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?

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

1 Answer 1


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.

The distribution of mass about the pivot, when considering the “resistance” of a body to changes in its angular speed, is important and the masses further from the pivot have a greater effect on the angular acceleration, proportional to distance from the pivot squared, than those closer to the pivot.


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