I Torque to velocity of shaft

1. Nov 13, 2016

fahraynk

I was trying to figure out the velocity a 6mm diameter shaft would spin given a torque of 4000 N-mm.

M =mass = .0227 Kg
i=moment of inertia = 1/2*MR^2 = 1.04E-7
T/R=Force = 1300N <---Am I correct that this is force experienced by shaft?
Bearing friction coefficient = .0015
iX'' + .0015X'=1300
X''+14423X'=1.25E10
X=e^st
S^2+14423S=0
S=14423
X=C1e^-14423t + c2 -> t=0, v=0, C1+C2 = 0. I just set both = 0 for homogeneous solution.
X=C3t
14423C3=1.25E10
X = 86671t

So, the velocity can NOT be 86671 m/s, because its ridiculous. Also after I did this I realized that the coeffiecient of friction was supposed to be for the force normal to the bearing. Still, it SORT of makes sense though, since the friction is ideally 0 for a bearing. So... was I correct here? Is most of the friction just aerodynamic drag on a rotating shaft for a motor? Otherwise motor speed would be ridiculously fast? Or did I make some fatal mistake?

2. Nov 13, 2016

Timakki

Hae can you define the terms iX"

3. Nov 13, 2016

Timakki

And also x"

4. Nov 14, 2016

Nidum

@fahraynk

I can't understand your calculations or even your purpose in doing them . Could you please just tell us in words what the actual problem is ?

Is this another homework problem ?

5. Nov 14, 2016

jbriggs444

4000 N-mm on a moment arm that is 3mm (diameter 6mm = radius 3mm) is 1300 N of force on the perimeter of the shaft, yes. One imagines that you are visualizing something like a chain drive: the force is tangential and applied at a fixed point on the outside of the shaft as the shaft rotates underneath.

Here, you seem to be setting up a differential equation. You appear to believe that the bearing friction coefficient relates friction to velocity rather than friction to normal force. As such, it is not a dimensionless coefficient. What are its units?

You bring in the moment of inertia, but are casual about making it clear whether X denotes tangential displacement or angular displacement.

Here you are solving the resulting differential equation.

6. Nov 14, 2016

CWatters

If you assume the friction torque is zero then the shaft will accelerate indefinitely.

If the friction torque is independent of velocity then then the shaft will accelerate indefinitely (because the friction torque is a lot less than the applied torque).

If the friction torque is proportional to velocity then at some velocity the frictional torque will equal the applied torque and it will stop accelerating. This is the only velocity that makes any sense. Any other velocity will be a function of time because it will be accelerating.

7. Nov 14, 2016

Cutter Ketch

There are many things wrong with this reasoning as others have pointed out. First let me reinforce one that jbriggs444 pointed out: it incorrect to convert to linear coordinates and just use the moment of inertia unchanged.

More to the point, why change to linear coordinates at all? You could leave it radial coordinates and convert the friction instead of the torque and inertia. I think that would be easier to follow.

I'm not sure you used the friction as you really intended, but proportional to velocity is one possibility, so that isn't crazy. The bearing friction may be proportional to the thrust force, or it may be proportional to velocity (or velocity squared, or side force, or torque perpendicular to the axis, etc etc) or it may be proportional to all of these. As others pointed out if it isn't proportional to speed the angular velocity will go to infinity. Net Torque is constant and > 0 so acceleration is constant and velocity increases without bound. Ok, let's say it is proportional to velocity, but if the coefficient is small, the steady state velocity may be huge. Not knowing what you really intend for the friction, I didn't work out the final velocity, but with the problem as stated a giant velocity, or in the case that friction is constant, an infinite velocity ARE actually the correct solution.

The reason this answer surprised you is that in the real world the driving torque almost always falls off with velocity. Engines, turbines, etc. can't keep applying torque up to infinite velocity, and the power curve is often as likely to be the limit as the load. A fun counter example would be a space ship with stuck maneuvering rockets. That would keep accelerating until the fuel ran out or the centrifugal force destroyed the ship. (yes I said "centrifugal". Sue me.) By the way this actually happened to Neil Armstrong on a Gemini mission. He achieved "ridiculous speed!" before he managed to shut the thruster down. However, here on earth if speed is your only purpose you can contrive to make things spin really insanely fast generally limited by the strength of materials, so that isn't a crazy result. Another thing that almost never happens in the real world is a shaft spinning with friction as the only load. Usually the shaft is being spun to accomplish some work, and generally that is much more important than the shaft friction (or an engineer needs to be fired)

8. Nov 17, 2016

fahraynk

Ha I got ripped apart here. Sorry you guys! I decided it was aerodynamic drag. Am I wrong?

X is position, i is moment of inertia. iX'' = moment of inertia times angular acceleration.
I had googled bearing friction and saw .0015 somewhere and assumed it should be proportional to velocity. But there is probably supposed to be a normal friction coefficient AND a velocity friction coefficient.

Basically I am trying to model a motor spinning a shaft, but the shaft is not translating power to anything. So, the only things that should slow it down are the bearing friction forces and aerodynamic drag. I am trying to estimate the max velocity the shaft will spin.

How can I set up and solve this equation? How would I set up friction on a bearing, 1 term for normal force 1 term for velocity friction? Or should I ignore bearing friction all together, just assume it is aerodynamic drag and wait until I can solve the Navier Stokes equations to answer the question in the future?

i * X'' -KFn -BX' = Torque/radius ( X'' is tangential distance. Does this make more sense? this is without a term for aerodynamic drag. Fn = force normal to bearing)

Cutter said leave it in radial coordinates. Out of curiosity, what do you mean ? Set it up for radions instead of distance in meters?

Last edited: Nov 17, 2016
9. Nov 17, 2016

Cutter Ketch

As I said before, as posed the answer is infinity. The torque is constant regardless of speed. The resisting torque is constant regardless of speed. The applied torque is larger than the resisting torque. So the net torque is constant regardless of speed. So the acceleration is constant regardless of speed. The shaft will accelerate forever.

To find the answer you seek you must correct one or all of the premises. Real torque usually is not constant regardless of speed. Practically all motors or turbines or pinwheels or gerbil wheels have a power curve where the torque falls off as the speed increases (the rocket thrusters being a counter example). Real friction is almost never constant regardless of speed (although it isn't uncommon to be close to that case). Failing that, all (?) real mechanical systems will break, melt, or catch fire at some speed.

You can, as you suggest include air resistance as the only thing that depends on speed. However the answer will still be unrealistic because that tiny contribution will result in a top speed that is still way beyond what any real motor can do. In fact you can put in more realistic bearing resistance with v dependence and it still won't be very realistic. In your no load or nearly no load condition it is the power curve of the motor that will give the largest contribution to determining the max speed. (Or the mechanical failure of some part). Imagine pedaling a stationary bicycle with no load on the wheel. Is it the resistance that limits you, or the fact that you just can't possibly move your legs any faster?

10. Nov 17, 2016

fahraynk

Torque * angular velocity = Current * Back emf (motor power equation)
So... I guess the limit is in the back emf * current, whenever that reaches values that melt the copper wire.
Thanks so much. Don't know why I did not see that.

11. Nov 17, 2016

Cutter Ketch

Ha! No. Electric motors have a very definite power curve. The torque drops off at high angular velocity because the inductance makes it hard to increase the rate at which the current in the coils reverses. The back emf as you sugggest, bu not because the coils melt. (Well not right away anyway). It will reach a limiting speed before the coils melt (probably!) Google "power curve electric motor" and then select images to see the typical torque curve. The speed will increase until the torque falls to a level equal to your small load.

12. Nov 17, 2016

fahraynk

Cool. I see this "power band" information I need to study. Thanks, you set me in the right direction.

13. Nov 18, 2016

CWatters

Typically a DC permanent magnet motor with no load will accelerate until the back emf is close to the applied voltage. So if you know the motor constant (in Rpm/Volt) you can work the no load rpm. At least it's easier than trying to guess the friction and rotor air drag.

14. Aug 2, 2017

Scott hunter

Torque is current*bemf as previously stated. Bemf is proportional to speed, the faster you go the more bemf generated. Current isn't figured on just the voltage applied to the coils . The voltage the motor coils actually sees is the applied voltage - bemf voltage. Therefore the faster you go the fewer volts the coils see. At some speed bemf will become >= applied voltage and it will not be able to accelerate anymore. If there is no voltage difference you can't get any current into the motor coils.