Pendulum

[Soaring Crane]

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.

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.

[profdh53]

Physics is not my field, but I'll give this a shot.
If you draw a right-angle triangle, using the change in height (D) as one side, and the length L where it is horizontal to D as the Hypotenuse, then you are looking for x, which is the other side. So, L^2= D^2 + x^2
Now, you have 3g=v^2/2, so g=v^2 (3/2), so L= 3/2
You have D= 3/5L, so D= 9/10
L^2 = D^2 + x^2
1.5^2= .9^2 + x^2
2.25 - .81 = x^2
x^2 = 1.44
x=.12

I hope this is true, and I hope it helps!

P

[profdh53]

My typo - sorry.
x is not .12
x is 1.2

Greetings,
P

[Soaring Crane]

Why do you solve for x?

Why is 3g = (v^2)/2? Wouldn't g then = (v^2)/6?