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In a conical pendulum the bob of mass ' $\mathrm{m}$ ' moves in a horizontal circle of radius ' $r$ ' with uniform speed ' $V$ '. The string of length ' $L$ ' describes a cone of semi vertical angle ' $\theta$ '. The centripetal force acting on the bob is ( $\mathrm{g}=$ acceleration due to gravity)
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The correct answer is:
$\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$
$\begin{aligned} & \mathrm{T} \cos \theta=\mathrm{mg} \\ & \sin \theta=\frac{\mathrm{r}}{\mathrm{L}} \\ & \cos \theta=\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{~L}}\end{aligned}$
$\begin{aligned} \therefore \quad \mathrm{T} & =\frac{\mathrm{mg}}{\cos \theta}=\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}} \\ \mathrm{mr} \omega^2 & =\mathrm{T} \sin \theta \\ \frac{\mathrm{T} \times \mathrm{r}}{\mathrm{L}} & =\mathrm{m}^2 ...(\sin \theta= \frac{r}{L}) \\ \omega^2 & =\frac{\mathrm{g}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\end{aligned}$
$\therefore \quad$ The centripetal force is
$\mathrm{F}=\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$
$\begin{aligned} \therefore \quad \mathrm{T} & =\frac{\mathrm{mg}}{\cos \theta}=\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}} \\ \mathrm{mr} \omega^2 & =\mathrm{T} \sin \theta \\ \frac{\mathrm{T} \times \mathrm{r}}{\mathrm{L}} & =\mathrm{m}^2 ...(\sin \theta= \frac{r}{L}) \\ \omega^2 & =\frac{\mathrm{g}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\end{aligned}$
$\therefore \quad$ The centripetal force is
$\mathrm{F}=\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$
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