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Nice. That's the accelerating cycloid as I called it. This gets at my
question:
There's a string that ultimately causes that motion of the bob. At some
point in time, it seems like the tension in the string would be high
enough to break the string. Meanwhile, in the elevator frame, the bob is
merely executing UCM forever...
Does IP still exist any more or still run in modern Windows? I thought
it was no longer developed.
Stefan Jeglinski
On 1/6/25 9:45 AM, Philip Keller via Phys-l wrote:
I was able to simulate this in Interactive Physics. In the elevatorframe,
it is just a vertical circle, as expected. But I would not have been ablehttps://docs.google.com/document/d/1SCH65itX6r2ltxAczJL1eJAkpgIYudhbU-noDJ-XXSQ/edit?usp=sharing
to guess the path in the stationary frame! I took a picture of it here:
--
(Btw, this is where Interactive Physics stands out from Phet and others
it took just a few minutes to get this running.)don't
On Mon, Jan 6, 2025 at 9:10 AM Carl Mungan via Phys-l <
phys-l@mail.phys-l.org> wrote:
UCM about a quadratically increasing y coordinate of the center of the
circle. You could probably graph it in Excel.
I don't anyone who makes as big and frequent blunders as myself, so
90-degworry about that! It's a neat problem; where did you get it from?
And yes, we're assuming elevator is so massive compared to the bob (and
assuming a massless string) that elevator does not deflect from its
vertical fall due to motion of bob.
On Mon, Jan 6, 2025 at 9:05 AM stefan jeglinski<jeglin@4pi.com> wrote:
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
=mark to calculate v at the bottom, and of course the value of T in thatthe
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 ofamplitude (for small oscillations). The speed of the bob at the instant
cable breaks is thus unknown. Call it V.
V is given by sqrt[2gL(1-cosA)] where A is the (unknown) angularamplitude before cutting the cable. Here the length of the string is L
tog
(T/2*pi)^2 which is known.the
After cutting, the bob will go around in a circle (relative to theelevator) with period 2*pi*L/V.
problem as posed: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
"An elevator cab is suspended from a steel cable. A pendulum inside
cab hangs from the ceiling of the cab on a string. The pendulum is set
theswinging 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
theelevator and also of an observer on the ground. What is the period of
stringpendulum as the cab falls?"and
allows the pendulum to swing outside of the elevator while it's falling(We imagine that the elevator roof has a slot or some opening which
then back in. The pivot point is some frictionless shaft bearing thathorizontal
allows the pendulum to swing in a complete plane)
The bob drops like every other part of the elevator but it has aMe:
Elevator:
When the cable breaks everything goes into free fall (“no gravity”).
speed at the moment of the break. The bob moves to the side as it dropsin
such a way that the original tension at the break is intact and thestring
stays taut at length L (if the bob was/freely/moving its distance fromits
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
follows acan’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
choosevertical cycloid that accelerates down. Punchline: an acceleratingcycloid
means the bob is under a non-uniform tension.
This is my key issue: if this analysis is correct then we could
elevatora value of g for which the pendulum string doesn’t break for the
isobserver but could break for the ground observer. Or maybe my analysis
stillincorrect.
period" T ~ sqrt(L/g) goes away (infinity for g = 0), the pendulumPS the “period” question is interesting. Although the “restoring
21402-1363has a period due to circular (or cycloidal) motion.21402-1363
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--
Carl E. Mungan, Professor of Physics, 410-293-6680
U.S. Naval Academy Mailstop 9c, 572C Holloway Rd, Annapolis MD
_______________________________________________http://usna.edu/Users/physics/mungan/_______________________________________________
_______________________________________________
Forum for Physics Educators
Phys-l@mail.phys-l.org
https://www.phys-l.org/mailman/listinfo/phys-l
Forum for Physics Educators
Phys-l@mail.phys-l.org
https://www.phys-l.org/mailman/listinfo/phys-l
Forum for Physics Educators
Phys-l@mail.phys-l.org
https://www.phys-l.org/mailman/listinfo/phys-l