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In a circular pendulum

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?

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closed as off-topic by ACuriousMind, Brandon Enright, Prahar, John Rennie, JamalS Nov 30 '14 at 9:13

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  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Brandon Enright, Prahar, John Rennie, JamalS
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You've got a very nice free body diagram there...have you tried applying the rules for uniform circular motion to the appropriate lengths and resultant forces? – dmckee May 2 '12 at 21:15
It seems the homework tag applies even if it is not actual homework, cf. the tag description – Qmechanic May 3 '12 at 8:49
up vote 1 down vote accepted

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

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

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Do NOT give full solutions to [homework] problems. Only hints. See… – Manishearth May 3 '12 at 3:39
@Manishearth There aren't homeworks. – Garmen1778 May 3 '12 at 6:10
@Garmen1778: It falls under the [homework] tag, regardless of it being homework. See the link above. – Manishearth May 3 '12 at 9:33

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)

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