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 2.66.

Step-by-step Explanation:

We are given a rigid uniform annular disc with inner radius r and outer radius R such that:

$r^2=\frac{R^2}{2}\Rightarrow r=\frac{R}{\sqrt{2}}$

As shown in the diagram, the disc is pivoted on a knife edge A located at the top of the inner boundary (at a distance of r from the center of mass G). The disc executes small amplitude oscillations in its own plane.

1. Moment of Inertia of the Annular Disc:
The mass moment of inertia of a uniform annular disc of mass m about the axis passing through its center of mass (G) and perpendicular to its plane is given by:

$I_G=\frac{1}{2}m\left(R^2+r^2\right)$

Substituting the given relation $r^2=\frac{R^2}{2}$ into the equation:

$I_G=\frac{1}{2}m\left(R^2+\frac{R^2}{2}\right)=\frac{3}{4}m{R}^2$

2. Moment of Inertia about the Pivot Point A:
Using the parallel axis theorem, we can determine the mass moment of inertia about the pivot point A (which is at a distance of r from G, perpendicular to the plane of the disc):

$I_A=I_G+m{r}^2$

Substituting the expressions for $I_G$ and $r^2$:

$I_A=\frac{3}{4}m{R}^2+m\left(\frac{R^2}{2}\right)=\frac{5}{4}m{R}^2$

3. Equation of Motion for Small Amplitude Oscillations:
When the disc is angularly displaced by a small angle $\theta$ about the pivot A, the restoring torque due to the weight of the disc acting at the center of mass G is:

$\tau =-mgrsin\theta$

Applying the small-angle approximation where $sin\theta \approx \theta$, the restoring torque becomes:

$\tau \approx -mgr\theta$

Applying the equation of rotational motion $\tau =I_A\frac{d^2\theta }{d{t}^2}$:

$I_A\frac{d^2\theta }{d{t}^2}+mgr\theta =0$

Substituting the values of $I_A$ and $r$:

$\left(\frac{5}{4}m{R}^2\right)\frac{d^2\theta }{d{t}^2}+mg\left(\frac{R}{\sqrt{2}}\right)\theta =0$

Simplifying the differential equation by dividing by $\frac{5}{4}m{R}^2$:

$\frac{d^2\theta }{d{t}^2}+\left(\frac{4g}{5\sqrt{2}R}\right)\theta =0$

4. Time Period of Simple Harmonic Motion:
Comparing this with the standard equation for simple harmonic motion $\frac{d^2\theta }{d{t}^2}+{\omega }_n^2\theta =0$, we find the angular frequency ${\omega }_n$:

${\omega }_n^2=\frac{4g}{5\sqrt{2}R}\Rightarrow {\omega }_n=\frac{2}{\sqrt{5\sqrt{2}}}\sqrt{\frac{g}{R}}$

The time period T of the oscillation is given by:

$T=\frac{2\pi }{{\omega }_n}=2\pi \left(\frac{\sqrt{5\sqrt{2}}}{2}\sqrt{\frac{R}{g}}\right)=\sqrt{5\sqrt{2}}\pi \sqrt{\frac{R}{g}}$

Comparing this with the given format $T=\beta \pi \sqrt{\frac{R}{g}}$:

$\beta =\sqrt{5\sqrt{2}}$

Evaluating this numerically using $\sqrt{2}\approx 1.4142$:

$\beta =\sqrt{5\times 1.4142}=\sqrt{7.071}\approx 2.659$

Rounding off to two decimal places, we get:

β = 2.66