# flawed logic, math or both.

joe shmo
Science 18 Aug '10 03:16
1. joe shmo
Strange Egg
18 Aug '10 03:161 edit
So my mother, who likes to swing alot, bent the S hook whilst swinging, thus sending her into the air. I just wanted to see what kind of force she applied to it, but somewhere I feel that my mathematics are inconsistent, can anyone help me find the flaw in my mathematics/logic/model?

I started with conservation of energy

-mgh = (1/2)mv^2

m= mass
g= acceleration due to gravity
v= linear velosity
h = height

so I start with her at rest at 0 P.E., and allow her to be lifted to a certain height whitch as a function of the angle subtended "A" and the radius "R" is as follows:

h = R - R*cosA

subbing this in to the energy equation and solving for v^2

v^2 = 2gR(cosA -1)

since there are two chains im assuming the force is split equally amongst them

such that

F = mv^2/(2R) = mg(cosA-1)

here by inspection i see the force is max at A = pi/2

but i should expect the force to be maximized when her velocity is maximized at A =0

and dF/dA = -mg*sinA = 0; A=0

which minimizes the force function i just derived. ๐

where am I being inconsistent in my analysis? I realize due to a lack of a diagram this may be a tricky to explain.
2. 18 Aug '10 10:30
I don't understand the problem, can you give the initial and boundary conditions?
3. 18 Aug '10 15:06
Originally posted by joe shmo
So my mother, who likes to swing alot
You really shouldn't be telling us things like that, you know.
Baby Gauss
18 Aug '10 15:07
Originally posted by mtthw
You really shouldn't be telling us things like that, you know.
Ahahah!
Baby Gauss
18 Aug '10 15:112 edits
Originally posted by joe shmo

and dF/dA = -mg*sinA = 0; A=0

which minimizes the force function i just derived. ๐
I have to say that I didn't actually read al of your post with that much attention but this bit I'm quoting isn't necessarily true.

Having a 0 derivative at a point doesn't imply that the function at that point reaches a minimum. It just means that a stationary point was found. To know if it is a minimum, a maximum or a saddle point you need the second derivative...

Sorry for the lazy reply but I hope it is of any help anyhow.

Ps:

but i should expect the force to be maximized when her velocity is maximized at A =0

Why do you say this? In the case of the harmonic oscillator the velocity is maximum when the force is zero and the force is maximum when the velocity is zero.
Thus maximizing the force doesn't imply maximizing the velocity (and the opposite is also true).
6. AThousandYoung
West Coast Rioter
18 Aug '10 16:49
Originally posted by mtthw
You really shouldn't be telling us things like that, you know.
I'm really curious what bending an S hook is such that it sends her into the air...

(yes I know this is about playgrounds and the little metal S that attaches the seat to the chain)
7. joe shmo
Strange Egg
18 Aug '10 17:52
Originally posted by mtthw
You really shouldn't be telling us things like that, you know.
๐
8. joe shmo
Strange Egg
18 Aug '10 18:14
but i should expect the force to be maximized when her velocity is maximized at A =0

"Why do you say this? In the case of the harmonic oscillator the velocity is maximum when the force is zero and the force is maximum when the velocity is zero.
Thus maximizing the force doesn't imply maximizing the velocity (and the opposite is also true)."

That was apparently false intuition that led me down that path of thought.

What you exclaimed above is consistent with the equation, so I guess it comes down to poor logical interpretation of the model.

and why did i say it?

F=(m/R)*v^2

m/R is constant, so the force would be largest in its valid range, when velocity is largest in its domain? Thats how I interpret it, why thats apparently incorrect i'm not sure?
9. joe shmo
Strange Egg
18 Aug '10 18:17
Originally posted by AThousandYoung
I'm really curious what bending an S hook is such that it sends her into the air...

(yes I know this is about playgrounds and the little metal S that attaches the seat to the chain)
If this is serious question, she eccentially straightened the S hook on the bottom causing the chain to come free of the crossbar.
Baby Gauss
18 Aug '10 18:491 edit
What you exclaimed above is consistent with the equation, so I guess it comes down to poor logical interpretation of the model.

and why did i say it?

F=(m/R)*v^2

m/R is constant, so the force would be largest in its valid range, when velocity is largest in its domain? Thats how I interpret it, why thats apparently incorrect i'm not sure?

"-mgh = (1/2)mv^2"

What this equation expresses is that you're considering the point of highest height (hence with zero velocity) and the point of 0 height (hence with maximum speed ).

Your next equation: h = R - R*cosA tells me that your A angle is measured starting from the axis that represents maximum height to the axis that contains the height we are currently considering.
If I'm interpreting you right I think you could just consider h = RsinA which is a much simpler expression.
Correct me if I'm wrong on this.

But now comes the real tricky part

subbing this in to the energy equation and solving for v^2

v^2 = 2gR(cosA -1)

You can't really do this because your first expression only makes sense when h is maximum and you're expression of h is a general one.

Incidentally you're last expression could also point out that you've made another error thus far.
Since cosA < 1 (except when A=0) the expression for v^2 gives negative values almost every time. Which obviously can't happen. But it could be that you're sing the convention that g is a negative number (and that would explain the minus sign in your first equation).
I want to comment on the rest of your post but I want to be sure I really understood the scenario you're describing.
11. joe shmo
Strange Egg
19 Aug '10 22:452 edits

"-mgh = (1/2)mv^2"

What this equation expresses is that you're considering the point of highest height (hence with zero velocity) and the point of 0 height (hence with maximum speed ).

Your next equation: h = R - R*cosA tells me that your A angle is measured starting from the axis that represents maximum your post but I want to be sure I really understood the scenario you're describing.
As for the "RsinA"...For some reason I just dont see that simplification.

|\
|.\
|..\
|...\
|

The line far left; 0 P.E.
Far right max P.E
"A" is subtending these 2 lines of equal (diagram obviously not to scale) length "R"

ok, I see the simplification now, but I must say that took a bit of messing around with sum and difference formulas, and the law of sines to arrive at...which probably indicates that I am overlooking the obvious...or it indicates that your willing to work harder for the pupose of simplification!๐

So where do we go from here, I also am having trouble wrapping my head around how/why my starting equation pigeon holes the height to be a maximum

and that negative had was not because of g (wether right or wrong) it came from
final PE minus Intital PE, final is 0, and final KE - Initial KE, this side inital =0
12. Bosse de Nage
Zellulรคrer Automat
20 Aug '10 00:28

Bring it on!
Baby Gauss
03 Sep '10 14:23
Originally posted by joe shmo
So where do we go from here, I also am having trouble wrapping my head around how/why my starting equation pigeon holes the height to be a maximum

and that negative had was not because of g (wether right or wrong) it came from
final PE minus Intital PE, final is 0, and final KE - Initial KE, this side inital =0
I also am having trouble wrapping my head around how/why my starting equation pigeon holes the height to be a maximum

Because by conservation of energy we know that:

1/2mv^2+mgh=1/2mu^2+mgz where v and u represent velocities and h and z represent heights. If you write -mgh = (1/2)mv^2 you're dropping terms from the full equation and the meaning of dropping those two terms is that you're saying the left hand side of your equation has zero velocity (thus maximum height) and the right side of your equation has zero height (thus maximum velocity).

and that negative had was not because of g (wether right or wrong) it came from
final PE minus Intital PE, final is 0, and final KE - Initial KE, this side inital =0

I think that I might have mislead you here. The thing is that v^2 = 2gR(cosA -1) is a wrong equation if you consider g to be positive. Since g>0 implies that 2gR(cosA-1 )<0 (except when A=0 and then it is v^2=0) what your equation states is that v^2<0 and this can't be since v is a real number.

Sorry for the delay.
14. joe shmo
Strange Egg
03 Sep '10 21:00
I also am having trouble wrapping my head around how/why my starting equation pigeon holes the height to be a maximum

Because by conservation of energy we know that:

1/2mv^2+mgh=1/2mu^2+mgz where v and u represent velocities and h and z represent heights. If you write -mgh = (1/2)mv^2 you're dropping terms from the full equation and ...[text shortened]... tion states is that v^2<0 and this can't be since v is a real number.

Sorry for the delay.
"1/2mv^2+mgh=1/2mu^2+mgz where v and u represent velocities and h and z represent heights. If you write -mgh = (1/2)mv^2 you're dropping terms from the full equation and the meaning of dropping those two terms is that you're saying the left hand side of your equation has zero velocity (thus maximum height) and the right side of your equation has zero height (thus maximum velocity)."

ahhhh...๐ณ
and about the "g" I did realize that it must be negative to make sense, however as you can see rather than redflagging the result for further analysis( which may have led me to my inconsitency) I blew it off...something I won't take lighty anymore.