Question

A thin rod is to be used as physical pendulum. The rod has a length of l -1.26m and a mass of m = 682 g. a) Where the pivot point should be located so that the oscillation frequency will be the same as if the pivot point was at one of the rod ends (recall the experiment we have done). (1P) b) Where needs the pivot point to be located, to maximize the oscillation fre- quency and what is its value? (1P)

Answer 1

Lets quickly derive the formula for general plane geometrical object. Suppose the moment of inertia(MOI) about the pivot point is $I_P$ and the distance between the pivot point and the center of mass(COM) is D, then the equation of motion for small oscillation is

$I_P\ddot{\theta}=-mgD\sin\theta\approx-mgD\theta$

so that the oscillation frequency is given by

$\omega^2=\frac{mgD}{I_P}$

$\omega^2\propto\frac{D}{I_P}$

(a) Let the distance from the COM of rod (midpoint in this case) be $x$. By parallel axis theorem

$I_P=I_{COM}+mx^2=\frac{1}{12}ml^2+mx^2$

What is required

$\frac{x}{\frac{1}{12}ml^2+mx^2}=\frac{l}{\frac{1}{3}ml^2}$

This is Quadratic equation. One solution is of course $l/2$ but we want the other one , which is  

$l/6=0.21\textup{ m}$

from the midpoint.

(b) Let the distance from the COM of rod (midpoint in this case) be $x$. Just maximise the function

$\omega=\sqrt{\frac{mgx}{\frac{1}{12}ml^2+mx^2}}=\sqrt{\frac{gx}{\frac{1}{12}l^2+x^2}}$

or minimise

$f(x)=\frac{l^2}{12x}+x$

Simply apply the following inequality, for arithmetic and geometric mean.

$\textup{A.M.}\geq\textup{G.M.}$

$\frac{\frac{l^2}{12x}+x}{2}\geq\sqrt{\frac{l^2}{12x}\cdot x}$

$f(x)\geq\frac{l}{\sqrt{3}}$

$\textup{min}f(x)=\frac{l}{\sqrt{3}}$

so that

$\omega_{\textup{max}}=\sqrt{\frac{\sqrt{3}g}{l}}$

where equality hold when quantities are equal, that is,

$\frac{l^2}{12x}=x\Rightarrow x=\frac{l}{2\sqrt{3}}$