In a demo, a pendulum was used. (See my pic.) It showed that the object experiences 3 g's at the bottom of the swing, according to F = (mv^2)/L. (where L, length of string = r) A peg is located a distance D directly below the point of attachment of the cord.(adsbygoogle = window.adsbygoogle || []).push({});

O_____L________o

+----------------|

-+---------------|

--+--------------|D

---+-------------|----+

----+------------*peg--+

------+----------|----+

---------+----+-O--+

+ = path of circular motion

O = object

First off, for a pendulum that starts at rest at a 90 degree angle to the vertical, energy conservation requires the speed to be consistent, so (mv^2)/2 = mgL or v^2 = 2gL. Now the force experienced at the bottom of the string is F = mg + (mv^2)/L, and after plugging in v^2 = 2gL, you get 3 mg (3 g).

Now my teacher gave us mg + T = (mv^2)/L. Is this for the force experienced at the top of the string? I don't understand how this is correct or what to do with it.

We are supposed to derive the relationship between D and L, where D ends up as (3/5)L. I read somewhere that the max. acceleration an object can endure is a = 5g. How can this be proven and how can I tie in this fact so the g's cancel out and I am left with D = (3/5)L?

Please help me. Thanks.

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# Homework Help: Pendulum force experienced at the top of the string?

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