Period of a Pendulum gave $T = 2\pi\sqrt{L/g}$ for a small bob on a string. A swinging bar, a leg or a door knocker is a rigid body: its mass is spread out, so we need The Parallel-Axis Theorem, the torque form of Newton's law, and the small-angle trick from Linear Approximation and Differentials. It builds on The Equation of Simple Harmonic Motion. See pendulums keep time in the Pendulums and Resonance lab.
Why it works
A rigid body hangs from a pivot with its center of mass a distance $d$ below. Tilted by $\theta$, gravity acting at the center of mass gives a torque $\tau = -mgd\sin\theta$ about the pivot. Newton's law for rotation: $$I\ddot{\theta} = -mgd\sin\theta.$$ For small angles (in radians), $\sin\theta \approx \theta$, the linear approximation near 0, so $\ddot{\theta} = -\dfrac{mgd}{I}\theta$: simple harmonic motion with $\omega^2 = \dfrac{mgd}{I}$.
Key idea
$$T = 2\pi\sqrt{\dfrac{I}{mgd}},$$ with $I$ about the pivot and $d$ from the pivot to the center of mass. A simple pendulum is the special case $I = mL^2$, $d = L$. The formula holds for small swings; larger swings take slightly longer.
Worked example: A swinging rod
A uniform meter stick swings from one end. Find its period for small swings.
- $I = \tfrac{1}{3}mL^2$ about the end, and $d = L/2$.
- $$T = 2\pi\sqrt{\dfrac{mL^2/3}{mgL/2}} = 2\pi\sqrt{\dfrac{2L}{3g}} = 2\pi\sqrt{\dfrac{2(1)}{3(9.8)}} = 1.64\ \text{s}$$
- A simple pendulum of length $\tfrac{2}{3}L$ has the same period: some of the rod's mass is close to the pivot and swings in a smaller arc.
Answer: $1.64\ \text{s}$.
Worked example: A disk on a nail
A uniform disk of radius $0.20\ \text{m}$ hangs on a nail through a point on its rim. Find its period.
- About the rim: $I = \tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2}MR^2$; $d = R$.
- $T = 2\pi\sqrt{\dfrac{3R}{2g}} = 2\pi\sqrt{\dfrac{3(0.2)}{2(9.8)}} = 1.10\ \text{s}$. The mass cancels.
Answer: $1.10\ \text{s}$.
Common mistake
Using the whole length of a swinging rod as the pendulum length, as in $2\pi\sqrt{L/g} = 2.01\ \text{s}$ for the meter stick. A rod isn't a bob on a string: its mass is spread from the pivot to the tip. Check: put $I$ about the pivot and $d$ from the pivot to the center of mass into $T = 2\pi\sqrt{I/(mgd)}$; for a rod pivoted at its end, $d = L/2$.
The small-angle period is very close for modest swings: at a $30^\circ$ amplitude the true period is only about 1.7% longer, and at $10^\circ$ about 0.2%. That near-independence from amplitude is why pendulums made good clocks.