds you must remember
- Optimum speed (no friction needed): \(v_0=\sqrt{Rg\tan\theta}\).
- Maximum safe speed (friction down the slope at limit): \(v_{\max}=\sqrt{Rg(\mu_s+\tan\theta)/(1-\mu_s\tan\theta)}\).
- Minimum safe speed (friction up the slope at limit; only if \(\tan\theta>\mu_s\)): \(v_{\min}=\sqrt{Rg(\tan\theta-\mu_s)/(1+\mu_s\tan\theta)}\).
Applications: railways, aircraft, cyclists
The same principle reappears across many systems. NCERT mentions roads and NIOS §4.4.1 extends the idea to rails and aircraft.
Railway tracks on a curve
Trains have flanged wheels that grip the rails; lateral friction is small. To supply the centripetal force around a curve, the outer rail is raised above the inner rail by a height \(h\), so that the track itself is banked at angle \(\theta\) with \(\sin\theta = h/L\), where \(L\) is the gauge (distance between rails). The condition for no lateral thrust on the rails is the same as the optimum-speed condition: \(\tan\theta = v^2/(Rg)\). Indian Railways calls this cant or superelevation.
Aircraft in a turn
An aeroplane in level flight has lift \(L\) acting perpendicular to its wings, balancing weight. To turn, the pilot banks the aircraft by an angle \(\theta\); the lift vector tilts with it. The vertical component \(L\cos\theta\) still has to balance \(mg\), while the horizontal component \(L\sin\theta\) supplies the centripetal force \(mv^2/R\). Dividing, \(\tan\theta = v^2/(Rg)\) — the same formula as for the optimum speed on a banked road, with the wing playing the role of the tilted ground.
Cyclist leaning on a turn
A cyclist cannot tilt the road, so they tilt the bike instead. By leaning inward by an angle \(\theta\) from the vertical, the line of action of the normal force passes through the centre of mass at an inclination. Taking torques about the centre of mass, equilibrium requires \(\tan\theta = v^2/(Rg)\) — the same relation, again. This is why cyclists on a sharp turn at 5 m/s in a 10 m radius bend lean by about 14° from vertical.