JEE Main & Advanced Physics Two Dimensional Motion Conical Pendulum

This is the example of uniform circular motion in horizontal plane.

A bob of mass m attached to a light and in-extensible string rotates in a horizontal circle of radius r with constant angular speed \[\omega \] about the vertical. The string makes angle \[\theta \] with vertical and appears tracing the surface of a cone. So this arrangement is called conical pendulum.  

The force acting on the bob are tension and weight of the bob.

From the figure \[T=\sin \theta =\frac{m{{\upsilon }^{2}}}{r}\]   ...(i) 

and \[T\cos \theta =mg\]                     ...(ii)  

(1) Tension in the string : \[T=mg\sqrt{1+{{\left( \frac{{{\upsilon }^{2}}}{rg} \right)}^{2}}}\]   ...(i)

\[T=\frac{mg}{\cos \theta }=\frac{mgl}{\sqrt{{{l}^{2}}-{{r}^{2}}}}\]

[As \[\cos \theta =\frac{h}{l}=\frac{\sqrt{{{l}^{2}}-{{r}^{2}}}}{l}\]]              

(2) Angle of string from the vertical :

\[\tan \theta =\frac{{{\upsilon }^{2}}}{rg}\]   ...(ii)

(3) Linear velocity of the bob : 

\[\upsilon =\sqrt{gr\tan \theta }\]

(4) Angular velocity of the bob :     

\[\omega =\sqrt{\frac{g}{r}\tan \theta }=\sqrt{\frac{g}{h}}=\sqrt{\frac{g}{l\cos \theta }}\]

(5) Time period of revolution :  

\[{{T}_{p}}=2\pi \sqrt{\frac{l\cos \theta }{g}}=2\pi \sqrt{\frac{h}{g}}\]

\[=2\pi \sqrt{\frac{{{l}^{2}}-{{r}^{2}}}{g}}=2\pi \sqrt{\frac{r}{g\tan \theta }}\]     ...(iii)

If \[\theta ={{90}^{o}}\], then pendulum becomes horizontal & from equations (i), (ii) and (iii) we get \[\upsilon =\infty ,\,\,T=\infty \] and \[{{T}_{p}}=0\] which is practically not possible.