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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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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

2 Answers 2

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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