← Mechanical Systems

4b Vertical circular motion and apparent weight

A recurring context every year: a loop, a roundabout, or an orbit, where gravity does not always point the same way relative to the centripetal direction — this changes which forces combine to supply F_c at different points.

At the top vs the bottom of a vertical loop

  • At the TOP: gravity and the normal force both point toward the centre (downward), so N + mg = mv²/r.
  • At the BOTTOM: gravity points away from the centre (down) while the normal force points toward it (up), so N − mg = mv²/r.
  • MINIMUM speed at the top is when N = 0 and gravity alone supplies the centripetal force: mg = mv²/r, giving v_min = √(gr).
  • Below v_min, the object cannot maintain contact with the track at the top — it falls away from the circular path.

Over a bump and through a dip

  • On TOP of a bump the centre is below the object: gravity points towards it and the reaction force away, so mg − N = mv²/r. The faster it goes, the smaller N gets.
  • It leaves the surface when N = 0, so the MAXIMUM speed over the top is v = √(gr). (At the top of a loop the same √(gr) is a minimum: there N points towards the centre.)
  • At the bottom of a dip the centre is above: N − mg = mv²/r, so N is bigger than the weight, as at the bottom of a loop.
Real paper — 2023 Feeling heavier at the bottom, lighter at the top Explain why a rider feels heavier at the bottom of a vertical loop and lighter at the top. At the BOTTOM, N − mg = mv²/r, so N = mg + mv²/r — the normal force (what the rider feels pressing on them) is LARGER than their weight, so they feel heavier. At the TOP, N + mg = mv²/r, so N = mv²/r − mg — the normal force is SMALLER than their weight, so they feel lighter. What you "feel" as weight is the normal force the seat exerts on you, not gravity itself — that is why it changes around the loop even though gravity is constant. ✓ Apparent weight = the normal force, which is larger than mg at the bottom and smaller than mg at the top of a vertical circle. Real paper — 2025 Why orbiting astronauts appear weightless yet stay in orbit Explain why astronauts on the ISS appear weightless yet stay in orbit. Gravity has NOT disappeared — it is still acting on the ISS and everything in it, providing exactly the centripetal force needed to keep them in orbit. Because gravity alone supplies F_c (there is no separate normal force from a floor pushing up on them), the astronauts and the station are both in continuous freefall together around the Earth. With nothing pushing back on them (no normal force), they experience no apparent weight — this is why it is "apparent weightlessness", not an absence of gravity. ✓ Gravity supplies the centripetal force for the whole orbit; with no normal force acting, apparent weight is zero even though real gravity is still very much present. ✕

Confusing "weightless" with "no gravity"

An orbiting astronaut is NOT beyond the reach of gravity — gravity at ISS altitude is only slightly weaker than at the surface. What is actually true is that gravity is providing the ENTIRE centripetal force, with no normal force from a floor to push back against them — that absence of a normal force is what "weightlessness" means, not an absence of gravity itself.

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