The conical pendulum

Suppose that an object, mass $m$, is attached to the end of a light inextensible string whose other end is attached to a rigid beam. Suppose, further, that the object is given an initial horizontal velocity such that it executes a horizontal circular orbit of radius $r$ with angular velocity $\omega$. See Fig. 60. Let $h$ be the vertical distance between the beam and the plane of the circular orbit, and let $\theta$ be the angle subtended by the string with the downward vertical.

Figure 60: A conical pendulum.

The object is subject to two forces: the gravitational force $m g$ which acts vertically downwards, and the tension force $T$ which acts upwards along the string. The tension force can be resolved into a component $T \cos\theta$ which acts vertically upwards, and a component $T \sin\theta$ which acts towards the centre of the circle. Force balance in the vertical direction yields

$$T \cos\theta = m g.$$ (262)

In other words, the vertical component of the tension force balances the weight of the object.

Since the object is executing a circular orbit, radius $r$, with angular velocity $\omega$, it experiences a centripetal acceleration $\omega^2 r$. Hence, it is subject to a centripetal force $m \omega^2 r$. This force is provided by the component of the string tension which acts towards the centre of the circle. In other words,

$$T \sin\theta = m \omega^2 r.$$ (263)

Taking the ratio of Eqs. (262) and (263), we obtain

$$\tan\theta = \frac{\omega^2 r}{g}.$$ (264)

However, by simple trigonometry,

$$\tan\theta = \frac{r}{h}.$$ (265)

Hence, we find

$$\omega = \sqrt{\frac{g}{h}}.$$ (266)

Note that if $l$ is the length of the string then $h=l \cos\theta$. It follows that

$$\omega = \sqrt{\frac{g}{l \cos\theta}}.$$ (267)

For instance, if the length of the string is $l=0.2 {\rm m}$ and the conical angle is $\theta = 30^\circ$ then the angular velocity of rotation is given by

$$\omega = \sqrt{\frac{9.81}{0.2\times \cos 30^\circ}} = 7.526 {\rm rad./s}.$$ (268)

This translates to a rotation frequency in cycles per second of

$$f = \frac{\omega}{2 \pi} = 1.20 {\rm Hz}.$$ (269)