Every circle so far has been travelled at a steady speed — uniform circular motion, with the inward force staying the same size all the way round. But swing a ball on a string in a vertical loop and that neat picture breaks. Gravity now helps the ball on the way down and fights it on the way up, so the speed changes, and the force needed to keep it on the circle changes too. This is non-uniform circular motion, and the key is to track how the forces line up at each point of the loop.
📘 What you need to know
In non-uniform circular motion the speed changes as the object goes round, because a resultant force acts along the motion
The classic case is a vertical circle, where weight acts downward everywhere while tension changes direction
At the bottom, tension and weight oppose — tension is at its maximum
At the top, tension and weight both point inward — tension is at its minimum
The object moves fastest at the bottom and slowest at the top
The minimum speed at the top (before it falls out of the circle) is vmin = √(gr)
Why the speed changes
In uniform circular motion the only force was directed straight to the centre, perpendicular to the motion, so it did no work and the speed stayed constant. In a vertical circle that’s no longer true. Weight always points down, so part of it usually lies along the direction of travel — and a force along the motion does work, changing the object’s speed.
On the way down, gravity has a component in the direction of motion, so the object speeds up. On the way up, gravity opposes the motion, so it slows down. The result: fastest at the bottom, slowest at the top.
Around a vertical loop the velocity (red) grows from top to bottom: longest arrow at the bottom (fastest), shortest at the top (slowest). Gravity speeds the ball up going down and slows it going up.
This is the whole difference from the last few pages. “Uniform” meant the inward force was constant and the speed never changed. “Non-uniform” means there’s now a bit of force pointing along the path as well as towards the centre — and that along-the-path bit is what changes the speed. Gravity is the culprit in a vertical circle.
Forces at the top and bottom
The cleanest way to handle a vertical circle is to look at the two special points — the very top and the very bottom — where the tension and weight both lie along the vertical, pointing straight towards or away from the centre. At every point the resultant of tension and weight must supply the centripetal force mv2/r towards the centre.
At the top, tension (teal) and weight (purple) both point down towards the centre, so they add. At the bottom, tension points up to the centre while weight points down, so they oppose — the tension has to be much bigger there.
At the bottom
Tension points up (towards the centre) and weight points down (away from the centre). The resultant towards the centre is T − mg, and this provides the centripetal force. So the string has to pull hard enough to both hold the weight up and bend the path:
At the bottom — maximum tensionTmax = mv2/r + mg
At the top
Now both tension and weight point down — both towards the centre. Gravity is already helping to bend the path, so the string needs to pull much less. The resultant towards the centre is T + mg:
At the top — minimum tensionTmin = mv2/r − mg
That’s why the string is tightest at the bottom of a swing and slackest at the top — and why the ball is most likely to fall out of its circular path near the top.
WE 1
A ball of mass 0.25 kg is swung on a string of length 0.80 m in a vertical circle. At the lowest point it is moving at 4.0 m s−1. Find the tension in the string there. (Take g = 9.81 m s−2.)
Step 1 — at the bottom, tension up, weight down
T − mg = mv² ÷ r
Step 2 — rearrange for T
T = mv²/r + mg
Step 3 — substitute the valuesT = (0.25 × 4.0²) ÷ 0.80 + (0.25 × 9.81)T = 5.0 + 2.45T = 7.5 NThe string must pull hard here — enough to hold the weight and bend the path.
WE 2
The same ball (mass 0.25 kg, string 0.80 m) is moving at 3.0 m s−1 at the highest point of the circle. Find the tension in the string there.
Step 1 — at the top, tension and weight both point down
T + mg = mv² ÷ r
Step 2 — rearrange for T
T = mv²/r − mg
Step 3 — substitute the valuesT = (0.25 × 3.0²) ÷ 0.80 − (0.25 × 9.81)T = 2.81 − 2.45T = 0.36 NFar smaller than at the bottom — gravity is doing most of the centripetal work up here.
The minimum speed at the top
Push this idea to its limit. As the ball slows near the top, the tension drops. If it goes too slowly, the string would need a negative tension — impossible, since a string can only pull, not push. At that point the string goes slack and the ball falls out of its circular path.
The slowest it can go while still on the circle is the moment the tension just reaches zero. Then gravity alone supplies the whole centripetal force:
At the top, when T = 0mg = mv2/r
The mass cancels, and rearranging gives the minimum speed at the top:
Minimum speed at the topvmin = √(gr)
Notice the mass cancelled — a heavy bucket and a light one need the same minimum speed at the top of a loop of the same size. This is the classic “bucket of water swung overhead” trick: swing it fast enough and the water stays in, because the bucket accelerates downward faster than the water would fall on its own. Go too slow, and you get wet.
🛠️ Solving a vertical-circle problem
Choose the point — top, bottom, or somewhere in between — that the question asks about.
Draw the forces there: weight always down, tension along the string towards the centre.
Find the resultant towards the centre and set it equal to mv2/r.
For the minimum top speed, set the tension (or normal force) to zero, so mg = mv2/r.
WE 3
A rollercoaster runs round a vertical loop of radius 8.0 m. Find the minimum speed the car must have at the top of the loop for the passengers to stay in contact with their seats. (Take g = 9.81 m s−2.)
Step 1 — at minimum speed the seat force is zero
gravity alone gives the centripetal force: mg = mv²/r
Step 2 — rearrange (mass cancels)
vmin = √(gr)
Step 3 — substitute the valuesvmin = √(9.81 × 8.0) = √78.5vmin = 8.9 m s⁻¹About 32 km/h — any slower and the passengers would start to lift out of their seats at the top.
💡 Top tips
Weight is always down. Only the tension (or normal force) changes direction round the loop — draw weight downward at every point.
Add at the top, subtract at the bottom. At the top both forces point inward (T + mg); at the bottom they oppose (T − mg).
Minimum speed means zero tension. Set T = 0 at the top and let gravity be the whole centripetal force.
The mass cancels in vmin = √(gr) — it depends only on g and the radius.
Quick recap: In a vertical circle the speed varies — fastest at the bottom, slowest at the top — because gravity has a component along the motion. Tension is greatest at the bottom (T = mv2/r + mg) and least at the top (T = mv2/r − mg). The slowest safe speed at the top is vmin = √(gr).
⚠ Common mistakes
Adding weight at the bottom when it should be subtracted (weight opposes the inward tension there)
Subtracting weight at the top when it should be added (both point inward there)
Drawing the tension pointing outward — a string always pulls towards the centre
Thinking the speed is constant round a vertical loop — it isn’t; gravity changes it
Forgetting that minimum speed comes from setting the tension to zero, not the weight
That wraps up the whole “Forces & Momentum” section — from free-body diagrams and Newton’s laws, through momentum and collisions, all the way to circular motion in and out of the vertical. You’ve now got the full toolkit for how forces make things move, turn and interact. Well done getting here — the next section builds straight on it.
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