H2 Physics Conical Pendulums: Resolve the Forces Before Finding the Radius

In the 17 November 2025 H2 Physics Paper 3 discussion, students asked how to find the radius in a circular-motion question when tension was unknown. One exchange identified a missing tension term. The useful lesson is to draw the real forces first: knowing the circular-motion formula does not yet tell you which force or component belongs on its left-hand side.

The official 2026 H2 Physics 9478 syllabus includes angular velocity, centripetal acceleration and the relationships F = mrω² and F = mv²/r. This guide uses a new idealised conical-pendulum example. It does not supply an official answer to the older thread’s examination question.

Decide which direction has acceleration

Imagine a small bob attached to a light string and moving steadily in a horizontal circle. The upper end of the string is fixed. The string makes an angle θ with the vertical. Assume negligible air resistance and a constant height.

The bob’s vertical acceleration is zero, so the upward component of tension balances its weight. Its horizontal acceleration points towards the centre of the circle. The horizontal component of tension supplies the required resultant force in that direction.

Draw only tension along the string towards the support and weight vertically downwards. “Centripetal force” describes the inward resultant required by the motion. Adding another inward force arrow labelled mv²/r, on top of the actual force components, would count the same effect twice.

Write the two component equations

Let tension be T and mass be m. Because θ is measured from the vertical, the vertical component is T cos θ and the horizontal component is T sin θ. Write T cos θ = mg and T sin θ = mω²r.

Divide the horizontal equation by the vertical one. Both tension and mass cancel, giving tan θ = ω²r/g. Therefore r = g tan θ/ω². The cancellation explains why a missing numerical mass or tension need not prevent you finding the radius.

Keep the angle definition beside the equations. If an angle is measured from the horizontal, the component expressions change. Reusing the same sine and cosine pattern without checking the diagram can produce tidy working for the wrong geometry.

Work through an original numerical example

Suppose the period is 2.50 s and the string is 20.0° from the vertical. Take g = 9.81 m s⁻². One complete revolution is 2π radians, so ω = 2π/2.50 = 2.51327… rad s⁻¹.

Substitute without rounding the intermediate angular velocity: r = 9.81 tan 20.0°/(2π/2.50)² = 0.56526… m. To three significant figures, the radius is 0.565 m. Use degrees for the supplied angle when evaluating the tangent; angular velocity remains expressed in radians per second.

Check the answer using the string geometry

If the string length is L, the horizontal-circle radius is L sin θ. The calculated example therefore requires L = 0.56526…/sin 20.0° = 1.65273… m, or about 1.65 m. This is a consistency check on the idealised setup, not an extra measurement supplied by the question.

The radius is the distance from the bob to the vertical axis, not the sloping string length. On a drawing, highlight these lengths in different colours. For a non-zero angle below 90°, the radius must be smaller than the string length. A calculation that violates that geometric condition needs review.

Explain work without confusing force and motion

During the ideal steady horizontal motion, the bob’s instantaneous displacement is tangential to the circle. Tension has no component along that tangent, and weight is vertical. Neither force does work on the bob in this model. Zero work does not mean zero force: the velocity direction is still changing.

Practise with appropriate items from the JC H2 Physics collection. Before calculating, label the circle’s centre, mark the radius and angle, draw the actual forces, and identify which components balance. Check each older item against your own syllabus and its stated assumptions.

For a final self-check, cover the algebra and explain why the vertical forces balance while the horizontal forces do not. Then recover the two equations from the diagram. If that explanation is clear, the radius formula follows from reasoning you can reuse in a different setup.