During a reaction if addition on one of the $\pi$ bonds of an allene (containing even number of $\pi$ bonds like $\ce {H2C=C=CH2}$) takes place then a carbocation is formed. To decide the position of the positive charge, we look at the stability of the carbocation. Example: $$\ce {H2C=C=CH2 ->[\ce{H3O+}]} \quad?$$
In the above reaction the proton ($\small{\ce{H+}}$) attacks the electron rich $\pi$ bond (say the left one) and can form one of the following carbocation: $$\ce{H2C+\bond{-}CH=CH2}\quad\quad \text{or }\quad\quad \ce{CH3\bond{-}C+=CH2}$$ $$ (\text{I})\quad\qquad\qquad\qquad (\text{II})$$
Now the major product (acetone) is formed as a result of the carbocation on the right (II), which apparently is less stable than the one on the left (I) as the positive charge in I is in resonance/conjugation with the $\pi$ bond. But our teacher said that since in allenes the overlapping p-orbitals (say $\ce{p_z}$) of one $\pi$ bond are perpendicular to the overlapping p-orbitals of the other (say $\ce{p_y}$; which makes $\ce{p_x}$ in the direction of internuclear axis $x$) and hence the positive charge cannot resonate with the $\pi$ bond (formed by overlap of $\ce{p_z}$) as the positive charge is effectively an empty p-orbital (the $\ce{p_y}$) and hence the carbocation (II) is relatively more stable.
But once addition has taken place the $\ce{C_1-C_2}$ $\pi$ bond no longer exists and the $\sigma$ bond is free to rotate so the empty p-orbital/positive charge can freely rotate and become coplanar with the p-orbitals of the $\pi$ bond. But why does this not happen? The reason cannot be steric hindrance as the two groups on $\ce{C_1}$ are the smallest, $\ce{H}$.