Conical pendulumsAQA A-Level Further Maths: Revision notes

A conical pendulum is a particle on a string that moves in a horizontal circle, the string tracing out a cone. Only two forces act on the particle: its weight $mg$ downwards and the tension $T$ along the string. If the string has length $L$ and makes angle $\theta$ with the downward vertical, the radius of the circle is $r=L\sin\theta$ and the particle is a depth $L\cos\theta$ below the fixed point. The particle has no vertical acceleration, so the vertical forces balance, while the resultant horizontal force provides the acceleration $\omega^2r=\frac{v^2}{r}$ towards the centre of the circle.

Resolve vertically and horizontally: $T\cos\theta=mg,\qquad T\sin\theta=m\omega^2r=\frac{mv^2}{r}.$ Dividing gives $\tan\theta=\frac{\omega^2r}{g}=\frac{v^2}{rg}$. Since $r=L\sin\theta$, the horizontal equation also gives $T=m\omega^2L$, and so $\cos\theta=\frac{g}{\omega^2L},\qquad T=\frac{mg}{\cos\theta}.$ The period is $\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L\cos\theta}{g}}$. A faster rotation gives a larger $\theta$ and a bigger tension.

A particle of mass $0.2$ kg is on a string of length $0.6$ m, making $60^\circ$ with the vertical. Take $g=9.8$. $T=\frac{mg}{\cos60^\circ}=\frac{0.2\times9.8}{0.5}=3.92$ N. $\omega^2=\frac{g}{L\cos\theta}=\frac{9.8}{0.3}=32.7$, so $\omega=5.72$ rad s⁻¹. $r=0.6\sin60^\circ=0.520$ m, so $v=\omega r=2.97$ m s⁻¹ and the period is $\frac{2\pi}{\omega}=1.10$ s.

Because $\cos\theta=\frac{g}{\omega^2L}$ and $\cos\theta<1$ for a string that is not vertical, the particle can move in a horizontal circle only if $\omega^2>\frac{g}{L}$. As $\omega$ increases, $\cos\theta$ falls and $\theta$ approaches $90^\circ$, but the string can never be horizontal, because the vertical component of tension must balance the weight. The tension $T=m\omega^2L$ rises without limit, so a string with a maximum tension sets a maximum $\omega$.

A particle may be attached by two strings to points $A$ and $B$ on the same vertical line, with $A$ above $B$, and rotate about that line with both strings taut. Let $AP$ make angle $\alpha$ and $BP$ make angle $\beta$ with the vertical, with tensions $T_A$ and $T_B$. The horizontal components both point towards the axis; the vertical component of $T_B$ points down: $T_A\sin\alpha+T_B\sin\beta=m\omega^2r,\qquad T_A\cos\alpha=T_B\cos\beta+mg.$ Example: $AP=BP=0.5$ m, $AB=0.8$ m, $m=0.4$ kg, $\omega=6$ rad s⁻¹. Then $r=0.3$, $\sin\alpha=0.6$, $\cos\alpha=0.8$, so $T_A+T_B=\frac{0.4\times36\times0.3}{0.6}=7.2$ and $T_A-T_B=\frac{0.4\times9.8}{0.8}=4.9$. Hence $T_A=6.05$ N and $T_B=1.15$ N. Both strings stay taut while $T_B\ge0$; at the limit $T_B=0$ and the case reduces to a single string, with $\omega^2=\frac{g\tan\alpha}{r}$.