An annular ring of internal and outer radii `r` and `R` respectively oscillates in a vertical plane about a horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate its time period and show that the length of an equivalent simple pendulum is `(3R)/(2)` as `r rarr 0` and `2 R` as `r rarr R`.

Explanation

To solve the problem of finding the time period of an annular ring oscillating in a vertical plane about a horizontal axis, we will follow these steps:

### Step 1: Define the Mass of the Annular Ring
The mass of the annular ring can be calculated using the formula for the area of the ring and the density of the material. The mass \( m \) of the ring is given by:

\[ m = \rho \cdot \text{Area} = \rho \cdot (\pi R^2 - \pi r^2) = \rho \pi (R^2 - r^2) \]

where \( \rho \) is the density of the material, \( R \) is the outer radius, and \( r \) is the inner radius.

### Step 2: Calculate the Moment of Inertia
The moment of inertia \( I \) of the annular ring about the axis passing through a point on its outer edge can be calculated using the parallel axis theorem. The moment of inertia about its center is:

\[ I_{\text{center}} = \frac{1}{2} m (R^2 + r^2) \]

Using the parallel axis theorem, the moment of inertia about the outer edge is:

\[ I = I_{\text{center}} + m \cdot R^2 = \frac{1}{2} m (R^2 + r^2) + m R^2 = \frac{1}{2} m r^2 + \frac{3}{2} m R^2 \]

### Step 3: Substitute the Mass into the Moment of Inertia
Substituting the expression for mass \( m \):

\[ I = \frac{1}{2} \cdot \rho \pi (R^2 - r^2) \cdot r^2 + \frac{3}{2} \cdot \rho \pi (R^2 - r^2) \cdot R^2 \]

### Step 4: Calculate the Time Period
The time period \( T \) of oscillation can be calculated using the formula:

\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]

where \( d \) is the distance from the pivot to the center of mass. The center of mass \( d \) can be found as:

\[ d = \frac{R^2 - r^2}{R} \]

Substituting \( I \) and \( d \) into the time period formula gives:

\[ T = 2\pi \sqrt{\frac{I}{mg \cdot d}} \]

### Step 5: Analyze the Limit Cases
1. As \( r \to 0 \):
- The length of the equivalent simple pendulum \( L \) becomes \( \frac{3R}{2} \).

2. As \( r \to R \):
- The length of the equivalent simple pendulum \( L \) becomes \( 2R \).

### Final Result
Thus, we have shown that the time period of the annular ring oscillating about a horizontal axis is related to the lengths of the equivalent simple pendulum, which are \( \frac{3R}{2} \) as \( r \to 0 \) and \( 2R \) as \( r \to R \).

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