Torque of a vibrator shaft rotating at constant rpm with eccentric mass attached 1

[Jaydeep_G]

Hello,

I am stuck at the procedure to find the required torque to rotate the shaft with eccentric masses attached with it. Following are the inputs that I have
Mass of rotating system= 14kg
rpm= 4000
M.I.(through axis of rotation)=0.008 Kg-m^2
Eccentricity from CG to axis of rotation= 4mm
As shaft rotates at constant rpm, so angular acceleration is zero.

I calculated as follows, but I don't know this method is right or wrong.
As the torque is rate of change of angular momentum, T=(Iw2-Iw1).......(w2 and w1 are final and initial angular speeds respectively)
And at start, w1=0 hence T=Iw1
Am doing this correct or not? Please suggest the correct approach to get to the answer. Any advice would be greatly appreciated.

Thanks and regards,
Jaydeep G

 

[BrianPetersen]

"Constant RPM" means angular acceleration is 0.

The non-zero and potentially significant forces on the bearings to constrain the bearings of the shaft (relatively) fixed in space are irrelevant to this.

Of course, the center of gravity of the shaft being away from the centerline of the shaft means that it matters a little bit if the orientation of the shaft is horizontal, vertical, or in outer space.

 

[Compositepro]

Rotating an eccentric weight does not use use energy or require torque at a steady state, other than friction. It is the absorption of the vibrational energy by the surrounding structure that requires energy and torque. Most vibrating structures, like vibratory conveyors, are designed to operate at a resonant frequency so that an empty conveyor uses little energy and most of the energy is used in moving the load.

 

[BrianE22]

Is the shaft horizontal such that gravity causes a moment about the shaft axis? If so, then you will have a small amount of torque ripple and speed change per revolution. They cancel out over a full turn though - gravity will slow the motor a bit (very small amount) when lifting the load but you get it back when dropping the load.

 

[rb1957]

is the question how much work to add to a stationary shaft to get it to rotate at 4000 rpm ?

another day in paradise, or is paradise one day closer ?

 

[BrianPetersen]

The statement "As shaft rotates at constant rpm, I'm unable to find angular acceleration" means it's not about acceleration ... and the "constant RPM" contains the answer to the angular acceleration right within that statement: 0.

 

[hydtools]

Torque = I x alpha not Iw1 - Iw2.

Torque x delta time,t2-t1 = deltaIw = Iw2 - Iw1. Impulse=change in momentum

Ted

 

[desertfox]

Hi Jaydeep

Not sure what you are really asking however if you are trying to balance the shaft during rotation this site might be of use:-

http://www.freestudy.co.uk/dynamics/balancing.pdf

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Jaydeep_G]

I think I took a wrong approach totally.

But I want to calculate the torque and then ultimately power consumed by that system to rotate with eccentric mass. The shaft is horizontal supported by two bearing 6205. This system is used in a vibration compactor.
Those inputs are the same as above. Please suggest the way around this or reference material to get the required result.

 

[MintJulep]

https://ocw.mit.edu/courses/physics...-fall-2016/syllabus/MIT8_03SCF16_Text_Ch2.pdf

 

[BrianPetersen]

The power consumed will merely be that due to friction. Those are deep-groove ball bearings, so the torque is going to be the friction in the seals and some agitation losses for the oil or grease inside the bearing and perhaps aerodynamic losses for the shaft and counterweight itself. For most practical applications, all of this can be neglected.

Angular acceleration averaged over a revolution = 0, therefore torque averaged over a revolution = 0, therefore net work consumed = 0, therefore power consumption = 0.

The fact that the eccentric is wibblywobbling around as the shaft rotates is not important to this when viewed in isolation with the assumption that the bearings are being rigidly held.

If that wibblywobbly shaft is making the rest of the machine surrounding it bounce up and down and in and out, then this is a WHOLE different and much more complex ball game, because the assumption of the bearings being rigidly held in space is no longer true, and the forces that they are applying in conjunction with a finite distance that they are moving while doing so means that they are performing work. Figuring out how much requires much more knowledge about the entire system, and it's going to depend on what your vibratory compactor is compacting. It's going to be different when sitting on a rigid surface than when sitting in mud, because of the damping (energy-absorbing) effect of the mud.

In other words ... this problem as a whole is waaayyyy more complicated than what you are seeing on the surface.

 

[hydtools]

Calculate centrifugal force.

https://www.engineersedge.com/physics/centrifugal_force.htm

Then work = energy

Force x displacement = work

Ted

 

[onatirec]

As pointed out above: if the bearings provide a rigid reaction, there will be no radial displacement from centrifugal force and therefore no work done nor energy lost.
Of course, even bearings have stiffness and compliance. And of course the eccentricity will produce vibrations and noise - which is effectively radiated energy, so the system is not closed.

hydtools does have the appropriate correction to the torque formula further above.

It's still not clear whether you are asking about the torque required to maintain a constant angular speed (which is 0 in an idealized system) or the energy required to bring the shaft up to a certain speed (which is easily calculated from already given formulas - for an idealized system).

Assuming this is driven by a motor - If you are interested in the actual power consumed (as your last post clarifies) you should measure the current and voltage going in to calculate power, since there are numerous losses from both the electromagnetic side (inverter switching, i^2R, core losses, etc) and mechanical (friction, vibration, noise, air resistance, etc)...

Asking about torque is somewhat nonsensical with the information given, since torque depends on acceleration, not final speed. Are you trying to reach 4000rpm in the span of milliseconds or hours?

 

[electricpete]

To summarize part of what has already been said:

If the rotating parts are constant speed and frictionless and the vibrating parts have no damping, then the average torque required to sustain that motion is zero.

If there is power dissipated by friction or damped vibrating components, then conservation of energy tells us that this will have to be provided by the average shaft power which is average torque times radian speed.

I think that's all you need if you're talking about average torque and power for example if you were sizing a motor (and by the way in that case friction within the motor itself isn't counted as shaft power.. the motor is rated by output power).

...but in the event you're also interested in instantaneous torque. There may be some variations above / below the average load torque associated with oscillations for example if the eccentric part of the mass is moving vertically on earth where there's gravity. Torque oscillations may lead to speed oscillations if motor output torque oscillations do not instantaneously match any load torque oscillations, the magnitude of such speed oscillations in a simplistic analysis depends on rotating inertia delta-w = delta-T / J although the dynamics of oscillations may be more complex (for example the assumption that everythign acts like a single J may not be valid if inertia is separated into two lumps with torsional spring between defining a 2DOF resonant frequency which can alter the dynamics depending on relation between that resonant frequency and the exciting frequency presumably which is related to rotation rate)

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(2B)+(2B)' ?