# Physical pendulum: moment of inertia vs center of mass

This has been confusing me for a while. Consider a solid, homogeneous rod of mass $$m$$ and length $$l$$, hanging from a fixed pivot. Its center of mass is located at $$\frac{1}{2} l$$, and its moment of inertia is $$I = \frac{1}{3}ml^2$$.

To find the frequency of small oscillations, we can find the torque from the equation $$\tau \theta = -I \ddot{\theta}$$ $$mg \left(\frac{1}{2} l\right) \theta = -\frac{1}{3}ml^2 \ddot{\theta}$$ and obtain $$\omega = \sqrt{\frac{3g}{2l}}$$ which, from my research, seems to be the correct answer. However, if we consider it as a point particle, we obtain $$\omega = \sqrt{\frac{g}{\frac{1}{2}l}} = \sqrt{\frac{2g}{l}}$$ For a point mass, these 2 answers give the same result. But why don't they agree in this case? What is causing the discrepancy?

• A rod and a particle have different moment of inertias for rotating about the hinge. Commented Oct 13, 2018 at 6:07

Using the same method of derivation for a point mass one gets $$mg \left(\frac{1}{2} l\right) \theta = -m\left (\frac{l}{2} \right ) ^2\ddot{\theta}$$ which results in the expected value of $$\omega$$ for a simple pendulum.