Need help understanding the Torque produced by a Engine...and how prvent engine stall

i am little confused about Torque which is measured in Newton meter...

Aussume,i have a engine which produces a 11Nm of torque at 2500RPM
the shaft of the engine is connected to a steel pipe which weighs 1kg..

so as far as i know,11Nm of torqe is the capacity of engine to turn something which weighs 11Newton and is connected to the center of the shaft at 1 meter distance...

so 11Nm is around 1.1kg-f ,so in the below image i have drawn the illustration,which shows a steel pipe of 1 meter long connected to the shaft,and weighs 1kg...

so the engine which produces over 1.1kg-f of force,can it turn the shaft when it is at rest? or will the engine stall and wont fire up??

I'll take a chance and show my ignorance, since I haven't done this kind of stuff in years, but I think you've got a slight misunderstanding of torque. Torque is the rotational equivalent of linear force. Thus, torque will determine how fast you can accelerate your rod up to 2500 RPM. Further, the fact that the motor is rated at 11Nm of torque at 2500 RPM doesn't tell you what its starting torque is. And further yet, you need to use Moment of inertia in your analysis, not just mass and length of the rod. FYI the moment of inertia (I) of a 1Kg rod 1m long is .833 Kg-m^2 (I looked it up).

I'm interested to see the proper answer to this.

@Barry
Well,i agree with what you said,i am new to the field of mechanics,well,it would be tough for some one from the field of coding and logic to real world Mechanical physics,at-least for me....

Well,my question might be kind of wrong in my #1 post,now i just a simple question to understand how it works

This is hypothetical question for my better understanding of torque and HP...

As you can see in the below 3D image,there's a motor connected to a 3 blades which are geometrically twisted...

Assumptions the motor is a 6.5 HP @ 3500RPM ,11Nm @ 2500 Torque gasoline small engine

the blades weigh 1 kg each ,so thats total of 3kg of weight of the blades

I am not considering the aerodynamic drag encountered by the blades when the revs go up...


So my simple yet confusing question to me is, with 3 kg of load at its shaft ,can this motor start the blades running... and how how would i calculate how at what weight of blades the motor will be unable to turn it and stall and turn off...


Please explain me,i am putting much effort to understand as much possible as i can,thank you

I think this is a little more complex than you (or I) understand. I think part of the equation is what kind of motor it is. (I'm not much more expert on motors than I am in physics ) I would think that if the motor is capable of "X" amount of stall torque, you could calculate how fast the motor will be able to accelerate your rotor to speed, (or whether it will just burn out). There is also 'stiction' that needs to be considered.

Well,as far as i know,the Engine with 6.5HP and 11Nm torque connected to that 3 Kg of blades will not be able to start the engine,so the blades need to be rotated manually or engine should be cranked manually,until the piston reaches a certain RPM from where the engine can run on its own power...

The reason for this as per me ,may be, this small engine is unable to generate enough torque to turn the blades in very low RPM... to turn this kind of load,may be the Torque generated by the engine should be much much more....

[I'm not sure about the inertia part below]

so if the 3 blades are say 1 meter long n 1kg each,then

the inertia of a blade,would be





so that's 0.33 Kg of inertia

0.33Kg multiplied by 3 for each blades = 1kg...

so the Engine needs to produce about 1Kg-f(9.82 N) of torque at 0 RPM to rotate the blades which are 1 meter in length,so that's like engine needs to produce about 9.82Nm @ 0 RPM theoretically,which the small gasoline engine surely cant...

So the Blades/Shaft should be cranked so that shaft reaches the minimal RPM for the engine to power by itself,by increasing the fuel flow,the revs can be increased,reaching the 2500RPM ,when the engine produces about 11Nm of torque,which would be quite enough to keep the momentum of the blades constant....


Well,i know somewhere i might i have made some wrong in my theory,please do correct me if i am wrong ...that will help me a lot ,thank you

For the definition and understanding of torque, you don't need to refer to inertia.

Starting with the post #1 scenario, 11N * 1 m = 11 Nm. How has 1m lenght and 1.1 kg "weight" to be arranged to generate 11 Nm?

Well 1 m is a lever arm in the simple case, perpendicular to the shaft. Imagine an horizontal, inflexible rod (with no weight) and a 1.1 kg weight at it's end. To lift the weight, you have to apply 11 Nm to the shaft. The weight is unbalanced, in contrast to your blades example.

That's how torque is usually introduced in mechanics text books, as a primarly static force. Inertia and related angular acceleration and deceleration torque is the next thing, but a bit more complicated because speed time differential comes in to play, which is completely missing from your previous attempts.