I am doing some calculations for a practical project of mine. Basically I have a levitating sphere with gravity countered and close to 0 friction when the sphere rotates(due to airdrag). Inside the sphere is a servo spinning at $.12sec/60 degrees$, or $2\pi /.72sec$(8.73 rad/s). The spinning servo on the inside has masses attached to both arms to allow it to have enough moment to make the globe spin. The arm's radius is 1 or .0254m.
I have the following equation:
JA=sum of torques on globe due to servo
J is the moment of inertia for the sphere and many other things inside, I did a rough calculation and got .5kg·m²
A is the angular acceleration, for now i have it be $\pi /6$ radians, I assume that this acceleration is good enough for the globe to turn at a reasonable rate.
Finally to find the torques, there are two torque motion on the servo, since its servo arm has two arms. All I need to do is calculate the torque on one arm and multiply by 2. $\omega $ is the angular velocity of the servo, r is the radius of the arm. m is the mass attached to the end of the servo arm, and v is the velocity. Assume the servo arm itself is massless.
$F=m\omega \omega r$
$T=m\omega \omega rr$
Now to solve:
$m=JA/(\omega ^2r^2)=.5(\pi /6)/(8.73^2*.0254^2)=5.324grams$
Thus, to make my globe rotate(at $\pi /6$ angular acceleration) in reaction to the servo on the inside spinning, the servo must hold $5.3 grams$ of weight on both ends.
Does this calculation make sense?