My question about the parallel theorem for the rotational inertia in the following cases (see detail)?

I = I(cm) + Md(d to square). It is right in the case that a body does not turn about its own center-of-mass axis, but about the other axis. If while a body turns about the other axis, it turns about its own center-of-mass axis. Then, I = ( I(cm) + Md (d to square) ) + I(cm). Is it correct?My question about the parallel theorem for the rotational inertia in the following cases (see detail)?
It is not correct.





To find the I(cm), use the same formula.





I = I(cm) + Md(d to square.





In this case, d = 0.





Therefore, I = I(cm) + M x 0 = I(cm)My question about the parallel theorem for the rotational inertia in the following cases (see detail)?
You have 1 extra I(cm) that should not be there. While turning about an axis off the CM by a distance d, the inertia about that axis is given by


Ia = I(cm) + md虏





That's all
no because this this is unreal situation.


as the body can not rotate at the same time about to Parrnell axis