Does center of mass affect how an object falls? Suppose you drop an object which has two ends, of which one is heavy and the other is pretty light. Will the object fall with its heavier end downward or with the lighter one? Why does it happen? 
 A: Drop a piece of paper and it glides sideways as well as flips. So aerodynamics (and hence the shape) affect the way things fall. 
Specifically aerodynamic forces have a center of pressure, which when ahead of the center of mass the body would rotate and flip, but if behind it will swing and stabilize at this orientation. This is the reason arrows, darts and rockets have fins.
A: The object's center of mass will fall straight downward, accelerated at g. All objects accelerate at the same rate in a vacuum, regardless of their weight, so neither end will accelerate faster than the other. That said, with air resistance, both ends of the object will feel the same drag force, so the end of the object with less mass will feel a greater deceleration from drag. This will cause a torque around the center of mass, spinning the object around the CM until the object is falling heavy end down (assuming no other drag cross-section effects).
A: $\newcommand{norm}[1]{\lVert #1 \rVert}
\renewcommand{vec}[1]{\pmb{#1}}$Let's take a mathematical point of view. Let the mass of the objects be $m_i$ for $i=1,2$ and the length of the rod $l$. The total torque with respect to the center of mass (COM) is given by:
$$\vec \tau \cdot  \vec e_z = \vec e_z \cdot \left[ \vec r_1 \times \vec F_{g1} + \vec r_2 \times \vec F_{g2} \right] = \norm{\vec r_1} \cdot \norm{\vec F_{g1}} - \norm{\vec r_2} \cdot \norm{\vec F_{g2}}$$
where $\vec r_i$ is the position vector of the $i$ th object measured from the COM and $\vec e_z$ is the unit vector in $z$-direction. Note that the force is perpendicular to the position vectors so $\sin \theta = 1$ and the minus sign in the second term arises because $\vec r_1 \times \vec F_{g1}$ and $\vec r_2 \times \vec F_{g2}$ show at opposite directions as you can show by your favorite right hand rule. Furthermore $\vec \tau \cdot \vec e_\xi = 0$ for $\xi = x,y$.
Now we know that $\norm{\vec r_i} = \frac{m_i}{m_1 + m_2} l$ and $\norm{\vec F_{g_i}} = g m_i $. In the end you get:
$$\vec\tau  \cdot \vec e_z= g\,l\cdot \left(\frac{m_1 m_2}{m_1+m_2} - \frac{m_1 m_2}{m_1+m_2}\right) = 0$$
Thus $\vec \tau \cdot \vec e_\xi = 0$ for $\xi =x,y,z$ thus $\vec \tau = \vec 0$. 
We conclude that there is no torque on the object with respect to the center of mass thus it can't rotate.
A: In a vacuum the rod will keep its orientation because of the equivalence principle, but with air resistance the heavier part will tilt toward the ground because its mass and therefore force is higher.
The force depends not only on the mass of the planet but also on the mass of the falling object, and since air resistance is also a force that acts on the surface the net acceleration is higher on the denser object.
A: The side with more mass has more momentum given that p = mv and gravity accelerates all masses equally. Therefore the heavier end is more resistant to changes in momentum given the concept of inertia (more mass = more resistant to changes in momentum = more force required to change momentum). Taking this into account, air resistance has a greater effect on the lighter side than the heavier side. While air resistance applies the same force to both ends, the heavier side has more inertia, so it is the lighter side's momentum that gets affected more in the direction of air resistance's force, which will be upward.
