Demonstration 1- I have seen this demonstration, but an excellent accompanying problem (Ch 8, number 72) is in Serway and Faughn’s 5th edition of College Physics.  The upper of two hinged boards has a small cup placed at C and a ball holder (a soda bottle top) at B. If the upper board is released at a small enough angle q (the problem is to find the minimum q), B falls with an acceleration greater than g and a ball placed in the holder lags behind the upper board and lands in the cup at C (the correct distance HC must also be determined).

First, find the minimum value of q for which this will happen. The outline of the solution is first to apply Newton’s second law for rotation to the upper board to find its angular acceleration, then use aT = ar to find the tangential acceleration of B, and last to find the vertical acceleration of B, ay, which must be greater than g. With the mass of the upper board being m, t = Ia leads to .

Solving,

.

Then, applying aT = ar , with r = L, the tangential acceleration aT of the board is

aT = .

From the triangle at B, ay = aTcosq. Using this and the fact that ay > g leads to

, or

.

Since the cosine is a decreasing function, q < cos-1() = 35.3°.

If the board is released when, for example, q = 30°, since x = HC must equal HA, x/L = cos30° and then x = 0.87L. The proper release angle may be gotten by inserting a 30° wedge between the boards, using a stick of the proper length between A and B, or using a large protractor.
 

Demonstration 2- A board or stick is held horizontally with end A on the edge of a table and end B is released. If a small object is placed a distance x from A, the problem is to show that if  x > 2L/3 the object will lag behind the board. That is, the initial downward acceleration of a point 2L/3 from A is g, and any points beyond that will fall with an acceleration greater than g.

Applying t = Ia leads to

, or .

Since aT = ar, aT = . Substituting aT = g and r = x, x =2L/3 .

Thus any point farther from A than 2L/3 will initially fall with an acceleration greater than g.

While this demonstration is much easier to do than the first, it happens so quickly that unless the object is close to B it is difficult to see that it lags behind the board.