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A particle of mass $m$ is suspended from a ceiling through a string of length $L$. If the particle moves in a horizontal circle of radius $r$ as shown in the figure, then the speed of the particle is
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The correct answer is:
$r \sqrt{\frac{g}{\sqrt{L^2-r^2}}}$
From the diagram,
$\begin{aligned} & \frac{T \sin \theta}{T \cos \theta}=\frac{F_{\text {Centripetal }}}{\text { weight }}=\frac{m v^2}{r} \times \frac{1}{m g} \\ & \Rightarrow \tan \theta=\frac{v^2}{r g} \\ & \text { Also, } \tan \theta=\frac{t}{\sqrt{L^2-r^2}} \\ & \therefore \frac{v^2}{r g}=\frac{t}{\sqrt{L^2-r^2}} \\ & \text { or, } v=r \sqrt{\frac{g}{\sqrt{L^2-r^2}}}\end{aligned}$
$\begin{aligned} & \frac{T \sin \theta}{T \cos \theta}=\frac{F_{\text {Centripetal }}}{\text { weight }}=\frac{m v^2}{r} \times \frac{1}{m g} \\ & \Rightarrow \tan \theta=\frac{v^2}{r g} \\ & \text { Also, } \tan \theta=\frac{t}{\sqrt{L^2-r^2}} \\ & \therefore \frac{v^2}{r g}=\frac{t}{\sqrt{L^2-r^2}} \\ & \text { or, } v=r \sqrt{\frac{g}{\sqrt{L^2-r^2}}}\end{aligned}$
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