Conical Pendulum

A tiny bob (a point mass) attached to a long, flexible, effectively massless and inextensible string, suspended from a rigid support and made to revolve so that the string sweeps out the surface of a right circular cone (rather than swinging back and forth in one vertical plane, which would instead be a simple pendulum) is called a conical pendulum.

The bob itself performs a practically uniform HORIZONTAL circular motion. Let L be the string length, $\theta$ the (constant) semi-vertical angle the string makes with the vertical at the support, and $r=L\sin\theta$ the radius of the bob's horizontal circular path.

Only two forces act on the bob: its weight mg, vertically downward, and the string tension $T_0$, directed along the string towards the support. Since the bob's actual motion is horizontal, the net force must be horizontal, so the VERTICAL component of tension balances the weight: $T_0\cos\theta=mg$ --- (1.6). The HORIZONTAL component is then the resultant, and hence centripetal, force: $T_0\sin\theta=mr\omega^2$ --- (1.5). Dividing (1.5) by (1.6), $\omega^2=\frac{g}{r}\tan\theta=\frac{g}{L\cos\theta}$ (using $r=L\sin\theta$, so $\tan\theta/r=1/(L\cos\theta)$). This gives the period of revolution $T=2\pi\sqrt{\frac{L\cos\theta}{g}} \qquad \text{--- (1.7)}$ and the frequency $n=\frac{1}{2\pi}\sqrt{\frac{g}{L\cos\theta}} \qquad \text{--- (1.8)}$ Both depend only on the string length L, the angle $\theta$, and g -- crucially, NEITHER depends on the mass of the bob, which cancels out of the ratio of Eqs. (1.5) and (1.6) entirely.

The same result is obtained (equally validly) by working in the NON-inertial frame that co-rotates with the bob: here the bob is at rest, and in addition to weight and tension a centrifugal pseudo-force $mr\omega^2$ must be introduced, acting horizontally outward; for equilibrium in this frame, $T_0\sin\theta$ must balance this centrifugal force, giving back the identical Eq. (1.5). …