A uniform rod of mass $M$ and length $L$ swings as a pendulum on a pivot at distance $L/4$ from one end of the rod. Find an expression for the frequency $f$ of small-angle oscillations.

1. A physical pendulum is an extended item that may be designed to travel in simple harmonic motion around a pivot that does not pass through the centre of mass for tiny angular displacements. This motion occurs on a regular basis.

$f=\frac{1}{2\pi }\sqrt{\frac{Mgd}{I}}$

Where $I$is the object's moment of inertia about an axis through the pivot, and $d$is the distance between the pivot and the object's centre of mass.

2. The Parallel-Axis Theorem: It is a relationship between rotational inertia, $I$comparing the rotation of a body around any axis to the rotation of the same body around a parallel axis through its centre of mass:

$I=I_{\text{com}}+M{d}^2$

The rotational inertia of the body about the axis through the centre of mass is$I$ and $d$is the perpendicular distance between the two axes. The real rotation axis has been displaced from the rotation axis via the centre of mass, which we can describe as $I_{\text{com}}$.

1. The uniform rod has a mass of: $M$.

2. The length of the rod is: $L$.

3. The rod swings as a pendulum on a pivot at distance from one end of the rod.

Because the pivot point is a quarter of the rod's length and the (uniform) rod's centre of mass is half its length, the distance $d$between the pivot and the centre of mass is:

$d=\left(L-\frac{L}{4}\right)-\frac{L}{2}$

$=\frac{L}{4}$

A uniform rod's moment of inertia around a pivot through its centre of mass is calculated:

$I_{\text{com}}=\frac{1}{12}M{L}^2$

Using Equation (2)'s parallel-axis theorem, we can calculate the rod's rotational inertia around the pivot point:

$I=I_{\text{com}}+M{d}^2$

$=\frac{1}{12}M{L}^2+M{\left(\frac{L}{4}\right)}^2$

$=\frac{1}{12}M{L}^2+\frac{1}{16}M{L}^2$

$=\frac{7}{48}M{L}^2$

The rod is modelled after a physical pendulum that moves in a basic harmonic pattern. Equation (1) is used to calculate the rod's oscillation frequency:

$f=\frac{1}{2\pi }\sqrt{\frac{Mgd}{I}}$

$=\frac{1}{2\pi }\sqrt{\frac{Mg\left(\frac{\backslash \mathrm{AA}}{4}\right)}{\frac{7}{48}MM{L}^2}}$

$=\frac{1}{2\pi }\sqrt{\frac{48g}{28L}}$

$=\frac{1}{2\pi }\sqrt{\frac{12g}{7L}}$

$=\frac{1}{\pi }\sqrt{\frac{3g}{7L}}$