On the nature of torque ....

[wuzak]

Apologies for making a late edit to my last post. I had it on the iPad and it was unsend. Then I just send it when I came back without checking what had been written. Sorry for that.

So lets repeat the look at the experiment. My view is it will not work as planned by Wuzak.

It will work just fine if the weight is adjusted. For half the angular elastical displacement, which is reached at double stiffness you must use twice the weight. It will travel half the height. I do not have to do the numbers because it is obvious for everybody that the work applied to the weight is the same in both cases.

And this is the proper demonstration that the energy of an elastic deformation in a torsional system is equivalent to the toque applied. The stiffness or the angular displacement is not relevant as we have seen. Only the torque defines the energy. You can use any stiffness or any angular displacement and you always keep the same torque, provided the stiffness and angular displacement are properly matched.

The same is not true for force in a linear system. It is not sufficient to define the energy of a linear elastic deformation because the linear deformation would be required as well to define the energy.

For case C, where mass = 2kg, length of shaft = 500mm, diameter of shaft = 5mm, etc, the deflection is 1.9985 radians, the mass moves 1.9985m and the change in potential energy of the mass is 39.210J - that is, the same angular deflection, twice the PE change.

FWIW my experiment won't work without support along the length of the shaft which still allows the shaft to twist. This is because unsupported it would bend - a lot! Also, I haven't verified that the load wouldn't cause the shaft to fail in shear. The size was made to deliberately make large deflections from a small load.

Basically, in a torsion spring, which you are describing, the torsional coefficient is described as K = T/w, where w = angular deflection in radians, and torque T is in NM.

That looks frighteningly similar to a linear spring. which has the spring stiffness K = F/x, where x is the linear displacement.

[WhiteBlue]

So the curious thing is that it is possible to have two measurement as Nm, one for energy and one for moment/torque that can't be added.

They can be added if you are working in the appropriate physical model. If your energy balance is about the torsional spring energy you can immediately take the torque. It is a true representation of how much energy is stored in your torsional system at any time.
If you are balancing a dynamic system where the shaft is in motion you are not so much interested in the internal energy absorption of the system but in what the system transmits in terms of energy. The inner absorption due to torque changes is very small compared to the energy transmitted by the system so you usually neglect it.

[WhiteBlue]

Basically, in a torsion spring, which you are describing, the torsional coefficient is described as K = T/w, where w = angular deflection in radians, and torque T is in NM.

That looks frighteningly similar to a linear spring. which has the spring stiffness K = F/x, where x is the linear displacement.

But the difference is not in the elasticity theory. Elasticity works for linear cases as for torsional cases. The difference is in the energy.

In a torsion spring the energy is completely defined by the torque. No other expression is needed if you have the torque. You do not have to check what the angular displacement happens to be by chance.

In a linear spring force is not sufficient to define the energy!!!! You have to know the linear displacement and multiply it with the force to arrive at the energy.

[Blanchimont]

Linear

energy = force x length
power = force x length/ time

Rotational (for 1 revolution)

power = torque x angular velocity = force x radius x 2 Pi x frequency = force x radius x 2 PI / period
energy = force x radius x 2Pi

The radius x 2 Pi is the same as the length in the linear formula. Without the 2 Pi the radius simply describes the distance between the pivot point and the force. IMO the multiplication of 2 Pi allows you to speak of energy.

[WhiteBlue]

@ blanchimont

There are different rotational energies. It is demonstrated by two different systems.

System A is a Williams kinetic energy storage. It works rotational. It's stored energy is kinetic and not elastic. The formula for the stored energy has nothing to do with the torque. It is only dependent of the angular velocity and the mass involved.

System B is the torsion bar suspension of a tank. The stored energy is directly depending of the torque in the torsion bars. In fact no other parameter is necessary to define that stored energy.

It follows that rotation can have different energies involved depending of the system design.

[Richard]

So you are saying that if I have a weight hanging from a spring then the energy transferred is force * displacement. That energy comes from the movement of the weight resulting in a loss of potential energy.

However if that rope is hanging from a torsion bar then no movement is required? If so where has the energy come from? The weight hasn't moved so hasn't lost any potential energy.

[Blanchimont]

@ blanchimont

There are different rotational energies. It is demonstrated by two different systems.

System A is a Williams kinetic energy storage. It works rotational. It's stored energy is kinetic and not elastic. The formula for the stored energy has nothing to do with the torque. It is only dependent of the angular velocity and the mass involved.

System B is the torsion bar suspension of a tank. The stored energy is directly depending of the torque in the torsion bars. In fact no other parameter is necessary to define that stored energy.

It follows that rotation can have different energies involved depending of the system design.

I just wrote the formulas down as i have the impression that you still believe that torque can be considered as a kind of energy. Did i get this wrong?

Some thoughts on your systems A and B.

The stored energy for the case A can also be described through torque as

torque = moment of inertia x angular acceleration

and

work = torque x angle = moment of inertia x angular acceleration x angle

And in case B the energy is dependent of the torque AND the angle!

Maybe we should discuss this during a "white sausage breakfast" and twist some of them to visualize the problem.

[wuzak]

Energy in a linear spring:



https://en.wikipedia.org/wiki/Hooke's_law#Spring_energy


Energy in a torsion spring:



http://en.wikipedia.org/wiki/Torsion_sp ... oefficient

Remarkably similar equations, don't you think?

[langwadt]

Linear

energy = force x length
power = force x length/ time

Rotational (for 1 revolution)

power = torque x angular velocity = force x radius x 2 Pi x frequency = force x radius x 2 PI / period
energy = force x radius x 2Pi

The radius x 2 Pi is the same as the length in the linear formula. Without the 2 Pi the radius simply describes the distance between the pivot point and the force. IMO the multiplication of 2 Pi allows you to speak of energy.

add an ideal rack and pinion and rotational becomes linear or vice versa, the radius x 2Pi is then just a "gearing"

[rjsa]

Energy in a linear spring:

https://upload.wikimedia.org/math/3/1/1 ... bc293b.png

https://en.wikipedia.org/wiki/Hooke's_law#Spring_energy


Energy in a torsion spring:

http://upload.wikimedia.org/math/5/4/2/ ... b7af22.png

http://en.wikipedia.org/wiki/Torsion_sp ... oefficient

Remarkably similar equations, don't you think?

Expected being those the first integral of f(x) dx in both cases, but quite misleading since K is newton/meter in the first case while it is newton-meters/radian for the second.

[wuzak]

Energy in a linear spring:

https://upload.wikimedia.org/math/3/1/1 ... bc293b.png

https://en.wikipedia.org/wiki/Hooke's_law#Spring_energy


Energy in a torsion spring:

http://upload.wikimedia.org/math/5/4/2/ ... b7af22.png

http://en.wikipedia.org/wiki/Torsion_sp ... oefficient

Remarkably similar equations, don't you think?

Expected being those the first integral of f(x) dx in both cases, but quite misleading since K is newton/meter in the first case while it is newton-meters/radian for the second.

Units a different because one is in a linear system and the other in a rotational system.

[WhiteBlue]

Energy in a linear spring:

https://upload.wikimedia.org/math/3/1/1 ... bc293b.png

https://en.wikipedia.org/wiki/Hooke's_law#Spring_energy


Energy in a torsion spring:

http://upload.wikimedia.org/math/5/4/2/ ... b7af22.png

http://en.wikipedia.org/wiki/Torsion_sp ... oefficient

Remarkably similar equations, don't you think?

Of course you can express it that way. You use the force and the elastic properties of the torsion spring. But it is not necessary to express the energy that way if you already have the torque.

[piast9]

There's a lot of confusion in this thread. In simple cases of linear motion the the work or energy is the scalar and it is the scalar products between the force and the displacement in the direction of the force. The force and displacement are parallel.

But the torque is a vector and it is the vector product between the force and the radius from the centre of roration at which the force is applied. In simple cases, when the force is tangential to the rotation the radius and the force are perpendicular.

Units may be the same but the nature of both these quantities are completely different.

Certain equivalent of the linear motion kinetic energy may be the moment of inertia but not the torque.

[wuzak]

Certain equivalent of the linear motion kinetic energy may be the moment of inertia but not the torque.

Mass moment of inertia is therotational equivalent of mass, not energy.

Energy and work are that whether they are in a linear or rotational system.

[WhiteBlue]

Torque is a potential energy, not a kinetic energy.