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A thin uniform circular disc of radius a and centre A, with density p, has a circular hole cut in it of radius b and centre B, where $AB = c < a−b$. The disc is free to oscillate in a vertical plane about a smooth fixed horizontal circular rod of radius b passing through the hole. Show that the period of small oscillations is $2\pi\sqrt{l/g}$, where $l = c + (a^4 -b^4)/(2a^2c)$.

I have the moment of inertia to be $I =p/2 pi(a^4 -b^4 +2a^2c^2)$, which looks correct. I don't know how to find the period of the oscillations - is there a formula I should know? Should I use conservation of energy of forces??

mvw
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guest10923
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1 Answers1

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The oscillating disk is a compound pendulum. The period is $2\pi \sqrt{\frac{I}{mgh}}$ where $I$ is moment of inertia about the axis of rotation and $h$ is the distance of the centre of mass from the axis.

You do not state the axis about which you have calculated moment of inertia. You might need to use the Parallel Axis Theorem if the axis you used is not the axis of rotation.

sammy gerbil
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