Zero equilibrium 0 x a force?

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The discussion centers on calculating the torque needed to spin a 3-foot, 10-pound bar attached to an axle at 30 or 60 RPM. While the bar is in equilibrium, it requires force to initiate movement. The conversation highlights the importance of torque in overcoming inertia and maintaining motion against friction and air drag. Nema class stepper motors with various gear ratios are being considered for this application. Understanding the relationship between torque, angular acceleration, and the forces involved is crucial for achieving the desired RPM.
j1ceasar
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Okay - a classic trap . multiplication times zero .
I have a 3 foot bar weighing 10 lbs . its attached along its mid point to an axle . how much torque / force - do I need along its axle ( motorized ) to get it to spin 30 rpm or 60 rpm ?

its in equilibrium , but of course it needs FORCE to move it

I am considering Nema class stepper motors / geared 5/1 - 16 to one or 27/1 - usually these motors go to 700- rpm ...
 
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j1ceasar said:
Okay - a classic trap . multiplication times zero .
I have a 3 foot bar weighing 10 lbs . its attached along its mid point to an axle . how much torque / force - do I need along its axle ( motorized ) to get it to spin 30 rpm or 60 rpm ?

its in equilibrium , but of course it needs FORCE to move it

I am considering Nema class stepper motors / geared 5/1 - 16 to one or 27/1 - usually these motors go to 700- rpm ...
Uh ... WHAT multiplication by zero?
 
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Just like F = m*a for linear motion, there is an analog for rotational motion as well. A torque will be required to accelerate the bar, but once it's moving only air drag and friction will slow it down.

See here: https://en.wikipedia.org/wiki/Angular_acceleration
 

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