Worked exercise — Compound pendulum

Problem

A homogeneous rigid rod of length $L = 1{,}0$ m and mass $m = 0{,}40$ kg is hinged at one end, free to oscillate in the vertical plane. Calculate: (a) the moment of inertia about the pivot, (b) the period of small oscillations, (c) a comparison with a simple pendulum of the same length.

Solution

(a) Moment of inertia. For a homogeneous rod hinged at one end: $I = \tfrac{1}{3}mL^2 = \tfrac{1}{3}\cdot 0{,}40\cdot 1^2 \approx \ev{0{,}133 \ \text{kg m}^2}$

(b) Period of the physical pendulum. We have $T = 2\pi\sqrt{I/(mgd)}$, where $d = L/2$ is the distance of the centre of mass from the pivot: $T = 2\pi\sqrt{\frac{0{,}133}{0{,}40\cdot 9{,}81\cdot 0{,}5}} = 2\pi\sqrt{0{,}0679} \approx \ev{1{,}64 \ \text{s}}$

(c) Comparison with the simple pendulum of $L = 1$ m: $T_s = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1}{9{,}81}} \approx 2{,}01 \ \text{s}$

The rod oscillates faster than a simple pendulum of the same length (about $20\%$ shorter period) because its centre of mass is closer to the pivot: the mass distributed along the rod has a smaller average lever arm than a mass concentrated at the far end.

Collegamenti

Argomenti: Oscillations and simple harmonic motion Concetti: Simple pendulum · Moment of inertia · Period and frequency Oggetti: Rod