Puzzle
Puzzle
(OP)
Defyiing law of angular momentum?
Here is a fun puzzle for you purists.
How is it that a kid sitting on a swing starting at rest can, without any help, always cause the swing to oscillate?
Please explain.
Here is a fun puzzle for you purists.
How is it that a kid sitting on a swing starting at rest can, without any help, always cause the swing to oscillate?
Please explain.





RE: Puzzle
B.E.
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Regards,
SNORGY.
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B.E.
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Regards,
SNORGY.
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Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
If you think there are no torques then try swinging in a swing that hasn't been firmly staked to the ground.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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Be careful.
The moment of inertia changes. Abiding by the conservation of the moment of momentum.
Fe
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In the absence of a rocket, and starting from rest, where does the linear horizontal force come from?
If there is no angular momentum to start with changing the moment of inertia will make no difference.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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Fe
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Where does the linear momentum come from?
Once the swing is started there is no problem magnifying it, but the mechanism for starting the swing is not that obvious. I'm not saying that it's impossible, obviously it isn't, but so far no-one has given an adequate explanation of how to start a swing from complete rest with no direct contact with the ground.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
1. Assign a coord sys on center of mass CM, somewhere near the kid's belly button. +X = forward, +Y = aligned with gravity field.
2. Kid leans back and kicks legs forward. CM is now (??) slightly rearward of chains by virtue of pivot point where kid is holding chains. This is change in momentum dM/dt = accel & applied at CM. Causing a -RZ rotation moment. Resolve forces in X,Y, kid moves slightly forward.
3. Repeat opposite in rearward direction -X.
4. Repeat in something near natural oscillation frequency of swing/pendulum to amplify motion.
Can it be done by kid sitting still? I don't see how. I would think replacing kid with a static mass would prove this. Kid would always be making small balance corrections and thus impart momentum change. However, a static weight would also start to oscillate due to (?) force imparted by action of Earth's rotation. Check out "Foucault Pendulum" for this effect, I think it may apply to the swing with static mass.
TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering
www.bluetechnik.com
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d(p)/dt=F, where p=m*v {ie. adding a force changes momentum}
Fe
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- Steve
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so, could a shuttle propel itself through space by this method?
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In a vacuum chamber, so gusts of wind don't acidentally get them started and convection currents set up by their body heat don't accidentally trigger micro-oscillations? Suppose that might also help with the "sitting perfectly still long enough" problem - even if it does put a bit of a mocker on the subsequent "get it going" attempts.
* Assuming an exponential decay, I guess this would take for ever - it would certainly feel that way to a hyperactive toddler.
A.
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Lets put a toddler in space and let him/her kick like crazy and see if he/she moves at all
Fe
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sit down move body back.
You have just created a (time separated) torque, equal to the difference in CG height times the acceleration times the mass/2.
My rocket example was solely to demonstrate that an indisputably linear force can create a torque.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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not at all.. you need a gravitational pull to generate it.. in space there isnt gravity
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thanks, I needed a good laugh. what do you suppose keeps the earth moving around the sun, anyway? rapid thrusts of its legs?
RE: Puzzle
The gravitational acceleration at the orbital height of a space shuttle is not that much less than on the surface of the Earth. That's what keeps it in orbit, rather than flying off into space.
The difference is that on a swing you can get the rope inclined to vertical, by standing and leaning forward for instance, but your centre of gravity is still directly vertical under the pivot. If you now change the tension in the rope, by squatting down for instance, this will generate a torque which will set up a small swing which can be amplified.
You can't do this in a shuttle because the line of action of the gravitational force is always exactly through the centre of gravity of the craft. The same applies to a swing with a vertical rope. You need to get the rope off vertical before changes in the elevation of your centre of gravity will have any effect.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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I think the reason is that the swing is constructed of a rope (or chain) that is of finite thickness. By leaning forwards and backwards, the kid is actually creating higher tension in the fore, followed by the aft fibers of the member, and in so doing, he (or she) is able to get tiny oscillations started in the system, at which point it's off we go. In a perfect system where the attachment to the rigid structure is frictionless and meets at an infinitely small point, the kid goes nowhere.
It's the only thing that I can think of.
Engineering is not the science behind building. It is the science behind not building.
RE: Puzzle
may raise more questions than it answers
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
I may be some time :)
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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Accordingly, no one is incorrect...
Fe
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- Steve
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Then it has a little rest. The swing keeps swinging.
Then it uses a control algorithm - legs forward when moving forward.
There's no friction in the model. The chain on the swing is a rigid link. The seat/body is fastened to the 'chain' via a rotational spring, this doesn't seem to make much difference but stops some mathematical problems.
Why, since you ask, yes I did have the day off.
I wonder if replacing the rigid chain by a series of links makes a difference?
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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I recall force being applied where my hands held the chains. And motion too, therefore energy being shifted. Only on the forward part of the swing though. Tucking the knees under and leaning forward on the back swing I can't yet explain in simple terms.
- Steve
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Is that software ADAMS?
Fe
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That's why most kids get a push, or push off with their feet on the ground to get it moving first.
IMO, all that's really going on is the kid grabs the chain or rope a couple feet above the seat. Holds that portion and what's above it in place, then kicks his feet out until his body is parallel to the ground or nearly upside down, this makes the portion of the chain below his hands start swinging a little. When you relax that grip, the whole chain/rope swings back rather than just the portion below your hands, then repeat that step and the swinging momentum gets larger and larger with each leg kick.
Cool question thuogh.
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- Steve
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I hear many good things about it. Think I will try to get my hands on a trial version.
I know that could be done in say Matlab by solving the DE and linking it with a real-time plot, but I suspect ADAMS makes this less time consuming. Not to mention its probable 3D multi-dynamics capabilities.
Fe
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Does it do impact dynamics as well? I was imagining an asteroid a little too close to Mars
Fe
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Connecting ODEs and DAEs in a causal fashion is difficult and prone to errors. Softwares like ADAMS are acausal, so the response you get isn't always what you expect. It's normally much more interesting than what you'd expected. If your firm has a real need for 3D multi-body dynamics, get in there.
- Steve
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A hard problem for all dynamic solvers. I used a function that I coined "atanlash" for my ADAMS work. Arctangents are like step functions, only they are continuous in all differentials - something that ODE solvers like.
- Steve
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Fe
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Where is the proof that supports that "solution".
?
And if one exists, then what respectable journal has published it
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Does greg's model prove, or dis-prove what I said in my reply?
B.E.
RE: Puzzle
energy (= work) is added to the system (boy on a swing) by the boy moving his legs (changing the inertia of the system and adding torque to the system). in-phase additions augment.
personally, the kids got bored with swings when they found they had to do some work, much preferred the thumb-based "fun" ... it's a different world.
RE: Puzzle
Zekeman -there's no proof, that is just a full physics model of a possible solution. That is, if you built what I modelled in the real world I would expect it to behave in a roughly similar fashion. And if it didn't then I would get very interested. It may not be a good model of how a real person swings, it is a model of a mechanism that exhibits swing-like behaviour.
Notice that it takes a long time to build up a respectable amplitude - if after a minute of swinging the swing was only moving by a couple few degrees the average child would give up. Therefore the REAL method you use to start swinging is different to this, but it does confirm the point that driving a resonant system at its resonant frequency, especially with the addition of even small non linearities (in this case geometry), allows energy to be put into the system.
fex32 Yes, collisions and contact are modelled in both ADAMS and WM3D. Analytically it is a pretty horrible problem, to get realistic behaviour it is often necessary to fine tune some pretty arbitrary numbers. Here's a contact model, of one of those big container gantries smacking into an obstruction on one rail. There's some pretty neat stuff going on inside that model, I think it is one of the best 'simple' models I've done.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
Yes, it would make a big difference! With a rigid rod we can apply a torque to the rod, which results in a horizontal force at the pivot, which is transmitted to the ground, and that gives us the required horizontal reaction to get a horizontal accelleration at the swing.
Now consider a weightless rope, with zero flexural stiffness, hanging exactly vertical in a vacuum.
Look at the free body diagram of the would be swinger and the swing from the level of the position of the hands on the rope downwards. Because no moment can be transmitted up the rope, the centre of gravity of the swinger + swing (+ breathing apparatus) will remain exactly below the position of the hands on the rope, whatever the swinger does. Similarly this point will remain exactly below the swing pivot point. The swinger can kick legs and lean back to his or her hearts content, there can be no accelleration of the centre of gravity, because there is zero horizontal reaction force.
Of course in reality the rope is not weightless or zero stiffness, and (probably most importantly) there is air resistance to provide some reaction, so a swing can get started, but it is much more difficult than when you start with a small initial movement.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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Fe
RE: Puzzle
Greg,
This is NOT physics.
It' star wars.
Without proof this is pure conjecture.
I wish somebody could show me how you could get the first increment of motion from rest. Energy alone won't do it.
Prove to me that throwing out your legs or moving your body will result in sustained motion.
If, at the start, you conserve linear horizontal momentum, there is zero motion of the CM.
Yes, or No?
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Fe
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Fe
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Fe
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Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
Didn't some old Greek bloke say something about being able to move the Earth, given a fulcrum and a lever?
In answer to your question:
If you have a rigid rod: no*
If you have a flexible rope or chain: yes
*or to be more precise, yes but the Earth is included in your system, so you can increase the angular momentum of the swing by reducing the angular momentum of the Earth.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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IDS, I'm not sure I can get on board with the idea that there's a difference between a flexible member vs. a rigid member. I read your post above, but I think that flexible vs. rigid makes no difference. Here's my thinking: any moment applied to the rigid member is only internalized inside the member. The moment is high at the bottom of the member, but zero at the top joint. Assuming there is a frictionless joint at the top, the net reaction is still only downwards -- i.e., no horizontal motion applied to the joint.
Engineering is not the science behind building. It is the science behind not building.
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- Steve
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Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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WM2D was always a better program, and it still lives
http://www.design-simulation.com/wm2d/demo.php
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
http://www.justice.gov/atr/cases/indx346.htm
- Steve
RE: Puzzle
If I replace the torsional spring between the seat and the rigid chain, what difference will it make?
The system is now
earth
revolute joint
rigid masless link to replace chain
rotational bush
high inertia seat
motorised revolute joint
leg pendulum
with gravity as the other constraint.
Also, try and predict what will happen if i replace the rigid chain by 5 massless links.
One thing you might like to think about is a 2dof system that consists of earth/ spring/ mass /linear motor /small mass, and what that system does as a function of the linear motor's activities, both in theory and then in practice.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
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- Steve
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Perhaps that is a better way to phrase it
Given a simple 2dof system, why would you expect to not be able to excite a resonant response, given control over one dof?
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
How can you accellerate the centre of mass of the system (everything below the pivot) if there is nothing to provide a horizontal reaction?
Can they?
Where does the horizontal force come from?
There's no problem setting up a resonant response in the vertical direction, but in the absence of any horizontal forces, the movement will be on a vertical line through the centre of mass.
Sorry to be so repetitive, but as far as I can see no-one has addressed this point.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
- Steve
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newton and several posts here, has already covered your question. a body at rest remains at rest untill acted upon by an external force. in the case under question, the kid on the swing moves his legs, and energy (work) input into the system and the rest is Newtonian motion.
greg,
your model starts very slowly, and you suppose kids would get bored, and so the real world is different. last time i was on a swing, i backed up so i wasn't under the suspension point of the swing, (still at rest zeke, supporting my weight on my legs) and let go ... and went from a position of rest to one of motion ...
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Cheers HM
No more things should be presumed to exist than are absolutely necessary - William of Occam
RE: Puzzle
Let's assume that you model the swing as a pendulum with a large bob, enough to enclose the kid. Now tell me how it is possible to move the pendulum from its vertical rest position.
My answer is that , in the absence of friction, you can't, since you are starting out with zero angular momentum with respect to the pivot and there are no external forces that can produce a torque .
So preserving angular momentum and it starts at zero it remains at zero and the CM won't move.
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- Steve
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All the kid has to do is start moving his/her body, and the swing will start moving.
So, the problem is that the model doesn't accurately reflect reality.
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let me say this simlpy, the kid is adding work to the system. if the kid was a quadruplegic, he'd have a very hard time of it (not only starting a swing but most other things as well). if the kid (with legs) sat there and did nothing; well, nothing would happen. by swinging his legs and pulling back on the swing ropes, he gets the swing going.
btw, is this a wind up ? (no pun intended)
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The kid, by moving his legs with respect to the rest of his body, changes the position of his center of mass in space. The swing moves, because the c.o.m. has shifted.
"Let's assume that you model the swing as a pendulum with a large bob, enough to enclose the kid. Now tell me how it is possible to move the pendulum from its vertical rest position."
If the kid can move within your plumb bob, then he can shift the position of the center of mass of the system, either by moving parts of his body (legs, arms), or by moving himself relative to the rest of the swing/bob.
If those concepts give you trouble, then tell me how a rider on a bicycle can generate forward motion...after all, angular momentum must be conserved, and all he's doing is rotating a crank.
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So preserving angular momentum and it starts at zero it remains at zero and the CM won't move. "
and another post stating "there is no horizontal force that can act" on the pendulum.
Again, as previously stated, angular momentum of a system is conserved IF no external forces (torques) act on the system.
Yes, there are horizontal components of force that can, and do act, on a pendulum. The external force acting is that of gravity, and the horizontal reaction forces transmitted through the pivot point of the pendulum (the reaction force is equal and opposite to the force that the kid exerts to move his legs to-and-fro). The forces are developed as soon as the c.o.m. is shifted from the nominal vertically-aligned or zero position. Once some angle theta exists between local gravity vector and the line between hinge and center of mass of the pendulum, the horizontal restoring force acting on the "bob" of the pendulum is exactly
mg*sin(theta).
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Now you are thinking my way,
""The forces are developed as soon as the c.o.m. is shifted from the nominal vertically-aligned or zero position. Once some angle theta exists between local gravity vector and the line between hinge and center of mass of the pendulum, the horizontal restoring force acting on the "bob" of the pendulum is exactly mg*sin(theta). ""
B.E.
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"So preserving angular momentum and it starts at zero it remains at zero and the CM won't move. "
...E pur si muove!
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If a rigid mass is used for the ballast then when fired, as well as the vertical load taken by the main structure, there is also a horizontal load due to the CoG of the mass moving horizontally as it inscribes an arc.
It is for this reason that typically they either have wheels to allow some horizontal motion, or (and probably better for accuracy) they have the ballast hanging from a pivot.
In the case where the ballast is rigid, and the trebuchet is on wheels, the fundamental system is not that different from a kid on a swing lifting and lowering their legs. The CoG/CoM of the kids legs inscribe an arc a bit like the mass on the arm of the trebuchet. This gives you a horizontal force to get you swinging.
The big difference is in then repeating this swinging motion at somewhere around the natural frequency of the swing with the impulse in the right direction at the right time.
What is Engineering anyway: FAQ1088-1484: In layman terms, what is "engineering"?
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Trueblood,
So you are saying that a rope attached to the bob can sustain a horizontal component of force at the pivot point.
Think again. The only forces on the rope are tensile.
The so-called reaction force you mention is internal and moves the legs and remainder of the body in opposite directions such that the CM must remain at the rest position.
RE: Puzzle
Let's summarise what's agreed:
Everyone agrees that with a rigid rod it's possible to generate horizontal movement. A bicycle operates in a similar way in principle.
Everyone agrees that a real child on a real swing can set it in motion from rest, although it's considerably more difficult than starting with a small amount of movement.
The question where people still differ is what happens on an idealised swing in a vacuum with a frictionless rope of zero flexural rigidity.
For those people who say that the reaction force comes from the momentum of the kids moving their centre of mass, how do the kids move their centre of mass with no reaction force to start with?
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
The chain now has an angle, and the cg has raised.
Now exploit that geometrical non linearity somehow.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
yes they did, pushing against friction of the pivot and possible air resistance
"Everyone agrees that with a rigid rod it's possible to generate horizontal movement. A bicycle operates in a similar way in principle."
a bicycle would be like pushing off the ground on a swing, you are fixed to the pivot in all axis and pushing the pedal/swing. Not touching ground on a swing you are only fixed in the vertical axis.
""The question where people still differ is what happens on an idealized swing in a vacuum with a frictionless rope of zero flexural rigidity.""
My instinct says you can't start the pendulum.
"" how do the kids move their centre of mass with no reaction force to start with?""
they don't
RE: Puzzle
You're sitting in the middle of a frozen lake. Now this is a special frozen lake. It is frictionless.( not a new problem) and you have to get to the shore.
Tell me that you can move by extending your legs.
Seems to be very much the same problem of the swing starting from rest.
(I know the other ways but they are not allowed for this exercise)
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Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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Does he now get to the shore. I don't think so.
He pushes (impulse Fdt) the ball, m2, achieving velocity v2 and the guy , m1,goes opposite with velocity v1.System Momentum=0 before push. System momentum just after
initial push =0. After the chain becomes taut both masses are stopped internal impulse same as before, final system momentum=0. Final position of the, CM of 2 mass system is at the starting point. Poor guy is stuck!
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IMO, the cg analogy pointed out makes perfect sense, from start to finish.
Fe
RE: Puzzle
No argument with that, but some are arguing (I now think correctly) that you don't need friction and/or air resistance.
OK, the bicycle analogy was a bit of a stretch, but the point I was making was that there is a fundamental difference between working on a rigid rod (to which you can apply a torque) and a chain or rope.
This is the question I wanted answered, and I think that Greg has now answered it.
With the no friction/ no air/ flexible chain and no initial movement, whatever you do the centre of mass of swing + rider will stay exactly where it started; under the top pivot, but the bottom end of the rope will not necessarily stay under the pivot, and as soon as you have some inclination in the rope you have a source of a horizontal reaction, and the ability to move your centre of mass.
I was arguing (to myself) that any offest between the bottom of the rope and the centre of mass will result in a rotation, and this will keep the bottom of the rope below the top, and above the centre of mass. What I missed was that will take time, so if some part of the swinger is moved sufficiently quickly, this will generate the required inclination of the rope, even though the centre of mass doesn't initially move.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
If you had a swing with a frictionless pivot and rope, in a vacuum, supported on a perfectly frictionless surface, starting from perfect rest, I don't think you could get that swinging.
Could you?
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
Maybe with a little flatulence you could
Fe
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Did anyone actually try and read that paper I posted? I didn't.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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Fe
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""any offest between the bottom of the rope and the centre of mass will result in a rotation,...that will take time, so if some part of the swinger is moved sufficiently quickly, this will generate the required inclination of the rope, even though the centre of mass doesn't initially move.""
dang, what you just said made me flash on it. With a flexible rope you create a second pendulum by grabbing the rope above you, just like the model. Now you are anchored in all axis to a pendulum pivot that you can push off of. You raise your weight up one side of the arc, let gravity swing it back and you have your horizontal force.
good thread...lots of brainstorming
RE: Puzzle
draw a free body diagram of the pendulum, it's something you probably did in school. the pivot point reacts radial load (zero bending stiffness in the cable), weight acts down, the horizontal component of the cable load is balanced be inertia force (the body in motion) or by a restraining force (if the body is at rest away from the neutral position).
RE: Puzzle
In air, as others have said, pushing against air will yield an external force and thus be capable of moving the swing and, taking advantage of resonance, it could make the swing work as designed. It also could help that fellow on the frozen lake.
However,friction cannot help for a rope swing since you cannot transmit the friction torque through the rope. The rod swing may be another story but my inclination is that it can't help but can only slow things down.
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Fe
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consider you put a small rocket on the swing. thrust from the rocket is tangential about the pivot. the reaction is inertia, the motion is circular about the pivot. consider you pulse the rocket adding an increment of thrust each time it crosses under the pivot.
really, this is a wind up (no pun intended), isn't it ?
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Interesting.
Fe
RE: Puzzle
Think again. The only forces on the rope are tensile."
Right. But if the rope is inclined at some angle to vertical, what are the reactions relative to horizontal? If no horizontal forces can be generated at the pivot point, then the swing will never swing, and any push given the rider will just send him sailing away over the horizon (this un-restrained swing is the equivalent of a guy on a frictionless surface). There are no forces reacted by the rope at the bottom of the arc of the swing - in this you are correct. But it is exactly that imbalance of forces that causes the center of mass to shift out-of-perpendicular, and thus start the oscillation of the swing.
"My take is that in a vacuum with and without friction there is no net reaction force and no starting the swing."
...
IF the kid can change the location of his c.o.m., then the swing will move, right?
When the kid's muscles contract, he accelerates the c.o.m. of his legs, and the muscles/bones react that (acceleration x mass = force) against what - the c.o.m. of the rest of his body. That force acts for a very short period relative to the natural frequency of the pendulum, but nevertheless, the integral of force over time generates a small displacement of the center of mass of the system. Granted, the displacement generated is small, and he has to repeatedly pump those legs at both the proper frequency and the proper phase in order to increase the amplitude of the swing over time, but friction need not be invoked to generate the forces and displacements.
You can talk about his arms bending the rope, and that technique works too for the same reasons as just moving his legs (he is shifting his weight about the pivot created by his hands holding the rope, thus shifting his c.o.m.).
There is a type of swing used by circus performers that has a long platform in place of the swing seat, and rigid links to the pivot points. You can watch a performer start from the swing perfectly at rest, and he then begins to shift his body (running, leaping, and stopping, then reversing direction) from one end of the platform to the next, eventually pumping the swing enough to go over-the-top (loop the loop).
But I'm coming to a point where I agree with rb1957.
RE: Puzzle
http://www.jstor.org/pss/2687680
http://www
RE: Puzzle
Call the top pivot H1, and the point where the arms hold onto the chain H2.
create a swing seat hanging below H2, CG obviously directly below h1 and H2. This has a high moment of inertia.
Now add a mass on a horizontal slider initially under H2.
Slowly move the mass away, the seat tips up, and the CG of the seat+mass, H2 and H1 remain colinear. No oscillation.
OK, that's the statics.
Now reset the mechanism, and this time move the mass very quickly.
The seat moves in the opposite direction, horizontally, as before, but due to its large MoI does not rotate, so H2 moves in the same direction. The chain, h1 to h2 is now off the vertical and the tension in it applies a torque about the CG of the seat mass system, which slowly rotates to bring H1 and H2 and the CG colinear, and then overshoots, and just sits there jiggling around the static location. Note that work was done on the first motion of the mass, this is what supplies the initial rotational kinetic energy of the seat/mass system. Once there is an oscillation then the mass can be moved in such a way as to feed more energy into the system.
So, that works. Is it how a real child gets a swing moving? No probably not. Swinging your legs is a far more obvious way of setting up a rotational motion. However it is robust (it needed no fine tuning to get it to work), obeys the laws of physics, doesn't rely on friction, and works equally well with chains or rigid links.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: Puzzle
There is no point in looking at how kids actually go about swinging because no-one disputes that a swing motion can be amplified by moving the centre of mass relative to the pivot point, once there is some movement, or the centre of mass has some accelleration. The question is, where does the initial accelleration come from in a system starting from rest with the centre of rest exactly under the pivot?
The answer is not is not that a rapid movement of the legs will cause the centre of mass of the system to move, even momentarily. It will not. If there is no external horizontal force the centre of mass stays precisely where it is horizontally. Certainly you can move the centre of mass vertically along the line of the rope, but that won't generate any horizontal force.
The answer is that although you can't move the centre of mass horizontally starting from rest, you can move the position of a point on the rope relative to the pivot (and the centre of mass). This position is unstable, and rotation of the body will tend to bring the two points back into line, but it does provide a temporary horizontal force (the horizontal component of the tension in the rope), which will accellerate the centre of mass, and this accelleration can then be amplified as discussd at length earlier in the thread.
Maybe you think this is a pedantic point, but I think it is important to recognise that if you have a system with no external force in a given direction nothing you do will accellerate the centre of mass in that direction, even momentarily, unless and until some external force is applied.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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I did not finish my problem I started earlier. Although what Greg described is almost exactly what I did, except he finished the thought, that I did not.
Fe
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I absolutely agree with you.
Which is why I'm gonna give Greg a star.
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Greg's link explained how swings work but did not address what we are discussing. What I learned from it though is that it is the shift in mass to and from the pivot point during the swing that pumps it. This works only when there is already a swinging motion.It cannot start a swing. The COM cannot be moved tangentially to the swing of the rope without an external force. The rider can only move his COM along the axis of the rope by swinging the legs or by standing and squatting on the seat. But this cannot start the swing.
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read the links in Bestwrench's post. The force is internal. It starts from rest when you move the small pendulum from the pivot where you grab the rope. You swing your legs back and forth and they gain angular momentum. Through timing and phase this is transferred to the larger pendulum and increased by either "parametric oscillation" (spinning skater centralizing mass) or "driven oscillation", moving back at the peak of back swing and forward at peak of forward swing, which increases the distance of oscillation.
I think the reason it's so hard to start the swing is that when you transfer the angular momentum from the short pendulum to the long pendulum in increases the moment of inertia by the multiple distance of long pendulum over short. Thus slowing the rotation rate by that multiple.
I'm just wondering what the kids are going to think when I tell them we're going to the park to play on the double pendulum mechanism and see if we can test the conservation of angular momentum through parametric oscillation.
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Those are great videos
None show any static start.
My take on the dynamics is seen from the first video where the key is the bob rising up the rope near the end of travel, thus reducing the length, r and since angular momentum is conserved (m*v*r= constant), then v increases as r decreases; the increase of v is due to the work done in climbing up the rope near the end of the travel, so we have an increase in energy, v and when the bob returns to the center position we have a higher velocity than it had previously; this process ,repeated, will cause sustained oscillation growing until perhaps friction or running out of height limits it.
The kid pumping his feet essentially mimics that motion since raising the feet causes the CM to rise
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The key difference is not the motion, it is the inclination of the rope or rod. If you have an inclined rope you do have something to push off (or in the case of a rope, pull on). When the swing is in motion then the rope is inclined (except momentarilly at the bottom of the swing). When it is not in motion then it is necessary to get some inclination in the rope, and that can be done by moving some part of the rope relative to the centre of mass.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
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There are two external forces. Gravity is one. The fixed pivot (the rope tied to the tree branch) is the second. This is not a free system, but a grounded system, with the grounding occurring at two seperated (but connected, via the swing ropes) points.
A star for tygerdawg, who made a good stab way up there ^ at a text description of the ideas. A star for Bestwrench, although you need a subscription to read either of his two references. Greg already has his star both for his model and for his paper reference.
Better are these, which describe the phenomena in (exhaustive!) detail, including stepwise derivation of the equations, and :
htt
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Please read the conclusion of the second paper at least, where it notes that the angular motion of the seated rider has an additive effect to the amplitude of swing motion, independent of the swing's amplitude (i.e. even if the initial angle of the swing is zero). This is shown in the plots in Greg's reference, but the explicit statement of the initial conditions is not made. They do also mention that their analysis assumes rigid links, not ropes. But I can show at least two ways where that wouldn't matter (one: the kid is a rigid link between the mass of his lardy butt and his legs). Both papers describe, and the first one graphs, how the driven angular oscillation motion of the suspended barbell (simulating the rider) can force oscillations of the entire swing, even from (please see the detailed description in the first paper) initial conditions with upper swing angle at zero, and first derivative of upper angle w.r.t. time of zero.
It is interesting that angular momentum is conserved, and in fact is essential (the transfer of that momentum from the lower oscillating system to the upper or global system is the mechanism at work), but - energy is clearly not conserved, as the energy in the system (provided by the suspended forcing function, i.e. rider) grows over time.
The best example of conservation of angular momentum does not mean that rotations of a multiple-jointed body cannot occur:
http://www.ocf.berkeley.edu/~barneye/kitty.html
(ok, that was just for fun :)
Whoops - nope, on third reading, Greg's paper does state (p. 465, bottom right) "When started from rest, the swing's amplitude will grow linearly...". Presumably, "at rest" means no initial motion, which also implies an initial angle of zero.
The stake has been well and fully driven into this one, the horse is long dead.
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Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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- Steve
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Bestwrench(BR) model videos.
The guy stands up at the peak of the swing in BR model and sits at the o position which is opposite to TB's guy.
It is obvious to me that it does not matter where in the cycle you stand up, and sit down you will get amplification.
Standing up increases the the total energy of the system,since work is done against gravity, while sitting down will not remove all of that energy.
Just think during sitting down at the end of the stroke you suddenly drop your body most of the energy increment is preserved; although, dropping the body at the 0 angle position may lose some energy through tension in the rope.
So, both models make the clear case that by standing you increase the total energy and dropping to the sitting position hardly changes the total energy, you have increasing velocity at the 0 position for each cycle with increasing amplitude.
I'm totally puzzled by the conclusion of TB's authors who limited the analysis to a rod rather than an ideal rope.
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JMW
www.ViscoAnalyser.com
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"I'm totally puzzled by the conclusion of TB's authors who limited the analysis to a rod rather than an ideal rope."
Their comment is regarding the idea that the perfect pin joint at the seat (junction of upper swing rope/link to the center of the dumbell) cannot support a torque, if the upper member is a rope (ropes don't support moments).
But, if a kid grabs the rope above the pivot, he creates a second rigid link between his arms and torso, which can create moments relative to other parts of his body. I.e. the system modelled in the references would add one more link and rigid member. This has been said by others, but it may be difficult to visualize.
Let's simplify it by re-sketching the first paper's "Figure 1", see the attached sketch Figure "A". Of course, I've now invoked a third link and third angle to the system...
After it's all said and done, think about the mathematical ideal of a massless rope with zero moment of inertia...ok, maybe it can't be started from rest...I can draw the free body diagram for such a case (figure B1 thru B3), and (I think) argue convincingly that the upper link cannot be displaced relative to the fixed pivot unless the rope has inertia...
So, ok, I will allow the hypotheses that a swing suspended from massless rope might not start from a rest position...please send me some of that rope so we can prove the theory.
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< why forget him. I was referring to him>
"I'm totally puzzled by the conclusion of TB's authors who limited the analysis to a rod rather than an ideal rope."
> my quote from that authors comment>
"Their comment is regarding the idea that the perfect pin joint at the seat (junction of upper swing rope/link to the center of the dumbell) cannot support a torque, if the upper member is a rope (ropes don't support moments).'
>A perfect pin has nothing to do with a rod that replaces an ideal rope>
"But, if a kid grabs the rope above the pivot, he creates a second rigid link between his arms and torso, which can create moments relative to other parts of his body. I.e. the system modelled in the references would add one more link and rigid member. This has been said by others, but it may be difficult to visualize."
> why add a another complication to an already complicated system. And, moreover, by adding another link you are still left with the rope above his hand where I suppose you get another ideal pivot, but what is the point>
Let's simplify it by re-sketching the first paper's "Figure 1", see the attached sketch Figure "A". Of course, I've now invoked a third link and third angle to the system...
"After it's all said and done, think about the mathematical ideal of a massless rope with zero moment of inertia...ok, maybe it can't be started from rest...I can draw the free body diagram for such a case (figure B1 thru B3), and (I think) argue convincingly that the upper link cannot be displaced relative to the fixed pivot unless the rope has inertia..."
> Those are the strangest free body diagrams I have ever seen.I don't have a clue.
Also you say an inertialess rope won't support the "model" or the physics; which is it?
"So, ok, I will allow the hypotheses that a swing suspended from massless rope might not start from a rest position...please send me some of that rope so we can prove the theory"
> Look in your neighbor's backyard.You don't really think its a rod or it has weight significant enough to change a well developed model. The models that do not allow for a flexible rope are somewhat.>
I do appreciate the math models and remember the solution is
often limited by the assumptions made.
I have looked at the standing- sitting method and am satisfied that the amplitude grows ( assuming a finite starting energy)since every time the guy stands he adds energy to the system, and there is no easy way to remove it (eg sitting down)
If you carry that out for the kid pumping, even without raising his CM, then there is a valid case that the energy of pumping (1/2Iw^2),will eventually manifest itself into increasing the energy of the system, and thus increase the amplitude.
So I believe that you don't need the math to understand this phenomenon.
I now firmly believe that any method of increasing the energy of the system,standing or thrashing will increase the velocity at the 0 position and thus increase the amplitude. Absent friction it will runaway.
Prove this wrong.
* http:
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You need to focus on change of momentum, not change of energy.
You can jump up and down all you like, if your centre of mass is exactly below the pivot point (which it will be if you start from rest) and the rope is vertical, you won't move anywhere. You will just convert a lot of chemical energy into potential energy, and then into kinetic energy, and then into heat.
Starting from rest you need to displace the rope horizontally at some point, and you can do that by pushing another part of the rope in the opposite direction. Once there is even a small deviation from the vertical you have a source of a horizontal reaction, and can start to move your centre of mass in the horizontal direction.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Puzzle
Didn't say forget him, I said forget the section that discusses standing/sitting motions, as they point out that this motion will not start a swing from the rest position.
>A perfect pin has nothing to do with a rod that replaces an ideal rope>
Taken out of context, the following statement clarifies.
> why add a another complication to an already complicated system. And, moreover, by adding another link you are still left with the rope above his hand where I suppose you get another ideal pivot, but what is the point>
To model the physical reality.
"> Those are the strangest free body diagrams I have ever seen."
Thanks. Yours, of course, are invisible to me.
"I don't have a clue."
Given. ;)
I'll admit the fbd's are less than complete, merely a method to try and describe the physics. Fig. 2b tries to describe the motion of the center of mass, ignoring any force input from the upper hinge (pure inertial reaction). The com must shift relative to the upper pivot for this case. Fig. 2c tries to describe the condition where the upper pivot must remain in-line with the center of mass as the rods change their angle relative to each other. The two cases are obviously different. I am trying to figure out why - but the realization that Fig. 2c only applies if the angle is changed very slowly, so that static equilibrium applies, and that fig 2b is the condition when the angle is changed very rapidly, makes me think twice. Like I said, I'd have to grind out the math to prove one way or the other.
> Look in your neighbor's backyard.You don't really think its a rod or it has weight significant enough to change a well developed model. The models that do not allow for a flexible rope are somewhat.>
My neighbor's swing has rods from two pivot points (glider swing). Are somewhat what? I dunno either.
"I now firmly believe that any method of increasing the energy of the system,standing or thrashing will increase the velocity at the 0 position and thus increase the amplitude. Absent friction it will runaway."
Ok, we disagree again I think - standing and sitting won't start a swing from the rest position, only the rocking motion will.
IDS: "Starting from rest you need to displace the rope horizontally at some point, and you can do that by pushing another part of the rope in the opposite direction. Once there is even a small deviation from the vertical you have a source of a horizontal reaction, and can start to move your centre of mass in the horizontal direction."
Yup. It's that first displacement that has me wondering again, but I'm satisfied with that statement if you are.
Good night.
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If you bothered to read my posts, I have repeatedly said that it is impossible to start a system at the rest position (assuming no air ) since you cannot generate an external horizontal force
My take on standing and thrashing assumes the motion has started.
So, since we are almost all in agreement about the starting problem,I guess we can put this to bed or have another marathon on ropes and models and math solutions.
But, I will now stand ( no pun intended)_on the standing/sitting and pumping means of making the amplitude grow until friction will limit it, with or without an ideal rope.
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Fe
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Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
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Fe
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Does anyone disagree that it is possible for a kid to do this on a swing that is hanging purely vertical?
Did someone say something about conservation of angular momentum?
-handleman, CSWP (The new, easy test)
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This thread makes my head hurt.
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Fe
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nah greg, simple harmonic ...
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Fe
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Fe
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Let's consider another example. The Rings in men's gymnastics is a swing and maybe easier for visualization and degrees of movement of the person. Now clearly there are lots of movements and rotations that will not produce produce swinging, as that is the point of the sport. However, the "swinger" (lots of google hits, btw) can perform a kip where, while hanging by his arms, he brings his legs over his head with is body folded at the hips. He then trusts his legs in a direction that is 45 degrees up from horizontal. In free-space this would of-course have no effect on his center of mass, but he is on a swing. The rings swing one way and the tension on the rope increases over the weight of the swinger. So during the leg acceleration there is a horizontal component of force on the center of mass (COM). When the kip is finished and the legs are decelerating, the rings will be on the other side of vertical from rope pivot point, along with the center of mass, which has been slightly displaced. During deceleration of the legs the rope tension is again higher than the swingers weight and there is a horizontal component of force applied to the COM which adds more energy to the swinger.
I vaguely remember doing this as kid. I guess that angular momentum is involved here but is not need for explanation (and would probably just confound understanding). It seems that the key is finding a way to create a small horizontal impulse to the swing that is not canceled by an equal and opposite reaction. The technique I described assures that the equal and opposite horizontal reactions naturally occur on opposite sides of vertical. So, it appears that it is possible to start a motionless swing.
Now I need to think if something similar is possible while seated.
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Here's an idea, based on the rocket... perhaps you can model this one, Greg. Picture a guy at the bottom of a perfect rope. Anytime he sticks his legs out, the rest of him goes the other way to keep the cg perfectly below the pivot. This guy is a clever fellow, though, and has someplace to be in the afternoon. He unravels his shirt into a long thread, ties it to his wallet, and throws the wallet as far as he can. This little push gets him off of dead center and starts him oscillating, at which point he can "pump" conventionally to build momentum. He has a really good arm, so he builds momentum While the wallet is still airborne. Eventually it comes to the end of the thread and stops, and he reels it back in before it ever hits the ground. He gets to keep a whole lot of momentum.
Maybe rigid bodies behave much differently than very flexible (or separable) ones?
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Or just a motor and a disc with a hole in it. Can it start oscillation of the rope.....
p.s. tune the frequency of the actuator to "her frequency of osculation".
Fe