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h3. Part A

Assuming that the rod is thin and uniform and that the bob can be treated as a [point particle], what is the approximate period of Big Ben's pendulum?

h4. Solution

{toggle-cloak:id=sys} *System:* {cloak:id=sys}Rod and pendulum bob together as a single rigid body.{cloak}

{toggle-cloak:id=int} *Interactions:* {cloak:id=int}Both components of the system are subject to [external influences|external force] from the earth ([gravity (near-earth)]).  The rod is also subject to an [external influence|external force] from the axle of the pendulum.  We will consider [torques|torque (single-axis)] about the axle of the pendulum.  Because of this choice of axis, the [external force|external force] exerted by the axle on the pendulum will produce no [torque|torque (single-axis)], and so it is not relevant to the problem. {cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod}[Single-Axis Rotation of a Rigid Body] and [Simple Harmonic Motion].{cloak}

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

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

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We begin with a [force diagram]:

!forcediagram.png!

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

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Looking at the [force diagram], we can see that the total [torque|torque (single-axis)] from [gravity (near-earth)] about the [axis of rotation] is given by:
{latex}\begin{large}\[ \tau = -m_{\rm rod}g\frac{L}{2}\sin\theta - m_{\rm bob}gL\sin\theta \]\end{large}{latex}

The [moment of inertia] of the composite pendulum is the sum of the [moment of inertia] of the thin rod rotated about one end plus the [moment of inertia] of the bob treated as a point particle:
{latex}\begin{large}\[ I_{\rm tot} = \frac{1}{3}m_{\rm rod}L^{2} + m_{\rm bob}L^{2}\]\end{large}{latex}

With these two pieces of information, we can write the [rotational version of Newton's 2nd Law|Single-Axis Rotation of a Rigid Body] as:
{latex}\begin{large} \[ \left(\frac{1}{3}m_{\rm rod}L^{2} + m_{\rm bob}L^{2}\right)\alpha = - \left(m_{\rm rod}g\frac{L}{2} + m_{\rm bob}gL\right)\sin\theta \]\end{large}{latex}
We can now perform some algebra to isolate α:
{latex}\begin{large} \[ \alpha = -\left(\frac{\frac{1}{2}m_{\rm rod}+m_{\rm bob}}{\frac{1}{3}m_{\rm rod}+m_{\rm bob}}\right)\frac{g}{L} \sin\theta\]\end{large}{latex}

This equation is not yet of the form required by the [Simple Harmonic Motion] model, since α is not directly proportional to θ.  To achieve the form required by the [Simple Harmonic Motion] model, we must make the standard [small angle approximation] which is generally applied to pendulums.  In the [small angle approximation], the sine of θ is approximately equal to θ.  Thus, we have:
{latex}\begin{large} \[ \alpha \approx -\left(\frac{\frac{1}{2}m_{\rm rod}+m_{\rm bob}}{\frac{1}{3}m_{\rm rod}+m_{\rm bob}}\right)\frac{g}{L}\theta \] \end{large}{latex}
which is of the proper form for [simple harmonic motion] with the [natural angular frequency|natural frequency] given by:
{latex}\begin{large}\[ \omega = \sqrt{\left(\frac{\frac{1}{2}m_{\rm rod}+m_{\rm bob}}{\frac{1}{3}m_{\rm rod}+m_{\rm bob}}\right)\frac{g}{L}} \]\end{large}{latex}

We are asked for the [period] of the motion, which is related to the [natural angular frequency|natural frequency] by the relationship:
{latex}\begin{large} \[ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\left(\frac{\frac{1}{3}m_{\rm rod}+m_{\rm bob}}{\frac{1}{2}m_{\rm rod}+m_{\rm bob}}\right)\frac{L}{g}} = 4.06 s\]\end{large}{latex}

{tip}The website of [Parliament|http://www.bigben.parliament.uk] claims that the "duration of pendulum beat" is 2 seconds.  This seems to contradict our calculation.  Can you explain the discrepancy?  Check your explanation using the video at the top of this page.{tip}

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