In a circular pendulum, where does the equation $v=\sqrt{rg\tan{\alpha}}$ come from?

Question

In a circular pendulum

the $v$ of the particle is $$v=\sqrt{gr\tan{\theta}}$$ where $r$ is the radius and $g$ is the gravity(positive sign), which is equal to $$v=\sqrt{gL\sin{\theta}\cos{\theta}}$$But where it comes from?

Answer 1

Actually it comes from Euler's Laws of Motion applied to a point mass instead of a rigid body (or as they are commonly known as Newton's laws).

Force equals mass times acceleration of the center of gravity. In a pendulum with steady motion, as it is rotating about the $+y$ with speed $\omega$, the tangential speed of the particle is $v=\omega\,r$. The acceleration of the particle, is that of circular motion with radius $r$ or $a = \omega^2 r = v^2/r$.

Since the forces in the vertical direction are balanced then $T\cos\theta = m g$ } $ T = \frac{m g}{\cos\theta}$

In the radial direction we have our equations of motion

$$ T \sin(\theta) = m v^2/r $$ $$ \frac{m g}{\cos\theta} \sin(\theta) = m v^2/r $$ $$ g \tan(\theta) = v^2/r $$ $$ v = \sqrt{ g\,r\,\tan\theta } $$

Answer 2

Looks like someone has forgotten what centripetal force is :)

Whenever you have circular motion, always use CPF as the resultant radial force.

(If you don't know everything behind CPF yet, you ought to learn that first)