Question

In Fig., the point D is chosen such that we have to extend OC for the perpendicular DN to fall on it. What will happen to the final expression of I_o, if point D is so chosen that the perpendicular DN falls directly on OC?

Answer 1

If point D is chosen such that the perpendicular DN falls directly on OC, then the geometrical construction changes, but the final result of the theorem of parallel axes remains unchanged.

For parallel axes,

I_O = I_C + Mh²

where:

I_O = moment of inertia about the axis through O

I_C = moment of inertia about the parallel axis through C

h = OC = perpendicular distance between the two axes

Choosing D differently only changes the intermediate geometrical relation used in the derivation. The cross-term still vanishes because C is the centre of mass.

Hence, the final expression is still:

I_O = I_C + Mh²

So, the final expression remains unchanged.