Answer
$ v= \sqrt{ \frac{gLcos^2\theta}{sin\theta}}$
Work Step by Step
We first find the two components:
x: $ F_tcos\theta = \frac{mv^2}{Lcos\theta}$
y: $F_tsin\theta=mg \\ F_t = \frac{mg}{sin\theta}$
Using the y result for tension, we use the x-equation to find v:
$ F_tcos\theta = \frac{mv^2}{Lcos\theta}$
$ \frac{mg}{sin\theta}cos\theta = \frac{mv^2}{Lcos\theta}$
$ v^2= \frac{gL}{sin\theta}cos^2\theta$
$ v= \sqrt{ \frac{gLcos^2\theta}{sin\theta}}$
