Question

Calculate the torque on the pivot of a simple pendulum of length $R$ and mass $m$, when the mass is at an angle $\theta$ from the equilibrium position?

• A. $m\frac{g}{R}\sin \theta$
• B. $m\frac{g}{R}\cos \theta$
• C. $mgR\sin \theta$
• D. $mgR\cos \theta$
• E. $mgR\tan \theta$

Answer 1

Answer: (C)

$mgR\sin \theta$

At equilibrium position, tension $\left(T\right)$ in the string balances the component of weight $mgcos\theta$.

From figure, torque about point O ${\tau }_o=rF=R\times mg$ $sin\theta$