[ponder] Driving torque for variable inertia 2

[mikeJW]

I have two expressions for torque with variable inertia. Will someone check out the maths for me please?

Deriving Torque from energy change

Work T.dq = Energy change = d(1/2 w².J ) q = rads, w = rad/S


d(1/2 w².J ) d( w².J )
Torque T = ------------ = ½ ---------
dq dq

= ½ [ w².dJ/dq + J.dw²/dq ] = ½ w².dJ/dq + ½ J.dw²/dq

We now need to re-arrange the second part to prove it equals J.a

dw²/dq = d(w.w)/dq = w.dw/dq + w.dw/dq = 2. w.dw/dq

= 2. (dq/dt).(dw/dq) = 2.dw/dt = 2 a

Thus ½ J.dw²/dq = ( ½ J).(2 a) = J . a

So T = J.a + ½ w².dJ/dq

This agrees with Pro-Mechanica results

Deriving Torque from rate of change of momentum

Torque T = d(Jw)/dt = J.dw/dt + w.dJ/dt

= J.a + w.(dJ/dq).(dq/dt)

= J.a + w.(dJ/dq).w = J.a + w².dJ/dq

Why is this different?

 

[quark]

Are you sure about dJ/dq????

 

[mikeJW]

Hello quark
If you mean should there be any DJ/dq then I can only point out that both equations for T use dJ/dq and that the first equation agrees with our solid modeller software.
If you mean that one of the maths steps involving dJ/dq is wrong can you elaborate please?

 

[zekeman]

I think your math is ok but the problem is that I believe the assumption of energy conservation is incorrect. The force derived from the momentum equation is the one I would go with.
By the way,is this an academic problem or if not, could you elaborate on how the moment of inertia is added or subtrated during the process.

 

[sreid]

Perhaps you can find the correct answer by applying the generalization of Newton's second law.

http://hyperphysics.phy-astr.gsu.edu/hbase/limn2.html#ln22

Replace the linear motion terms with rotational motion terms.

 

[mikeJW]

Thanks to zekeman and sreid.
Ok my second method listed is Newtons Second Law applied to rotating inertia and this was my first solution when I started the model.
However we have a solid modelling package which can also determine torque for a variable inertia provided the inertia is strictly defined.
Comparison of the results between my model and a simple solid model showed by inspection, that my w^2.DJ/dj term was twice the solid model value.
So I looked again, this time starting from energy and the result agreed with the solid model.
That's my dilema I think it should be momentum, you think it should be momentum but I am using the energy version just because it agrees with the solid model.

For your interest, I am involved in concept studies of high speed machines. They major torque component is from driving a reciprocating mass via cams and rocking levers.
These torques are typically 500 - 1500Nm.
The next significant source of torque is from a rotating mechanism whose centre of mass changes cyclically, driven by cams. This torque is up to 500Nm.
So you can see the problem is not trivial.

 

[quark]

I am not an expert in these issues and left the topic long back. My question is how the polar moment of inertia changes with angular displacement?

Regards,

 

[mikeJW]

Hello quark.
Think of a governor for steam engine where two balls are mounted on opposite corners of a parallelogram and the central rotating shaft goes through the other two corners.
The inertia can be varied by contracting or extending the distance between the two balls. The radius of rotation can be driven by closing or opening the gap between the two corners on the shaft.
If there is no energy input then the mechanism would change speed, like an ice skater.
However if a change in radius occurs on a constant speed system then a torque is produced instead.

In our case the radius of rotation (of the centre of mass) of the revolving parts are also variable. It completes one cycle for one revolution of the shaft. Hence we would know exactly the radius at any time in the cycle and thus the instantaneous J and dJ/dq.

In fact I am trying to determine the locus of the radius of rotation required for a desired torque cycle. But that's another story.
Regards.

 

[prex]

Sorry for Pro-Mechanica guys, but their equation appears to me as incorrect.
In your first derivation you apply the law of conservation of energy, but of course this is valid only for a closed system. Now your system is not closed because, to obtain the displacement of masses that cause a change in moment of inertia, you need to do work by means of an axial force (using your example of a governor). Hence you need to enter this work in the energy balance.
Momentum conservation (or if you prefer the Newton's 2nd law) is on the contrary valid for any of the three displacements and three rotations of a mechanical system, when the external forces and moments are correctly accounted for.
So your 2nd equation is correct, because the axial force used to displace the two balls has no moment around the axis of rotation.
Energy is not necessarily conserved in a system where momentum is conserved: the classical example is the inelastic collision of two masses, one moving, the other one initially at rest: half of the initial mechanical energy is lost (for two equal masses).
By comparing the two equations one may infer that the force needed to displace the balls is F=w(dJ/dx)/2 (w as defined by you and x being the axial displacement of the force): it would be interesting to obtain this result independently to confirm my reasoning.


prex

http://www.xcalcs.com

Online tools for structural design

 

[mikeJW]

Many thanks prex.
Yes everyone agrees that the momentum approach is correct.
Rather than blame Pro-mechanica, I will go to my colleague and see how he set his solid model up.
There may be an assumption in his model that it is a conservation of energy system.
I had included in my model determination of the work done to change the inertia as I need to know the forces here too.
So I am very happy to stay with conservation of momentum.
Regards,

 

[dford]

mike:

It turns out that the energy equation you used is not valid for a variable mass system.

For fixed mass translational system d(KE) = F dx , like you tried to use. KE is kinetic energy 1/2mvv.

For variable mass system d(m Ke) = m F dx

Found reference to a similar problem in Goldstein's "Classical Mechanics."

You can derive this relation via impulse/momentum arguments.

Doug

 

[mikeJW]

Eureka! It seems everyone is correct.
The momentum equation applies.
What was declared as 'Drive torque for the variable inertia' in the solid model, is in fact the 'System torque' as it includes the control means for varying the inertia.
In this case a set of cams was used to vary the radius of the rotating mass.
Since the work done by the control cams has the opposite sign to dJ/dq then the total 'system torque' is less than the torque to drive the variable inertia alone.
Probably the control torque will equal -1/2 w^2.dJ/dq and so completely explain the difference.
I am on another project but will conclude this topic asap.

Many thanks to all who responded for helping me keep my sanity.
Regards.

 

[GregLocock]

Great thread.

And the fundamental reason is that there IS a law called the conservation of momentum, and there is NOT a law called the conservation of kinetic energy.

In the simplest case take T=0, and final inertia=2*initial inertia=2*J, and initial velocities are w and 0 for the two inertias, and the final velocities are equal

at t=0

KE=1/2J*w^2

at t=final

by momentum Jw=2*J*(wfinal)

KE=1/2*2*J*(wfinal)^2

which is half as much as we started with.

On a slight tangent if it were possible to do this then you could break the second law of thermodynamics. The penalty for that is quite severe.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[mikeJW]

Disclaimer I believe the expressions are correct but use at your own risk.

T₁ is the general expression for driving a variable inertia load.
T₂ is a particular expression for controlling the variable inertia AND using a constant speed shaft.


Torque for variable inertia load

Torque T₁ = d(Jw)/dt = J.dw/dt + w.dJ/dt = J.a + w.(dJ/dq).(dq/dt) = J.a + w.(dJ/dq).w = J.a + w².dJ/dq


Torque required to change the variable inertia Shaft driven at constant speed only !

Centrifugal force F = Mw².R

dF = F1 – FO = M.w².R₁ – M.w²R_O = M.w².dR

Work done = ½ dF.dR = M.w².dR.dR

Equating work done

T₂.dq = ½.w².dJ and T₂ = ½ w².dJ/dq


Now the direction of work is opposite to T₁ so T₂ = - ½ w².dJ/dq


Thus Total System Torque

T = J.a + ½.w².dJ/dq

If shaft speed is trully constant then a = 0, but it is left in to deal with instantaneous torque pulses.
They are short lived (in my case) so constant speed is still valid.

Where J = instantaneous rotating inertia
q = shaft angle
w = shaft speed
a = shaft acceleration

 

[zekeman]

Not to beat a dead horse.... but the proof of your case does not depend on the constancy of w:
consider the following:
Newton's second law yields(since J=mr^2)
T=d/dt(Jw)=wJdw/dq+ w^2dJ/dq=mwr^2dw/dq+2mw^2rdr/dq
The energy equation including the internal energy expended
(and stored inside the system) in moving the mass from r to r+dr is
Tdq=d(1/2Jw^2)+mw^2rdr where seconfd term contains the centrifugal force, and expanding
Tdq =mwr^2dw+mw^2rdr+ mw^2rdr
where the third term is the internal energy expended
Therefore,combining, we get
Tdq=mwr^2dw+2mw^2rdr= wJdw+2mw^2rdr
dividing by dq yields precisely the equation above derived from momentum.
Since we are not usually concerned with the internal energy per se, the conclusion of this exercise is that the torque you want is obtained from the momentum equation as everyone has suggested.