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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
{Solution:=} *System:* {:=} {cloak}
{:=} *Interactions:* {:=} [ |external force] [gravity (near-earth]) [ |external force] [|torque (single-axis)] [external force| ] [torque| (single-axis)] {cloak}
{:=} *Model:* {:=mod}[Single-Axis Rotation of a Rigid Body] and [Simple Harmonic Motion].{cloak}
{:=} *Approach:*
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{: Diagrammatic Representation: We begin with a force diagram:  Image Added
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: | Latex |
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=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:
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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 |
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\begin{large}\[ I_{\rm tot} = \frac{1}{3}m_{\rm rod}L^{2} + m_{\rm bob}L^{2}\]\end{large}{latex}
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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:
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}\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 |
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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}
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This equation is not yet of the form required by the [Simple Harmonic Motion ] model, since α 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 |
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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}
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which is of the proper form for [simple harmonic motion ] with the [natural angular frequency |natural frequency] given by:
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}\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:
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}\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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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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