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An open merry go round rotates at an angular velocity $ω$ . A person stands in it at a distance $r$ from the rotational axis. It is raining and the rain drops falls vertically at a velocity $v_{0}$ .How should the person hold an umbrella to protect himself from the rain in the best way. Angle made by umbrella with the vertical is

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$\cot α = \frac{V_{0}}{r ω}$

$\tan α = \frac{V_{0}}{r ω}$

$\cot α = \frac{ω}{r V_{0}}$

$\tan α = \frac{ω}{r V_{0}}$

answer is A.

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Detailed Solution

Let the velocity of the drops above the person $r e l$ . to the merry go round be at an angle $α$ to the vertical. This angle can be determined from the velocity triangle $V_{0} = V_{r e l} + V_{m . g . r}$ ,

$V_{r e l} = V_{0} - V_{m . g . r}$

$A s V_{m . g . r} = r ω$

$\cot α = \frac{V_{0}}{r ω}$

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