Physics 9702 · Paper 4

Angled Forces: Cones, Pendulums & Banking

The current flagship question: a ball on a string, in a cone, or a banking aircraft, where one tilted force does two jobs at once.

The two-line resolve

The two-line resolve the mark scheme wants: the vertical part of the tension balances the weight (T cos θ = mg); the horizontal part is the resultant, pointing at the centre (T sin θ = mv²/r).

Split the tilted force (tension, contact force or lift) into components. The mark scheme wants exactly two lines:
Vertically: the vertical component balances the weight, so there is no vertical resultant: $T\cos\theta = mg$ .
Horizontally: the horizontal component is the resultant, pointing at the centre: $T\sin\theta = \dfrac{mv^2}{r}$ (9702/42/F/M/25 Q1(b)).
Divide the two equations and the force size cancels: $\tan\theta = \dfrac{v^2}{gr}$ , or $a = g\tan\theta$ .

Worked example

Exam version: a ball inside a smooth cone

A steel ball rolls in a horizontal circle of radius 0.15 m inside a frictionless cone whose wall makes 52° with the horizontal, so the contact force is tilted at 52° from the vertical direction of the resolve. Show that the speed is about 1.4 m s⁻¹, then find ω. (9702/42/F/M/25 Q1)

Two lines: vertical component of contact force = weight; horizontal component = $mv^2/r$ . Dividing: $a = g\tan 52^\circ = 12.6 \text{ m s}^{-2}$ .
$v = \sqrt{ar} = \sqrt{12.6 \times 0.15} = 1.4 \text{ m s}^{-1}$ .
$\omega = \dfrac{v}{r} = \dfrac{1.4}{0.15} = 9.3 \text{ rad s}^{-1}$ .

Answer

v = 1.4 \text{ m s}^{-1}, \; \omega = 9.3 \text{ rad s}^{-1}

The follow-up one-marker: with the same surface angle the acceleration is fixed, so going faster needs a bigger radius ( $v^2 \propto r$ ).

The same resolve, different actors: a conical pendulum on a string ( $T\cos\theta = mg$ , radius = $L\sin\theta$ ) (9702/41/M/J/23 Q2); an aircraft banking (vertical part of lift balances weight, horizontal part turns the plane); a banked road (the normal contact force leans inward).

Common mistake

Check which line the angle is measured from. A cone wall at 52° to the horizontal and a string at 27° to the vertical put the sin and cos in different places. Draw the triangle before writing the two lines.

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