Oscillations of a disc with a smaller disc removed (Feynman ex. 17.23)

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The discussion focuses on the oscillations of a disc with a smaller disc removed, analyzing the center of mass and moment of inertia. The center of mass is calculated to be at a distance of 5a/6 from the edge, while the moment of inertia about the center is derived as 13πa^4ρ/32. The kinetic and potential energy expressions are formulated, leading to an energy equation that does not exhibit simple harmonic motion for finite angles. After some corrections and verification, the moment of inertia about the center of mass is confirmed as 37πa^4ρ/96. The final discussion highlights a discrepancy in the period of oscillation, prompting a review of the parallel axis theorem.
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Homework Statement
Determine the period of small oscillations, given that the disc rolls without slipping
Relevant Equations
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A disc of radius ##a## has a smaller disc of radius ##a/2## removed. The resulting object has mass ##m##:

1630700847115.png
The centre of mass ##G## is a distance ##h = \dfrac{\pi a^3 - \dfrac{3\pi a^3}{8} }{\dfrac{3\pi a^2}{4}} = \dfrac{5a}{6}## from the edge. The moment of inertia of the shape about the centre ##C## is\begin{align*}
I_C = \dfrac{1}{2}(\pi a^2 \rho)a^2 - \left(\dfrac{\pi a^2 \rho}{4} \right) \left( \dfrac{1}{2} \left(\dfrac{a}{2} \right)^2 + \left(\dfrac{a}{2} \right)^2 \right) = \dfrac{13\pi a^4 \rho}{32}
\end{align*}therefore the moment of inertia about ##G## is $$I_G = I_C + \dfrac{3\pi a^2 \rho}{4} \left(\dfrac{a}{6} \right)^2 = \dfrac{41 \pi a^4 \rho}{96}$$If the angle through which the disc has rotated is denoted ##\varphi##, then the centre of mass ##G## has velocity\begin{align*}
\mathbf{v}_G &= (a\dot{\varphi})\mathbf{\hat{x}} + \boldsymbol{\omega} \times \overrightarrow{CG} = \left(a\dot{\varphi} - \dfrac{a \dot{\varphi}}{6} \cos{\varphi} \right)\mathbf{\hat{x}} + \left(\dfrac{a\dot{\varphi}}{6} \sin{\varphi} \right)\mathbf{\hat{y}} \\

\implies v_G^2 &= \frac{a^2 \dot{\varphi}^2}{36}\left(37 - 12\cos{\varphi} \right)
\end{align*}Therefore the kinetic energy of the body is\begin{align*}
T = \dfrac{1}{2} m v_G^2 + \dfrac{1}{2}I_G \dot{\varphi}^2 = \pi a^4 \rho \dot{\varphi}^2 \left( \dfrac{1}{96}(37 - 12\cos{\varphi}) + \frac{41}{192} \right) = \pi a^4 \rho \dot{\varphi}^2 \left( \dfrac{115}{192} - \dfrac{1}{8} \cos{\varphi} \right)
\end{align*}Meanwhile, the height of the centre of mass ##G## when the body has rotated by ##\varphi## is\begin{align*}
y_G = a(1-\cos{\varphi}) + \dfrac{5a}{6} \cos{\varphi} = a(1- \dfrac{1}{6} \cos{\varphi})
\end{align*}therefore the potential energy is ##U = \dfrac{3\pi a^3 \rho g}{4}\left(1- \dfrac{1}{6} \cos{\varphi} \right)##. Overall I write the energy ##E## as\begin{align*}
\dfrac{1}{\pi a^3 \rho} E &= a \dot{\varphi}^2 \left( \dfrac{115}{192} - \dfrac{1}{8} \cos{\varphi} \right) + \dfrac{3g}{4}\left(1- \dfrac{1}{6} \cos{\varphi} \right) \\ \\

\implies \dot{E} &= 2a \dot{\varphi} \ddot{\varphi}\left( \dfrac{115}{192} - \dfrac{1}{8} \cos{\varphi} \right) + \dfrac{a\dot{\varphi}^3}{8} \sin{\varphi} + \dfrac{g \dot{\varphi}}{8} \sin{\varphi} \overset{!}{=} 0
\end{align*}The motion does not look simple harmonic, at least for finite ##\varphi##. To investigate we can put ##\cos{\varphi} = 1## and ##\sin{\varphi} = \varphi##, and I assume also ##\dot{\varphi}^2 = 0##, which gives\begin{align*}
\ddot{\varphi} + \dfrac{12g}{91a} \varphi = 0 \implies \Omega &= \sqrt{\dfrac{12 g}{91 a}} \\

T &= \pi \sqrt{\dfrac{91a}{3g}}
\end{align*}However in the solution manual they write ##T = \pi \sqrt{29a/g}##?
 
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State the parallel axis theorem. Did you use it correctly?
 
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haruspex said:
State the parallel axis theorem. Did you use it correctly?
ahhhh, kill me
ought to be ##I_G = I_C - \dfrac{3\pi a^2 \rho}{4} \left(\dfrac{a}{6} \right)^2##, right?

[Edit: Yep, I checked it through with ##I_G = \dfrac{37 \pi a^4 \rho}{96}## and it works. Thanks haru! ☺️]
 
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