A ball of mass ' $\mathrm{m}$ ' is attached to the free end of a string of length ' $l$ '. The ball is…

A ball of mass ' $\mathrm{m}$ ' is attached to the free end of a string of length ' $l$ '. The ball is moving in horizontal circular path about the vertical axis as shown in the diagram. The angular velocity ' $\omega$ ' of the ball will be [ $\mathrm{T}=$ Tension in the string. $]$

1. $\sqrt{\frac{\mathrm{T} l}{\mathrm{~m}}}$
2. $\sqrt{\frac{\mathrm{Tm}}{l}}$
3. $\sqrt{\frac{\mathrm{m} l}{\mathrm{~T}}}$
4. $\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}}$

Solution

The tension in the string can be resolved in two components along the perpendicular axis. The gravitational force is acting downwards and the centrifugal force is acting in $-\mathrm{x}$ direction $\begin{aligned} & \mathrm{T} \sin \theta=\mathrm{mr} \omega^2 \\ \therefore \quad & \omega^2=\frac{\mathrm{T} \sin \theta}{\mathrm{mr}} \end{aligned}$ $\therefore \quad \omega=\sqrt{\frac{\mathrm{T} \sin \theta}{\mathrm{mr}}}$ From figure, $\sin \theta=\frac{\mathrm{r}}{l}$ $\begin{aligned} & \therefore \quad \omega=\sqrt{\frac{\mathrm{Tr}}{\mathrm{mr} l}} \\ & \therefore \quad \omega=\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}} \\ & \end{aligned}$ ~

Asked in: MHT CET 2023 (11 May Shift 1)

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