{composition-setup}{composition-setup}
{table:border=1|frame=void|rules=cols|cellpadding=8|cellspacing=0}
{tr:valign=top}
{td:width=350|bgcolor=#F2F2F2}
{live-template:Left Column}
{td}
{td}
{HTMLcomment}The header material sets up the left navigation column.{HTMLcomment}


{excerpt:hidden=true}Compares the simple pendulum model with a slightly more detailed one.{excerpt}

|!179px-Grandfather_clock_pendulum.png!|
|Photo Courtesy of Wikimedia Commons|
|Downloaded 2009-01-16 from Charles H. Henderson & John F. Woodhull (1901) The Elements of Physics, D. Appleton & Co., New York, p.59, fig.21|


Adding detail to the model of the [pendulum].


{HTMLcomment}The "Solution" header and the *bold* items below should NOT be changed.  Only the regular text within the {cloak} macros should be altered.{HTMLcomment}

h4. Solution 

{toggle-cloak:id=sys} *System:* {cloak:id=sys}A model of a [pendulum], simply supported and free to swing without friction about a supporting axis under the [torque|torque (single-axis)] due to [gravity|gravity (near-earth)].{cloak:sys}

{toggle-cloak:id=int} *Interactions:* {cloak:id=int}[torque|torque (single-axis)] due to [gravity|gravity (near-earth)] and the upward force exerted against gravity by the axis.{cloak:int}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod}[Angular Momentum and External Torque about a Single Axis]{cloak:mod}

{toggle-cloak:id=app} *Approach:*

{cloak:id=app}

{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diag}

We consider first the usual *Simple Model of a Pendulum*

|!Basic Pendulum 01.PNG!|


And then a slightly more detailed model

|!Basic Pendulum 02.PNG!|

With two variations.


{cloak:diag}

{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

{cloak:id=math}

The simple model has the virtue that it is extremely simple to calculate the[moment of inertia], *I*, of the pendulum about the axis of rotation. We assume a massless stick of length *L* and a [point mass|point particle] *m* at the end. The moment of inertia is simply
\\
{latex}\begin{large} \[ I = mL^{2} \]\end{large}{latex}
\\
If we pull the pendulum away from its vertical [equilibrium position] by an angle {*}θ{*}, then the [restoring force] *F{~}res{~}* is given by 
\\
{latex}\begin{large} \[ F_{res} = m g sin(\theta) \]\end{large}{latex}
\\  
And the natural frequency, as noted in the vocabulary entry on [pendulum], is given by

\\
{latex}\begin{large} \[ \omega = \sqrt{\frac{g}{L}} \]\end{large}{latex}
\\
Now consider the second diagram above. Now, instead pf a [point particle], the mass of the pendulum ihas real extent. It is a disc of radius *r*. We can calculate the real [moment of inertia] by using the [parallel axis theorem]. According to that, we can calculate the moment of inertia about a point other than the center of mass (about an axis parallel to one running through the center of mass) by simply adding *md{^}2{^}* to it, where *d* is the distance between the center of mass and the axis of rotation. For a uniform disc, rotating about its center, the moment of inertia is
\\
{latex}\begin{large} \[ I = \frac{1}{2}mr^{2} \]\end{large}{latex}
\\
In this case *d* is the length of the pendulum *L*, so the moment of inertia is
\\
{latex}\begin{large} \[ I = \frac{1}{2}mr^{2} + mL^{2} \]\end{large}{latex}
\\
and the angular frequency is given by
{latex}\begin{large} \[ \omega = \sqrt{\frac{mLg}{I}} = \sqrt{\frac{mLg}{\frac{1}{2}mr^{2} + mL^{2}}} \]\end{large}{latex}
\\
After some algebra, this reduces to:
\\
{latex}\begin{large} \[ \omega = \sqrt{\frac{g}{L + \frac{r^{2}}{2L}}} \]\end{large}{latex}
\\

{cloak:math}
{cloak:app}

{HTMLcomment}The footer completes the page setup and attaches the license and acknowledgments.{HTMLcomment}
{td}
{tr}
{table}