Question

A pendulum was kept horizontal and released. Find the acceleration of the pendulum when it makes an angle $\theta$ with the vertical.

• A. $g\sqrt{1+3{\cos }^2\theta }$
• B. $g\sqrt{1+3{\sin }^2\theta }$
• C. $g\sin \theta$
• D. $2g\cos \theta$

Answer 1

The correct option is A $g\sqrt{1+3{\cos }^2\theta }$

$\frac{{mv}^2}{2}$ = $mgl\cos \theta$ or
$\frac{v^2}{l}$ = $2g\cos \theta$
and $\sqrt{a_t^2+a_r^2}=\sqrt{{\left(g\sin \theta \right)}^2+{\left(2g\cos \theta \right)}^2}$
= $g\sqrt{1+3{\cos }^2\theta }$