What's actually happening is something more like this:
Here, $x_0$ and $v_0$ are the initial position and velocity of the moon, $a_0$ is the acceleration experienced by the moon due to gravity at $x_0$, and $\Delta t$ is a small time step.
In the absence of gravity, the moon would travel at the constant velocity $v_0$, and would thus move a distance of $v_0 \Delta t$ during the first time step, as shown by the arrow from the green circle to the red one. However, as it moves, the moon is also falling under gravity. Thus, the actual distance it travels, assuming the gravitational acceleration stays approximately constant, is $v_0 \Delta t + \frac12 a_0 \Delta t^2$ plus some higher-order terms caused by the change in the acceleration over time, which I'll neglect.
However, moon's velocity is also changing due to gravity. Assuming that the change in the gravitational acceleration is approximately linear, the new velocity of the moon, when it's at the blue circle marking its new position $x_1$ after the first time step, is $v_1 = v_0 + \frac12(a_0 + a_1)\Delta t$. Thus, after the first time step, the moon is no longer moving horizontally towards the gray circle, but again along the circle's tangent towards the spot marked with the second red circle.
Over the second time step, the moon again starts off moving towards the next red circle, but falls down to the blue circle due to gravity. In the process, its velocity also changes, so that it's now moving towards the third red circle, and so on.
The key thing to note is that, as the moon moves along its circular path, the acceleration due to gravity is always orthogonal to the moon's velocity. Thus, while the moon's velocity vector changes, its magnitude does not.
Ps. Of course, the picture I drew and described above, with its discrete time steps, is just an approximation of the true physics, where the position, velocity and acceleration of the moon all change continuously over time. While it is indeed a valid approximation, in the sense that we recover the correct differential equations of motion from it if we take the limit as $\Delta t$ tends towards zero, it's in that sense no more or less valid than any other such approximation, of which there are infinitely many.
However, I didn't just pull the particular approximation I showed above out of a hat. I chose it because it actually corresponds to a very nice method of numerically solving such equations of motion, known as the velocity Verlet method. The neat thing about the Verlet method is that it's a symplectic integrator, meaning that it conserves a quantity approximating the total energy of the system. In particular, this means that, if we use the velocity Verlet approximation to simulate the motion of the moon, it actually will stay in a stable orbit even if the time step is rather large, as it is in the picture above.
Conceptually you can "replace" the force on the left side with a weight of 100 N, then the force downwards on the ceiling is 100 N plus 100 N $\times\cos \theta$, plus the weight of the pulley, where $\theta$ is the angle of the rope with the vertical.
The reason why the downward force on the ceiling is bigger than the weight of the object is that an extra 100N is needed to stop the pulley wheel from rotating.
Imagine there were two differently sized pulley wheels which rotate about the same axle. First of all we glue the string from the 100N weight on to the smaller pulley wheel, then we glue the string from a much smaller weight on to the pulley wheel with a much larger diameter. If the weights and diameters are in inverse proportion, we stop it from rotating, and we get a downward force on the ceiling which is smaller than 200N.
For example, if the large pulley wheel is twice as big as the first one, then to get equal torque you would only need 50N of force to stop the rotation, and then you would only have 150N of force acting on the ceiling due to the pulley.
Conversely the 100N could be attached to the larger wheel, then we would need a 200N weight on the other pulley, and we would get 300N total force downwards. That is to balance the torques on the pulley wheels.
And if the pulley bearing was seized, then you wouldn't need any extra weight at all. The extra force is to stop the rotation of the pulley wheel.
Best Answer
I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like:
The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity.
To write down the equation of motion for the rod we use the rotational analogue to Newton's second law:
$$ T = I\frac{d^2\theta}{dt^2} $$
where $T$ is the torque and $I$ is the moment of inertia, and rearranging this gives:
$$ \frac{d^2\theta}{dt^2} =\frac{T}{I} \tag{1} $$
If the length of the rod is $\ell$, then the torque is:
$$\begin{align} T &= mg\frac{\ell}{2}\cos\theta + Mg\ell\cos\theta \\ &= (\frac{m}{2} + M)g\ell\cos\theta \end{align}$$
The moment of inertia of the rod is:
$$ I_{rod} = \frac{m\ell^2}{3} $$
and assuming our weight can be approximated as a point mass its moment of inertia is:
$$ I_{weight} = M\ell^2 $$
and the total moment of inertia is just the sum of these two:
$$ I = \frac{m\ell^2}{3} + M\ell^2 $$
And to get our equation of motion we just substitute our expressions for $T$ and $I$ into equation (1) to get:
$$ \frac{d^2\theta}{dt^2} = -\frac{(\frac{m}{2} + M)g\ell\cos\theta}{\frac{m\ell^2}{3} + M\ell^2} $$
There's a minus sign because as I've drawn the diagram the angle $\theta$ decreases with time so the angular acceleration is negative. With some rearranging we get:
$$ \frac{d^2\theta}{dt^2} = -\left(\frac{m + 2M}{m + 3M}\right) \frac{3g\cos\theta}{2\ell} \tag{2} $$
Now actually solving this equation of motion would be hard, but we don't need to solve it to answer your question. The left hand side of equation (2) is the angular acceleration, and if we increase the magnitude of the angular acceleration the rod falls faster while if we decrease it the rod falls slower. So your question simplifies to:
On the right hand side everything outside the brackets is independent of $M$, so we just need to answer whether the term in the brackets increases or decreases if we change $M$.
This is easy to answer because we have a term $2M$ on the top of the fraction and a term $3M$ on the bottom, and obviously $3M \gt 2M$. So if we increase $M$ the fraction in the brackets decreases, and therefore the magnitude of the angular acceleration decreases.
So the answer is that attaching a mass to the top of the rod does indeed make it fall more slowly.