# Conical Pendulum > Conical pendulum in JEE Physics: tanθ = v^2/rg, period T = 2π√(Lcosθ/g), tension components, banking analogy and speed-dependent cone angle. - Canonical URL: https://prepelephant.com/topics/jee/physics/conical-pendulum - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Conical Pendulum", PrepElephant, https://prepelephant.com/topics/jee/physics/conical-pendulum ## Direct answer Whirling a bob so that it traces a horizontal circle while the string sweeps out a cone gives the conical pendulum, defined by two force equations: T cosθ = mg and T sinθ = mv²/r, which combine into tanθ = v²/rg with r = L sinθ the circle radius. The period is T = 2π√(L cosθ/g) — the pendulum behaves like a simple pendulum whose length is only the vertical projection L cosθ, called the height of the cone. Mass cancels everywhere: a heavier bob at the same angle needs proportionally more tension but the same speed and the same period. The identical mathematics governs banking of a frictionless curved road, where tanθ = v²/rg gives the banking angle. ## What you must remember - **Force balance:** vertical equilibrium T cosθ = mg, radial Newton's law T sinθ = mv²/r; dividing gives tanθ = v²/rg, the master equation. - **Period and frequency:** T = 2π√(L cosθ/g); frequency n = (1/2π)√(g/L cosθ); both depend on L, θ and g but never on the mass of the bob. - **Speed:** v = √(rg tanθ); centripetal acceleration a = g tanθ, directed horizontally towards the axis, supplied entirely by the horizontal component of tension. - **Limiting case:** as θ → 0 the period tends to 2π√(L/g), the small-oscillation simple pendulum value; as θ → 90° the period → 0 but the required tension → ∞ (horizontal string is impossible). - **Cone height:** period depends only on h = L cosθ, the vertical height of the bob below the suspension point — pendulums of different lengths but the same h keep time together. - **Banking connection:** a vehicle on a frictionless banked road at angle θ is the same free-body diagram, tanθ = v²/rg; the optimum speed needs no friction at exactly this angle. - **If the string snaps:** the bob becomes a projectile launched horizontally with the instantaneous tangential velocity — a classic follow-up option in JEE Main. ## How to attack a conical pendulum problem Start with the string length and angle as the given data, because everything else follows. Take L = 2 m and θ = 30°. The circle radius is r = L sinθ = 1.0 m and the cone height is h = L cosθ = 1.732 m. Period first: T = 2π√(1.732/9.8) = 2π × 0.42 ≈ 2.64 s. Speed next, from v = 2πr/T = 2π × 1/2.64 ≈ 2.38 m/s, or equivalently v = √(rg tan30°) = √(9.8 × 0.577) = 2.38 m/s — computing it both ways is a fifteen-second self-check worth building as habit. Tension last: T = mg/cosθ = 1.155 mg, always larger than the weight whenever the bob is moving. The subtle behaviour JEE probes is the speed dependence. Spin faster at fixed L: tanθ grows, the cone opens, h = L cosθ shrinks, and the period falls — a faster bob actually revolves more times per second partly because it travels a bigger circle at disproportionately higher speed. Spin towards v → ∞ and θ → 90°, tension → ∞: no finite string can hold a truly horizontal circle. These two limit statements, slow limit recovering the simple pendulum and fast limit being forbidden, answer most reasoning questions set on this device. ## Examiner's framing The standard error is writing the period as 2π√(L/g) with the full string length instead of L cosθ; options are engineered so that this slip yields the "attractive" wrong answer, usually the middle one. A second favourite: asking for the angle when the tension equals the weight — set T = mg, and since T = mg/cosθ, this forces cosθ = 1, meaning θ = 0; students who mechanically solve tanθ = v²/rg without checking often miss that tension equals weight only in the trivial non-whirling case. JEE Advanced dresses the same physics as a bob on a string inside a rotating dome, or as a car on a banked track where the friction case (tanθ ± μ = v²/rg form) extends the diagram, so treat the banked road as this topic's twin rather than a separate chapter. ## Frequently asked questions ### What supplies the centripetal force in a conical pendulum? The horizontal component of string tension, T sinθ = mv²/r; gravity has no horizontal component and only balances the vertical component T cosθ = mg. ### Why is the period of a conical pendulum independent of the bob's mass? Mass cancels between the force equations — both the required centripetal force and the weight scale with m — leaving T = 2π√(L cosθ/g) with no m anywhere. ### Can the string of a conical pendulum ever become horizontal? No; θ = 90° would demand zero vertical force balance since cosθ = 0, which cannot support the weight, so the angle always stays below 90°. ### How does the period change if the bob is whirled faster in the same string? The cone opens (θ increases), L cosθ decreases, and the period drops; formally n = (1/2π)√(g/L cosθ) rises with speed. ### What happens to the bob the instant the string is cut? It moves as a projectile with horizontal velocity equal to its instantaneous tangential speed, following a parabola under gravity alone.