Line joining the centre of path and cylinder is having angular velocity $\omega$ and angular acceleration $\alpha$ at the given instant:

In this

$V_{cm} = (R - r)\omega$

Total kinetic energy = $(1/2)mV_{cm}^2 + (1/2)I\omega ^2$

In this will we take moment of inertia (I) about point O or point C .

Also Acceleration of centre of cylinder = $(R-r)\alpha \hat{i} + (R-r)\omega ^2 \hat{j}$

But how could we find acceleartion of point of contact ?