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Springs and Vertical Circles: Energy Conservation Explained

Quick answer: Spring energy and vertical circles made simple: the two classic energy stages every JEE and NEET aspirant must master, with worked examples and quick revision points.

In one line: Springs and Vertical Circles — exam-ready notes in one glance.

In one line: JEE/NEET Physics · Work, Energy & Power series · Part 7 of 8 · All parts →✪ Key points — the 30-second versionSprings obey Hooke's law: pull force =.

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
In fact, 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. Moreover, 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.

Hooke’s law, simply.
Therefore, the vertical circle’s top point: the weak link.
The √(gR) rule, derived.
Energy connects the levels.
Solved examples.
Common mistakes.
Meanwhile, this physics in your daily life.
Practice set.
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)

Letter.	What it means (plain words).	Value / unit.
F.	the force the spring pulls back with.	N.
k.	stiffness — newtons per metre of stretch.	N/m.
x.	stretch (or squeeze) from natural length.	m.

Combined with energy: stretch stores ½kx², and releasing converts it to KE. As a result, 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 other words, in a vertical circle. Meanwhile, the top is the danger point — gravity pulls you toward the centre (helping the circle) and speed is lowest there (energy spent on climbing). In fact, the question: how slow can you go at the top and still keep the circle?

The √(gR) Rule, Derived.

Notably, at the top, gravity pulls down — straight toward the centre. Meanwhile, 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). Indeed, at or above it, the track/rotation holds. Meanwhile, 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. Specifically, at the circle’s top, gravity points toward the centre — it’s an ally. Indeed, at the bottom, it’s opposition (the track must push extra).
Centimetres in ½kx². 5 cm = 0.05 m, always — the eternal spring trap.
Getting the tension wrong at the top. Similarly, 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.

Overall, 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.
Consequently, 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)

One idea, three doors — open whichever clicks for you

Same concept (why spring energy is ½kx² and loops trade height for speed), three different ways of seeing it. If one door confuses you, try the next — at least one will stick.

Door 1 · The story way

A trampoline catch: the deeper you sink it, the HARDER it pushes back — resistance grows with stretch. That growing resistance is why spring energy isn’t just force × distance: the average force over the stretch is only half the final force.

Door 2 · The numbers way

Stretch a spring (k = 100 N/m) by 0.2 m: final force 20 N, average force 10 N, energy = 10 × 0.2 = 2 J = ½kx². The ½ is the triangle’s area — force ramping from 0 to 20 over the stretch. Vertical loop: at the top, mg = mv²/r; solve and v_top = √(gr) — the minimum loop speed.

Door 3 · The picture way

Draw force versus stretch: a straight diagonal line from zero. The stored energy is the shaded triangle under it — base x, height kx, area ½kx². For the loop: a circle with speed arrows — long at the bottom, shortest at the top — and a height ledger trading between them.

Why is this happening at all? Why the ½? Because spring force ramps linearly from zero, so the average over the pull is half the peak — energy is the area under the force curve, and a ramp to a peak only fills half the rectangle. Why do loops work? Because gravity is conservative: every metre of height lost is refunded exactly as speed, so only the difference matters.

▶ 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

← Part 6: Collisions: The Great Sorting — What Survives, What Dies Part 8: The Finale: Energy in the Real World, and the Complete Formula Card →

Frequently Asked Questions.

What should you know about Hooke's Law, Simply?

What should you know about The Vertical Circle's Top Point: The Weak Link?

What should you know about The √(gR) Rule, Derived?

Furthermore, at the top, gravity pulls down — straight toward the centre. Meanwhile, in the most desperate case, gravity alone supplies the entire inward push the circle demands : Below √(gR). Gravity wants more inward pull than the circle’s path can provide — the object leaves the circle (water leaves the bucket). In other words, at or above it, the track/rotation holds. Meanwhile, one number, universal: bucket, plane, rollercoaster, satellite (whose ‘circle never fails’ because it’s always in free fall).

What should you know about Energy Connects the Levels?

What should you know about Solved Examples?

Note: force linear (20 N), energy quadratic — different books. ✔ Using √(gR) as the BOTTOM speed. It’s the TOP minimum. Bottom needs √(5gR). Forgetting gravity helps at the top. Likewise, at the circle’s top, gravity points toward the centre — it’s an ally. Meanwhile, at the bottom, it’s opposition (the track must push extra).

Examiner’s Corner: The Three Traps

Trap 1: Forgetting the pivot is a point, not a support

In vertical-circle problems the top of the circle is the stress point: the minimum speed there is sqrt(gR) for a string or a track contact, because gravity alone must supply the centripetal force. Students who draw the free-body diagram at the bottom and never check the top lose the whole question. At the bottom, the tension or normal force is maximum: T = mv²/r + mg; at the top, minimum: T = mv²/r − mg. The difference between the two readings is 6mg for a full swing at critical speed — a favourite numerical.

Trap 2: Mixing spring energy with gravitational potential

A spring released from natural length while a mass falls does not convert all gravitational potential into spring energy unless the question says equilibrium. At maximum stretch the mass is momentarily at rest but acceleration is not zero — it is (kx minus mg)/m. Equilibrium stretch, where net force vanishes, is mg/k, exactly half the maximum stretch when dropped from rest. That factor-of-two distinction generates endless single-digit questions.

Trap 3: SHM disguised as circular motion

Uniform circular motion projected on any diameter is SHM. So a spring-block oscillation problem can be solved by energy conservation or by the SHM toolkit; both must agree. When an exam gives a spring constant and a mass and asks for the time period, T = 2*pi*sqrt(m/k) is the one-line route — do not re-derive from force equations under time pressure.

Worked Mini-Set

A 0.5 kg mass on a spring (k = 200 N/m) is pulled 5 cm and released. Find T and maximum speed. (T = 2*pi*sqrt(0.5/200) about 0.31 s; v = A*omega = 0.05 * sqrt(200/0.5) about 1 m/s.)
A stone on a 1 m string just completes a vertical circle. Speed at the top? (sqrt(gR) about 3.1 m/s; at the bottom sqrt(5gR) about 7 m/s.)
A block falls onto a spring from 20 cm above. Maximum compression versus equilibrium compression? (Maximum is where all gravitational energy is stored; equilibrium is mg/k — the falling case gives the larger, by the factor relation above.)

Sources and further reading

References & authoritative sources

Source: compiled from official notifications, standard textbooks and our own mock-test analytics; last reviewed September 2026.

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
In fact, a pail, a plane looping, a satellite: one rule, √(gR)
Hooke’s law, simply.

Series · Part 7 of 8

Physics: Work, Power and Energy

1Work Done: When a Force Actually Achieves Something
2Kinetic Energy and the Work-Energy Theorem
3Potential Energy: Stored Work, Ready to Strike
4Conservation of Energy: The Universe’s Perfect Bookkeeping
5Power and Efficiency: How FAST You Can Do the Work
6Collisions: The Great Sorting — What Survives, What Dies
7Springs and Vertical Circles: Energy in Two Classic Stages
8Energy in the Real World: The Formula Card

Next: Part 8 — Energy in the Real World: The Formula Card →

Download the whole series as PDF →

Have a doubt on this topic?

Sources & official references

NEET (UG) official site - NTA https://jeemain.nta.nic.in | JEE Main official site - NTA https://www.pib.gov.in | Press Information Bureau https://www.indiacode.nic.in | India Code - central acts

External references for fact-checking and further reading.

Springs and Vertical Circles: Energy in Two Classic Stages