In uniform circular motion the speed stays constant. In non-uniform circular motion — like a ball swinging in a vertical circle — the speed and the tension change as the object moves around. The key is applying Newton’s second law at each position, where gravity and tension combine differently depending on where you are on the circle.
📘 What you need to know
In a vertical circle, the speed and tension both vary because weight always acts downward while the required centripetal force changes direction.
At the bottom: tension and centripetal force point up; weight points down → Tmax = mv² / r + mg.
At the top: tension and weight both point down (towards the centre) → Tmin = mv² / r − mg.
Minimum speed at the top: set T = 0 → vmin = √(gr).
The object is fastest at the bottom and slowest at the top — energy conservation explains why.
Why the speed and tension change
In a vertical circle, gravity acts downward at every point. As the object rises, gravity does negative work and slows it down; as it falls, gravity does positive work and speeds it up. This is unlike a horizontal circle, where gravity is perpendicular to the motion throughout.
The tension in the string must provide the centripetal force after accounting for whichever component of gravity already acts towards (or away from) the centre. So the tension — and therefore the net inward force — varies continuously around the circle.
Weight (red, mg) always points down. Tension (blue) always points towards the centre. At the top both are inward — so tension is smallest. At the bottom tension fights weight — so it is largest.
Newton’s second law at the top and bottom
The trick is to take inward as positive at each position, then apply Fnet = mv² / r.
At the bottom (tension is maximum)
Tension points up (inward), weight points down (outward). The net inward force is T − mg:
Bottom of the circleTmax − mg = mv² / r
→
Tmax = mv² / r + mg
At the top (tension is minimum)
Both tension and weight point down (inward). The net inward force is T + mg:
Top of the circleTmin + mg = mv² / r
→
Tmin = mv² / r − mg
Minimum speed at the top
For the object to maintain contact with the circular path at the top, the string must remain taut. The critical condition is when the tension just reaches zero — at that point, gravity alone supplies the centripetal force:
Minimum speed at the top (T = 0)mg = mv² / r
→
vmin = √(gr)
Below this speed, gravity would need to be supplemented by an inward tension — but the string can only pull, not push, so it goes slack and the object leaves the circular path.
🛠 Solving a vertical-circle problem
Identify the position (top, bottom, or side).
Draw the forces: tension inward along the string, weight downward.
Take inward as positive and write: Fnet (inward) = mv² / r.
At the bottom: T − mg = mv² / r.
At the top: T + mg = mv² / r.
For minimum speed at the top, set T = 0 and solve.
Quick reference: bottom → Tmax = mv²/r + mg • top → Tmin = mv²/r − mg • minimum speed at top → vmin = √(gr). Object is fastest at bottom, slowest at top.
Worked examples
WE 1
Minimum speed at the top of a loop
An 8.0 kg bucket of water is swung in a vertical circle on a 0.50 m string. Find the minimum speed at the top so the water stays in the bucket. (Take g = 9.81 m s⁻².)
At minimum speed, tension T = 0gravity alone provides the centripetal forcemg = mv² / r → v = √(gr)v = √(9.81 × 0.50)vmin ≈ 2.21 m s⁻¹
WE 2
Tension at the bottom of the swing
A 0.40 kg ball on a 0.60 m string swings in a vertical circle. At the bottom its speed is 4.0 m s⁻¹. Find the tension in the string at that point. (Take g = 9.81 m s⁻².)
At the bottom: T − mg = mv² / rT = mv² / r + mg= (0.40 × 4.0²) / 0.60 + (0.40 × 9.81)= (0.40 × 16) / 0.60 + 3.924= 10.67 + 3.924T ≈ 14.6 N
WE 3
Tension at the top
The same ball (0.40 kg, 0.60 m string) has speed 2.5 m s⁻¹ at the top. Find the tension at the top.
At the top: T + mg = mv² / rT = mv² / r − mg= (0.40 × 2.5²) / 0.60 − (0.40 × 9.81)= (0.40 × 6.25) / 0.60 − 3.924= 4.167 − 3.924T ≈ 0.24 Njust above zero — the ball barely stays on its circular path
💡 Top tips
Always take inward as positive and write the net inward force equal to mv² / r.
At the top, both T and mg point inward (down) → they add. At the bottom, T points inward but mg points outward → they subtract.
Minimum speed means T = 0 — gravity alone provides centripetal force, giving vmin = √(gr).
Fastest at the bottom, slowest at the top — energy is being traded between kinetic and gravitational potential.
If T comes out negative, the required centripetal force is less than mg alone can provide — the object can’t maintain circular motion at that speed.
⚠ Common mistakes
Using T − mg = mv²/r at the top. At the top both T and mg point inward, so it’s T + mg = mv²/r.
Using T + mg = mv²/r at the bottom. At the bottom mg points outward, so it’s T − mg = mv²/r.
Thinking tension is constant around the circle — it isn’t; it changes at every point because speed changes too.
Forgetting to check the sign of T. If T < 0, the object can’t complete the loop at that speed.
That’s the full Forces & Momentum unit covered. The golden rule for vertical-circle questions is always the same: draw the forces, pick inward as positive, and set the net inward force equal to mv²/r. Get that habit right and every variation of this problem becomes routine.
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