# 6.2.2 Conical Pendulum

Consider the conical pendulum: a pendulum bob of mass m traveling along a circular path in the horizontal plane at a constant speed v, with the string of length L making an angle θ with the vertical.

We realize that there are two and only two forces acting on the bob: the tensional force T and the weight mg. There isn’t a third force acting in the centripetal force, as many students may think. Circular motion does not require any physical force to be in the centripetal direction. It only requires the resultant force to be in the centripetal direction. Therefore,

vertically:

\begin{aligned}(\sum {{F}_{y}}&=0)\\T\cos \theta &=mg\quad \cdots (1)\end{aligned}

horizontally:

\displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\T\sin \theta &=m\frac{{{{v}^{2}}}}{r}\quad \cdots (2)\end{aligned}

(2)÷(1):

$\displaystyle \tan \theta =\frac{{{{v}^{2}}}}{{rg}}\quad \cdots (3)$

You will see equation (3) appear in many other circular motions e.g. an airplane turning in a horizontal circle, and a car turning on a banked curve. They all share the similar dynamics of having the vertical component of some force balancing the weight, while the horizontal component provides the required centripetal force. Note that not a single force is in the centripetal direction. But the resultant force is.

Vertically:

$\displaystyle L\cos \theta =mg\text{ }\cdots \cdots \cdots (1)$

Horizontally:

$\displaystyle L\sin \theta =m\frac{{{{v}^{2}}}}{r}\text{ }\cdots \cdots \cdots (2)$

(2)÷(1):

$\displaystyle \tan \theta =\frac{{{{v}^{2}}}}{{rg}}$

Vertically:

$\displaystyle N\cos \theta =mg\text{ }\cdots \cdots \cdots (1)$

Horizontally:

$\displaystyle N\sin \theta =m\frac{{{{v}^{2}}}}{r}\text{ }\cdots \cdots \cdots (2)$

(2)÷(1):

$\displaystyle \tan \theta =\frac{{{{v}^{2}}}}{{rg}}$

Video Explanation

Conical Pendulum, Flying Chairs, Aeroplane Turning and Banked Curve