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

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abhi@eda said:
... the Mi of the shaft was calculated including the blades right? how can i calculate the Mi of the blade again???
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Seems this was a misunderstanding on my side about the shaft: I thought with shaft you thought of the axis of the motor. I don't think you have to recalculate the Mi value.



Right. It's just analog to the equation for the covered distance of linear movement (s) of a mass (m) accelerated by the acceleration (a) during a time (t):
s = (a/2)*t2 , where a = F/m (F=force [N])

If you need a more exact explanation, I'd recommend to study thoroughly this Wiki explanation - which I've linked already in my first answer to you.
 
Sorry,im back after some gap,

Another simple doubt was raised,the SI unit of Moment of Inertia is kg.m² not Nm² right??


thank you
 
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abhi@eda said:
Another simple doubt was raised, the SI unit of Moment of Inertia is kg.m² not Nm² right??
Click to expand...

Yes, you are right: the original error was here in your very first question of this thread (and I didn't notice it either, before now):
the weight of the each blade is 1.5 kg/14.7 Newtons
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It is not allowed to use the acceleration of gravity for the conversion of [kg] into [N*s2/m] (not simply [N]!), because not the gravity's acceleration will be used, but the motor's.

Hence 1kg = 1N*s2/m

So your above calculation seems correct:
T = 3.375kg.m² * 17.46 rads/sec² = 58.92 Newton Meter of Torque
Click to expand...

[kg.m2 * [rads]/s2] =[(N*s2/m)*m2 * [[rads]/s2] = [Nm]

[rads] is a dimensionless unit.

Merry Christmas! - erikl
 
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