6.0 Conical Pendulum

If a small particle of mass $m$ tied to an inextensible massless string is whirled in a horizontal circle as shown in fig. This arrangement is called as conical pendulum.

In conical pendulum, the vertical component of tension $(T)$ balances the weight $(W=mg)$ while the horizontal component of tension provides necessary centripetal acceleration $\left( {{a_c}} \right)$.

Therefore, $$\begin{equation} \begin{aligned} T\cos \theta = mg\quad ...(i) \\ T\sin \theta = \frac{{m{v^2}}}{r}\quad ...(ii) \\\end{aligned} \end{equation} $$

Dividing the equation $(ii)$ by $(i)$ we get, $$\begin{equation} \begin{aligned} \tan \theta = \frac{{{v^2}}}{{rg}} \\ v = \sqrt {rg\tan \theta } \\\end{aligned} \end{equation} $$ Also, $$\omega = \frac{v}{r} = \sqrt {\frac{{g\tan \theta }}{r}} $$

Time period $\left( \tau \right)$ of pendulum, $$\begin{equation} \begin{aligned} \tau = \frac{{2\pi }}{\omega } \\ \tau = 2\pi \sqrt {\frac{r}{{g\tan \theta }}} \\ \tau = 2\pi \sqrt {\frac{{l\sin \theta .\cos \theta }}{{g\sin \theta }}} \\ \tau = 2\pi \sqrt {\frac{{l\cos \theta }}{g}} \\\end{aligned} \end{equation} $$

where,

$\tau $: Time period of pendulum

$\theta $: Angle made by the string with the vertical

$v$: Tangential velocity of particle

Squaring and adding equation $(i)$ and $(ii)$ we get,

$$\begin{equation} \begin{aligned} {T^2} = {\left( {mg} \right)^2} + {\left( {\frac{{m{v^2}}}{r}} \right)^2} \\ T = \sqrt {{{\left( {mg} \right)}^2} + {{\left( {\frac{{m{v^2}}}{r}} \right)}^2}} \\\end{aligned} \end{equation} $$

Question 8. A carousel is rotating with angular velocity $\omega $ about its center as shown in the figure. Find:

(a). tension $T$ in the rod

(b). angle $\theta $ which the rod makes with the vertical

(c). time period $\left( \tau \right)$ of the rotation

(Assume rod to be massless).

Solution:

The horse is rotating in a circular motion with angular velocity $\omega $ about the center of carousel with radius $(L+r)$.

So, the components of tension will provide necessary centripetal force and balance weight of the horse.

Therefore, $$\begin{equation} \begin{aligned} T\cos = mg\quad ...(i) \\ T\sin \theta = m{\omega ^2}\left( {L + r} \right)\quad ...(ii) \\\end{aligned} \end{equation} $$

Squaring and adding equation $(i)$ and $(ii)$ we get,

Dividing equation $(i)$ by $(ii)$ we get, $$\tan \theta = \frac{{{\omega ^2}(L + r)}}{g}\quad ...(iii)$$

From equation $(iii)$ we get, $$\omega = \sqrt {\frac{{g\tan \theta }}{{L + r}}} $$

Time period is given by, $$T = \frac{{2\pi }}{\omega }$$