Consider a rod of uniform mass $m$, and length $l$, placed vertically on a frictionless surface. Now if it is displaced slightly it will start falling. As there is no net horizontal force the trajectory of the center of mass of the rod would be vertical, and this the acceleration of the center of mass would all at times be vertically downwards, whereas the acceleration of the end of the rod in contact with the floor will always be horizontal (let this be point $B$) Therefore at any angle $\phi$ with the vertical, the instantaneous center of rotation would be vertically above point $B$, and would be horizontally besides the center of mass. Therefore we can conclude that the trajectory of a the instantaneous center (call it $O$) would be a straight line that connects the initial position of the COM (at a height $l/2$) and the final position of point $B$. Therefore, at any given angle with vertical point $O$ will have both a vertical and horizontal component of force. Now, if we are given the angular acceleration and velocity as $\alpha$ and $\omega$ respectively when the rod makes and angle $\phi$ with the vertical, then we can express the acceleration of the center of mass as the product of the angular acceleration and the distance between point $O$ and the center of mass. This gives the acceleration as $\alpha\frac{L}{2} \sin\phi$ However this result doesn't match with the answer given in my workbook which also includes a term with $\omega^2$ in it. This led me to believe that maybe the value of acceleration obtained using the above procedure would be the relative acceleration of the center of mass with respect to point $O$, but if this was the case then the center of mass would have a horizontal component of acceleration also which we know is not possible. I am not able to understand how to approach the problem or what the issue in my original line of thought is.
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