Not every loop is travelled at a steady speed. When gravity has a say in the matter — like a ball on a string swinging over the top of a circle — the resultant force keeps changing, and so does the speed. This is non-uniform circular motion.
📘 What you need to know
Non-uniform circular motion happens when the resultant force on an object changes as it moves round the circle — a vertical circle is the classic example
Weight always points straight down, so it adds to or subtracts from the string tension depending on where the object is
Tension (or normal force) is at its maximum at the bottom and its minimum at the top of the loop
Because the resultant force changes, the object’s speed changes too — fastest at the bottom, slowest at the top
There is a critical minimum speed at the top of the loop, below which the string goes slack (or a car loses contact with the track)
Why the Motion Isn’t Uniform
In uniform circular motion, the centripetal force always points towards the centre and stays constant in size — think of a puck on a frictionless table tied to a fixed post. But swing an object in a vertical circle, and its weight mg never changes direction — it always points down towards the ground, no matter where the object is on the loop.
This means the string (or track) has to do a different amount of work depending on position:
At the bottom, tension must support the weight and supply the centripetal force, so it is largest here
At the top, weight already points towards the centre, so tension only needs to make up the rest — it is smallest here
At the sides, weight acts at 90° to the string, contributing nothing to the centripetal force at that instant
Tension (teal) always points towards the centre; weight (red) always points straight down. Their combination is what changes as the ball goes round.
Tension at the Top and Bottom
At the bottom of the loop, tension and the centripetal force requirement point the same way (towards the centre, i.e. upwards), while weight pulls the opposite way. Tension has to overcome weight and provide the centripetal force:
Tension at the bottom
Tmax = mv2 ÷ r + mg
At the top of the loop, weight already points towards the centre (downwards), so it does part of the job for you. Tension only has to supply what weight doesn’t:
Tension at the top
Tmin = mv2 ÷ r − mg
Because v also changes around the loop (energy is conserved, so the object slows down as it climbs and speeds up as it falls), the difference between Tmax and Tmin is even larger than these two equations suggest on their own.
The Critical Condition at the Top
A string can only pull — it can’t push. So tension can never be negative. Looking at the formula for the top of the loop, if the speed drops too low, Tmin would have to become negative to keep the maths balanced. In reality, this can’t happen: the string simply goes slack, and the object stops following a circular path.
The slowest the object can go at the top and still keep the string taut is when tension has dropped to exactly zero — at that point, weight alone supplies all the centripetal force needed:
Minimum speed at the topvmin = √(gr)
The same idea applies to a car on the inside of a vertical loop track, except the normal force from the track takes the place of tension — if the car goes too slowly, it loses contact with the track before reaching the top.
🧭 Solving a vertical circular motion problem
Sketch the object at the position you’re interested in, and mark on the weight (always straight down) and the tension or normal force (always towards the centre)
Identify whether the two forces act in the same direction or opposite directions at that point
Write the resultant, taking care with signs, and set it equal to mv2 ÷ r
Rearrange for whatever you need — tension, speed, or radius
If a “just barely makes it round” scenario is described, set tension (or normal force) to zero before rearranging
Quick recap: in a vertical circle, weight never changes direction — so tension is greatest at the bottom, least at the top, and the string can only go slack at the top, never the bottom.
WE 1
A 0.45 kg ball is attached to a string of length 0.80 m and swung in a vertical circle. At the top of the circle, the ball has a speed of 4.0 m s⁻¹.
(a) Calculate the tension in the string at the top.
(b) Determine the minimum possible speed at the top for the string to stay taut.
Part (a)
At the top: T = mv²/r − mg
T = (0.45 × 4.0²) ÷ 0.80 − (0.45 × 9.81)T = 9.0 − 4.41T ≈ 4.6 NPart (b)
Slack string means T = 0, so weight alone provides the centripetal force
vmin = √(gr) = √(9.81 × 0.80)vmin ≈ 2.8 m s⁻¹Since 4.0 m s⁻¹ is comfortably above 2.8 m s⁻¹, the string does stay taut here.
WE 2
A roller-coaster car of mass 600 kg enters a vertical circular loop of radius 10 m. Assume the track is frictionless.
(a) Calculate the minimum speed needed at the top of the loop for the car to stay on the track.
(b) Use conservation of energy to find the minimum speed needed at the bottom of the loop.
Part (a)vtop = √(gr) = √(9.81 × 10)vtop ≈ 9.9 m s⁻¹Part (b)
Height gained from bottom to top = 2r, so:
½vbottom² = ½vtop² + g(2r)vbottom² = 98.1 + 4 × 9.81 × 10 = 490.5vbottom ≈ 22 m s⁻¹Notice how much faster the car must be moving at the bottom — that’s the height gain (energy) as well as the circular motion requirement.
💡 Top tips
Always draw weight pointing straight down, no matter where the object is on the circle — it’s tension (or normal force) that changes direction to stay pointed at the centre
“Just completes the loop” or “string stays taut” is exam code for tension (or normal force) = 0 at the top
Speed isn’t constant around a vertical circle — don’t reuse the same v from the bottom of the loop when working at the top
Pair this topic with conservation of energy when a question asks you to compare speeds at different heights around the loop
⚠ Common mistakes
Assuming tension is the same all the way round — it isn’t, because weight’s direction relative to the string keeps changing
Forgetting that tension can’t be negative, and getting a “negative tension” answer without realising the string must have already gone slack
Mixing up the top and bottom equations — check whether weight is helping or opposing the centripetal force before writing the resultant
Using the same speed value at every point on the loop instead of accounting for the object slowing down near the top
Up next: with forces and momentum now covered, we move into circular motion’s close cousin — rotational quantities and the language of angular velocity showing up in fields, before Theme B picks up with thermal energy transfers.
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