Forces in a vertical circle
Key idea: In a vertical circle, combine tension and weight towards the centre at each position. At the top, a string just stays taut when gravity alone supplies mv²/r.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Identify the inward direction and resolve the weight.
- Write a force equation at each position.
- Find the bottom resultant force and tension.
- Find the side tension and explain the effect of weight.
- Find the top speed at which the string just stays taut.
- Calculate the minimum top speed and its dependence on mass.
- Calculate the tension and decide whether the string stays taut.
- Use forces and energy to find the minimum bottom speed.
- Explain your force equations and energy calculation.
- Find the tension ratio and compare bottom and top tensions.
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ContinueWhirl a bucket of water in a vertical circle fast enough and the water stays in, even upside down at the top. In a vertical circle the weight sometimes points towards the centre and sometimes away from it, so the tension in the string changes round the circle. This lesson writes the force equation at the top, bottom and side of a vertical circle, and finds the slowest speed at which a string stays taut at the top.
Forces at the top and bottom
Model a small mass on a light string, swung in a vertical circle of radius r about a fixed centre, with no air resistance. Call the tension in the string FT. At every point the string pulls towards the centre, while the weight mg always acts straight down.
At any instant, the resultant force towards the centre must be mv²/r, using the speed at that instant. In a vertical circle the speed changes as the mass rises and falls, so you write the equation separately at each point:
- at the top, the centre is below, so tension and weight both point towards it: FT + mg = mv²/r;
- at the bottom, the centre is above, so tension points towards it and weight points away: FT-mg = mv²/r;
- at the side, the weight is along the tangent, so it has no component towards the centre: FT = mv²/r. Here the weight changes the speed instead.
So the tension is largest at the bottom, where it must hold up the weight as well as provide the centripetal force.
FT + mg = mvₜₒₚ²/r
FT - mg = mv_bottom²/r
FT = mv_side²/r
O is the fixed centre. Dashed arrows show tension FT; solid arrows show weight mg. Draw only these real forces. Arrow lengths are schematic and do not compare magnitudes or speeds between positions.
Worked example 1
Tension at the top and at the bottom
Problem
A 0.300 kg mass is swung in a vertical circle on a string of length 0.800 m. For two separate swings, find the tension (a) when it passes the top at 5.00 m s⁻¹, (b) when it passes the bottom at 7.00 m s⁻¹. Take g = 9.81 m s⁻².
Show full solution
(a) At the top
Method
Take downwards, towards the centre, as positive.Reason
Tension and weight both point towards the centre, so together they provide mv²/r.Working
Answer
FT = mv²/r-mg = 0.300(5.00)²/0.800-0.300(9.81) = 9.375-2.943 = 6.43 N
(b) At the bottom
Working
Answer
Upwards is towards the centre and the weight points away, so FT = mv²/r + mg = 0.300(7.00)²/0.800 + 2.94 = 18.4 + 2.94 = 21.3 N
Quick check
For these two separate swings, the bottom tension is more than three times the top tension. In a single swing with no energy loss, the top and bottom speeds must also obey energy conservation: v_bottom² = vₜₒₚ² + 4gr. The two stated speeds are not the top and bottom speeds of that one swing.
Guided practice 2
Tension at the bottom
Problem
A 0.250 kg mass on a string passes the bottom of a vertical circle of radius 0.500 m at 4.00 m s⁻¹. Find the required inward resultant and the tension. Give the tension to three significant figures. Take g = 9.81 m s⁻².
Find the resultant and tension
Show the steps Optional working
Hints
Hint 1
which way is inwards
At the bottom, the centre is above, so the weight points away from it.
Show full solution
Tension
Working
Answer
FT-mg = mv²/r, so FT = 0.250(4.00)²/0.500 + 0.250(9.81) = 8.00 + 2.45 = 10.5 N
Check your understanding 1 At the side of the circle
A 0.200 kg mass on a string of length 0.800 m passes the side of a vertical circle, level with the centre, at 2.00 m s⁻¹. Find the tension. What is the weight doing at this point?
Show answer
The weight is vertical, along the tangent, so it has no component towards the centre. So FT = mv²/r = 0.200(2.00)²/0.800 = 1.00 N. The weight gives the mass a tangential acceleration of 9.81 m s⁻², changing its speed.
The slowest speed at the top
A string can pull but it cannot push, so the tension can never be negative. At the top,
At the slowest possible speed, v = square root of gr, the tension is zero and the weight alone provides the centripetal force. Any slower and the string goes slack: the weight is more than the centripetal force needed, so the mass falls inside the circle. This is why the water stays in a fast-whirled bucket: at the top, the bucket and the weight together push the water round the circle.
Check your understanding 2 The slowest speed
Find the slowest speed at which a mass on a string of length 1.20 m can pass the top of a vertical circle with the string taut. Would a heavier mass need a higher speed? Take g = 9.81 m s⁻².
Show answer
v = square root of gr = square root of 9.81(1.20) = 3.43 m s⁻¹. No: the mass cancels from mv²/r = mg, so the slowest speed is the same for any mass.
Check your understanding 3 Too slow at the top
A 0.500 kg mass on a string of length 1.20 m is meant to pass the top of a vertical circle at 3.00 m s⁻¹. Can it? Take g = 9.81 m s⁻².
Show the steps Optional working
Show answer
FT = 0.500(3.00²/1.20-9.81) = -1.16 N. A string cannot push, so this is impossible: 3.00 m s⁻¹ is below the slowest speed of 3.43 m s⁻¹, so the proposed taut circular motion at the top is impossible. The string must go slack if the motion would otherwise require negative tension.
Check your understanding 4 Will the string stay taut for a complete loop?
A mass starts at the bottom of a vertical circle of radius 0.800 m with just enough speed to keep a light string taut at the top. Only weight and tension act. Take g = 9.81 m s⁻². Find the minimum bottom speed, then justify your method with an energy equation and the top force equation. You may work on paper.
Show the steps Optional working
Show answer
At the top, limiting tension is zero, so vₜₒₚ² = gr. Energy conservation gives (1/2)mv_bottom² = (1/2)mvₜₒₚ² + mg(2r), hence v_bottom = square root of 5gr = 6.26 m s⁻¹. Checking only square root of gr at the bottom would omit the energy needed to rise. For the full method, check the zero-tension condition, the rise of 2r and the factor 5.
Check your understanding 5 Explain the two conditions
Show your force and energy equations for the complete-loop calculation. Explain why square root of gr applies at the top rather than the bottom. You may compare your paper working without typing it all.
Show answer
Compare: at the top, T + mg = mvₜₒₚ²/r with T = 0 gives vₜₒₚ² = gr; from bottom to top, (1/2)mv_bottom²-(1/2)mvₜₒₚ² = 2mgr gives v_bottom² = 5gr. The mass loses kinetic energy as it rises, so its initial bottom speed must be higher than its limiting top speed.
Common mistakes
- Using the same force equation at the top and bottom of the circle.
- Using the same speed all the way round, when the mass slows as it rises.
- Adding a separate “centripetal force” arrow to the diagram.
- Treating a negative calculated tension as an outward push from the string.
Before you move on
Check your understanding 6
A mass on a light string passes the top of a vertical circle at twice the minimum speed needed to keep the string taut. Express the tension as a multiple of the weight. Is the tension at the bottom larger or smaller, if mechanical energy is conserved?
Show answer
At the top, v = 2 square root of gr, so mv²/r = 4mg. Thus FT + mg = 4mg and the tension is 3mg, three times the weight. The mass gains kinetic energy as it falls to the bottom, so its speed is larger there. At the bottom the tension must also overcome the weight: FT = mv²/r + mg. Both effects make the bottom tension larger.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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