Conical Pendulum

A conical pendulum is a pendulum consisting a bob suspended by a massless thread which moves in a horizontal circle. The massless thread is only an idealization. No thread is massless in the world. In conical pendulum the bob does not oscillate back and forth but it moves in a circle.

This article deals with the derivation of time period of conical pendulum and speed of rotation of the bob. We consider a bob of mass $m$ suspended by a massless thread of length $L$. The bob moves along a horizontal circle of radius $R$ with constant speed $v$.

In conical pendulum (definition) the bob moves in a horizontal circle and the thread sweeps out the surface of a cone and hence it is called conical pendulum. We first determine the time period of one complete rotation in conical pendulum.

The bob does not show any vertical motion (moves in horizontal circle only). The bob experiences the centripetal (radial) acceleration in horizontal direction and we consider its motion as the uniform circular motion.

Figure 1 Conical pendulum

In Figure 1 the weight of the bob acts vertically downwards whose magnitude is $w = mg $. The conical pendulum tension ${{T}_{F}}$ in the thread has two components. The vertical component of $T_F$ is $T_F \cos \theta $ and the horizontal component is $T_F \sin \theta $. The vertical component balances the weight and the horizontal component provides the centripetal force on the bob. Here we find the period of the pendulum and how it is related to the angle $\theta$ the thread makes with the vertical line. You can get,

\[\begin{align} {{T}_{F}}\cos \theta &=mg \tag{1} \label{1}\\ \text{and,}\quad {{T}_{F}}\sin \theta & =\frac{m{{v}^{2}}}{R} \tag{2} \label{2} \end{align}\]

If $T$ is the time period of the bob to complete one rotation along the circle, the bob travels a distance of $2\pi R$ in time $T$. So the constant speed is, $v = 2\pi R / T$. Also $R = L\sin \theta $. Now putting the values of ${T_F} = mg/\cos \theta$ from Equation \eqref{1} and $v = 2\pi R/T$ in Equation \eqref{2} and solving you get,

\[T = 2\pi \sqrt {\frac{R}{{g\tan \theta }}} \tag{3} \label{3}\]

You know $R = L\sin \theta $, so replacing $R$ in Equation \eqref{3} by $L\sin \theta $ you get the expression (conical pendulum equation) for the time period of rotation, that is,

\[T = 2\pi \sqrt {\frac{{L\cos \theta }}{g}} \tag{4} \label{4}\]

Equation \eqref{4} shows that time period of pendulum is related to the length of the thread, angle $\theta$ between the thread and the vertical line, and the acceleration due to gravity. When $\theta$ increases, the value of $\cos \theta$ decreases and hence the time period decreases. As the angle $\theta $ increases, the tension $T_F = mg/\cos \theta$ increases. Notice that the angle $\theta$ can never be $90^\circ$. If it is $90^\circ$, the time period would be $T = 0$, the tension would be $T_F = \infty$ and the speed would be $v = \infty$.

Again we can determine the expression of speed $v$ of the bob in conical pendulum. To do so, divide Equation \eqref{2} by Equation \eqref{1} and,

\[\tan \theta = \frac{v^2}{Rg} \tag{5} \label{5}\]

Solving for $v$ in Equation \eqref{5} gives $v = \sqrt{ Rg\tan \theta}$. You know from above Figure 1 that, $R = L\sin \theta$, and we get the conical pendulum equation for speed,

\[v = \sqrt {Lg\sin \theta\, \tan\theta} \tag{6} \label {6}\]

The above equation shows that the speed $v$ of the bob is proportional to the length of the thread, the acceleration due to gravity, sine of the angle and tangent of the angle. The above equation also shows that the speed does not depend on the mass of the bob. You can also verify from the above equation that if $\theta$ is $90^\circ$, $v = \infty$ (tangent of $90^\circ$ is $\infty$) which is not possible.

In conical pendulum problems you often need to solve the time period of rotation. Since the time period directly depends on the angle between the thread and the vertical line, you can not make it a good clock!

Mechanics
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