Pendulums — Question 2

Question 2

A uniform rod is not a point mass at its center

A uniform slender rod of length $L=1.20\,\mathrm{m}$ pivots freely about a horizontal axis through one end. Its inertia about that axis is $ML^2/3$. Use $g=10\,\mathrm{m/s^2}$ and small angles.

1. Derive and evaluate its period.

2. Compare it with a point mass at the rod's center of mass, distance $L/2$ from the pivot. Why does that replacement give the wrong result?

Question 2 - Solution

A uniform rod is not a point mass at its center

(a) Derivation and calculation

For a uniform rod pivoted at an end,$$I_P=\frac13ML^2,\quad d=L/2,\quad I_P\ddot\theta=-Mgd\sin\theta.$$$$\omega^2=\frac{Mgd}{I_P}=\frac{3g}{2L},\quad\boxed{T=2\pi\sqrt{\frac{2L}{3g}}}.$$$$L=1.20,\ g=10:\quad\boxed{T=1.77715\,\mathrm s}.$$

(b) Result and interpretation

Replacing the rod by a point mass at its center gives$$T_{\mathrm{point}}=2\pi\sqrt{L/(2g)}=1.53906\,\mathrm s.$$$$\boxed{\frac{T}{T_{\mathrm{point}}}=\sqrt{\frac{2L/3}{L/2}}=\sqrt{4/3}=1.15470}.$$The center of mass fixes the gravitational moment, but does not supply the rod's full rotational inertia.

Physical check

The distributed rod has greater inertia about the pivot than a point mass at the same center of mass.