Question

Show that the angle made by the string with the vertical in a conical pendulum is given by $\theta =\frac{g}{L{\omega }^2}$ where L is the length of the string and $\omega$ is the angular speed.

Answer 1

Let $T$ be the tension in the string of conical pendulum.

From figure $OB=r=Lsin\theta$

Horizontal direction : $Tsin\theta =mr{w}^2$

$\therefore$ $Tsin\theta =m\left(Lsin\theta \right)w^2$ $⟹T=mL{w}^2$

Vertical direction : $Tcos\theta =mg$

$\therefore$ $\left(mL{w}^2\right)cos\theta =mg$ $⟹\theta =co{s}^{-1}(\frac{g}{L{w}^2})$