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Drive fast over a hump in the road and your stomach lifts. Go fast enough and the wheels leave the ground. Both effects come from one equation, and exam questions on humps, dips and domes all test whether you write it with the right signs. This post sets up that equation, finds the lift-off speed, compares a hump with a dip, and solves the classic block on a smooth dome.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "dynamics of uniform circular motion, centripetal force and its applications" and, under Work, Energy and Power, "motion in a vertical circle". The NEET (UG) 2026 syllabus lists centripetal force and its applications too. In NCERT Class 11 (2026-27 reprint), the Laws of Motion chapter covers circular motion in Section 4.10 with the level and banked road, and the Work, Energy and Power chapter works a bob in a vertical circle in Example 5.7. A hump is not a named NCERT example; it uses the same centripetal idea. Check the current year's syllabus.
What equation governs the top of a hump?
At the top of a hump of radius r, the centre of the circle is below the car. Two vertical forces act: weight mg downward (towards the centre) and the normal force N upward (away from it). The net force towards the centre must equal mv²/r:
mg − N = mv²/r, so N = m(g − v²/r)
Three things follow. N is less than mg whenever the car is moving. N falls as speed rises. And N reaches zero at
v = √(gr)
Above that speed the road would have to pull the car down, which it cannot do. The car loses contact.
Worked example 1: a car on a hump
A 1000 kg car crosses a hump of radius 20 m. Take g = 9.8 m/s².
At 10 m/s (36 km/h): v²/r = 100/20 = 5.0 m/s². N = 1000 × (9.8 − 5.0) = 4800 N, less than half the car's weight of 9800 N.
Lift-off speed: v = √(9.8 × 20) = √196 = 14 m/s, about 50 km/h.
The driver: a 60 kg driver at 10 m/s presses on the seat with 60 × 4.8 = 288 N instead of 588 N. That drop in apparent weight is the stomach-lifting feeling.
A smaller normal force also means less friction is available, so braking or steering on the crest of a hump is harder at speed.
Worked example 2: the same car in a dip
Now the road curves the other way, a dip of radius 20 m, and the centre is above the car. The normal force points towards the centre, so
N − mg = mv²/r, giving N = m(g + v²/r)
At 10 m/s, N = 1000 × (9.8 + 5.0) = 14,800 N. The car is pressed into the road, its suspension compresses, and the driver feels heavier. There is no lift-off speed in a dip; instead the load on tyres and springs grows with speed.
Worked example 3: past the crest
Suppose the car keeps a steady 12 m/s on a circular hump of radius 20 m. At an angle θ from the top, only the component of gravity along the radius, mg cos θ, points towards the centre:
N = mg cos θ − mv²/r
At the very top, N = 1000 × (9.8 − 7.2) = 2600 N, so the car is still in contact. As it moves past the crest, cos θ falls. N reaches zero when cos θ = v²/(gr) = 144/196 ≈ 0.735, that is θ ≈ 42.7°. So a car can stay on the crest and still leave the road on the way down.
The classic variant. A small block starts from rest at the top of a smooth hemisphere of radius R and slides down. Here the speed is not fixed: energy conservation gives v² = 2gR(1 − cos θ). Set N = 0, so g cos θ = v²/R. Then cos θ = 2(1 − cos θ), and cos θ = 2/3, about 48.2° from the top. The block leaves the surface at a height 2R/3 above the centre.
Why is this easy to get wrong?
Uniform circular motion is a known difficulty, even for future teachers. Mutsvangwa (2020) tested 41 second-year physical sciences student teachers with a 20-question multiple-choice test and found that factual misconceptions and conceptual misunderstandings about uniform circular motion were prevalent. For hump problems, stay in the ground frame: draw only the real forces, weight and the normal force, and set their net towards the centre equal to mv²/r. Adding an outward "centrifugal force" as well counts the same effect twice.
Which mistakes should you avoid?
Writing N = mg at the top. That is only true when the car is at rest.
Using the same sign for hump and dip. Decide where the centre is first.
Treating mv²/r as a force on the diagram. It is what the net force must equal.
Fixing the speed in the dome problem. A sliding block speeds up, so use energy.
For level and banked roads and the vertical circle on a string, see circular motion and banking problems. For speed changing along a curve, read non-uniform circular motion.
A practice routine
Redo Example 1 with r = 40 m and find the new lift-off speed.
Do the dip version of the same numbers.
Solve the dome problem without notes, then for a block given a small push at the top.
Draw the free-body diagram first, every time, and mark where the centre is.
If you are not sure where the problem starts, Fermi's AI tutor can work out what you already know and build a plan for the session, so you sit down to a clear next step.
Before your next mechanics test, find the lift-off speed for a hump of radius 10 m and check it gives N = 0.
