How would the angular velocity of the rod change if it slipped on the table? I wanted to consider a second case of my homework assignment. We were asked to solve the question:
A uniform rod of length b stands vertically upright on a horizontal plane in a position
of unstable equilibrium. The rod is then given a small displacement at the top and
it tips over. What is the rod's angular velocity when it hits the plane, assuming the rod does not slip?
I managed to get an answer from conservation of energy. However, I wanted to also understand what difference there would be if we considered if the base of the rod could slip. How would I go about understanding that situation? Any suggestions would be much appreciated! :)
 A: If there is no friction on the surface, the rod will slide and rotate at the same time. In this case, you should analyze the motion of the center of mass. Since there is no horizontal force acting on the rod, its center of mass has vertical motion only. Therefore the rod will move such that its centroid falls vertically as shown in this figure

A: If it doesn't slip, which means the base of the rod is fixed on the plane.
Let the mass of rod be $m$, the potential energy would be $E_p=mg\frac{b}{2}\mathrm{sin}\theta$, when the angle between rod and plane is $\theta$. Let the angular velocity of the rod be $\omega$, then the kinetic energy is $E_k=\frac{1}{2}I\omega^2$, which $I=\frac{1}{3}mb^2$. Then as the conservation of energy, we have$$E_p+E_k=E_{p0}=mg\frac{b}{2}$$
thus,$$\frac{1}{6}mb^2\omega^2=mg\frac{b}{2}(1-\mathrm{sin}\theta)$$ then,
$$\omega=\sqrt{\frac{3g(1-\mathrm{sin}\theta)}{b}}$$
this is how $\omega$ changes by $\theta$.
If it could slip on the plane, we assume the plane is absolutely smooth.
Let the velocity of the center of mass be $v_c$, we could easily know the horizontal component of $v_c$ must be $0$, because the conservation of the horizontal component of momentum.
We have the velocity of the base of the rod is $\vec{v_b}=\vec{v_c}+\vec{\omega}\times\vec{b}/2$. We noticed that $v_b$ must be horizontal, so the vertical component of $v_b$  is $v_{by}=v_c-\frac{1}{2}\omega b\cos \theta=0$, so,$$v_c=\frac{1}{2}\omega b\cos\theta$$
then the conservation of energy would be,$$\frac{1}{2}mv_c^2+\frac{1}{2}I_c \omega^2=mg\frac{b}{2}(1-\mathrm{sin}\theta)$$where $I_c=\frac{1}{12}mb^2$.Let's Put  $v_c$ and $I_c$ in, we get,
$$\omega=\sqrt{\frac{12g(1-\sin\theta)}{b(3\cos^2\theta+1)}}$$
There is an assumption that the rod wouldn't leave the plane, which is true. But I'm not gonna prove it for brevity.
I guess this might be the answer you want. If you have any more questions, let me know.
A: If it doesn't slip, you can model it as rotating around the end of the rod (where it touches the ground).  When it strikes the ground, this restriction means that you know both the angular speed of rotation and the speed of the center of mass.
If there is no friction, then the rod will rotate around the center of mass.  This changes the moment of inertia for the rod, and it decouples the rotation and the fall.  At the beginning of the fall, the two will be related.  But during the fall it could happen that both ends lift from the ground.  That makes determining the energy due to rotation difficult.
