PhysicsISC Class 11 · Laws of Motion

Dynamics of Circular Motion: Level Roads, Banking and Vertical Circles

Updated 17 October 2026 · Fact-checked

Circular motion needs a net inward (centripetal) force mv²/r. On a level road friction supplies it, giving v_max = √(μrg). On a frictionless banked road tanθ = v²/rg. In a vertical circle the minimum speed at the top is √(gr) and at the bottom √(5gr).

At the top only weight supplies the centripetal force (T = 0, v = √(gr)); reaching the bottom needs √(5gr) and the string then pulls with 6mg.

Illustration drawn for QuestionBuddy.

Understand Circular Motion, Banking and Vertical Circle

Centripetal force. A body moving in a circle of radius r at speed v needs a net inward force mv²/r. It is not a new force: it is the net inward force from tension, friction, gravity or the normal reaction.

Level road. Friction supplies the centripetal force, so the safe speed is v ≤ √(μ r g). It does not depend on the mass of the vehicle.

Banked road. On a frictionless road banked at θ, the horizontal component of the normal reaction supplies the centripetal force: N sinθ = mv²/r and N cosθ = mg, so tanθ = v²/(rg). With friction μ, the maximum speed is given by v² = rg(μ + tanθ)/(1 − μ tanθ), and the minimum safe speed by v² = rg(tanθ − μ)/(1 + μ tanθ) when tanθ > μ; otherwise it is zero.

Vertical circle. For a body on a string of length r, tension and weight vary. At the top: T + mg = mv²/r, so the minimum speed (T = 0) is √(gr). At the bottom: T − mg = mv²/r. Energy conservation gives v_bottom² = v_top² + 4gr, so the minimum speed at the bottom is √(5gr), where T = 6mg. On a convex bridge top, N = mg − mv²/r, and the body leaves the surface when v = √(gr).

Conical pendulum. A bob on a string of length L moving in a horizontal circle with the string at angle θ to the vertical has tanθ = v²/(rg), r = L sinθ, and period T = 2π√(L cosθ/g).

Circular Motion, Banking and Vertical Circle: formulas and key facts

Centripetal force: $F = m v^{2} / r$; Directed to the centre.
Level road: $v_{m} a x = \sqrt (μ r g)$; Independent of mass.
Banked road: $t an θ = v^{2} / r g ∣ v^{2} = r g (μ + t an θ) / (1 - μ t an θ)$; Frictionless; with friction (maximum speed).
Vertical circle: $v_{t} o p \geq \sqrt (g r) ∣ v_{b} o tt o m \geq \sqrt (5 g r)$; T_bottom = 6mg at the minimum speed.

Circular Motion, Banking and Vertical Circle in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Bucket whirled in a vertical circle

Suppose a child whirls a bucket of water, total mass 2 kg, in a vertical circle of radius 0.8 m. Take g = 10 m s⁻².

Find the minimum speed at the top and the tension at the bottom if the speed at the top is minimum.

Try it first, then show the solutionSolution

At the top (T = 0): mg = mv²/r → v = √(gr) = √8 = 2.83 m s⁻¹.
At the bottom: v² = 5gr = 40.
T = mg + mv²/r = 20 + 2 × 40/0.8 = 20 + 100 = 120 N (6mg).

Answer: 2.83 m s⁻¹ at the top; 120 N at the bottom.

Memory hook: Water stays in the bucket if the net force at the top is still enough to curve the path.

Tricks to remember Circular Motion, Banking and Vertical Circle

Short memory aids. Say each one aloud once, then use it on a question the same day.

Top: √(gr), bottom: √(5gr)
Add 4gr to v² between top and bottom.
Use it when: Vertical circle minimum speeds.
Bank without friction: tanθ = v²/rg
Quote v² over rg for the banking angle.
Use it when: Any "ideal banking" question.

Circular Motion, Banking and Vertical Circle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Safe speed on a level curve · 1 mark · MCQBoard level

A car goes round a level curve of radius 100 m with μ = 0.4 (g = 10 m s⁻²). The maximum safe speed is:
(a) 10 m s⁻¹
(b) 20 m s⁻¹
(c) 30 m s⁻¹
(d) 40 m s⁻¹

Try it first, then show the solutionSolution

v = √(μrg) = √(0.4 × 100 × 10) = √400 = 20 m s⁻¹.

Answer: (b) 20 m s⁻¹

Example 2: Banking angle · 3 marks · Short answerBoard level

Find the angle of banking for a frictionless curve of radius 100 m for vehicles at 72 km/h (g = 10 m s⁻²).

Try it first, then show the solutionSolution

72 km/h = 20 m s⁻¹. (1 mark)
tanθ = v²/(rg) = 400/(100 × 10) = 0.4. (1 mark)
θ = tan⁻¹0.4 = 21.8°. (1 mark)

Answer: θ = 21.8°

Example 3: Stone in a vertical circle · 4 marks · Short answerHarder board level

A 0.5 kg stone is whirled on a 1.2 m string in a vertical circle (g = 10 m s⁻²). Find the minimum speed at the top, the minimum speed at the bottom, and the tension at the bottom in that case.

Try it first, then show the solutionSolution

v_top = √(gr) = √12 = 3.46 m s⁻¹. (1 mark)
v_bottom = √(5gr) = √60 = 7.75 m s⁻¹. (1 mark)
T = mg + mv²/r = 5 + 0.5 × 60/1.2 = 5 + 25 = 30 N. (1 mark)
Check: 6mg = 6 × 5 = 30 N. (1 mark)

Answer: 3.46 m s⁻¹, 7.75 m s⁻¹, 30 N.

Example 4: Banked road with friction · 4 marks · Short answerHarder board level

A road of radius 50 m is banked with tanθ = 0.2, and μ = 0.4 (g = 10 m s⁻²). Find the maximum safe speed and the minimum safe speed.

Try it first, then show the solutionSolution

v_max² = rg(μ + tanθ)/(1 − μ tanθ) = 500 × 0.6/(1 − 0.08) = 300/0.92 = 326.1. (2 marks)
v_max = 18.1 m s⁻¹, about 65 km/h. (1 mark)
Since μ > tanθ, the vehicle cannot slide down even when at rest, so v_min = 0. (1 mark)

Answer: v_max = 18.1 m s⁻¹; v_min = 0.

Example 5: Wall of death · 5 marks · Long answerHOTS / competency

Suppose a motorcyclist rides along the inner wall of a vertical cylinder of radius 4 m, with μ = 0.5 between tyres and wall (g = 10 m s⁻²). Find the minimum speed at which he does not slip down, and for a total mass of 200 kg at that speed the normal reaction and friction.

Try it first, then show the solutionSolution

Vertical: friction f = mg. Horizontal: N = mv²/r. (1 mark)
Limiting: μN = mg → μ mv²/r = mg. (1 mark)
v² = gr/μ = 10 × 4/0.5 = 80, v = 8.94 m s⁻¹. (1 mark)
N = 200 × 80/4 = 4000 N. (1 mark)
Friction = μN = 2000 N = mg = 200 × 10 (consistent). (1 mark)

Answer: v_min = 8.94 m s⁻¹; N = 4000 N; f = 2000 N.

Where students lose marks in Circular Motion, Banking and Vertical Circle

The slips examiners see most often, and what to write instead.

Mistake: Calling centripetal force an extra force
Drawing both friction and centripetal force on the free-body diagram.
Do this: Show real forces only; the centripetal force is their net inward resultant.
Usually costs: 1 to 2 marks.
Mistake: Using T = mg at the top of a vertical circle
Forgetting that the tension adds to the weight in the inward direction.
Do this: At the top T + mg = mv²/r.
Usually costs: 1 to 2 marks.
Mistake: Mixing banking formulas
Using tanθ = v²/rg when friction is present.
Do this: Use the frictionless formula only if friction is stated to be absent.
Usually costs: 1 mark.

Exam tips

Mark the direction towards the centre and write net inward force = mv²/r.
Convert km/h to m/s before substituting.
In vertical circles, check the answers with energy conservation.
State g and any approximation you use.

Practice questions on Circular Motion, Banking and Vertical Circle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A particle moves in a circle at constant speed. Its acceleration is:
(a) zero
(b) along the tangent
(c) directed towards the centre
(d) directed away from the centre
Hint
Direction of change of velocity.
Answer
(c) directed towards the centre
Q2 · 2 marks · Short answerBoard level
A 1000 kg car moves at 10 m s⁻¹ on a level circular track of radius 50 m. Find the friction force needed.
Hint
Friction = mv²/r.
Answer
f = 1000 × 100/50 = 2000 N.
Q3 · 3 marks · Short answerHarder board level
A motorcycle rider takes a flat bend of radius 24 m at 12 m s⁻¹ (g = 10 m s⁻²). Find the angle his machine makes with the vertical and the least μ needed between tyres and road.
Hint
tanθ = v²/(rg), and the same ratio is the least μ.
Answer
tanθ = 144/(24 × 10) = 0.6, so θ = 31.0°. Friction supplies the centripetal force, so μ ≥ v²/(rg) = 0.6.
Q4 · 3 marks · Short answerHarder board level
A bob of a conical pendulum has a string of length 1 m at 30° to the vertical (g = 10 m s⁻²). Find the speed and the period.
Hint
r = L sinθ; tanθ = v²/(rg).
Answer
r = 0.5 m. v² = rg tan30° = 0.5 × 10 × 0.5774 = 2.887, v = 1.70 m s⁻¹. T = 2π√(L cosθ/g) = 2π√(0.0866) = 1.85 s.
Q5 · 3 marks · Short answerHarder board level
An 800 kg car crosses the crest of a convex bridge of radius 40 m at 10 m s⁻¹. Find the normal reaction and the speed at which it would leave the road (g = 10 m s⁻²).
Hint
N = mg − mv²/r.
Answer
N = 8000 − 800 × 100/40 = 8000 − 2000 = 6000 N. It leaves the road when N = 0: v = √(gr) = √400 = 20 m s⁻¹.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a pilot of mass 70 kg flies a loop of radius 400 m at a constant 60 m s⁻¹ (g = 10 m s⁻²).
(i) Find the net force needed to keep the pilot in the circle. (1 mark)
(ii) Find the seat force on the pilot at the lowest point. (1 mark)
(iii) Find the force at the highest point. (1 mark)
(iv) At what speed would the pilot feel weightless at the top? (1 mark)
Hint
At the bottom N − mg = mv²/r; at the top N + mg = mv²/r.
Answer
(i) mv²/r = 70 × 3600/400 = 630 N. (ii) N = 630 + 700 = 1330 N. (iii) N = 630 − 700 = −70 N, so the seat cannot push; the straps must pull down with 70 N. (iv) √(gr) = √4000 = 63.2 m s⁻¹.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Safe speed on a level curve?Show answer
√(μ r g).
2.Banking angle without friction?Show answer
tanθ = v²/(rg).
3.Minimum speed at the bottom of a vertical circle?Show answer
√(5gr).
4.Tension at the bottom for minimum speed?Show answer
6mg.

Circular Motion, Banking and Vertical Circle: frequently asked questions

Why do racing tracks bank their curves?

Banking lets the normal reaction supply part of the centripetal force, so vehicles can take the bend at higher speed without relying only on friction.

Does a heavier car have a lower safe speed on a level curve?

No. In v = √(μrg) the mass cancels, as both friction and the required centripetal force are proportional to the mass.

Why does water not fall from a whirled bucket at the top?

If the speed is at least √(gr), the weight alone is not more than the centripetal force needed, so the water presses against the bucket and stays in.