Time period of conical pendulum!

Classical Mechanics Level 2

A small object of mass $m$ is tied to string of length $l$ and is whirled round in a horizontal circle of radius $r$ at a constant speed. The centre of the circle is vertically below the point of support. What is the time period of the rotations in terms of $g$ , $r$ and $\theta$ (where $\theta$ is the angle made by the string with the vertical)?

2\pi\sqrt{\dfrac{r}{g\,\tan\theta}}

2\pi\sqrt{\dfrac{g}{r\tan\theta}}

2\pi\sqrt{\dfrac{\tan\theta}{gr}}

2\pi\sqrt{\dfrac{g\, \tan\theta}{r}}

1 solution

Skanda Prasad
Oct 30, 2016

The solution is very simple. A diagram would have been helpful. But as I'm not sure, from where the diagram has to be taken, I'm giving the solution theoritically. Anyone having the $FBD$ of this problem is requested to post it please.

If we make a $FBD$ , we get $Tsin\theta=\dfrac{mv^2}{r}$

And, $Tcos\theta=mg$ (where $T$ is the tension in the string).

Dividing the above equations and solving for $v$ , we get

$v=\sqrt{rgtan\theta}$

Now $\omega=\dfrac{v}{r}=\dfrac{\sqrt{rgtan\theta}}{r}=\sqrt{\dfrac{gtan\theta}{r}}$

We know, Time period, $t=\dfrac{2\pi}{\omega}$

$\implies$ $t=2\pi\sqrt{\dfrac{r}{gtan\theta}}$

Also, a quick dimension checking shows that no other option works.

Agnishom Chattopadhyay - 4 years, 7 months ago

Well, yeah...that would work too...

Skanda Prasad - 4 years, 7 months ago

Time period of conical pendulum!

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