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A person with his hand in his pocket is skating on ice at the rate of $10 \mathrm{~m} / \mathrm{s}$ and describes a circle of radius $50 \mathrm{~m}$. What is his inclination to vertical:
$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)$
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$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)$
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Verified Answer
The correct answer is:
$\tan ^{-1}(1 / 5)$
Since surface (ice) is frictionless, so the centripetal force required for skating will be provided by inclination of boy with the vertical and that angle is given as $\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}$ where $\mathrm{v}$ is speed of skating $\&$
i is radius of circle in which he moves.
i is radius of circle in which he moves.
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