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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 321 kg and a length of 4.48 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: Rod and pendulum bob together as a single rigid body.
Interactions: Both components of the system are subject to external influences from the earth ([gravity]). 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.
Model: Single-Axis Rotation of a Rigid Body and Simple Harmonic Motion.
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}} = 3.92 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
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. How far from the axis of rotation must the penny be placed to have this effect?
Solution
System: The rod and the pendulum bob plus one old English penny.
Interactions: As in Part A.
Models: As in Part A.
Approach:
Diagrammatic Representation:
The penny can very reasonably be treated as a point mass.
Mathematical Representation:
The penny adjusts the net [torque] on the pendulum
\begin
[ \tau = -m_
g\frac
\sin\theta - m_
gL\sin\theta - m_
gd\sin\theta ]\end
and also the moment of inertia of the pendulum
\begin
[ I_
= \frac
m_
L^
+ m_
L^
+ m_
d^
]\end
Following the same procedures as in Part A, we can then arrive at a formula for the period of the adjusted pendulum:
\begin
[ T = \frac
= 2\pi\sqrt{\left(\frac{\frac
m_
+m_
+m_
\frac{d^{2}}{L^
}}{\frac
m_
+m_
+m_
\frac
{L}}\right)\frac
{g}}]\end
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:
\begin
[\frac
\times\frac
\times\frac
=
]\end
which means that the period of the pendulum is decreased by:
\begin
[ \frac
= 1.9\times10^{-5} s]\end
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
[ T = 2\pi\left(\frac
m_
+m_
+m_
\frac{d^{2}}{L^{2}}\right)^
\left(\frac
m_
+m_
+m_
\frac
\right)^
\frac
{g}}]\end