Find the point that falls at exactly g.

A uniform ruler of length L is hinged at one end, held horizontal and released. At the moment of release, which point of the ruler starts falling with an acceleration of exactly g?

Answer: The point 2L/3 from the hinge.

At release the ruler has angular acceleration α = 3g/2L about the hinge, and a point at distance x from the hinge accelerates downward at xα. Points near the hinge fall slower than g, and points near the end fall faster.

x α = g

\Rightarrow x \cdot \frac{3 g}{2 L} = g

\Rightarrow x = \frac{2 L}{3}

The point two thirds of the way from the hinge falls at exactly g. The middle, which is the centre of mass, falls at 0.75g, which is why the hinge must push up on the ruler.

Original practice question · JEE Physics Decoded

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