You are currently viewing Springs and Vertical Circles: Energy in Two Classic Stages
Engineering Exams6 min readAug 30, 2026

Springs and Vertical Circles: Energy in Two Classic Stages

Springs and Vertical Circles: Energy in Two Classic Stages
6 min read · 1,013 words

JEE/NEET Physics · Work, Energy & Power series · Part 7 of 8 · All parts →

✪ Key points — the 30-second version

  • Springs obey Hooke’s law: pull force = stiffness × stretch (F = kx)
  • Vertical circles: minimum top speed = √(gR) (gravity supplies the whole inward push)
  • Energy solves both: ½kx² ↔ KE ↔ mgh trades
  • Water in a rotating bucket doesn’t fall — the circle’s demand holds it
  • A pail, a plane looping, a satellite: one rule, √(gR)

Swing a bucket of water over your head — the water stays in. Loop a plane — passengers are pushed into seats, not belts. Both are one rule about the minimum speed at the top of a vertical circle. Part 7 of the Work, Energy & Power series: two classic energy stages every exam loves.

In this card

  1. Hooke’s law, simply
  2. The vertical circle’s top point: the weak link
  3. The √(gR) rule, derived
  4. Energy connects the levels
  5. Solved examples
  6. Common mistakes
  7. This physics in your daily life
  8. Practice set
  9. Recap

Hooke’s Law, Simply

The vertical circle: the TOP is the weak link — gravity alone must supply the whole inward push, giving the minimum speed √(gR)

TOP: min speed √(gR) gravity pulls toward centre (helper) BOTTOM: needs √(5gR) gravity opposes (must climb out) release height h = 2.5R

pull-back force = stiffness × stretch  (F = kx)double the stretch, double the pull — and the stored energy squares (½kx², Part 3)
LetterWhat it means (plain words)Value / unit
Fthe force the spring pulls back withN
kstiffness — newtons per metre of stretchN/m
xstretch (or squeeze) from natural lengthm

Combined with energy: stretch stores ½kx², and releasing converts it to KE. This pairing (F = kx to find forces; ½kx² to find energy) solves every spring question.

The Vertical Circle’s Top Point: The Weak Link

In a vertical circle, the top is the danger point — gravity pulls you toward the centre (helping the circle) and speed is lowest there (energy spent on climbing). The question: how slow can you go at the top and still keep the circle?

The √(gR) Rule, Derived

At the top, gravity pulls down — straight toward the centre. In the most desperate case, gravity alone supplies the entire inward push the circle demands:

mg = mv²/R  →  v_top = √(gR)the absolute minimum top speed — any slower, the circle fails and you drop

Below √(gR), gravity wants more inward pull than the circle’s path can provide — the object leaves the circle (water leaves the bucket). At or above it, the track/rotation holds. One number, universal: bucket, plane, rollercoaster, satellite (whose ‘circle never fails’ because it’s always in free fall).

Energy Connects the Levels

To find the minimum launch speed at the BOTTOM for a full loop: bottom speed must be enough to climb 2R and still have √(gR) at top. Energy: ½mv_b² = ½m(gR) + mg(2R) → v_b = √(5gR) — the famous √5, sibling of Part 4’s 2.5R height rule (they’re the same statement, one in speeds, one in heights).

Solved Examples

✎ Easy — Hooke. A spring (k = 400 N/m) stretched 5 cm. Pull-back force and stored energy?

Force: 400 × 0.05 = 20 N. Energy: ½(400)(0.05²) = 0.5 J.

Note: force linear (20 N), energy quadratic — different books. ✔

Answer: F = 20 N; E = 0.5 J

✎ Exam level — the bucket. Minimum speed at the top of a 1 m vertical circle (g = 10)?

√(gR): √(10 × 1) ≈ 3.16 m/s.

Feel it: one full turn per ~2 seconds — that’s why you swing a bucket briskly, not lazily. ✔

Answer: ≈ 3.16 m/s

✎ JEE level — the √5 launch. A bead on a frictionless vertical loop (R = 0.8 m). Minimum bottom speed to complete the loop (g = 10)?

Energy route: ½v_b² = ½(gR) + 2gR → v_b = √(5gR) = √(5×10×0.8) = √40.

v_b ≈ 6.32 m/s.

Cross-check with Part 4: release height needed = 2.5R = 2 m → v from 2 m drop = √(2×10×2) = √40 ✔ — speeds and heights tell the same story.

Answer: v_b = √(5gR) ≈ 6.32 m/s

⚠ Mistakes students make — and how to avoid them

  • Using √(gR) as the BOTTOM speed. It’s the TOP minimum. Bottom needs √(5gR).
  • Forgetting gravity helps at the top. At the circle’s top, gravity points toward the centre — it’s an ally; at the bottom, it’s opposition (the track must push extra).
  • Centimetres in ½kx². 5 cm = 0.05 m, always — the eternal spring trap.
  • Keeling the tension wrong at the top. At minimum speed the track/string pushes (or pulls) with ZERO extra force — gravity does it all. That’s the meaning of √(gR).

This Physics in Your Daily Life

◎ This physics in your daily life

  • The bucket trick works exactly when your hand-side speed beats √(gR) — feel it fail as you slow: water falls from the top.
  • Rollercoaster loops are engineered above √(5gR) with safety margin — the screams at the top are physics holding you in.
  • Washing machine spin cycles: the drum spins clothes at speeds where water ‘can’t stay’ in the fabric — it leaves through the holes tangentially. √(gR) logic, laundry edition.
  • Pilots looping aircraft feel ‘g-force’ at the loop’s BOTTOM (extra push needed) and lightness at the top — the vertical circle’s asymmetry, worn as body weight.
  • Every trampoline bounce is F = kx catching you and ½kx² returning you — this card’s two halves in one mattress.

Practice set (answers hidden — try first)

(NEET-level) Spring k = 200 N/m, x = 10 cm. Force:
F = 200 × 0.10 = 20 N.
(JEE Main-level) Minimum top speed in a 0.4 m vertical circle (g = 10):
√(10 × 0.4) = 2 m/s.
(JEE Main-level) Minimum bottom speed for the same loop:
√(5 × 10 × 0.4) = √20 ≈ 4.47 m/s.
(Concept) At the top at minimum speed, the string’s tension is:
Zero — gravity supplies the entire inward push.
(NEET-level) Doubling a spring’s stretch multiplies its stored energy by:
4 (x²).
🧠 Memory tricks & everyday anchors — the 20-second revision

  • 🧠 Chant: ‘top is √gR, bottom is √5gR, height is 2-and-a-half R’.
  • 🧠 Gravity flips roles: helper at the top, opponent at the bottom of a vertical circle.
  • 🏠 Daily: swing a bucket briskly — you’re personally verifying √(gR).
  • 🏠 Daily: the washing machine’s spin cycle is water failing to keep the circle.
  • 🔁 F = kx (linear) vs ½kx² (squared)
  • 🔁 v_top(min) = √(gR): gravity alone supplies the push
  • 🔁 v_bottom(min) = √(5gR)
▶ Recap card — save for revision week

  • Hooke: F = kx (force linear); energy: ½kx² (squared)
  • top of vertical circle: minimum v = √(gR)
  • bottom launch for a full loop: √(5gR)
  • height equivalent: 2.5R (same rule, in heights)
  • at minimum top speed, gravity supplies the whole inward push

Quick revision

  • Springs obey Hooke’s law: pull force = stiffness × stretch (F = kx)
  • Vertical circles: minimum top speed = √(gR) (gravity supplies the whole inward push)
  • Energy solves both: ½kx² ↔ KE ↔ mgh trades
  • Water in a rotating bucket doesn’t fall — the circle’s demand holds it
  • A pail, a plane looping, a satellite: one rule, √(gR)
  • Hooke’s law, simply
ShareTelegramX

Have a doubt on this topic?