Pendulum period: max velocity.

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Lloyd Burton
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Pendulum period: max velocity.

Post by Lloyd Burton »

Hello friends, I know the formulas but I am having trouble relating the period of a pendulum to the velocity at the bottom of the swing. The formula for the period involves calculus which is unfortunately beyond me but there is a handy calculator available, so that is not a problem, and I know the formula for a body in free fall. But I have got into a muddle and can't reconcile one with the other. Would some kind soul care to help a mathematically challenged member with a simple explanation?
Lloyd.
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Tarsier79
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re: Pendulum period: max velocity.

Post by Tarsier79 »

The way I used to calculate, as if the pendulum string was weightless.

So, mgh of pendulum bob at top = potential energy = "E".

At the bottom, all potential energy is converted to Kinetic energy. solve for "E": E =1/2 mv^2 or mvv/2
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ME
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Post by ME »

To add: The speed of the pendulum bob can be calculated from Ek=Ep for all the heights.
v = √(2·g·∆h)
∆h = (h₀-h)

This velocity is a tangent velocity at the rim, while the acceleration (g) is straight down.
Because it's tangent you can simply divide by the pendulum-radius to get the angular speed:
ω = √(2·g·∆h) /r [units: rad/second]

When you convert the position to some polar notation then you'll get something like this for the height difference:
∆h = r· (1-cos(θ))
Where:
r: pendulum radius
θ: angle of the pendulum (0 is straight down)
∆h: height above ground

The time-of-arrival is not simple.
As far as I'm able to understand:
While gravity is constant for the vertical, the bob does not follow this line (as like in free fall): thus the acceleration will depend on the angle of the bob, while the velocity depends on the pure height and (besides direction) is actually independent of the angle.
It looks like it has all the ingredients to be a simple plug-and-calculate operation, but it lacks the right symmetry to be simple and skews things up a bit.
Thus we get some elliptic function which is as difficult as calculating the arclength of an ellipse which is not simply stretching a circle. IOW: not simple.

So we plug our angle "θ" into an elliptic integral, which is luckily figured out by someone else, to make a correction to the simple "period" formula we know.

As can be seen in the formula series: a small theta (θ) only get smaller (θ²), hence we usually ignore this correction for small angles.
T = 2·π·√(r/g) · ( 1+ (θ²)/16 + (θ⁴)*11/3072 ... )

And divide by 4 if you only want the travel-time from some starting-angle to the bottom.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
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