First slide

Vertical circular motion

Question

A simple pendulum is oscillating with an angular amplitude of 90^o as shown in fig. (4). The value of θ for
which the resultant acceleration of the bob is directed horizontally is

Moderate

A

$0^{\circ}$
B

$90^{\circ}$
C

$\sin^{- 1} (1 / \sqrt{3})$
D

$\cos^{- 1} (1 / \sqrt{3})$

Solution

See fig. (11).
From figure $a_{t} = gsin θ$
and $a_{c} = \frac{v^{2}}{l} = \frac{2 glcos θ}{l} = 2 gcos θ$
Now, $\tan θ = \frac{a_{c} \sin 90^{\circ}}{a_{t} + a_{c} \cos 90^{\circ}}$
or $\tan θ = \frac{2 gcos θ}{gsin θ} = \frac{2}{\tan θ}$
or $\tan^{2} θ = 2 or \sec^{2} θ = 3$
$∴ \cos θ = 1 / \sqrt{3}$
$θ = \cos^{- 1} (1 / \sqrt{3})$

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