Worked example 11.6: Oscillating disk

Question: A uniform disk of radius $r = 0.8 {\rm m}$ and mass $M=3 {\rm kg}$ is freely suspended from a horizontal pivot located a radial distance $d=0.25 {\rm m}$ from its centre. Find the angular frequency of small amplitude oscillations of the disk.

Answer: The moment of inertia of the disk about a perpendicular axis passing through its centre is $I= (1/2) M r^2$. From the parallel axis theorem, the moment of inertia of the disk about the pivot point is

$$I' = I + M d^2 = \frac{3\times 0.8\times 0.8}{2} + 3\times 0.25\times 0.25 = 1.1475 {\rm kg m^2}.$$

The angular frequency of small amplitude oscillations of a compound pendulum is given by

$$\omega = \sqrt{\frac{M g d}{I'}} = \sqrt{\frac{3\times 9.81\times 0.25}{1.1475}} = 2.532 {\rm rad./s}.$$

Hence, the answer is $2.532 {\rm rad./s}$.