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:
 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 (near-earth) about the axis of rotation is given by:
\begin{large}\[ \tau = -m_{\rm rod}g\frac{L}{2}\sin\theta - m_{\rm bob}gL\sin\theta \]\end{large} |
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{large}\[ I_{\rm tot} = \frac{1}{3}m_{\rm rod}L^{2} + m_{\rm bob}L^{2}\]\end{large} |
With these two pieces of information, we can write the rotational version of Newton's 2nd Law as:
\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} |
We can now perform some algebra to isolate α:
\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} |
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{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} |
which is of the proper form for simple harmonic motion with the natural angular frequency given by:
\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} |
We are asked for the period of the motion, which is related to the natural angular frequency by the relationship:
\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} |
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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