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!
When the beam is barely balanced, the plank nearly looses contact with the pillar labelled "A", which is why the normal force at "A" is zero. Any more weight at "B" will result in a torque imbalance, causing the plank to tip over.
To mathematically prove that a mass of $45\ \text{kg}$ at "B" is the maximum mass that results in a normal force of zero (at "A"), model the plank as a rigid body in static equilibrium.
These are the two equations that form the basis of the rigid object in static equilibrium model:
$$ \sum \vec{F} = 0 \ \ \text{and} \sum \vec{\tau_z} = 0$$
The sum of torques can be made about any axis.
Call the normal forces at "A" and "C" as $\vec{n_A}$ and $\vec{n_C}$ respectively. Let the plank's mass be $M = 15 \ \text{kg}$, and the boy's weight be $\vec{F_g}$ = $(45 \text{kg}).\vec{g}$
The net torque about an axis perpendicular to point "A" is zero, so:
$$\sum \tau_A = 0 \rightarrow \ -Mg(\overline{\text{AO}}) \ + n_C(\overline{\text{AC}}) -F_g(\overline{\text{AB}}) = 0$$
Solving this equation should yield : $n_C = 588N$
The net force on the plank is also zero, so:
$$\sum F = 0 \rightarrow Mg \ + F_g - n_C \ -n_A = 0$$
which yields $\boxed{n_A = 0}$
As an exercise, try using this method to caluclate $n_A$ if the mass of the boy was $46 \ \text{kg}$. You will get a negative normal force at "A" which is unphysical.
Hope this helps.
Best Answer
Imagine a thin stick that you place on the floor and you want it stand. If you succeed to put the stick so as its weight, considered as acting on the center of mass, pass through the point $O$ of contact with the floor, the stick will stand. Otherwise it will fall.
Let's see why. Let's decompose the weight of the stick into a component along the stick, and one perpendicular to it. The latter component creates a torque around the point $O$. Let's calculate this torque.
$$\tau = g\int_A^B R \ \rho (R) \ \text dR \tag{1},$$
where $R$ is the distance from the point $O$, $\rho$ is the density of the stick per unit length, $A$ and $B$ are the extremities of the stick.
Now let me express $R = R_0 + r$ where $R_0$ is a point whose meaning will appear below. We get
$$\tau = g R_0\int_{R_A - R_0}^{R_B - R_0} \rho (R_0 + r) \ \text dr \ + g \int_{R_A - R_0}^{R_B - R_0}r \ \rho (R_0 + r) \ \text dr. \tag{2}$$
Well, the 1st integral gives
$$\tau = g R_0 M = GR_0 \tag{3}$$
where $G$ is the weight of the stick, and this is the final result if we choose the point $R_0$ in such a way that the 2nd integral in (2) be zero. The point $R_0$ for which the 2nd integral in (2) vanishes is the center-of-mass. As to the result (3), it says that the torque imposed by the weight of the stick is equal to the distance to the center-of-mass, $R_0$, times the weight of the stick as if it were concentrated, all of it, in the center-of-mass.
Two simple examples If the stick has uniform density, then the 2nd integral yields the expression $g\ \rho \frac { (R_B - R_0)^2 - (R_A - R_0)^2}{2}$ which is clearly zero for $R_0$ chosen in the middle of the stick. But if, say, the lower third of the stick is has four times the linear density of the rest, the 2nd integral is zero for $R_0$ chosen at 1/3 height, i.e. $R_A - R_0 = -L/3$, where L is the stick length, and $R_B - R_0 = 2L/3$. Indeed, $g\frac {1}{2}[-4\rho \frac {L^2}{9} + \rho \frac {4L^2}{9}]$.