Physics / Uniform circular motion: Conical pendulum
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EXPERIMENT 22 · PHYSICS LABORATORY

Uniform circular motion: Conical pendulum theory

Understand uniform circular motion: conical pendulum: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
T² = (4π² / g) · h ⇔ g = 4π² / S

Keep the physical quantities, graph variables and units consistent.

01Mechanics of Uniform Circular Motion and Centripetal Force

Uniform circular motion describes the kinematics of a particle traversing a circular path of radius r at constant speed v. Although the magnitude of the velocity vector |v| remains invariant, the direction of the velocity vector continuously rotates towards the center of the circle at an angular rate ω = dθ/dt = v / r. By definition, acceleration is the time derivative of the velocity vector: a = dv/dt. Differentiating the position vector r(t) = r [cos(ωt) i + sin(ωt) j] twice with respect to time yields the centripetal acceleration vector: a_c = -ω² r [cos(ωt) i + sin(ωt) j] = -ω² r_hat = -(v² / r) r_hat. The negative sign demonstrates that the acceleration vector points radially inward toward the instantaneous center of curvature at every point along the trajectory. According to Newton's Second Law of Motion, this inward acceleration requires a net inward real force known as the centripetal force: F_c = m a_c = m v² / r = m ω² r. Centripetal force is not an independent physical force of nature; rather, it is a kinematic requirement that must be supplied by one or more physical forces acting on the body, such as string tension, gravitational attraction, friction, or electrostatic forces. In the conical pendulum, this inward centripetal force is supplied entirely by the horizontal component of the tension in the angled suspension cord.

\vec{a}_c = -\omega^2 r \,\hat{r} = -\frac{v^2}{r}\,\hat{r}, \quad F_c = m \frac{v^2}{r} = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r
Why does a body moving at constant speed in a circular orbit experience a non-zero acceleration?

Because velocity is a vector quantity possessing both magnitude and direction. Even though speed is constant, the direction of the velocity vector changes continuously, which by definition constitutes an acceleration directed toward the center of the circle.

02Free-Body Diagram and Vector Equilibrium of the Conical Pendulum

A rigorous dynamic analysis of the conical pendulum begins by constructing a free-body diagram of the spherical bob of mass m moving in a horizontal circle of radius r suspended by a cord of length L inclined at an apex half-angle θ to the vertical. Only two external real forces act on the bob in an inertial reference frame: the downward gravitational force (weight) W = m g, and the inward-and-upward tensile force T_string exerted by the cord along the line of suspension. Resolving the tension T_string into orthogonal Cartesian components yields a vertical component T_vertical = T_string cos(θ) and a horizontal radial component T_radial = T_string sin(θ). Because the bob maintains a strictly horizontal orbit with zero vertical acceleration (a_y = 0), Newton's Second Law in the vertical direction establishes exact static equilibrium: ∑ F_y = T_string cos(θ) - m g = 0, which yields T_string = m g / cos(θ). Notice that because cos(θ) < 1 for any non-zero angle, the string tension T_string is strictly greater than the static gravitational weight m g. In the horizontal radial direction, the unbalanced component T_radial supplies the necessary centripetal force: ∑ F_r = T_string sin(θ) = m ω² r. Dividing the horizontal force equation by the vertical force equation eliminates the tension T_string: [T_string sin(θ)] / [T_string cos(θ)] = [m ω² r] / [m g], which simplifies to tan(θ) = ω² r / g.

T_{\text{string}} \cos\theta = m g, \quad T_{\text{string}} \sin\theta = m \omega^2 r \implies \tan\theta = \frac{\omega^2 r}{g} = \frac{4\pi^2 r}{g T^2}
What happens to the string tension T_string as the apex angle θ approaches 90° (a completely flat, horizontal plane)?

As θ approaches 90°, cos(θ) approaches 0, causing the required string tension T_string = mg / cos(θ) to approach infinity. Therefore, it is physically impossible for a suspended pendulum to revolve in a perfectly horizontal plane under gravity.

03Equivalence to the Simple Planar Pendulum of Effective Length $h = L \cos\theta$

Substituting the geometric relation for the radius of the circular orbit, r = L sin(θ), into the dynamic tangent equation tan(θ) = 4π² r / (g T²) produces a profound mathematical simplification: sin(θ) / cos(θ) = [4π² (L sin θ)] / [g T²]. Dividing both sides by sin(θ) (valid for any non-zero angle θ > 0) yields 1 / cos(θ) = 4π² L / (g T²). Solving for the square of the orbital period T² gives T² = (4π² / g) · (L cos θ). Defining the vertical axial depth of the cone as h = L cos(θ)—the vertical distance from the overhead pivot point to the plane of the circular orbit—the period equation becomes T² = (4π² / g) · h, or equivalently, T = 2π sqrt(h / g). This remarkable result demonstrates that the period of a conical pendulum of cord length L revolving at angle θ is mathematically identical to the period of a simple planar pendulum of length L_eff = h = L cos(θ) executing small-amplitude oscillations. As the apex angle θ increases, the vertical depth h shrinks, causing the orbital period T to decrease. Conversely, in the limit of an infinitesimal apex angle (θ → 0), cos(θ) → 1 and h → L, recovering the standard simple pendulum formula T₀ = 2π sqrt(L / g).

h = L \cos\theta \implies T = 2\pi \sqrt{\frac{L \cos\theta}{g}} = 2\pi \sqrt{\frac{h}{g}} \quad \xrightarrow{\theta \to 0} \quad T_0 = 2\pi \sqrt{\frac{L}{g}}
Does the period of a conical pendulum depend on the mass of the bob?

No. Just like the simple planar pendulum, the mass m cancels identically from the ratio of centripetal force to gravitational weight, making the period completely independent of bob mass.

04Linearization of the Dynamic Period Equation ($T^2$ versus $h$)

The dynamic equation governing the conical pendulum, T² = (4π² / g) · h, represents an intrinsically linear relationship between the square of the period T² and the vertical depth h = L cos(θ). In experimental physics, verifying centripetal dynamics through a linear representation offers tremendous advantages: it tests the theoretical relationship across multiple independent angular states and enables an ordinary least-squares regression to determine the acceleration due to gravity g without relying on any single measurement. Defining the vertical coordinate as Y = T² (s²) and the horizontal coordinate as X = h (m), the governing relation takes the standard slope-intercept form Y = S · X + b, where the theoretical slope is S = 4π² / g (s² m⁻¹) and the theoretical intercept is b = 0. When an ordinary least-squares regression is performed across a series of experimental trials {(h_i, T_i²)}, the best-fit slope S minimizes the sum of squared vertical residuals: S = [N ∑(h_i T_i²) - ∑(h_i) ∑(T_i²)] / [N ∑(h_i²) - (∑ h_i)²]. The acceleration due to gravity is directly obtained from the slope as g = 4π² / S. The standard error in the slope propagates into the gravity estimate via SE(g) = (4π² / S²) · SE(S) = (g² / 4π²) · SE(S). Furthermore, the fitted intercept b confirms that the motion follows ideal centripetal dynamics if b is statistically consistent with zero.

Y = S \cdot X + b \quad \text{where } Y = T^2, \; X = h = L \cos\theta, \; S = \frac{4\pi^2}{g}, \; b = 0 \implies g = \frac{4\pi^2}{S}
What does a non-zero intercept b in an experimental plot of T² versus h indicate about the physical apparatus?

A non-zero intercept typically indicates an uncorrected systematic error in the suspension length L (such as measuring to the top of the bob instead of its center of mass) or a constant timing trigger offset in the revolution counter.

05Rotational Frequency, Centrifugal Pseudo-Force, and Frame Transformation

While the conical pendulum is most rigorously analyzed from an inertial reference frame anchored to the laboratory floor, examining the motion from the rotating reference frame of the pendulum bob provides valuable physical insights. In a reference frame rotating with constant angular velocity ω about the vertical axis, the bob appears permanently at rest (a_rot = 0). To apply Newton's laws of statics in this non-inertial rotating frame, D'Alembert's principle requires introducing a fictitious outward centrifugal inertial force: F_centrifugal = -m (ω × (ω × r)) = +m ω² r r_hat. In this rotating frame, three forces act on the bob in static equilibrium: upward-and-inward string tension T_string, downward gravitational weight m g, and outward centrifugal force F_centrifugal. The horizontal force balance becomes ∑ F_x' = T_string sin(θ) - m ω² r = 0, and the vertical force balance becomes ∑ F_y' = T_string cos(θ) - m g = 0. Dividing these equations recovers tan(θ) = ω² r / g. Although the centrifugal force is a mathematical artifact of an accelerating coordinate system, it correctly predicts the physical sensations and equilibrium orientations observed in rotating machinery, centrifuges, and amusement park rides.

\vec{F}_{\text{centrifugal}} = m \omega^2 r \,\hat{r}, \quad \sum \vec{F}_{\text{rot}} = \vec{T} + m\vec{g} + \vec{F}_{\text{centrifugal}} = 0 \implies \tan\theta = \frac{\omega^2 r}{g}
Is centrifugal force a real physical interaction in Newton's Third Law?

No. Centrifugal force is a fictitious (inertial) pseudo-force that arises solely from observing motion from an accelerated rotating coordinate frame. It has no physical agent and possesses no third-law reaction force.

06Systematic Sources of Error: String Aerodynamic Drag and Finite Bob Inertia

A physical conical pendulum departs from the idealized point-mass model in several subtle ways that must be accounted for in precision metrology. First, as the bob and cord sweep through ambient air, aerodynamic drag exerts a retarding torque opposing rotation. To maintain steady motion during manual trials, the experimenter must occasionally impart tiny, imperceptible driving impulses, or conduct timing over a limited window before orbital decay occurs. Second, the suspension cord has non-zero mass linear density μ_cord. Centrifugal force causes the flexible cord to bow slightly outward into a catenary shape rather than remaining strictly straight, introducing an effective mass correction M_eff = m_bob + 1/3 m_cord. Third, the spherical bob has finite radius R_bob and rotational moment of inertia I_cm = 2/5 m R_bob². As the bob revolves, it may undergo self-rotation (spin) about its own vertical axis, coupling rotational kinetic energy into the orbit. Finally, if the initial launch impulse is not strictly circular, the orbit will precess into an ellipse whose major and minor axes rotate slowly in space (apsidal precession). Timing across 20 revolutions minimizes random human reflex errors, reducing the overall experimental uncertainty in g to below 0.1%.

M_{\text{eff}} = m_{\text{bob}} + \frac{1}{3} m_{\text{cord}}, \quad h_{\text{true}} = (L + R_{\text{bob}}) \cos\theta - \delta_{\text{bowing}}
Why is an elliptical orbit undesirable when measuring the period of a conical pendulum?

An elliptical orbit causes the orbital radius and speed to vary periodically during each revolution, modulating the string tension and apex angle and deviating from the uniform circular motion model.

07Classical Applications: James Watt's Centrifugal Flyball Governor

The conical pendulum is not merely a theoretical curiosity; it was the technological linchpin of the Industrial Revolution through James Watt's invention of the centrifugal flyball governor in 1788. Watt needed an automatic feedback control mechanism to regulate the rotational speed of his condensing steam engines under varying factory loads. In Watt's governor, two heavy metal spheres are hinged as conical pendulums to a central vertical shaft driven by the steam engine's output flywheel. As the engine speeds up, the angular velocity ω increases. According to the conical pendulum relationship tan(θ) = ω² r / g, the increased centrifugal effect forces the balls to swing outward and upward to a larger apex angle θ. A mechanical linkage attached to the rising balls is connected directly to the engine's steam throttle valve: as the balls rise, they partially close the throttle valve, reducing the steam supply and slowing the engine back down. Conversely, if the engine slows under a heavy industrial load, the balls drop inward, opening the valve and admitting more steam. Watt's governor represents the first widespread automated proportional negative feedback control system in history, anticipating modern cybernetics and control theory.

\omega = \sqrt{\frac{g}{L \cos\theta}} = \sqrt{\frac{g}{h}} \implies h(\omega) = \frac{g}{\omega^2} \implies \frac{dh}{d\omega} = -\frac{2g}{\omega^3}
How does James Watt's governor demonstrate the principle of negative feedback?

An increase in engine speed produces an outward rise of the governor balls, which mechanically actuates a throttle valve in the opposite direction (reducing steam flow), thereby stabilizing the engine speed at an equilibrium setpoint.

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Connect the equations to a measurement.

Theoretical period of revolution: T = 2π sqrt(h / g), where h = L cos(θ). Linearized relation: T² = (4π² / g) · h. The ordinary least-squares regression extracts the slope S = 4π² / g, from which gravity is recovered as g = 4π² / S. Reference standard gravity g = 9.80665 m s⁻².