# Vertical Circle Dynamics

> Vertical circle dynamics in JEE Physics: minimum speeds √(gR) and √(5gR), tension T = mv^2/R + mg cosθ, rod vs string cases and slack-string motion.

- Canonical URL: https://prepelephant.com/topics/jee/physics/vertical-circle-dynamics
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Vertical Circle Dynamics", PrepElephant, https://prepelephant.com/topics/jee/physics/vertical-circle-dynamics

## Direct answer

A body looping a vertical circle of radius R must clear a speed floor set by gravity: for a string or the inner surface of a track, the minimum speed at the top is √(gR) and the corresponding minimum at the bottom is √(5gR), because energy conservation links the two points through a 2R climb. At any angle θ measured from the lowest point, the tension follows T = mv²/R + mg cosθ while the speed obeys v² = u² − 2gR(1 − cosθ). A rod changes everything: it can also push, so the topmost speed may drop to zero and the body still completes the circle. Across the critical loop, T_bottom − T_top = 6mg — a number worth carrying into the exam hall.

## What you must remember

- **Critical speeds (string/inner track):** v_top = √(gR) when tension just vanishes; v_bottom = √(5gR) for the same loop; at the horizontal diameter level v = √(3gR).
- **General tension:** T = mv²/R + mg cosθ with θ from the lowest point; at the bottom T = mv²/R + mg, at the top T = mv²/R − mg.
- **The 6mg result:** for the minimum-speed loop, T_bottom = 6mg and T_top = 0, so the tension difference between bottom and top is always 6mg whenever the body just completes the circle.
- **Speed at angle θ:** v² = u² − 2gR(1 − cosθ) from energy conservation; only gravity does work on the bob, so mechanical energy is conserved throughout.
- **Rod versus string:** a rod or bucket (two-sided constraint) sustains zero or even downward-forcing contact, so the top speed can be zero; a string needs v_top ≥ √(gR) or it goes slack.
- **Leaving the circle:** the instant T = 0 (string) or N = 0 (track), the body leaves circular motion and behaves as a projectile launched along the tangent.
- **Water in a bucket:** the same v ≥ √(gR) at the top keeps water pressed to the bucket; the classic demonstration value at R = 1 m is about 3.1 m/s.

## Worked loop: minimum speed at the top

Run a 1 kg bob on a 1 m string at the tightest possible loop. At the top, gravity alone must supply the entire centripetal force: mg = mv²/R, so v_top = √9.8 ≈ 3.13 m/s and T_top = 0. Energy conservation from bottom to top gives v_bottom² = v_top² + 4gR = 9.8 + 39.2 = 49, so v_bottom = 7 m/s. Tension at the bottom: T = mv²/R + mg = 49 + 9.8 = 58.8 N, exactly 6mg. The whole loop's numbers — 3.13, 7, 58.8, and the intermediate v = √(3gR) = 5.42 m/s at the sides — fall out of two equations only, and memorising this ladder for R = 1 m, g = 9.8 m/s² converts many numericals into recognition.

The deeper JEE skill is the slack-string case. Launch the same bob from the bottom at 6 m/s, less than 7. It rises; at angle θ the tension is T = m(v²/R + g cosθ) with v² = 36 − 19.6(1 − cosθ). Setting T = 0: 36 − 19.6 + 19.6cosθ + 9.8cosθ = 0 gives cosθ = −16.4/29.4 ≈ −0.56, so θ ≈ 124° — above the horizontal, the string slackens, and from that instant the bob is a projectile launched at 30-odd degrees above the horizontal at v ≈ 2.35 m/s.

## Classic traps in loop problems

The most-tested distinction is constraint type: "minimum speed at the top for a rod" is zero, but for a string it is √(gR); questions deliberately mix a bead on a wire (two-sided), water in a bucket (one-sided if open at top? — actually still needing v ≥ √(gR)) and a mass on a spring. Second trap: applying T = mv²/R − mg at the bottom; the sign flips with position, which is why the θ-from-bottom formula is safer. Third: after the string slackens, students keep applying circular dynamics to the projectile leg; the moment T = 0 is precisely the moment the chapter changes.

## Frequently asked questions

### What is the minimum speed at the top of a vertical circle for a body on a string?

√(gR), obtained by setting tension to zero so that gravity alone provides the centripetal force, mg = mv²/R.

### Why is the bottom speed for a complete loop √(5gR)?

Energy conservation across the 2R climb gives v_bottom² = gR + 4gR = 5gR when the top speed is the critical √(gR).

### How does a rod change the minimum speed condition?

A rod can push as well as pull, so it prevents the body from falling away inside the circle; the topmost speed may be zero and the loop is still completed.

### When does the body leave the circular path?

The instant the constraint force reaches zero — tension for a string, normal reaction for a track — after which the body moves as a tangent-launched projectile.

### Is the tension difference between bottom and top always 6mg?

Only for the minimum-speed loop; in general T_bottom − T_top = 6mg holds whenever energy conservation links the speeds across a 2R climb, as it does for any completed loop.