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Parliamentary copyright images |
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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.
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
System:
Interactions:
Model:
Approach:
Diagrammatic Representation:
We begin with a force diagram:
Mathematical Representation
Looking at the force diagram, we can see that the total torque from [gravity] about the axis of rotation is given by:
\begin
[ \tau = -m_
g\frac
\sin\theta - m_
gL\sin\theta ]\end
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:
\begin
[ I_
= \frac
m_
L^
+ m_
L^
]\end
With these two pieces of information, we can write the rotational version of Newton's 2nd Law as:
\begin
[ \left(\frac
m_
L^
+ m_
L^
\right)\alpha = - \left(m_
g\frac
+ m_
gL\right)\sin\theta ]\end
We can now perform some algebra to isolate α:
\begin
[ \alpha = -\left(\frac{\frac
m_
+m_{\rm bob}}{\frac
m_
+m_{\rm bob}}\right)\frac
\sin\theta]\end
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:
\begin
[ \alpha \approx -\left(\frac{\frac
m_
+m_{\rm bob}}{\frac
m_
+m_{\rm bob}}\right)\frac
\theta ] \end
which is of the proper form for Simple Harmonic Motion with the natural angular frequency given by:
\begin
[ \omega = \sqrt{\left(\frac{\frac
m_
+m_{\rm bob}}{\frac
m_
+m_{\rm bob}}\right)\frac
{L}} ]\end
We are asked for the period of the motion, which is related to the natural angular frequency by the relationship:
\begin
[ T = \frac
= 2\pi\sqrt{\left(\frac{\frac
m_
+m_{\rm bob}}{\frac
m_
+m_{\rm bob}}\right)\frac
{g}} = 4.06 s]\end
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.
Part B
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Parliamentary copyright images |
Parliamentary copyright images |
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.