A simple pendulum is oscillating with angular displacement $90^{0}$ . For what angle with vertical the acceleration of bob directed horizontally?

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Solution

$tan α = \frac{a_{n}}{a_{t}} = \frac{\frac{v^{2}}{R}}{g sin α} = \frac{\frac{2 g h}{R}}{g sin α}$

$= \frac{2 g (R cos α)}{R g (sin α)}$

$tan α = \sqrt{2}$

$cos α = \frac{1}{\sqrt{3}}$

$α = {cos}^{- 1} (\frac{1}{\sqrt{3}})$

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