PhysicsLaws of motion · circular motionClass 11

Circular motion: banking and vertical circles: interactive simulation

Anything moving in a circle needs a net force towards the centre, mv²/r. Circular motion problems are about finding which real forces — friction, the normal force, tension, gravity — add up to that.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 1, A-level Physics, IB Physics · 3 modes in one lab

What you can explore

  1. 1

    Banked road

    Design speed, maximum and minimum safe speeds with friction, and the direction of friction.
  2. 2

    Vertical circle

    Tension around the loop, the √(5gR) condition, and where the string goes slack.
  3. 3

    Conical pendulum

    Cone angle, period, speed and tension, with the force diagram.

Try this

Drive a car round a banked curve and find its safe speed range, whirl a ball in a vertical circle and watch the tension change, and set a conical pendulum spinning.

  1. 1

    Banked road: R = 100 m, θ = 15°, μ = 0.3.

    What you should see

    Design speed ≈ 58 km/h needs no friction; the car holds the curve up to about 89 km/h. Slower than that is fine because tan 15° < μ.

  2. 2

    Set μ = 0 and drive slower and faster than the design speed.

    What you should see

    Only exactly v₀ works: slower slides down and in, faster skids up and out.

  3. 3

    Vertical circle with R = 1 m: try 7.5 m/s at the bottom, then 6 m/s.

    What you should see

    The minimum is √(5gR) = 7 m/s. At 6 m/s the string goes slack in the upper half and the ball falls inside the circle.

  4. 4

    Try 4 m/s.

    What you should see

    v² < 2gR, so the ball never rises above the centre and simply swings like a pendulum.

  5. 5

    Conical pendulum with L = 1 m at 30°.

    What you should see

    Period ≈ 1.87 s and tension = mg/cos 30° ≈ 1.15 mg.

Formulas behind it

Centripetal acceleration

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

Banked road, no friction

v0=rgtan⁡θv_0 = \sqrt{rg\tan\theta}

Banked road, maximum safe speed

vmax=rg(μ+tan⁡θ)1−μtan⁡θv_{max} = \sqrt{\frac{rg(\mu + \tan\theta)}{1 - \mu\tan\theta}}

Vertical circle (string)

vbottom≥5gR,vtop≥gRv_{bottom} \ge \sqrt{5gR},\qquad v_{top} \ge \sqrt{gR}

Tension at angle θ from the bottom

T=mv2R+mgcos⁡θT = \frac{mv^2}{R} + mg\cos\theta

Conical pendulum

T=2πLcos⁡θgT = 2\pi\sqrt{\frac{L\cos\theta}{g}}

Common mistakes

  • Adding “centripetal force” as an extra force in the free body diagram. It is the net of the real forces.
  • Using v_min = √(gR) at the bottom of a vertical circle. That is the minimum speed at the top.
  • Forgetting that friction can act down the slope on a banked road when the car is too fast.
  • Treating a rod and a string the same in vertical circles. A rod can push, so the minimum speed at the top is zero.
Practise 210 questionsCheck your answer instantly, with a worked solution.Stuck on a circular motion: banking and vertical circles question?Snap it and get a whiteboard video solution.

FAQ

Why are roads banked on curves?

Tilting the road lets part of the normal force point towards the centre, so the car can turn without relying only on friction — safer on wet or icy roads.

Why does water stay in a bucket swung in a vertical circle?

At the top the water needs mv²/r ≥ mg to keep moving in the circle. If the bucket is fast enough, gravity alone is not enough to pull the water out of the circular path.

What is the difference between centripetal and centrifugal force?

Centripetal force is the real net inward force in an inertial frame. Centrifugal force is a pseudo-force that appears only when you describe the motion from the rotating frame.

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