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Composition Setup
 Excerpt
hiddentrue
Check Parliament's math by calculating the period of Big Ben's pendulum.
Image Added
Parliamentary copyright images are reproduced with the permission of Parliament.
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...
Parliamentary 
...
 video 
...
 of 
...
 the 
...
 pendulum 
...
 of 
...
 the 
...
 great 
...
 clock
The Great Clock of Parliament (Big Ben) uses a pendulum to keep time. The website of Parliament reports that the pendulum rod has a mass of 107 kg and a length of 4.4 m, and the bob attached to the rod has a mass of 203 kg. 
 Deck of Cards
idbigdeck
| 

The [Great Clock of Parliament|http://www.bigben.parliament.uk] (Big Ben) uses a pendulum to keep time.  The website of Parliament reports that the pendulum rod has a mass of 107 kg and a length of 4.4 m, and the bob attached to the rod has a mass of 203 kg.  

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

*System:* Rod and pendulum bob together as a single rigid body. 

*Interactions:* Both components of the system are subject to [external influences|external force] from the earth ([gravity]).  The rod is also subject to an [external influence|external force] from the axle of the pendulum.  We will consider [torques|torque (one-dimensional)] 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 (one-dimensional)], and so it is not relevant to the problem.  

*Model:* [Single-Axis Rotation of a Rigid Body] and [Simple Harmonic Motion].  

*Approach:*  

{color:red} *Diagrammatic Representation:* {color}

We begin with a [force diagram]:

{color:red} *Mathematical Representation* {color}

Looking at the [force diagram], we can see that the total [torque|torque (one-dimensional)] from [gravity] about the [axis of rotation] is given by:
{latex}
 Card
labelPart A
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?
Solution
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idsys
 System: 
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idsys
Rod and pendulum bob together as a single rigid body.
 Toggle Cloak
idint
 Interactions: 
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idint
Both components of the system are subject to external influences from the earth (). The rod is also subject to an external influence from the axle of the pendulum. We will consider torques about the axle of the pendulum. Because of this choice of axis, the external force exerted by the axle on the pendulum will produce no torque, and so it is not relevant to the problem. 
 Toggle Cloak
idmod
 Model: 
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idmod
 and .
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idapp
 Approach: 
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idapp
 Toggle Cloak
iddiag
  Diagrammatic Representation: 
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iddiag
We begin with a force diagram:
Image Added
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diag
diag
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idmath
  Mathematical Representation 
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idmath
Looking at the force diagram, we can see that the total torque 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}
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 
 &alpha;:
{latex}
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 
 &alpha; 
α is 
 
 not 
 
 directly 
 
 proportional 
 
 to 
 &theta;.  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 &theta; is approximately equal to &theta;.  Thus, we have:
{latex}
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}

h3. Part B

Fine adjustment of the pendulum is accomplished by adding old (pre-decimal) pennies to the pendulum.  According to the website of [Parliament|http://www.bigben.parliament.uk], each 9.4 g penny used to adjust the clock is added to the pendulum in such a way that the clock mechanism speeds up enough to gain two fifths of one second in 24 hours of operation.  How far from the axis of rotation must the penny be placed to have this effect?

h4. Solution

*System:* The rod and the pendulum bob plus one old English penny.

*Interactions:* As in Part A.

*Models:* As in Part A.

*Approach:*

{color:red} *Diagrammatic Representation:* {color}

The penny can very reasonably be treated as a point mass.  



{color:red} *Mathematical Representation:* {color}

The penny adjusts the net [torque|torque (one-dimensional)] on the pendulum

{latex}\begin{large}\[ \tau = -m_{\rm rod}g\frac{L}{2}\sin\theta - m_{\rm bob}gL\sin\theta - m_{\rm penny}gd\sin\theta \]\end{large}{latex}

and also the [moment of inertia] of the pendulum

{latex}\begin{large}\[ I_{\rm tot} = \frac{1}{3}m_{\rm rod}L^{2} + m_{\rm bob}L^{2} + m_{\rm penny}d^{2}\]\end{large}{latex}

Following the same procedures as in Part A, we can then arrive at a formula for the period of the adjusted pendulum:

{latex}\begin{large} \[ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\left(\frac{\frac{1}{3}m_{\rm rod}+m_{\rm bob}+m_{\rm penny}\frac{d^{2}}{L^{2}}}{\frac{1}{2}m_{\rm rod}+m_{\rm bob}+m_{\rm penny}\frac{d}{L}}\right)\frac{L}{g}}\]\end{large}{latex}

We are hoping to solve this equation for _d_, which implies we need to know _T_, the adjusted period.  Finding the absolute period is tricky, since we are given information about the _change_ in the period.  We know that the pendulum's period is adjusted such that the pendulum ticks out an extra 2/5 of a second per day.  We can use this information to estimate the new period.  Suppose we assume that the pendulum's original period is exactly 4 seconds to simplify the calculations.  The number of complete oscillations executed in one day is then:
{latex}\begin{large}\[\frac{\rm 24\:hours}{\rm 1\:day}\times\frac{\rm 3600\:seconds}{\rm 1\:hour}\times\frac{\rm 1\:oscillation}{\rm 4\:seconds} = {\rm 21600 oscillations/day}\]\end{large}{latex}
which means that the period of the pendulum is decreased by:
{latex}\begin{large}\[ \frac{\rm 2/5 second}{21600 oscillations} = 1.9\times10^{-5} s\]\end{large}{latex}

There are now two possible roads to the solution.  The most obvious way is to find the new total period of the pendulum by subtracting this adjustment from 4 seconds and then input the period into the equation we found above.  We can alternatively perform an approximation in the equation above that will allow us to directly use the _change_ in the period.  We will use this latter method as a means of showcasing the utility of series expansions.

The formula given above for the adjusted period can be written:

{latex}\begin{large} \[ T =  2\pi\left(\frac{1}{3}m_{\rm rod}+m_{\rm bob}+m_{\rm penny}\frac{d^{2}}{L^{2}}\right)^{1/2}\left(\frac{1}{2}m_{\rm rod}+m_{\rm bob}+m_{\rm penny}\frac{d}{L}\right)^{-1/2}\sqrt{\frac{L}{g}}\]\end{large}{latex}

We know for a fact that the mass of the penny is much, much smaller than the mass of the rod and the pendulum bob.  Thus, we can write:
{latex}\begin{large} \[ T = T_{\rm old}\left(1+\frac{m_{\rm penny}d^{2}}{\frac{1}{3}m_{\rm rod}L^{2} + m_{\rm bob}L^{2}}\right)^{1/2}\left(1+\frac{m_{\rm penny}d}{\frac{1}{2}m_{\rm rod}L + m_{\rm bob}L}\right)^{-1/2}\]\end{large}{latex}
where _T_~old~ is 4 seconds in our approximation.  The two binomials in the expression above are each of the form:
{latex}\begin{large} \[ (1+x)^{r}\qquad x \ll 1\]\end{large}{latex}
which allows the series expansion:
{latex}\begin{large}\[ (1+x)^{r} \sim 1+rx+\mathcal{O}(x^{2}) \]\end{large}{latex}
where the last term is an error term of order _x_^2^.  Since _x_ is so small, we can ignore errors of order _x_^2^.  Using this approximation procedure, we can write:

{latex}\begin{large}\[ T \sim T_{\rm old}\left(1+\frac{1}{2}\frac{m_{\rm penny}d^{2}}{\frac{1}{3}m_{\rm rod}L^{2}+m_{\rm bob}L^{2}} - \frac{1}{2}\frac{m_{\rm penny}d}{\frac{1}{2}m_{\rm rod}L + m_{\rm bob}L}\right)\]\end{large}{latex}

This is a quadratic equation in _d_:
{latex}\begin{large}\[ \frac{m_{\rm penny}}{\frac{2}{3}m_{\rm rod}+2m_{\rm bob}}\frac{d^{2}}{L^{2}} - \frac{m_{\rm penny}}{m_{\rm rod}+2m_{\rm bob}}\frac{d}{L} + \frac{T_{\rm old}-T}{T_{\rm old}} = 0\]\end{large}{latex}

which can be solved using the quadratic equation:
{latex}\begin{large}\[ \frac{d}{L} = \]\end{large}{latex}

 Tip
The website of Parliament 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.
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math
math
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app
app
 Card
labelPart B
Part B (Challenge)
Image Added
Image Added
Parliamentary copyright images are reproduced 
 with the permission of Parliament.
Parliamentary copyright images are reproduced 
 with the permission of Parliament.
Fine adjustment of the pendulum is accomplished by adding old (pre-decimal) pennies to the pendulum. According to the website of Parliament, each 9.4 g penny used to adjust the clock is added to the pendulum in such a way that the clock mechanism speeds up enough to gain two fifths of one second in 24 hours of operation. The placement of the coins on the pendulum can be estimated using BBC video available at http://news.bbc.co.uk/2/hi/science/nature/7792436.stm. Use the model of Part A plus the estimated location of the penny to predict the effect of the penny and compare to the reported effect.
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