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A thin long rod A revolves at 300 rpm about a vertical axis. Another rod B is held vertically with it's lower end just above the horizontal plane swept by rod A during it's rotation. Find the maximum length (in m) of rod B, if it should fall without being hit by the rotating rod after B is released from rest. (Assume $$g = 10\text{ m/s}^2$$)
Correct Answer: 0.2
At 300 rpm the rod makes 5 revolutions per second, so it sweeps past a given point every $$0.2\text{ s}$$. Released just after rod A passes, rod B must fall its whole length through the plane within 0.2 s. So $$L = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 0.2^2 = 0.2\text{ m}$$, i.e. 20 cm.
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