Conical Pendulum
Category : JEE Main & Advanced
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.
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