Algorithm to find out how much Torque is necessary by a DC motor to rotate a rotor...

[abhi@eda]

Simple question about on how much Torque is necessary by a DC motor to rotate a rotor arrangement..

This image below sums up my question more clear,its a hypothetical setup i know,but its just for my understanding



the length of each blade being 1.5 meter

As you can see,there's a electric motor,and it is connected directly to a shaft,and that shaft is connected to 3 blades ,where each blade weighs 1.5 kg,i know a gear can be used to increase the torque or increase RPM..but i want to understand it without confusion...

i want to calculate what amount of torque is to be produced by the motor to rotate the shaft,the torque reduces as the RPM goes up,but i want to find the initital torque needed to move it...

i want to know the steps or algorithm you could say

should i calculate the moment of inertia? or..?

please tell me the algorithm to calculate ,ill do it my self and try to find out the answer..thank you

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the MI of the each blade

the weight of the each blade is 1.5 kg/14.7 Newtons
and the length of blade is 1.5 meter

therefore MI = (14.7 X 1.5^2)/3=11.025 Newtons

and as there are 3 blades its 11.025 N * 3 = 33.075N of total Moment of inertia on the rotational Axis..


so now the DC motor has to produce more than 33 Newtons per ... ?

so that it rotates the shaft a 0 RPM ,and if i produces any torque less than 33 Newtons per ... then the motor stalls and will never be able to start the rotation by itself..


am i right about this?
please correct me,this is aching my head,i am not from physics background...

 

[godfreyl]

Small torque will give small angular acceleration. Larger torque will give larger angular acceleration.

There is no minimum torque. If there is no friction then even the tiniest torque will start it turning.

 

[crutschow]

As godfreyl noted, the motor torque is just that needed to overcome any static friction plus how fast to want to accelerate the rotor. As the rotor speeds up, then the motor need to provide enough torque to overcome the moving friction plus the air drag from the blades.

 

[erikl]

MI = (14.7 X 1.5^2)/3=11.025 Newtons

and as there are 3 blades its 11.025 N * 3 = 33.075N of total Moment of inertia on the rotational Axis..

so now the DC motor has to produce more than 33 Newtons per ... ?

so that it rotates the shaft a 0 RPM ,and if i produces any torque less than 33 Newtons per ... then the motor stalls and will never be able to start the rotation by itself..

am i right about this?

Not quite. To start the rotation, the motor torque just has to overcome the dry friction of the motor (i.e. of the rotor in its bearing(s)).
Then any minimum torque τ is enough to accelerate any Moment of Inertia (MI) with an angular acceleration α

τ = MI * α
_______________
PS: mind the units:
MI = (14.7N X 1.5^2m²)/3=11.025 N*m² (Newtons times square-meters)
... 3 blades is: 11.025 N*m² * 3 = 33.075 N*m² of total Moment of inertia on the rotational Axis. <- Right! (+ MI of rotor & axis)

 

[abhi@eda]

Thank u god n chow,i understood wat u explained n i agree with it,but my confusion is there will be some intertia,as the weight of the blades sum up to 4.5 kg,so the motor has to apply some amount of torque to put a rotating force more than the intertia of the shaft to over come the inertia n rotate the shaft

How do i find that torque???

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Not quite. To start the rotation, the motor torque just has to overcome the dry friction of the motor (i.e. of the rotor in its bearing(s)).
Then any minimum torque τ is enough to accelerate any Moment of Inertia (MI) with an angular acceleration α

τ = MI * α
_______________
PS: mind the units:
MI = (14.7N X 1.5^2m²)/3=11.025 N*m² (Newtons times square-meters)
... 3 blades is: 11.025 N*m² * 3 = 33.075 N*m² of total Moment of inertia on the rotational Axis. <- Right! (+ MI of rotor & axis)

U said any amount of torque can rotate any amount of MOI ,this where my confusion starts

Ill illustate a example to express my question
Ill take a small motor which are used in radios n toy cars and it is powered by 12 v dc ,n now ill connect a metal rectangular plate which weighs 1kg n the motor shaft is connected to the center of gravity of the plate

Its obvious the motor cant rotate it,it reaches its maximum stall current,unable to rotate it ,as it is unable produce enough torque to rotate the plate...

How is this ?? U said any amount of torque can rotate any amount of moi..

 

[godfreyl]

Its obvious the motor cant rotate it...

Yes it can. That's the point.

 

[FvM]

How do i find that torque?

A physical correct attempt of torque calculation must contain an expression for angular acceleration. It's till missing from your posts. Static friction, as mentioned by erikl is a differnt point, but not directly related to the problem. You can (in principle) design a drive with frictionless bearing, e.g. magnetic or air bearing, so there's a realistic method to rotate a shaft with arbitrarily small torque (and respective large acceleration time).

 

[crutschow]

.................................

Ill illustate a example to express my question
Ill take a small motor which are used in radios n toy cars and it is powered by 12 v dc ,n now ill connect a metal rectangular plate which weighs 1kg n the motor shaft is connected to the center of gravity of the plate

Its obvious the motor cant rotate it,it reaches its maximum stall current,unable to rotate it ,as it is unable produce enough torque to rotate the plate...

How is this ?? U said any amount of torque can rotate any amount of moi..

If the plate is truly mounted at the center of gravity (i.e. the plate will stay in any position it is manually rotated to), then the only reason the plate won't rotate is if the static (dry) friction is greater than the motor torque.

 

[abhi@eda]

Yes it can. That's the point.

So u mean that the 12v dc motor can rotate the 1kg metal plate wen it the shaft is connected to the CG but the angular velocity will be very low ? Did i get that right ??

So this 12v motor cannot say produce enough torque to rotate the metal plate at say 100rads /sec because the dry friction of shaft is more than the motor max torque?

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If the plate is truly mounted at the center of gravity (i.e. the plate will stay in any position it is manually rotated to), then the only reason the plate won't rotate is if the static (dry) friction is greater than the motor torque.

Ok so till today i guess i had thought that the torque of motor should be higher than the shaft load or the combined inertia of shaft to make it move,but the real force the motor has to over come by its torque is the dry friction of shaft ,right?

 

[crutschow]

...................

Ok so till today i guess i had thought that the torque of motor should be higher than the shaft load or the combined inertia of shaft to make it move,but the real force the motor has to over come by its torque is the dry friction of shaft ,right?

That's what we've all been trying to tell you. :wink:

 

[abhi@eda]

That's what we've all been trying to tell you. :wink:

thank you all for helping me understand this thing , my assumption on torque was wrong so far..


ill revisit the static friction part later,that is a hard one i guess so

so i will assume that,there is no friction in the rotor(hypothetically),

and now the shaft has 33 Newtons of Moment of Inertia,that is the 33 Newtons of resistance of shaft to change its motion..

so in order to rotate the shaft,the 33 N of MOI should be overcomed by the motor,but now i know the motor's torque has nothing to with it,
then what characteristic/attribute of a motor comes to play to overcome this MOI?

 

[crutschow]

........................
so i will assume that,there is no friction in the rotor(hypothetically),

and now the shaft has 33 Newtons of Moment of Inertia,that is the 33 Newtons of resistance of shaft to change its motion..

so in order to rotate the shaft,the 33 N of MOI should be overcomed by the motor,but now i know the motor's torque has nothing to with it,
then what characteristic/attribute of a motor comes to play to overcome this MOI?

That's not what was said. Please re-read the posts. The motor's torque (over and above that needed to overcome friction) will determine how fast the rotor accelerates. And, of course, there will be also be air friction to overcome as the rotor speeds up.

To summarize the motor torque:

1. Overcomes any bearing friction
2. Accelerates the rotor mass (the more torque, the faster the acceleration)
3. Overcomes air friction at speed

Make sense?

 

[godfreyl]

the shaft has 33 Newtons of Moment of Inertia

No. See post 4. The unit of Moment of inertia is N*m² not N.

In your example, MI = 33 N*m² (not 33 N)

so in order to rotate the shaft,the 33 N of MOI should be overcomed by the motor

No. Moment of Inertia is not a force, and it does not have to be "overcomed" before the shaft can rotate.

 

[FvM]

Torque definitely matters when you want to speed up the rotor in finite time. That's what the angular acceleration formula is talking about. https://en.wikipedia.org/wiki/Angular_acceleration

 

[abhi@eda]

That's not what was said. Please re-read the posts. The motor's torque (over and above that needed to overcome friction) will determine how fast the rotor accelerates. And, of course, there will be also be air friction to overcome as the rotor speeds up.

To summarize the motor torque:

1. Overcomes any bearing friction
2. Accelerates the rotor mass (the more torque, the faster the acceleration)
3. Overcomes air friction at speed

Make sense?

yes,thank you that makes sense...

the torque after overcoming the bearing friction,will accelarate the mass,if the torque is more,then it accelerates the shaft quicker,if the torque is low,the acceleration is slow..i understood it clearly..

but my confusion is, how do i know how much torque can overcome the bearing friction??

as i now know that the dry friction is the factor which stops the shaft from rotating,so as you pointed out

1)Overcomes any bearing friction

my main concern/problem is,i want to buy a motor,which is capable of rotating the shaft descibed in the picture,i want to make sure,the motor i buy is not underpowered and will fail to rotate the shaft

i dont want to buy a over powered motor,which will increase the weight...


i am stuck at this point... if i sort this out,then i can calculate the necessary gears to increase or decrease the torque/RPM(which i will not include in this thread)

as soon as i found out the solution to this problem,i will mark this thread solved once and for all n leave it alone! so please help me out :-(

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Torque definitely matters when you want to speed up the rotor in finite time. That's what the angular acceleration formula is talking about. https://en.wikipedia.org/wiki/Angular_acceleration

yes,i understood that,the torque determines the accelration rate,one can play with gears to increase or decrease torque to accelerate a shaft or wheel within desired time..

But i want to know what "power" motor will be able to overcome the friction of the rotor setup in my first post and rotate it rather than stalling...

so that i could make sure i am not buying a under rated motor or over powered motor..

please help me out on this , thank you

 

[Tunelabguy]

I don't think you are being honest with yourself about the real requirements of your project. You are asking for a motor that is just enough to overcome starting friction and begin turning the rotor. But what good is that to you? If you are only able to turn the rotor at 20 RPM, will that satisfy you? I don't know if you are building a helicopter or a personal fan or just a work of art. But whatever it is, you should take into account the reasonable acceleration and top speed necessary to make whatever it is you are doing a success. Once you do that, you will probably find out that you need a motor that is capable of much more torque than just what is needed to break friction.

 

[erikl]

... i want to know what "power" motor will be able to overcome the friction of the rotor setup in my first post and rotate it rather than stalling...

Any motor should be able to overcome its own dry friction - otherwise it wouldn't be worth to be called a motor.

You've probably already passed by a wind turbine which you thought to be in state of rest because you didn't see it rotate. If you looked again after a few seconds you might have seen that the position of the blades had changed: it actually rotated, but rather slowly, so you couldn't detect the rotation itself, but only its action after some time.

Now let's assume your motor's torque (τ) is enough to overcome the static (dry) friction of its bearings - as any motor will be able to achieve - , however is rather small compared to the mass of inertia (Mi) of the blades, so after the begin of moving it will start to rotate very slowly.

Let's try an example: Say your motor's torque is just 1Nm , and the total load of your blades has an Mi of 33Nm² - as calculated by you - then you achieve an angular acceleration of
α = τ/Mi = 1/33 m^-1 = 0.03 rad/s² .

One second after start, the blades will have rotated an
angle = (α/2)*t² = 0.015 rad = 0.87°
- a rotational movement not easily to be detected.

If now the dynamic friction of the blades in the air eats up a part (or all) of the motor's torque, the rotational velocity will increase even slower (or not at all) - its moving action practically invisible - just by the position of the blades after a while.

Without dynamic friction this acceleration would go on, and after 10 seconds the total rotational movement of the blades would reach 100*0.87°, or nearly a right angle difference - easily detected.

In any case the dynamic friction of the blades in the air will match the motor's torque at some rotational velocity, and so prevent further acceleration of its speed.

 

[abhi@eda]

Any motor should be able to overcome its own dry friction - otherwise it wouldn't be worth to be called a motor.

You've probably already passed by a wind turbine which you thought to be in state of rest because you didn't see it rotate. If you looked again after a few seconds you might have seen that the position of the blades had changed: it actually rotated, but rather slowly, so you couldn't detect the rotation itself, but only its action after some time.

Now let's assume your motor's torque (τ) is enough to overcome the static (dry) friction of its bearings - as any motor will be able to achieve - , however is rather small compared to the mass of inertia (Mi) of the blades, so after the begin of moving it will start to rotate very slowly.

Let's try an example: Say your motor's torque is just 1Nm , and the total load of your blades has an Mi of 33Nm² - as calculated by you - then you achieve an angular acceleration of
α = τ/Mi = 1/33 m^-1 = 0.03 rad/s² .

One second after start, the blades will have rotated an
angle = (α/2)*t² = 0.015 rad = 0.87°
- a rotational movement not easily to be detected.

If now the dynamic friction of the blades in the air eats up a part (or all) of the motor's torque, the rotational velocity will increase even slower (or not at all) - its moving action practically invisible - just by the position of the blades after a while.

Without dynamic friction this acceleration would go on, and after 10 seconds the total rotational movement of the blades would reach 100*0.87°, or nearly a right angle difference - easily detected.

In any case the dynamic friction of the blades in the air will match the motor's torque at some rotational velocity, and so prevent further acceleration of its speed.

so by taking the calculations you made and the theorem you said "Any minimum torque τ is enough to accelerate any Moment of Inertia (MI) with an angular acceleration α"
i think i need not worry about how much torque is necessary to just start the rotation of the shaft..

rather i need to calculate the Moment of inertia of the shaft and and determine what amount of Troque is necessary to accelerate the shaft to desired RPM in the given finite time..

so even if the Moment of inertia increases ,say 100Nm² with just 1Nm of torque produced by motor,the shaft will rotate but too too too slow,which is virtually like no rotation...

so if assume i have 12Nm of torque and MoI of 33Nm²,the the Angular acceleration would be 0.36 rads/sec²..

and this 0.36 rads/sec² of accelration acclerates the shaft until it reaches the same angular acceleration ,after the shaft reaches to the point where the shafts has the accleration equal to or more than 0.36 rads/sec²,the the torque produced by the motor cannot influence the shaft to increase its speed any further,but when by friction/air resistance,the shaft rotating acceleration reduces,the torque by the motor tries to maintian the acceleration to 0.36 rads/sec²..

am i right about this?

 

[erikl]

so by taking the calculations you made and the theorem you said "Any minimum torque τ is enough to accelerate any Moment of Inertia (MI) with an angular acceleration α"
i think i need not worry about how much torque is necessary to just start the rotation of the shaft..

Right!

rather i need to calculate the Moment of inertia of the shaft and and determine what amount of Troque is necessary to accelerate the shaft to desired RPM in the given finite time..

That's correct if you want to operate the "naked" motor alone, i.e. without the load of the blades.

Together with the blades, the total Moment of inertia is the sum of the M_i's of the rotor, the shaft, and the blades. I'd guess, however, the M_i's of the rotor and the shaft are insignificant compared to the blades' M_i.

so even if the Moment of inertia increases ,say 100Nm² with just 1Nm of torque produced by motor,the shaft will rotate but too too too slow,which is virtually like no rotation...

Yes, at least in the beginning. Due to such small acceleration it will take quite a while until the rotation is detectable. However the acceleration continues until the dynamic friction of the bearings + the dynamic friction of the blades in the air totally consumes the motor's torque. After this, there's no further acceleration, as the provided and consumed torque are balanced, so no more torque left for acceleration. Then, the rotational speed will stay constant.

so if assume i have 12Nm of torque and MoI of 33Nm²,the the Angular acceleration would be 0.36 rads/sec²..
am i right about this?

Exactly!

 

[abhi@eda]

Right!


That's correct if you want to operate the "naked" motor alone, i.e. without the load of the blades.

Together with the blades, the total Moment of inertia is the sum of the M_i's of the rotor, the shaft, and the blades. I'd guess, however, the M_i's of the rotor and the shaft are insignificant compared to the blades' M_i.

Well,the the Mi of the shaft was calculated including the blades right?

the weight of the each blade is 1.5 kg/14.7 Newtons
and the length of blade is 1.5 meter

therefore MI = ( (14.7N) X (1.5meter)^2)/3=11.025 Newtons

and as there are 3 blades its 11.025 N * 3 = 33.075N of total Moment of inertia on the rotational Axis..

how can i calculate the Mi of the blade again???

Thank you