A rigid uniform annular disc is pivoted on a knife edge A in a uniform gravitational field as shown, such that it can execute small amplitude simple harmonic motion in the plane of the figure without slip at the pivot point. The inner radius 𝑟 and outer radius 𝑅 are such that 𝑟2 = 𝑅2/2, and the acceleration due to gravity is 𝑔. If the time period of small amplitude simple harmonic motion is given by T= ßπ√(R/g), where 𝜋 is the ratio of circumference to diameter of a circle, then 𝛽= ________ (round off to 2 decimal places).

The correct answer is 2.65 (when the pivot point is located at the outer boundary of the disc) or 2.66 (when the pivot point is located at the inner boundary of the disc). Both values lie within the official range of 2.62 to 2.70.

1. Physical Pendulum Theory:
An annular disc pivoted about a knife edge executing small-amplitude simple harmonic motion behaves as a physical pendulum. The time period $T$ of a physical pendulum is given by:
$$T=2\pi \sqrt{\frac{I_A}{Mgd}}$$
where:
• $M$ is the mass of the annular disc,
• $g$ is the acceleration due to gravity,
• $d$ is the distance from the center of mass (the geometric center of the disc) to the pivot point $A$,
• $I_A$ is the moment of inertia of the annular disc about the pivot axis passing through $A$.

2. Moment of Inertia of the Annular Disc:
The moment of inertia of a uniform annular disc of inner radius $r$ and outer radius $R$ about its central axis passing through the center of mass is:
$$I_{\mathrm{CM}}=\frac{1}{2}M\left(R^2+r^2\right)$$
Using the Parallel Axis Theorem, the moment of inertia about the pivot axis $A$ is:
$$I_A=I_{\mathrm{CM}}+M{d}^2=\frac{1}{2}M\left(R^2+r^2\right)+M{d}^2$$

3. Case A: Pivot point A on the outer boundary ($d=R$)
If the knife edge $A$ is on the outer boundary of the disc, then the distance $d=R$.
Substituting $d=R$ and the given relation $r^2=\frac{R^2}{2}$ into the moment of inertia expression:
$$I_A=\frac{1}{2}M\left(R^2+\frac{R^2}{2}\right)+M{R}^2=\frac{3}{4}M{R}^2+M{R}^2=1.75M{R}^2$$
Now we substitute $I_A$ and $d=R$ back into the time period equation:
$$T=2\pi \sqrt{\frac{1.75M{R}^2}{MgR}}=2\sqrt{1.75}\pi \sqrt{\frac{R}{g}}$$
Comparing this with the given expression $T=\beta \pi \sqrt{\frac{R}{g}}$, we have:
$$\beta =2\sqrt{1.75}=2\sqrt{\frac{7}{4}}=\sqrt{7}\approx 2.6458$$
Rounding off to 2 decimal places yields:
$$\beta \approx 2.65$$

4. Case B: Pivot point A on the inner boundary ($d=r$)
If the knife edge $A$ is on the inner boundary of the annular disc, the distance $d=r=\frac{R}{\sqrt{2}}$.
Using the Parallel Axis Theorem:
$$I_A=\frac{1}{2}M\left(R^2+r^2\right)+M{r}^2=\frac{1}{2}M\left(R^2+\frac{R^2}{2}\right)+M\left(\frac{R^2}{2}\right)=1.25M{R}^2$$
Substituting $I_A$ and $d=\frac{R}{\sqrt{2}}$ into the time period equation:
$$T=2\pi \sqrt{\frac{1.25M{R}^2}{Mg\frac{R}{\sqrt{2}}}}=2\sqrt{1.25\sqrt{2}}\pi \sqrt{\frac{R}{g}}$$
Comparing this with the given expression:
$$\beta =2\sqrt{1.25\sqrt{2}}\approx 2\sqrt{1.7678}\approx 2.659$$
Rounding off to 2 decimal places yields:
$$\beta \approx 2.66$$