A uniform thin ring of radius R and mass m suspended in a vertical plane from a point in its circumference. Its time period of oscillation is

Options

A. `2pi sqrt((2R)/(g))`
B. `pi sqrt((2R)/(g))`
C. `(pi)/(2) sqrt(( R)/(g))`
D. `pi sqrt((R )/(2g))`

Answer

To find the time period of oscillation of a uniform thin ring of radius \( R \) and mass \( m \) suspended from a point in its circumference, we can follow these steps:

### Step 1: Understand the Setup
The ring is suspended from a point on its circumference and can oscillate in a vertical plane. The center of mass of the ring is at its geometric center.

### Step 2: Identify the Forces
When the ring is displaced by a small angle \( \theta \), the gravitational force \( mg \) acts downwards through the center of mass. The component of this force that acts to restore the ring back to its equilibrium position is \( mg \sin \theta \).

### Step 3: Calculate the Torque
The torque \( \tau \) about the point of suspension due to the weight of the ring is given by:
\[ \tau = -mg \sin \theta \cdot R \]
The negative sign indicates that the torque acts in the opposite direction of the displacement.

### Step 4: Use Small Angle Approximation
For small angles, we can approximate \( \sin \theta \approx \theta \) (in radians). Thus, the torque can be rewritten as:
\[ \tau \approx -mg \theta R \]

### Step 5: Relate Torque to Angular Acceleration
According to the rotational dynamics, torque is also related to angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the ring about the point of suspension.

### Step 6: Calculate the Moment of Inertia
Using the parallel axis theorem, the moment of inertia \( I \) of the ring about the point of suspension is:
\[ I = I_{cm} + md^2 \]
where \( I_{cm} = mR^2 \) (moment of inertia about its center) and \( d = R \) (distance from the center of mass to the point of suspension). Thus,
\[ I = mR^2 + mR^2 = 2mR^2 \]

### Step 7: Substitute into the Torque Equation
Now substituting \( I \) into the torque equation:
\[ -mg \theta R = 2mR^2 \alpha \]
Since \( \alpha = \frac{d^2 \theta}{dt^2} \), we can rewrite this as:
\[ -mg \theta R = 2mR^2 \frac{d^2 \theta}{dt^2} \]

### Step 8: Rearrange to Form a Simple Harmonic Motion Equation
Dividing through by \( mR \):
\[ -g \theta = 2R \frac{d^2 \theta}{dt^2} \]
Rearranging gives:
\[ \frac{d^2 \theta}{dt^2} + \frac{g}{2R} \theta = 0 \]
This is the standard form of simple harmonic motion.

### Step 9: Identify the Angular Frequency
From the equation, we can identify the angular frequency \( \omega \):
\[ \omega^2 = \frac{g}{2R} \]
Thus,
\[ \omega = \sqrt{\frac{g}{2R}} \]

### Step 10: Calculate the Time Period
The time period \( T \) of oscillation is given by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2R}{g}} \]

### Final Answer
The time period of oscillation of the ring is:
\[ T = 2\pi \sqrt{\frac{2R}{g}} \]

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