Is the period of a conical pendulum the same as a simple pendulum of the same length?

No. For a given length L, the period of a conical pendulum (T_rev = 2π/ω) depends on the cone angle θ. Using ω² = g/(L cosθ), we find T_rev = 2π√(L cosθ / g). This is shorter than the small-angle simple pendulum period T₀ = 2π√(L/g) by a factor of √(cosθ). They are equal only in the limit as θ approaches zero, where the conical motion becomes a vanishingly small horizontal circle.

What happens if I spin the bob too fast or too slowly?

The steady conical motion described by the equations exists only for a specific ω for a given L and θ. If you spin it faster than the required ω, the angle θ increases (the cone opens up and the bob rises) to provide the larger centripetal force needed. If you spin it too slowly, the angle decreases; if ω is less than √(g/L), a horizontal circular path is impossible and the bob will begin to oscillate in a complex, non-uniform motion.

Where do we see conical pendulums in the real world?

While not as common as the simple pendulum, the physics is directly applicable to any object undergoing steady horizontal circular motion under gravity and a single inclined support force. Examples include a tetherball winding around a pole, the motion of a car on a banked curve (where the normal force acts like tension), and certain amusement park rides. It's a fundamental model for analyzing forces in inclined circular paths.

Why does the tension force change with the cone angle?

Tension has two jobs: hold the bob up against gravity and provide centripetal force. From T cosθ = mg, we see T = mg/cosθ. As θ increases (a wider cone), cosθ decreases, so tension must increase. This makes sense—at a steeper angle, more of the tension force is directed horizontally to provide centripetal force, so its overall magnitude must be larger to still have a sufficient vertical component to balance weight.

Conical Pendulum

A conical pendulum consists of a mass (bob) attached to a string of fixed length L, moving in a horizontal circle at a constant angular speed ω. The string traces out a cone, hence the name. This simulator models the steady-state motion of such a pendulum, governed by Newton's second law applied to uniform circular motion. The two primary forces acting on the bob are its weight, mg (vertically downward), and the string tension T (acting along the string). Resolving these forces reveals that the vertical component of tension balances the weight (T cosθ = mg), while the horizontal component provides the necessary centripetal force for circular motion (T sinθ = mω² r). The radius of the circle is r = L sinθ. Combining these equations yields the fundamental relationship between angular speed and the cone's half-angle: ω² = g / (L cosθ). This shows that for a given length L, a specific angle θ corresponds to a specific ω to maintain steady motion. The simulator allows exploration of how changing ω or L affects the cone's geometry, the magnitude of the tension force, and the period of revolution T_rev = 2π/ω. A key comparison is made with the period of a simple pendulum undergoing small-angle oscillations, T₀ = 2π√(L/g). The model simplifies reality by assuming a massless, inextensible string, a point-mass bob, and the absence of air resistance and friction. By interacting, students solidify concepts of force resolution, centripetal force, and the conditions for uniform circular motion in a non-vertical plane.

Who it's for: High school and introductory university physics students studying Newtonian mechanics, specifically uniform circular motion and force decomposition.

Key terms

Conical Pendulum
Centripetal Force
Uniform Circular Motion
Angular Velocity (ω)
Period of Revolution
Tension Force
Force Resolution
Simple Pendulum Period

Live graphs

How it works

The bob moves on a horizontal circle at fixed polar angle θ from the vertical, tracing a cone. In steady motion the string tension balances weight and supplies the centripetal force. The angular speed is ω = √(g/(L cosθ)), so steeper cones (larger θ) require faster rotation. This is a different motion from the planar swing of a simple pendulum: compare T_rev here with the small-angle harmonic period T₀ = 2π√(L/g) — they are not the same physical period, but both scale with √(L/g).

Key equations

T cosθ = mg · T sinθ = mω²(L sinθ) ⇒ ω² = g/(L cosθ)

T_rev = 2π/ω = 2π√(L cosθ/g) · T_simple,small ≈ 2π√(L/g)

Frequently asked questions

Is the period of a conical pendulum the same as a simple pendulum of the same length?: No. For a given length L, the period of a conical pendulum (T_rev = 2π/ω) depends on the cone angle θ. Using ω² = g/(L cosθ), we find T_rev = 2π√(L cosθ / g). This is shorter than the small-angle simple pendulum period T₀ = 2π√(L/g) by a factor of √(cosθ). They are equal only in the limit as θ approaches zero, where the conical motion becomes a vanishingly small horizontal circle.
What happens if I spin the bob too fast or too slowly?: The steady conical motion described by the equations exists only for a specific ω for a given L and θ. If you spin it faster than the required ω, the angle θ increases (the cone opens up and the bob rises) to provide the larger centripetal force needed. If you spin it too slowly, the angle decreases; if ω is less than √(g/L), a horizontal circular path is impossible and the bob will begin to oscillate in a complex, non-uniform motion.
Where do we see conical pendulums in the real world?: While not as common as the simple pendulum, the physics is directly applicable to any object undergoing steady horizontal circular motion under gravity and a single inclined support force. Examples include a tetherball winding around a pole, the motion of a car on a banked curve (where the normal force acts like tension), and certain amusement park rides. It's a fundamental model for analyzing forces in inclined circular paths.
Why does the tension force change with the cone angle?: Tension has two jobs: hold the bob up against gravity and provide centripetal force. From T cosθ = mg, we see T = mg/cosθ. As θ increases (a wider cone), cosθ decreases, so tension must increase. This makes sense—at a steeper angle, more of the tension force is directed horizontally to provide centripetal force, so its overall magnitude must be larger to still have a sufficient vertical component to balance weight.