Spring Pendulum & Torque

Are these drawn upside down?

It is inverted spring pendulum. thanks.

In that case it's simple (maybe)!
At the end of it's travel the mass is stationary (just before it reverses direction), therefore treat the spring as a rigid lever the mass and the angle relative to the pivot should give the torque, which must be transmitted to the joint else the spring wouldn't have anything to react against to provide the restoring force.
You could always look upon it as a conventional rigid pendulum on a pivot, with a spring either side of the mass providing the restoring force when the mass is deflected.

Isn't the drawing wrong? Surely the base of the spring is rigidly mounted upright in the platform (or whatever it is). I'm assuming the spring is what's bending back and forth and making the mass oscillate, or is there something driving the bottom of the spring? or is this just another bonkers theoretical Q which on close inspection makes no sense at all?
Del
_{(look on it as a flow of water)}

" Isn't the drawing wrong? " - looks it to me. As far as I can tell, the result in either case will be the mass moving anti-clockwise around the pivot point until it either meets an obstruction or behaves as a normal pendulum (the one with the spring would also bob up and down a bit).

Is there something I'm missing?

non-slip condition.

thanks!

As a walking model it is more complex and I have to think it over since the forces are following the applied torque. Give me some time for a better analysis not only a superficial one.

The spring must be relatively stiff else it wouldn't support mass b.
Therefore treat it a rigid, in reality it would just reduce the start up torque to get b moving, but once b was moving it would overshoot when the torque was removed.
When in doubt build it and see, you must have an old biro lying about!
Remember the theory is just a tool to explain the reality. It's reality that actually matters.
Del

Only the first of many opinions you are sure to get and most sure to be better than mine;

It seems that if the spring is heavy enough to allow the hinged point to travel then yes there must be torque...

If it is not heavy enough, it would still transmit what ever energy the spring is capable of delivering as torque but at lower values than A or B with a heavy spring???

That 'hinge' is driving me mad.
a) If it's a real pivot the thing just flops over and the spring is irrelevant.
b) If it's a rigid connection, why is it shown as a pivot/hinge?
c) If it's driven by a torque on a shaft to which the pivot is mounted, it would be nice if that was made obvious.
Whatever the answer, I still think the spring is pretty irrelevant, after all the rigid pendulum is just the case of a very stiff spring, so we end up arguing about degree, at the other extreme case the spring is sooo floppy it doesn't support the darned thing.
Oh my furry brain hurts... can I be excused?
Del

Can I draw a torque in free body diagram or not ? is there any problem ?

What do you think about this ?

If you can write F x L, then you can describe the torque. My thinking isn't perfectly clear on this yet but for now I'm thinking that the presence of the spring makes L variable. In which case L takes on a form like L(t) = L₀*(1+ k*sin(ω*t)). Where ω is the period of the spring's oscillation and L₀ is the unstretched/uncompressed length of the pendulum. As the spring pendulum begins to move, inertia of the point mass causes the spring to stretch, so you'll have one period for the pendulum and another for the spring. It might be somewhat like the double pendulum, if you've studies that.

I should have said ω is the frequency, not the period.

It's still early in the morning for me. Need more coffee.

If your spring is compliant only in axial direction then it is possible to consider that the mass is "guided" on the rod and the spring makes only the axial connection between the mass and the rod. For your simulation you consider the mass "sliding" on the rod depending on the forces between mass and spring and the rod turning with respect to the bottom point. It MUST be said that there is no torque transmitted by a rotating bearing (at least in theory) only radial forces. If you want to go into details then the only torque possible to transmit in the way as you made the sketch is the friction in the bearing.

If you consider the inverted pendulum then the basic plate, in its movement, will only transmit a force, this force will act upon the mass in 2 ways : 1st as a force in the movement direction and 2nd as a torque due to the lever arm. You should move the torque arrow to the mass where a torque will be present but not put it at the bearing where there is none.

If it is not clear enough I shall make you a sketch, let me know.

Thanks,

[1] spring is compliant only in axial direction.

[2] Torque is a control input.

Do you mean that I can't give a control input to the system (B) ?

(in case of (B), "move the torque arrow to the mass" isn't possible ?)

* Is there any way to give the control torque to the system (B) ?

Would you please tell me more about that (with sketch if possible) ?

The figure was modified.

Non-slip condition.

If I assumed that the spring is compliant only in axial direction, Can I give a control torque to the system ?

(as above mentioned by someone, very stiff sping makes the system rigid pendulum...)

Still confused..

The introduction only of a torque will be an error since the model, to be valid, has to "copy" reality" even if in a simplified form.

I think that you should consider a force due to the "push" of the leg behind and the effect of the displacement of the COG due to the pendular movement of the leg passing from behing to the front for the next step.

It will be NOT a rigid pendulum since the spring will change the length (and the inertia in rotation) according to the axial component of resulting effects force+torque.

At the first contact of the foot with the ground at step start the weight will compress the joints which will have an influence on dynamics.

It is on my opinion a VERY complex "system" and you should also consider the muscles as compliant force generators (actuators).

I noticed that you are working in posturology and I would suggest you look at possible models developed by other labs and if you can refine them.

The model you think about seems too simplified.

With respect to the forces mentioned above, if you have a platform able to quantify the vertical and horizontal forces at the contact between foot and ground and corelate them to the step film you get the information you need for your model. If you do not have yet such a platform I can give you some hints how to build one efficient an simple at same time (a couple of years ago I had to do it for a lab). In fact you need 2 (or even better 3) to obtain the forces for the whole step.

Let me know if you need more support.

Nick

please check this link : http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2918273/

It's a simple walking model.

I am wondering why rigid pendulum is ok, but spring pendulum is not ok with torque..

please do not think about walking model. Just see the figure that I attached in forum.

so, you mean that, although i assumed that the spring is compliant only in axial direction, I can not give a torque to the system ??

I had a look at the paper, it is a very good one. I want to read it carefully but from the start I think that your model is not as adapted as their. I noticed that a "point contact" is not representative since the contact force between foot and ground changes versus position and time. this is the reason they used "rolling" contacts.

I let you know my comments to morrow.

thanks ! I will wait till then.

figure link : http://blog.naver.com/nkaier/50096159402

Your equation is correct. You can consider 2 situations: transient & steady state.

In the 1st 2 forces will appear : one radial Fr= R*(dθ/dt)^2 and a tangential Ft= M*R*(d/dt(dθ/dt)). In fact with a torque you never come to a steady state situation since in a steady state the only force is radial and gives no torque.

In the second case only the Fr will appear. Fx and Fy are the projections of the above mentioned vectors which depend on the angle. There is an effect you neglect: the weight which is always vertical and is generating either a positive or a negative torque depending on the position and makes the system unstable if only a leg is considered.

Now the introduction of a torque means that you introduce energy in the "system" and I have the feeling which I still have to make sure that the model is based only on a balance between potential and kinetic energy.

I have to read sevral other papaers before I can give you a solid answer.