On the nature of torque ....

[wuzak]

Torque is a potential energy, not a kinetic energy.

Torque is not energy at all.

torque (tôrk)
n.
1. The moment of a force; the measure of a force's tendency to produce torsion and rotation about an axis, equal to the vector product of the radius vector from the axis of rotation to the point of application of the force and the force vector.
2. A turning or twisting force.

http://www.thefreedictionary.com/torque

[Richard]

Torque is a potential energy, not a kinetic energy.

So how does holding a hammer by its handle (ie a turning moment = torque) have potential energy that is not there when holding it at the centre of mass (linear force)? Same hammer = same potential energy (ie gravity).

[WhiteBlue]

Torque is a potential energy, not a kinetic energy.

So how does holding a hammer by its handle (ie a turning moment = torque) have potential energy that is not there when holding it at the centre of mass (linear force)? Same hammer = same potential energy (ie gravity).

If torque is applied to an elastic system like a crank shaft you get a finite deformation. That much should be obvious from the various discussions we had on this board about harmonics of different engines. There are torsional oscillations and radial oscillations. So a crank shaft has to be considered as an elastic system. I don't know about your hammer though.

If you put 1000 Nm across the crank shaft it will show a certain angular displacement. If you reduce the torque to 500 Nm the displacement will go down to half of the original value. This follows from elastic theory. The crank shaft therefore is a torsion spring that absorbs more energy the more torque you load it up with. So we can probably agree at least that the torque is proportional to the stored energy in the crank shaft. I only go a little step further and say that the torque is exactly the energy stored and the reason is the identity of the physical dimension. It usually is a dead give away that you are dealing with the same thing if it has the same physical dimension.

[Richard]

So what happens if you double the length of the shaft with the same applied torque?

= L · T / (I · G)

The rotation deflection will double when L doubles and everything else is constant.



Hence the energy goes up fourfold, but the torque remains the same.

[amc]

Yes, torque is definitely not energy, in whatever form. I stand by my previous assertion that it is easier to consider torque as having units of Nm/rad, as it clears up the issues with units.

WB, you could say that torque is one of the things (the other being rotational speed) that defines how much potential something has to transfer energy. However, this is not analogous to potential energy. (Clearly, otherwise any system that increased or decreased torque (i.e. gears) would contravene the Conservation of Energy principle).
On an atomic level, the strain (elastic) energy in the torsion spring comes from the bonds between atoms being lengthened and thus moving into a higher energy state. More lengthening, or more bonds being lengthened, will increase the energy gained - not more force lengthening them.

Perhaps it would help to show the energy stored in a torsion spring as a product of force and distance, as we would traditionally define energy.
In the diagram below, consider a beam of length r, with a force F applied at the end. It deflects by angle ϴ; let x be any distance along the beam.

The force on any point is (Fr/x) and the extension is (xϴ). The product is therefore Frϴ; this can also be written as τϴ - which we know to be the strain energy stored in a torsional spring.

[langwadt]

Torque is a potential energy, not a kinetic energy.

So how does holding a hammer by its handle (ie a turning moment = torque) have potential energy that is not there when holding it at the centre of mass (linear force)? Same hammer = same potential energy (ie gravity).

If torque is applied to an elastic system like a crank shaft you get a finite deformation. That much should be obvious from the various discussions we had on this board about harmonics of different engines. There are torsional oscillations and radial oscillations. So a crank shaft has to be considered as an elastic system. I don't know about your hammer though.

If you put 1000 Nm across the crank shaft it will show a certain angular displacement. If you reduce the torque to 500 Nm the displacement will go down to half of the original value. This follows from elastic theory. The crank shaft therefore is a torsion spring that absorbs more energy the more torque you load it up with. So we can probably agree at least that the torque is proportional to the stored energy in the crank shaft. I only go a little step further and say that the torque is exactly the energy stored and the reason is the identity of the physical dimension. It usually is a dead give away that you are dealing with the same thing if it has the same physical dimension.

so you would claim that a torsion spring torqued up to 1000Nm regard less of angular displacement
has the energy to lift ~102kg(1000N) 1 meter?

[WhiteBlue]

So what happens if you double the length of the shaft with the same applied torque?

= L · T / (I · G)

The rotation deflection will double when L doubles and everything else is constant.

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

Hence the energy goes up fourfold, but the torque remains the same.

If you double the length of the shaft you just make the spring softer. It has the same effect as if you make the shaft slimmer. But is has no impact on the energy storage IMO. The elasticity k goes down. You cannot use the same elasticity if you change the properties of the spring.

[WhiteBlue]

WB, you could say that torque is one of the things (the other being rotational speed) that defines how much potential something has to transfer energy. However, this is not analogous to potential energy. (Clearly, otherwise any system that increased or decreased torque (i.e. gears) would contravene the Conservation of Energy principle)....

I do not look at the energy that is transferred. I'm only looking at the stored elastic energy that gets stored by increasing torque and is given back from the shaft as work when the torque is reduced. You must mentally separate the two things.

[WhiteBlue]

so you would claim that a torsion spring torqued up to 1000Nm regard less of angular displacement
has the energy to lift ~102kg(1000N) 1 meter?

It has the energy to move 1000N by 1 m. Yes. 1000Nm is quite some torque. There are few cars that make that much torque.

[wuzak]

so you would claim that a torsion spring torqued up to 1000Nm regard less of angular displacement
has the energy to lift ~102kg(1000N) 1 meter?

It has the energy to move 1000N by 1 m. Yes.

Not necessarily.

1000Nm torque means that it can exert a 1000N force at a radius of 1m. Or 100N at a radius of 10m. Or 10N at a radius of 100m. etc.

Whether or not it can move a 1000N weight 1m cannot be known.

If the deflection is small, then it might only be able to move the weight by 20mm. If the deflection is large it could move the weight by 10m.

You simply cannot know from the torque alone.

1000Nm is quite some torque. There are few cars that make that much torque.

The torque doesnt have to be at the engine power takeoff. Remember that there are two stages of reduction after that, which multiplies the torque.

Perhaps you could aks yourself this:
If, by your reasoning, an engine with 1000Nm can move a 1000N weight by 1m, how can it move the same weight by 2m after a 2:! reduction gearbox, and how can it move it by 6m after a 2:1 reduction gearbox and a 3:1 reduction differential?

[wuzak]

So what happens if you double the length of the shaft with the same applied torque?

= L · T / (I · G)

The rotation deflection will double when L doubles and everything else is constant.

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

Hence the energy goes up fourfold, but the torque remains the same.

If you double the length of the shaft you just make the spring softer. It has the same effect as if you make the shaft slimmer. But is has no impact on the energy storage IMO. The elasticity k goes down. You cannot use the same elasticity if you change the properties of the spring.

Well, let's look at that shall we?

The angle of defection of a shaft is given by the formula

= L · T / (Ip · G) as shown by Richard.

I have added the subscript p to indicate that the property is the polar moment of inertia.

Let's rearrange the formula to get T in terms of

So,

T = *(Ip · G) / L

The definition of the torsional spring constant k is T= k *

Therefore

k = (Ip · G) / L

As you have noted, doubling the length halves the value of k.

Now, the energy stored in a torsional sping is given by U = 1/2 * k * ^2.

Let's use k1 as the constant for the spring with double length. And 1 is the angle of deflection of that second spring.

Since we are using the same torque we can write

T= k * = k1 * 1

Thus, 1 / = k/k1 = k / (1/2 * k) = 2

Therefore 1 = 2*

The energy of the lengthened spring is:
U1 = 1/2 * k1 * (1)^2

Substituting:
U1 = 1/2 * 1/2 * k * (2*)^2

U1 = 1/4 * k * 4*^2

U1 = k *^2 = 2 * U

So, the more flexible spring stores twice as much energy as the original spring given the same torque amount and, funnily enough, twice the angular deflection.

[WhiteBlue]

It is very likely that you have made a wrong assumption somewhere. Probably in the elasticity constant.
I have never said that k goes to 50% if you double the shaft length. It probably also goes by the square.

[Rohit_Gupta]

Energy stored in the shaft will be equal to the work that you do.
Torque cannot be equal to the work that you put in.
The same torque can produce multiple angular displacements. In this case for the same torque angular displacement doubles.
Hence, work done = torque X angular disp. Since the angular disp doubles,the work done and the energy stored doubles.

[wuzak]

It is very likely that you have made a wrong assumption somewhere.

Why, because it doesn't agree with your thesis?

Probably in the elasticity constant.

Nope, I believe I have the correct formula (also used by Riachard Leeds) for angular deflection in torsion.

I used it consistently for both cases.

I have never said that k goes to 50% if you double the shaft length.

No, you probably didn't. But it does.

It probably also goes by the square.

No, it doesn't.

k is directly proportional to the Polar Moment of Inertia (Ip = 1/2 * pi * R^4) and the Modulus of Rigidity (G, a property of the material, which remains the same in both cases), and inversly proortional to the Length (L).

k = (Ip · G) / L

Double the length, halve the k.

[rjsa]

Take a 1m radius drum connected to a torsion spring and rotated pi.rad, giving 1000N.m torque like that. The drum has a line around it connected to a weight on a flat surface. No mass on the drum or line for the sake of simplicity

When you release the drum it will rotate pi.rad, with a initial torque of 1000N.m and final torque of 0N.m, varying linearly.

With half a turn, the line will have been pulled 2.1416 meters, with a force linearly varying from 1000 to 0 N.m. That will give an average of 500N.M over 2.1416 meters, or 1070.8 joules. Not 1000 joules.