Which moment of inertia do I need to calculate torque? I'm trying to calculate the torque needed to rotate an array of solar panels. I've found some formulas for calculating torque, but they require moment of inertia. I designed this array in Autodesk Inventor and it gives me multiple options for the moment of inertia calculation. I've tried asking on the autodesk forums but no one will help me. I would be fine calculating this manually if need be also.
The pivot point is in the center of the array and the dimensions are: 117" x 1.5" x 129 1/16". The weight is 470 pounds and the center of gravity is in the center of the axis of rotation. The pivot is on the center of the 117" side. (see drawing below)

Which of these moments of inertia would I want to use in this equation, should I use principal global or center of gravity? I'm currently on the principal tab and it gives me I1, I2, and I3. Are any of these the right one?
The speed of rotation is no more than 0.001 rpm. (0.00003333333 π rad/s)
Does anyone know what value I should use for the angular acceleration on a dc motor? I know very little about physics and I'm having trouble.
 A: The answer first: given your axis of rotation, which is the x-axis, you are looking for the moment of inertia associated with that axis, which is the first, smallest principal moment ($I_1$). It only becomes that simple because in your setup the coordinate axes and symmetry axes are identical.
Now for some comments: As @BowlOfRed also pointed out in the comments, there is some hints that suggest that the moment of inertia is of no or little significance when dimensioning your motor for this setup:
In a perfectly constructed assembly, that is slowly accelerated to such slow speeds (say you allow about 5-10s to accelerate from zero to final angular velocity), motor torque needed to achieve this acceleration is going to be negligible. In an ideal system, the constant rotation itself needs no torque to be sustained. What you do have to take care of is


*

*System asymmetry due to imperfect construction

*Friction

*Power of the elements (wind, precipitation, debris)


I expect the additional resistance introduced through those effects to be much higher than the "clean" moment of inertia.
As the forces will be highest at the centre of rotation, you may want to consider any drive mechanism that attaches to the outer area of the panel (using pistons, chains, rope or similar), as it will have to overcome much lower resistance that way. This (pistons attached to the outer area) is in fact the way the solar panels at my workplace are set up.
All the best for your project!
A: The torque needed at the pivot A to rotationally accelerate an object by $\ddot{\theta}$i s
$$ \tau_A = I_A \ddot{\theta} + c_x W $$ where


*

*$I_A = I_{com} + m c^2$ is the mass moment of inertia at the pivot

*$c$ is the total distance between the pivot and the center of mass

*$c_x$ the horizontal distance between the pivot and the center of mass

*$W$ is the weight of the part


In your case if the pivot is at the center of mass ($c=c_x=0$) then all you need is
$$ \tau_A = I_A \ddot{\theta}$$ and you need to measure the mass moment of inertia about the pivot axis. I don't how Inventor handles this, but you should be able to choose which point and what coordinate system you want inertia measured about.
A: 1.) if you could assign a new coordinate system to your model such that one of the axis aligns with the rotation axis then take inertia along that axis and multiply with the angular acceleration to get the torque.
