Why do the string and rod models have different critical speeds at the top?

A string can only pull. At the top gravity alone must at least supply the centripetal force, mg = mv²/R, so v_min = √(gR); any slower and the string goes slack and the ball leaves the circle early. A rod can pull or push, so the speed at the top can drop to zero with the rod pushing up with mg. Hence a string (or inside track) needs at least √(5gR) at the bottom, while a rod (or tube) needs only √(4gR).

When does the rod pull and when does it push at the top?

Taking the rod force downward as positive, F + mg = mv²/R, so F = mv²/R − mg. For v > √(gR) F is positive and the rod pulls (in a tube, the outer wall presses in). At v = √(gR) there is no rod force. For v < √(gR) F is negative and the rod pushes the ball up (the inner wall of a tube). Drag the bottom speed in the tool and watch the top force change direction at √(gR).

Why is the string tension always 6mg larger at the bottom than at the top?

At the bottom T₁ − mg = mv₁²/R and at the top T₂ + mg = mv₂²/R. Subtracting gives T₁ − T₂ = 2mg + m(v₁² − v₂²)/R, and energy conservation gives v₁² − v₂² = 4gR, so T₁ − T₂ = 6mg whatever the speed. The F–θ graph in the tool labels the values at 0° and 180° so you can check it directly.

Where does a ball that is too slow leave the circle?

If 2gR < v₀² < 5gR the ball rises above the centre but the tension drops to zero somewhere in the upper half. If the radius there makes angle β with the vertical, mg cosβ = mv²/R and ½mv₀² = ½mv² + mgR(1 + cosβ), giving cosβ = (v₀² − 2gR)/(3gR). After that it is a projectile, and the tool draws its parabola. If v₀² ≤ 2gR the ball just swings below the centre and never leaves.

With two blocks tied together on a turntable, which slides first and what is the critical ω?

The block farther out needs more centripetal force mω²r, so it reaches maximum static friction first and then the string starts to pull. Tied on one side, the outer block gives T + μm₂g = m₂ω²r₂ and the inner one μm₁g − T = m₁ω²r₁, so ω² = μg(m₁ + m₂)/(m₁r₁ + m₂r₂). Tied through the axis on opposite sides, the inner block’s friction reverses and the limit is reached when its outward friction is maximal: ω² = μg(m₁ + m₂)/(m₂r₂ − m₁r₁). The tool plots both frictions and the tension against ω.

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🎡Vertical Circular Motion - String vs Rod, Loop-the-Loop, Turntable, Banked Curves

Q: Why is the string tension always 6mg larger at the bottom than at the top?

At the bottom T₁ − mg = mv₁²/R and at the top T₂ + mg = mv₂²/R. Subtracting gives T₁ − T₂ = 2mg + m(v₁² − v₂²)/R, and energy conservation gives v₁² − v₂² = 4gR, so T₁ − T₂ = 6mg whatever the speed. The F–θ graph in the tool labels the values at 0° and 180° so you can check it directly.

Q: Where does a ball that is too slow leave the circle?

If 2gR < v₀² < 5gR the ball rises above the centre but the tension drops to zero somewhere in the upper half. If the radius there makes angle β with the vertical, mg cosβ = mv²/R and ½mv₀² = ½mv² + mgR(1 + cosβ), giving cosβ = (v₀² − 2gR)/(3gR). After that it is a projectile, and the tool draws its parabola. If v₀² ≤ 2gR the ball just swings below the centre and never leaves.

Q: With two blocks tied together on a turntable, which slides first and what is the critical ω?

The block farther out needs more centripetal force mω²r, so it reaches maximum static friction first and then the string starts to pull. Tied on one side, the outer block gives T + μm₂g = m₂ω²r₂ and the inner one μm₁g − T = m₁ω²r₁, so ω² = μg(m₁ + m₂)/(m₁r₁ + m₂r₂). Tied through the axis on opposite sides, the inner block’s friction reverses and the limit is reached when its outward friction is maximal: ω² = μg(m₁ + m₂)/(m₂r₂ − m₁r₁). The tool plots both frictions and the tension against ω.

This tool animates the classic circular-motion exam problems: a ball on a string (or on the inside of a track), a ball on a rod (or in a tube), a loop-the-loop entered from height h, blocks on a turntable, a conical pendulum, a banked railway curve and a car on a humped or dipped bridge. It uses energy conservation and Newton’s second law to find the speed and the string, rod or track force at every point, simulates where the ball leaves the circle and its projectile path afterwards, and writes out each step with the numbers substituted.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Vertical Circular Motion

1Pick the string, rod, loop-the-loop, turntable, conical pendulum, rail curve or bridge tab
2Choose a classic case or adjust mass, radius, speed, height, friction or angle
3Watch the force arrows and the F–θ graph, and read the results and step-by-step analysis
4Open Practice for random circular-motion problems with instant marking

FAQ

Why do the string and rod models have different critical speeds at the top?: A string can only pull. At the top gravity alone must at least supply the centripetal force, mg = mv²/R, so v_min = √(gR); any slower and the string goes slack and the ball leaves the circle early. A rod can pull or push, so the speed at the top can drop to zero with the rod pushing up with mg. Hence a string (or inside track) needs at least √(5gR) at the bottom, while a rod (or tube) needs only √(4gR).
When does the rod pull and when does it push at the top?: Taking the rod force downward as positive, F + mg = mv²/R, so F = mv²/R − mg. For v > √(gR) F is positive and the rod pulls (in a tube, the outer wall presses in). At v = √(gR) there is no rod force. For v < √(gR) F is negative and the rod pushes the ball up (the inner wall of a tube). Drag the bottom speed in the tool and watch the top force change direction at √(gR).
Why is the string tension always 6mg larger at the bottom than at the top?: At the bottom T₁ − mg = mv₁²/R and at the top T₂ + mg = mv₂²/R. Subtracting gives T₁ − T₂ = 2mg + m(v₁² − v₂²)/R, and energy conservation gives v₁² − v₂² = 4gR, so T₁ − T₂ = 6mg whatever the speed. The F–θ graph in the tool labels the values at 0° and 180° so you can check it directly.
Where does a ball that is too slow leave the circle?: If 2gR < v₀² < 5gR the ball rises above the centre but the tension drops to zero somewhere in the upper half. If the radius there makes angle β with the vertical, mg cosβ = mv²/R and ½mv₀² = ½mv² + mgR(1 + cosβ), giving cosβ = (v₀² − 2gR)/(3gR). After that it is a projectile, and the tool draws its parabola. If v₀² ≤ 2gR the ball just swings below the centre and never leaves.
With two blocks tied together on a turntable, which slides first and what is the critical ω?: The block farther out needs more centripetal force mω²r, so it reaches maximum static friction first and then the string starts to pull. Tied on one side, the outer block gives T + μm₂g = m₂ω²r₂ and the inner one μm₁g − T = m₁ω²r₁, so ω² = μg(m₁ + m₂)/(m₁r₁ + m₂r₂). Tied through the axis on opposite sides, the inner block’s friction reverses and the limit is reached when its outward friction is maximal: ω² = μg(m₁ + m₂)/(m₂r₂ − m₁r₁). The tool plots both frictions and the tension against ω.

Features

String vs rod: critical top speed √(gR) vs 0, and the rod force switching from push to pull at √(gR)
Simulates where a ball leaves the circle, its speed there and the projectile path that follows
F–θ graph of the string or track force, showing F_bottom − F_top = 6mg
Loop-the-loop from height h, with the h ≥ 2.5R condition
Turntable with one or two blocks (free, tied on one side, tied through the axis): critical ω and friction–ω graph
Conical pendulum, smooth bowl, rail design speed and flange forces, humped and dipped bridges, plus random practice

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