Why can't you use conservation of energy for MC3 on the midterm?

Energy is indeed conserved, but the argument, ``the speed at the bottom must be the same because in both cases speed is zero at the top, and because potential energy is the same at the top for both the rod and string cases, $A$ and $B$ must have equal amount of kinetic energy at the bottom'' isn't the right answer.

The condition for the ball to just barely reach the top is actually not that its speed must be zero there... rather the condition is that it must be undergoing circular motion at the top. The key here is that the particle must satisfy $F_r = m \frac{v^2}{R}$ at the top. In the string case, the sum of the radial forces (in the downward vertical direction) is $mg+T$. Because string tension can only be down (positive direction), this has a minimum value $mg$, corresponding to $T=0$. So the minimum speed for $A$ to be undergoing circular motion at the top is that for gravity providing the centripetal force. In contrast, the rod can provide an upward force at the top, canceling gravity, and the sum of the radial forces is $mg-T$. Then the minimum speed at the top for $B$ in the rod case is zero.

By conservation of energy, that means that in the rod case the speed at the bottom is lower, and so particle $A$ on the string will need greater speed at the bottom in order to have non-zero (horizontal) speed at the top.