Question
A car drives over a circular hump of radius . At the top of the hump, why does the apparent weight (normal force) decrease? And what is the maximum speed before the car loses contact? Many students confuse this with circular motion in a vertical loop and apply wrong sign conventions.
Solution — Step by Step
At the top of the hump, the centre of the circular path is below the car (the hump curves downward). So centripetal acceleration points downward, toward the centre.
This is the opposite of a car in a valley, where the centre is above.
Forces on the car at the top:
- Weight acting downward (toward the centre)
- Normal force acting upward (away from the centre)
Net inward (downward) force = centripetal force:
Solving:
So . The driver feels lighter — apparent weight is reduced.
The car loses contact when :
Apparent weight = . Maximum speed before losing contact = .
Why This Works
The centripetal direction always points toward the centre of the circular path. Whether forces help or oppose depends on this geometry. Over a hump: gravity helps (points toward centre), normal opposes. In a valley: gravity opposes (points away from centre), normal helps.
If the required centripetal force is more than gravity can provide alone, the surface must “pull” the car down — but a road surface can only push, not pull. So the car flies off at exactly .
For circular motion problems, always sketch the centre of the circle first, mark inward direction, and write . Sign confusion vanishes.
Alternative Method
Energy/free-fall view: at the critical speed, the car is in projectile motion with zero normal force, like an object thrown horizontally that just barely follows the curvature. Setting gravity equal to centripetal acceleration: , so . Same answer.
Common Mistake
Students often write (treating as inward) over a hump. That gives the wrong answer because it implies helps the centripetal force, when actually opposes it. Always draw the diagram and identify which forces point toward the centre before writing the equation.