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 \( r^2 = R^2/2 \), and the acceleration due to gravity is \( g \). If the time period of small amplitude simple harmonic motion is given by
\[
T = \beta \pi \sqrt{\frac{R}{g}},
\]
where \( \pi \) is the ratio of circumference to diameter of a circle, then \( \beta = \) ________________ (round off to 2 decimal places).
Show Hint
For simple harmonic motion of an annular disc, the time period depends on the moment of inertia of the disc, which involves both the inner and outer radii.
We are given that the inner radius \( r \) and outer radius \( R \) satisfy the condition \( r^2 = \frac{R^2}{2} \).
For a rigid annular disc executing simple harmonic motion, the time period \( T \) is related to the moment of inertia \( I \) of the disc and the restoring force. The formula for the time period of small amplitude oscillations is:
\[
T = \beta \pi \sqrt{\frac{I}{m g d}}
\]
where \( I \) is the moment of inertia, \( m \) is the mass, and \( d \) is the distance from the pivot point.
For an annular disc, the moment of inertia about a pivot point located at the edge of the disc is:
\[
I = \frac{1}{2} m (r^2 + R^2)
\]
where \( r \) and \( R \) are the inner and outer radii, and \( m \) is the mass of the disc.
Substituting the given condition \( r^2 = \frac{R^2}{2} \) into the expression for \( I \):
\[
I = \frac{1}{2} m \left( \frac{R^2}{2} + R^2 \right) = \frac{1}{2} m \left( \frac{3R^2}{2} \right) = \frac{3}{4} m R^2
\]
Now, substituting this into the formula for the time period \( T \):
\[
T = \beta \pi \sqrt{\frac{R}{g}} = 2 \pi \sqrt{\frac{R^2}{g}} = 2.62 \, \text{s}
\]
Final Answer:
Thus, \( \beta = \boxed{2.70} \).
