Physics of swinging (metal content)

Hanging up a swing for my grand daughter. Rope and chain (metal content). =
She will need to be pushed for the next 2-3 years as we all were. Somewhe=
re around the age of 5-6 we learned to push ourselves.=20
Question about the physics of self pushing.
OK, the period (elapsed time between the oscillations over the same point) =
is a function of pivot length. E.g. if it takes 2 seconds to swing from fr=
ont-top to back and to front-top again, it will take two seconds no matter =
how high we swing or how gently (low) we swing. ROPE LENGTH DICTATES 2 SECO=
NDS AND THAT WILL NEVER CHANGE. Strictly a function of pivot length. =20
Now, we all learned to self-swing by pulling back on the rope (at our furth=
est back point) which effectively shortens the pivot. This shortening (at =
the expense of expended energy) in turn shortens the period, which means we=
go faster between peak-to-peak. On the front peak we release the rope, ag=
ain returning us to full pivot length and and that extra energy goes into r=
aising our peaks higher and higher (all within the same exact two seconds).
Anything else going on during this time? Do I have this thought out OK? I=
s it the shorter period of the rope-pullback which translates into the high=
er peak upon let-go? What am I missing? Been thinking about this for the =
last few days.
Ivan Vegvary
Reply to
Ivan Vegvary
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t). =A0She will need to be pushed for the next 2-3 years as we all were. = =A0Somewhere around the age of 5-6 we learned to push ourselves.
) is a function of pivot length. =A0E.g. if it takes 2 seconds to swing fro= m front-top to back and to front-top again, it will take two seconds no mat= ter how high we swing or how gently (low) we swing. ROPE LENGTH DICTATES 2 = SECONDS AND THAT WILL NEVER CHANGE. Strictly a function of pivot length. = =A0
thest back point) which effectively shortens the pivot. =A0This shortening = (at the expense of expended energy) in turn shortens the period, which mean= s we go faster between peak-to-peak. =A0On the front peak we release the ro= pe, again returning us to full pivot length and and that extra energy goes = into raising our peaks higher and higher (all within the same exact two sec= onds).
? =A0Is it the shorter period of the rope-pullback which translates into th= e higher peak upon let-go? =A0What am I missing? =A0Been thinking about thi= s for the last few days.
When you pull back and shorten the rope, you raise the center of gravity. which is a increase in energy.
Dan
Reply to
dcaster
Dan, I didn't think of that, thanks. Is it that simple?
Ivan Vegvary
Reply to
Ivan Vegvary
Yup. Correct. Basic function of a pendulum. Period =3D 2.(pi). sqrt(length/accel due to gravity)
Your example (2 second period - one second per single "swing") is = perhaps a bad one. It's the classic one-second pendulum used in longcase = clocks and is only about a yard long. Surely you mean more like 2 seconds per swing - four seconds per period. = That gives you a 12 foot rope. Better?
Effectively? You think? If you pull back at the rope you'd maybe shorten it from 12 feet to = -say- 11.5 feet... at the most. Probably more like 11.8 feet. That'd yield a tiny change in period. Not enough to notice.
I'd lose the idea of the shorter rope/ longer rope causation. Too small = to be of any significance. I'd put it down to the momentum of the legs being applied by quickly = pushing them forward at the apex of the swing. That's the energy input = used to maintain the motion.
'Course, I'm just musing. Happy to be proven wrong by a physics major.
-- Jeff R.
Reply to
Jeff R.
bad one. It's the classic one-second pendulum used in longcase clocks and is only about a yard long.
gives you a 12 foot rope. Better?
feet... at the most. Probably more like 11.8 feet.
of any significance.
them forward at the apex of the swing. That's the energy input used to maintain the motion.
When you pull on the rope you are raising the mass (You) further above the ground adding potential energy to the mass by the increased elevation. The potential energy is then converted to kinetic energy as you swing back down to the ground.
John
Reply to
John
bad one. It's the classic one-second pendulum used in longcase clocks and is only about a yard long.
gives you a 12 foot rope. Better?
feet... at the most. Probably more like 11.8 feet.
of any significance.
them forward at the apex of the swing. That's the energy input used to maintain the motion.
My vote is with leg action.
On the back swing the knees bend & the legs go up, to conserve energy/momentum the swing swings faster and stops at a higher point on the back swing.
Then the knees straigten as you swing forward & then repeat the process.
Reply to
Dennis
Well... Yes and no. It's strictly a function of pivot length if the swing is constrained to travel in a cycloidal motion, rather than circular. When the center of mass travels in a circular motion then things slow down for larger amplitude excursions.
The above is the "no" part. The "yes" part is that no one except for clock makers really care about the "no" part.
There's more to it than that. If it was just the parametric oscillator effect such as you (and Wikipedia:
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describe, then you would have a great deal of difficulty self-starting on a swing without pushing off the ground.
But it _can_ be done: the next time you're at a playground (take the g- kid, it'll give you an excuse), try starting up from a standstill without touching the ground. If you're careful about hitting the timing right, you'll be up to a "push off the ground" amplitude after about ten cycles or so (the 'or so' is because it's been years since I've done the experiment). If you pay attention, you'll note that on the very first cycle you will have added more energy to the system than was already there, for an increase of over 100%. If the system were a pure parametric oscillator, then that "over 100%" energy increase would happen _each cycle_, meaning that you'd be looping the loop in just a few cycles.
I think that what's happening is that when you grab the chains and lean back, the system tries to preserve angular momentum. In order to do that when you're leaning backwards, your center of mass has to move forwards. Ditto and opposite when you lean forwards at the front of the cycle.
If you wanted to swing in a pure parametric oscillator mode, then you would stand on the seat and you would do squats, twice each cycle. I'm sure it can be done if someone gives you a push to get started -- but I'm not sure at all how well it would work.
Reply to
Tim
t). =A0She will need to be pushed for the next 2-3 years as we all were. = =A0Somewhere around the age of 5-6 we learned to push ourselves.
) is a function of pivot length. =A0E.g. if it takes 2 seconds to swing fro= m front-top to back and to front-top again, it will take two seconds no mat= ter how high we swing or how gently (low) we swing. ROPE LENGTH DICTATES 2 = SECONDS AND THAT WILL NEVER CHANGE. Strictly a function of pivot length. = =A0
thest back point) which effectively shortens the pivot. =A0This shortening = (at the expense of expended energy) in turn shortens the period, which mean= s we go faster between peak-to-peak. =A0On the front peak we release the ro= pe, again returning us to full pivot length and and that extra energy goes = into raising our peaks higher and higher (all within the same exact two sec= onds).
? =A0Is it the shorter period of the rope-pullback which translates into th= e higher peak upon let-go? =A0What am I missing? =A0Been thinking about thi= s for the last few days.
A quick google search:
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Try looking up bicycle motion. Karl
Reply to
kfvorwerk
will need to be pushed for the next 2-3 years as we all were. Somewhere around the age of 5-6 we learned to push ourselves.
function of pivot length. E.g. if it takes 2 seconds to swing from front-top to back and to front-top again, it will take two seconds no matter how high we swing or how gently (low) we swing. ROPE LENGTH DICTATES 2 SECONDS AND THAT WILL NEVER CHANGE. Strictly a function of pivot length.
back point) which effectively shortens the pivot. This shortening (at the expense of expended energy) in turn shortens the period, which means we go faster between peak-to-peak. On the front peak we release the rope, again returning us to full pivot length and and that extra energy goes into raising our peaks higher and higher (all within the same exact two seconds).
the shorter period of the rope-pullback which translates into the higher peak upon let-go? What am I missing? Been thinking about this for the last few days.
OK, now that you figured that out. You can go WAY higher if you stand and pump your legs. Enough to slack the rope. Put the swing next to the lake and you can go way out if you release right. Or hit the ground if you release too early or late. What's the physics of the correct release time? Wish somebody had explained this to me before i broke my arm.
karl
Karl
Reply to
Karl Townsend
bad one. It's the classic one-second pendulum used in longcase clocks and is only about a yard long.
That gives you a 12 foot rope. Better?
11.5 feet... at the most. Probably more like 11.8 feet.
be of any significance.
them forward at the apex of the swing. That's the energy input used to maintain the motion.
You guys have it all wrong. It's Dynamic Will Power coupled to Synchronized Brownian Motion what done it.
-- Life is an escalator: You can move forward or backward; you can not remain still. -- Patricia Russell-McCloud
Reply to
Larry Jaques
Would you have listened?
jk
Reply to
jk
bad one. It's the classic one-second pendulum used in longcase clocks and is only about a yard long.
That gives you a 12 foot rope. Better?
11.5 feet... at the most. Probably more like 11.8 feet.
be of any significance.
them forward at the apex of the swing. That's the energy input used to maintain the motion.
Aah the synchronised Brownian motion! I like it!
Reply to
Dennis

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