A hollow cylinder is free to rotate about it\'s horizontal axis. A hole is cut i

Question

A hollow cylinder is free to rotate about it's horizontal axis. A hole is cut into the side of the cylinder. T object is to drop a ball throught the hole whilst the hole is at it's topmost position , rotate the cylinder and have the ball exit from the same hole as it is in the bottom most position. the diameter of the cylinder is 0.50m and the question is at what revolutions per second can this event happen. Having spent some time on this my answer does not corespond to the given book answer of 1.6 revs per second. Is this answer correct? Please ,can you show your work? Thank you.

Explanation / Answer

The ball falls through 2R in time t=4R/g

So the hole should be at the bottom at the same time.
Hence the answer is =(2n+1)g/4R and
the minimum 0=6.95c/s

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