Conical Pendulum
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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.