41.3: The Conical Pendulum

A conical pendulum is also similar to a simple plane pendulum, except that the pendulum is constrained to move along the surface of a cone, so that the mass $m$ moves in a horizontal circle of radius $r$, maintaining a constant angle $\theta$ from the vertical.

For a conical pendulum, we might ask: what speed $v$ must the pendulum bob have in order to maintain an angle $\theta$ from the vertical? To solve this problem, let the pendulum have length $L$, and let the bob have mass $m$. A general approach to solving problems involving circular motion like this is to identify the force responsible for keeping the mass moving in a circle, then set that equal to the centripetal force $m v^{2} / r$. In this case, the force keeping the mass moving in a circle is the horizontal component of the tension $T$, which is $T \sin \theta$. Setting that equal to the centripetal force, we have

$$T \sin \theta=\frac{m v^{2}}{r} .$$

The vertical component of the tension is

$$T \cos \theta=m g$$

Dividing Eq. $\PageIndex{1}$ by Eq. $\PageIndex{2}$,

$$\tan \theta=\frac{v^{2}}{g r}$$

From geometry, the radius $r$ of the circle is $L \sin \theta$. Making this substitution, we have

$$\tan \theta=\frac{v^{2}}{g L \sin \theta}$$

Solving for the speed $v$, we finally get

$$v=\sqrt{L g \sin \theta \tan \theta} .$$