## Essential University Physics: Volume 1 (4th Edition)

$v= \sqrt{ \frac{gLcos^2\theta}{sin\theta}}$
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: $F_tcos\theta = \frac{mv^2}{Lcos\theta}$ $\frac{mg}{sin\theta}cos\theta = \frac{mv^2}{Lcos\theta}$ $v^2= \frac{gL}{sin\theta}cos^2\theta$ $v= \sqrt{ \frac{gLcos^2\theta}{sin\theta}}$