Re: [Phys-L] pendulum in free fall

[stefan jeglinski <jeglin@4pi.com>]

Thanks Carl yes that was an obvious blunder on my part. As I was 
thinking about this I began imagining the pendulum dropped at the 90-deg 
mark to calculate v at the bottom, and of course the value of T in that 
case doesn't go as √L/g. I should have left the question as just v at 
the bottom. Both the value of v and T are irrelevant to the question at 
hand, which is a description of the motion in the two reference frames. 
I think the elevator frame of reference seems pretty straightforward 
(UCM). It's the ground frame I'm still wondering about.

Stefan Jeglinski

On 1/6/25 7:13 AM, Carl Mungan via Phys-l wrote:
> I think the problem is indeterminate because period is independent of amplitude (for small oscillations). The speed of the bob at the instant the cable breaks is thus unknown. Call it V.
>
> V is given by sqrt[2gL(1-cosA)] where A is the (unknown) angular amplitude before cutting the cable. Here the length of the string is L = g (T/2*pi)^2 which is known.
>
> After cutting, the bob will go around in a circle (relative to the elevator) with period 2*pi*L/V.
>
> > On Jan 6, 2025, at 12:27 AM, stefan jeglinski<jeglin@4pi.com> wrote:
> >
> > Happy New Year, I need help sorting out my thinking on this one. The problem as posed:
> >
> > "An elevator cab is suspended from a steel cable. A pendulum inside the cab hangs from the ceiling of the cab on a string. The pendulum is set to swinging and has a period T while the elevator cab is stationary. Suddenly, the steel cable supporting the cab breaks (or is cut) at precisely the moment when the pendulum bob reaches its *maximum speed*. Describe the pendulum’s subsequent motion from the point of view of an observer in the elevator and also of an observer on the ground. What is the period of the pendulum as the cab falls?"
> >
> > (We imagine that the elevator roof has a slot or some opening which allows the pendulum to swing outside of the elevator while it's falling and then back in. The pivot point is some frictionless shaft bearing that allows the pendulum to swing in a complete plane)
> >
> > Me:
> >
> > Elevator:
> > When the cable breaks everything goes into free fall (“no gravity”). The bob drops like every other part of the elevator but it has a horizontal speed at the moment of the break. The bob moves to the side as it drops in such a way that the original tension at the break is intact and the string stays taut at length L (if the bob was/freely/moving its distance from its pivot would be > L); thus tension is maintained and the bob moves in uniform circular motion with a tangential speed v = sqrt(2gL).
> >
> > Ground:
> > Everything about the pendulum must/look/the same – the pendulum string can’t go slack for one observer and not the other (yes?); however, the bob’s path doesn’t look like a circle from the ground – the bob follows a vertical cycloid that accelerates down. Punchline: an accelerating cycloid means the bob is under a non-uniform tension.
> >
> > This is my key issue: if this analysis is correct then we could choose a value of g for which the pendulum string doesn’t break for the elevator observer but could break for the ground observer. Or maybe my analysis is incorrect.
> >
> > PS the “period” question is interesting. Although the “restoring period" T ~ sqrt(L/g) goes away (infinity for g = 0), the pendulum still has a period due to circular (or cycloidal) motion.
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