Angular Acceleration Under Gravity
Everybody should know by now that when you drop two different objects, they will fall down at the same rate if there is no air resistance. This can be shown quite simply by dropping a pencil and a book, and observing how they hit the ground at the same time. But I want to consider a bit more complicated system. Consider two rigid, strong sticks of different lengths that want to topple over, like two chimneys that are going to be destroyed as shown in the picture on the right. Are the centers of mass falling down at the same rate? After all, the only external force acting on it is gravity, just like with two free falling objects.
Let’s start with with finding the angular acceleration on the thin rod. I’m going assume that you know the formula for the rotational inertia of a uniform rod rotating about one of its ends is (I=1/3ML^2) and that the center of mass is at the point 1/2L
As you can see from the equation, length of the chimney is inversely related to it’s angular acceleration. So if we have a shorter chimney, then its angular acceleration will be much larger, and so its angular velocity will always be larger. This means that a shorter chimney will cover the 90 degree distance to fall down first, and so hit the ground faster. But does this mean that it’s center of mass is also falling down faster? We can find out by finding its linear acceleration, and using the same reasoning. If the angular acceleration is in radians, we can just multiply it by the radius (1/2L for center of mass).
It turns out that the downwards acceleration of the center of mass does not depend on the length of the rod, and so it does not matter you are examining a tall chimney and a short chimney. They will follow an idea analogous to Galaleo’s gravity, where objects fall down at the same rate regardless of their mass. This result is still consistent with the fact that the shorter chimney falls down first, because the center of mass of the shorter chimney has a shorter distance to fall down.