Banking, Friction and the Safe Speed Window on a Hill Curve
A 100 m radius highway curve on a hill road near Chikmagalur is banked at 15°; tan 15° = 0.268, sin 15° = 0.259 and cos 15° = 0.966. Tyre tests give a coefficient of static friction of 0.40 on dry asphalt and 0.25 when the surface is wet, and g may be taken as 10 m/s². The transport department is reviewing the posted limit after two accidents in which cars left the road on the outside of the curve. Local drivers say that on wet mornings their cars feel as if they are 'sliding inwards' when crawling near the hairpin, while taxi drivers regularly take the curve at 30 m/s (about 108 km/h). A junior engineer proposes raising the banking to 25° (tan 25° = 0.466) instead of resurfacing the road, arguing that extra banking 'costs nothing and always helps'. The chief engineer wants numbers for both ends of the safe speed range, dry and wet, before accepting or rejecting that proposal.
1. Find the maximum safe speed on the dry curve and decide, with numbers, whether 30 m/s is acceptable.
Read the worked answer
At the limit of sliding outwards friction acts up the slope, so the force balance along and perpendicular to the incline gives
N(cos θ − μ sin θ) = mg and N(sin θ + μ cos θ) = mv²/R
which combine to v_max² = Rg(tan θ + μ)/(1 − μ tan θ).
Dry road: numerator 0.268 + 0.40 = 0.668; denominator 1 − 0.40(0.268) = 0.8928.
v_max² = (100)(10)(0.668/0.8928) = 748 m²/s², so v_max = 27.3 m/s (about 98 km/h).
Cross-check at 30 m/s: required centripetal acceleration is v²/R = 900/100 = 9.0 m/s². Banking alone supplies g tan θ = 2.68 m/s², so friction must provide the rest. Required friction = m(9.0)(0.966) − m(10)(0.259) = 6.10m N, while N = m(10)(0.966) + m(9.0)(0.259) = 11.99m N, giving μ_required = 6.10/11.99 = 0.51.
Judgement: 0.51 exceeds the available 0.40, so a car at 30 m/s cannot hold the curve and will drift outward. The posted limit must be at or below roughly 27 m/s on dry road, which already rules out the taxi drivers' habit.
2. Repeat the calculation for the wet road and explain the drivers' report of sliding inward at very low speed.
Read the worked answer
With μ = 0.25, numerator = 0.268 + 0.25 = 0.518 and denominator = 1 − 0.25(0.268) = 0.933.
v_max² = 1000(0.518/0.933) = 555 m²/s², so v_max = 23.6 m/s (85 km/h) - about 3.7 m/s lower than dry.
The lower end of the window exists only if tan θ > μ, because a slow or stationary car needs friction to stop it sliding down the slope. At the point of sliding down, friction acts up the slope:
v_min² = Rg(tan θ − μ)/(1 + μ tan θ)
Dry: tan θ − μ = 0.268 − 0.40 < 0, so no minimum speed exists; a car can stop on the curve and friction alone holds it.
Wet: tan θ − μ = 0.268 − 0.25 = 0.018, denominator = 1 + 0.25(0.268) = 1.067, so v_min² = 1000(0.018/1.067) = 16.9 and v_min = 4.1 m/s (about 15 km/h).
Judgement: on a wet day the curve is safe only between roughly 15 km/h and 85 km/h. A car forced below 15 km/h - a queue, a stalled truck, a cyclist - will slide down the banking. The drivers' 'sliding inwards at crawling speed' is therefore a real phenomenon, not a misperception, and it is the lower limit that the junior engineer has ignored.
3. Evaluate the proposal to bank the curve at 25° with wet friction still 0.25, including its effect on the low-speed end, and give a recommendation.
Read the worked answer
v_max² = 1000(0.466 + 0.25)/(1 − 0.25 × 0.466) = 1000(0.716/0.8835) = 810, so v_max = 28.5 m/s (102 km/h). The top-speed problem is essentially solved even in the wet.
v_min² = 1000(0.466 − 0.25)/(1 + 0.25 × 0.466) = 1000(0.216/1.1165) = 193, so v_min = 13.9 m/s (50 km/h).
Because 0.466 > 0.40, a lower limit appears even on a dry day: v_min² = 1000(0.066/1.1864) = 55.6, giving v_min = 7.5 m/s (27 km/h).
Judgement: the proposal converts a top-speed failure into a large low-speed failure. At 25° banking nothing may travel below 50 km/h in the wet or 27 km/h when dry, so any stopped or slow vehicle becomes the hazard, and the change does nothing when μ approaches zero (oil, ice), where friction cannot be traded for geometry at all. Resurfacing to raise μ, say to 0.55, gives v_max = √[1000(0.818)/(1 − 0.147)] ≈ 31 m/s while keeping tan θ − μ negative, so no lower limit is created. Recommendation: reject the 25° proposal, improve the surface, and set the limit near 20 m/s.