How to determine when a body has only kinetic translational energy or only rotational kinetic energy or both?

Question

does the rod have translational kinetic energy when it falls flat? since I thought gravity acts on the body center of mass should have some downward linear velocity only but the solution given says that when the rod is almost flat .. it has only rotational kinetic energy and they found angular velocity using energy conservation is the solution right? I even had a doubt regarding the choice of center of rotation...the end of the rod has some linear velocity so can that be taken as center of rotation.

The thin rod is released from the vertical position as shown, and falls by itself, then angular speed of the rod just before it falls flat, will be

$K_f~=~mg{\frac{l}{2}}$ and ${\frac{1}{2} }m{\frac{l^2}{3}}ω^2~=~mg{\frac{l}{2}}$

Answer 1

The decomposition of KE into translation and rotation depends on the reference point. In general, only the center of mass allows for a neat separation, with no cross terms. So, if in doubt, use the COM. If you apply this decomposition here, you will get the formula you show in the OP. Note that the moment of inertia about the center of the rod has a factor of $\frac{1}{12}$ and not $\frac{1}{3}$ so the KE formula you wrote aboe includes both the rotation obout the COM and the translation of COM. In order to get it, you will need tofind a relationship between the velocity of the COM and the angular velocity around the ceneter of mas. Just write the vertical position of the COM (y) as a function of the angle of the bar with the vertical and take the derivative.

But, as you probably tought, this also look like a pure rotation about the end of the bar (with the 1/3 factor in the moment of inertia) and not about the COM. This is so because at the very last moment the velocity of the end of the bar in contact with the ground is zero so it has no translation at all. You can see this if you write the displacement of this end, x, as a function of the angle between the bar and the vertical, $x=(l/2) \sin \theta$. If you take the derivatives of both sides, you get $v_x={l/2}\cos\theta$. This is zero when the angle is $90^o$.