studiot

So is the original arrangement being abandoned as unworkable? The point about a couple is that you can apply one to the rod to get the rotation started, but there is no guarantee it will continue for very long. As a matter of interest, if you have a large m1 and a small m2 you have a situation that occurs in nature. There are quite a few molecules that have a heavy atom at one end of the bond and a very light one at the other. They perform a very peculiar sort of tumbling rotation. I believe there are macro toys based on this as well. Final comment. When you have rotations about 3 axes (known as Euler axes) , two rotations are stable but the middle one is wildly unstable. This can be demonstrated by trying to spin a matchbox/book/brick about each of the three axes and observing the behaviour.

I have some heavy curtains, weighted down at the bottom. They run on a very low friction track using brass bogie rollers. Now if I grab the bottom an give an impulse all that happens is the curtains move bodily sideways. They do not attempt rotation, even though the force of drawing the curtains has substantial mement about the rail. If you do the same with your rod ie grab m1 and give it a push the whole rod will simply slide sideways. I am still waiting for you to tell me if you understand the difference between a couple and a moment (of a force) A couple has the same moment about every point in the plane, including its point of application. A moment of a force has a different moment about nearly every point in the plane and zero moment about its point of application.

Thank you. I will have to look some facts and figures up. Did you understand my comment about the strut? The femur is most likely to fail in buckling or twisting, rather than straight compression.

these equations are very concerning since they imply that space is neither isotropic nor homogeneous. The situation is even worse when we extend it to 3D since in your model when we are aligning the x and x' axes and considering V directed along these, we have y= ct z=ct

Thermal agitation of the particles. I say particles because in the case of the steel bar the particles are the atoms bonded to each other, forming the steel alloy structure. In the case of the Sun the particles are in the plasma state and in sigfnificant motion because of their temperature. The Bohr spectra come from promoting electrons within an atom between discrete energy levels, just as you say. The thermal spectra come from vibration of the bonded atoms about the bond length/angle without breaking the bond.

duplicate thread Moderator please merge. see

Please tell us what you actually want to achieve. Then proper advice can be given. There is a whole lot more to stress analysis than perhaps you realise. In particular the femur is technically a 'slender strut' that is also subject to significant torsion and bending. This can 'magnify' the effects of direct compression dramatically.

The continuous emission spectrum is due to the surface temperature of the Sun. This is obviously hotter than that of a steel bar heated to red or white heat, but the principle is the same. see stephans law https://en.wikipedia.org/wiki/Stefan–Boltzmann_law The black lines are absorbtion lines. That means certain wavelengths this thermal luminescence are removed. This happens as the light travels outwards from the Sun's surface through its atmosphere. And yes they are specific to the elements that are in the very outer layers of the Sun and in its atmosphere.

the motion of m2 does not affect the KE of m1 the motion of m1 is the motion of m1, whatever that is, and its KE is also given by my equation. Not also that since the rod is free to pivot about m2 m1 has motion in both the x and y directions each with its corresponding KE, which both contribute to the total KE of m1. You appear to be trying to count something twice because you are jumping between frames again.

I had hoped this meant you were beginning to listen and understand and could perhaps even answer my earlier hint question about the difference between a moment and a couple.

Why is that ? KE = 0.5 mv2 If m is zero then KE = 0, whatever v is.

Wouldn't it be nice if this principle could be used in a bar? After all it only takes a small amount of money to buy one drink. But if you used the same amount lots of times think how many drinks you could get.

My older brother is 76 and has two boats. I dread the times I have to go and help him with one of them.

I did not say how the mass m1 would move, just that whatever energy was input by the initial impulse would remain and that the other parts of the mechanism would have zero KE at all times and that there were no dissipative or potential forces acting to change the PE of m1. The question as to whether the parts of the mechanism would act in the way seanie describes has been challenged by all of us. So I was hoping to dispel or remove the side issue of energy conservation. By the way do you not find calling the ends of the rod m1 and m2 confusing? I certainly do.

Would this be ac or (I presume dc) ? What minumum are you seeking ? What load are you driving ? The simplest way might be a battery and a high value rheostat (in relation to the load resistance) connected potentiometer. That would give a sensibly constant cuurent up to 200 microamps. If you can live with a range of fixed currents then you could use fixed Hiz resistors and a switch instead of the pot.

Bouyancy is the result of the balance of foces on a body totally immersed in a fluid that is subject to a body force field such as gravity. You don't fall to the core because you are standing on the deck of a ship in your sou'wester

How and in which frame of reference?

Then the kinetic energy of the system is fluctuating. Where is the excess energy stored when the kinetic energy drops? It is worth reviewing the energy situation. 1) As with any isolated mechancial system the total mechanical energy is constant. 2) The total energy of the system comprises that input at first setting in motion (whatever that is) of the system. 3) Macswell suggests the KE is fluctuating, but is it? The end m2 which is alleged to execute SHM has zero mass so zero KE at all times. 4) There are no dissipative forces as the system is frictionless and no energy is lost. 5) But no energy is produced either so there is no 'overunity'. Does this sum it up?

Thank you for the subsequent links. I note the BBC had none of it in the 1 o'clock news nor under 'What else did he say' on their website. I also wonder (tongue in cheek since it would be marvellous if they finally succeeded) if this is the same team that has been promising commercial fusion since the late 1950s when they were on the brink of......... with ZETA.

Note that it is the time rate of change of these fields that is important. It is this time rate that links the two of Maxwells equations that lead to the wave equation when combined. [math]\frac{{\partial {E_x}}}{{\partial z}} = - \frac{{\partial {B_y}}}{{\partial t}}[/math] [math] - \frac{{\partial {H_y}}}{{\partial z}} = \frac{{\partial {D_x}}}{{\partial t}}[/math] Where E, D are the electric vectors B and H are the magnetic vectors with subscripts denoting the direction relative to x, y, z axes and travel of the EM wave in the z direction.

Thank you for a response deserving serious consideration. It will certainly receive that. Meanwhile a couple of quick first impressions. This implies a discontinuity at x' = 0. unless we add these equations (equivalent to taking the average) 2x' = 2ct x' = ct This is, of course identical to the equation for observer1 viz x = ct You seem to have finally correctly identified the necessary second postulate of SRT. If this is your replacement postulate it doe not provide the means to numerically evaluate the apparent change of position. Where do your formulae come from?

This is in no way consistent either with the very clear and simple scenario I outlined, Nor is is a possible equation unless the light source was modulated by some sinusoidal signal, something I most definitely did not say. Furthermore I definitely specified the position of the second observer. I did not say that there was any separation between the observers. and everything that follows is completely and utter rubbish and thoroughly discouraging to any sensible discussion. Good night. Are you just being plain akward or what?

And yet when I ask for details I get answer like You can have no idea why I ask these questions, most have led to an improvement of your system description. I have asked several times how you set the system going in the desired motion and other responders have queried whether it will ever reach that motion yet you seem to think there is nothing in the concerns of all these others. Think about this very carefully. If you simply apply a single temporary force to m1, you have specified zero friction on the x axis guide rail. So why will the x component of that force not simply move the rod bodily along the guide rail when there is no resistance? Real world mechanisms all have friction. Applying a force with zero x component to the chosen starting configuration produces zero motion, but unstable equilibrium.

I am not suprise some Physicists won't help you if you take that attitude with them, and given the peremptory way you have dismissed my attempts to help you. Below I have set out a simple plan of the starting conditions in accordance with your instructions. Now please tell me how you impart a motion to your rod to set it rotating about m2. Are you aware of the difference between a moment and a couple ? Once you have demonstrated that it is possible to establish the rotating motion at all we can get down to the business of the geometry.

Work may be considered a form of energy. Energy is the capacity to perform work. Mechanically we have Potential Energy, which does not involve motion but is present because of the position of a body or the configuration of a system And we have kinetic energy which is present when bodies are in motion. Both of these can be converted into mechanical work and there is a controlling equation called the work - energy equation in mechanics. https://www.google.co.uk/search?source=hp&ei=zoiSXY-aM4OZlwT9sKfIAQ&q=work+-+energy+theorem&oq=work+-+energy&gs_l=psy-ab.1.0.0j0i22i30l9.1076.5750..10642...0.0..0.89.887.13......0....1..gws-wiz.......0i131.3Yo3mDVfoIk#spf=1569884379468 I should add to swansont's note about thermodynamics There is a form of energy called free energy (in fact there are two types Gibbs free energy and Helmholtz free energy) Loosely speaking this refers to the amount of energy available to do work after other necessary considerations have been satisfied.