$\begin{array}{1 1} \tan^{-1} v^2/gr \\ \cot^{-1} v^2/gr \\ \sin^{-1} v^2/gr \\\cos^{-1} v^2/gr \end{array} $

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Answer :$ \tan^{-1} v^2/gr$

By Newton's second law the force Fproducing this acceleration on a mass m is $F= ma= \large\frac{mv^2}{r}$

Taking moments about G gives $R_1 \large\frac{l}{2} $$= R_2 \large\frac{l}{2}$

Therefore $R_1=R_2-$the reaction forces are the same, we will call them both R from now on.

Vertically there is no motion so no acceleration so resolving forces in the vertical direction gives $2R \cos \theta =mg$ ----(i)

Resolving forces horizontally gives $\large\frac{mv^2}{r} $$=2R \sin \theta$ -----(ii)

Dividing (ii) by (i) gives

$\tan \theta = \large\frac{\sin \theta}{\cos \theta} =\frac{mv^2}{mgr}$

$\tan \theta =\large\frac{v^2}{gr}$

Therefore $\theta= \tan ^{-1} v^2/gr$

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