Moment of Inertia: uniform rigid rod on smooth plane Consider a rod of length $b$ and mass $m$ on a smooth horizontal plane. A force is applied to one end of the rod. 
What is the acceleration $a$ and angular acceleration $\alpha$ of the other end of the rod. 
At what distance should the force be applied such that the one end of the rod has initial acceleration of zero.
The essence of the question is, at what point does a rod rotate about given the force is applied a distance $x$ from one end of the rod?
 A: Let us denote $\ P$ the point corresponding to the end of the rod with no force applied, then we apply a force $\ \overrightarrow{F} $ on the other end of the rod, such that is parallel to the plane and perpendicular to the rod. 
Now, with respect to the center of mass, the only force whose torque is different from zero is $\ \overrightarrow{F}$:
$$\vec{M_{cm}}=\overrightarrow{R}\times\overrightarrow{F}$$
Now, projecting this equation on an axis with versor $\hat{a}$ perpendicular to the plane, we get:
$$\ M_a=RF=\frac{b}{2}F=I_a\dot{\omega}=I_a\alpha=\frac{mb^2}{12}\alpha$$
$$\ \implies \alpha=\frac{6F}{mb}$$
Where $\alpha$ is the angular acceleration around the center of mass.
So at distance $\frac{b}{2}$ from the center of mass, the acceleration is 
$$\ a=\alpha \frac{b}{2}-\frac{F}{m}=\frac{2F}{m}$$
Because for the center of mass, we have that $\sum \vec{F}=m\vec{a_{cm}}\implies a_{cm}=\frac{F}{m}$
And this is in the opposite direction.
The generic force, applied at distance $\ r$ from the center, gives a torque:
$$\ \ M=rF=\frac{mb^2}{12}\alpha\implies \alpha=\frac{12rF}{mb^2}$$
So, in order to have $\ a=0 $ for the point $\ P$, you must have:
$$\ a=\alpha \frac{b}{2}-\frac{F}{m}=\frac{6rF}{mb}-\frac{F}{m}=0$$
$$\ \implies 6\frac{r}{b}=1\implies r=\frac{1}{6}b$$
