Question

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

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

Answer 1

Answer: (A)

g$\sqrt{1 + 3\cos^{2}\theta}$

Answer 2

$\frac{mv^{2}}{2} = mgl\cos\theta or\frac{v^{2}}{l} = 2g\cos\theta$

and = $\sqrt{a_{t}^{2} + a_{r}^{2}} = \sqrt{\left( g\sin\theta \right)^{2} + \left( 2g\cos\theta \right)^{2}}$

= g$\sqrt{1 + 3\cos^{2}\theta}$

tan β = $\frac{g\sin\theta}{v^{2}/l} = \frac{\tan\theta}{2}$